WAVE DAMAGE. November 19, 2002

The leaking oil tanker Prestige sinks some 240km off Spain's north-western coast, taking more than 70,000 tonnes of fuel to the seabed 4 km below.

Strength of Materials (HE 306)

Prof. Dr. R. Chandra Sekaran Ph.D (C.Engg)

INTRODUCTION—CONCEPT OF STRESS 2 STRESS AND STRAIN—AXIAL LOADING 3 TORSION 4 PURE BENDING 5 ANALYSIS AND DESIGN OF BEAMS FOR BENDING

In the experiment, a wire of a given thickness and length was used to suspend a basket. Thebasket was filled slowly with sand, fed from an adjacently suspended hopper (Figure). When the wire suspending the basket breaks, a spring closes the hopper opening, and the basket falls a short distance into a hole, so as not to upset the basket. The sand in the basket was then weighed to establish the tensile strength of the wire.

What is Mechanics?• Mechanics is the science which describes and predicts the conditions of rest or motion of bodies under the action of forces.• Categories of Mechanics:- Rigid bodies – Statics and Dynamics- Deformable bodies – Strength of Materials- Fluids• Mechanics is an applied science - it is not an abstractor pure science but does not have the empiricism found in other engineering sciences.• Mechanics is the foundation of most engineering sciences and is an indispensable prerequisite to their study.

Fundamental Concepts• Space - associated with the notion of the position of a point P given interms of three coordinates measured from a reference point or origin.• Time - definition of an event requires specification of the time andposition at which it occurred.• Mass - used to characterize and compare bodies, e.g., response toearth’s gravitational attraction and resistance to changes in translationalmotion.• Force - represents the action of one body on another. A force ischaracterized by its point of application, magnitude, and direction, i.e.,a force is a vector quantity.

In Newtonian Mechanics, space, time, and mass are absolute concepts,independent of each other. Force, however, is not independent of theother three. The force acting on a body is related to the mass of the body and the variation of its velocity with time.

EQUILIBRIUMThis section will be limited to one- or two-dimensional systems, where all the forces and couples will be acting in on plane; such a system of forces is called a coplanar system.In two dimensions, equilibrium is achieved when the following laws are satisfied:

upward forces = downward forces , forces to the left = forces to the right and clockwise couples = counter clockwise couples.

Taking moments about Bclockwise couples = counter clockwise couples RAx4 = 3 x 6 + 1 0 x 2 RA = (18 + 20 )/4 RA = 9.5 kN

Resolving forces vertically, RA+RB = (3+10) or RB = (13 - 9.5) = 3.5 kN

Determine the reactions of RA and RB for the simply-supported beam

Tension and compression: direct stresses

One of the simplest loading conditions of a material is that of tension, in which the fibres of the material are stretched.

Consider, for example, a long steel wire held rigidly at its upper end and loaded by a mass hung from the lower end. If vertical movements of the lower end are observed during loading it will be found that the wire is stretched by a small, but measurable,amount from its original unloaded length. The material of the wire is composed of a large number of small crystals which are only visible under a microscopic study; these crystals have irregularly shaped boundaries, and largely random orientations with respect to each other; as loads are applied to the wire, the crystal structure of the metal is distorted.

For small loads it is found that the extension of the wire is roughly proportional to the applied load. This linear relationship between load and extension was first discovered by Robert Hooke in 1678; a material showing this characteristic is said to obey Hooke's law.

As the tensile load in the wire is increased, a stage is reached where the material ceases to show this linear characteristic; the corresponding point on the load-extension curve is known as the limit of proportionality.

If the wire is made from a high-strength steel then the load-extension curve up to the breaking point has the form shown.

Beyond the limit of proportionality the extension of the wire increases non-linearly up to the elastic limit and, eventually, the breaking point.

Elastic limit is important because it divides the load-extension curve into two regions. For loads up to the elastic limit, the wire returns to its original un-stretched length on removal of the loads; this properly of a material to recover its original form on removal of the loads is known as elasticity; the steel wire behaves as a still elastic spring.

When loads are applied above the elastic limit, and are then removed, it is found that the wire recovers only part of its extension and is stretched permanently; in this condition the wire is said to have undergone an inelastic, or plastic extension.

For most materials, the limit of proportionality and the elastic limit are assumed to have the same value.

In the case of elastic extensions, work performed in stretching the wire is stored as strain energy in the material; this energy is recovered when the loads are removed. During inelastic extensions, work is performed in making permanent changes in the internal structure of the material. Not all the work can be recovered : balance of this energy reappears in other forms, mainly as heat.

When tensile tests are carried out on steel wires of the same material, but of different cross sectional area, the breaking loads are found to be proportional approximately to the respective cross-sectional areas of the wires.

This is so because the tensile strength is governed by the intensity of force on a normal cross-section of a wire, and not by the total force. This intensity of force is known as stress, denoted by “σ”. Suppose δA is an element of area of the cross section of the bar, if the normal force acting on this element is δP, then the tensile stress at this point of the cross-section is defined as the limiting value of the ratio (δP/ δA) as δA becomes infinitesimally small. Thus,

When the forces P in Figure are reversed in direction at each end of the bar they tend to compress the bar; the loads then give rise to compressive stresses.

Tensile and compressive stresses are together referred to as direct (or normal) stresses, because they act perpendicularly to the surface.

A steel bar of rectangular cross-section, 3 cm by 2 cm, carries an axial load of 30 kN. Estimate the average tensile stress over a normal cross-section of the bar.

A steel bolt, 2.50 cm in diameter, carries a tensile load of 40 kN. Estimate the average tensile stress at the section a and at the screwed section b, where the diameter at the root of the thread is 2.10 cm.

Tensile and compressive strains

In the steel wire experiment we discussed the extension of the whole wire.

If we measure the extension of, say, the lowest quarter-length of the wire we find that for a given load it is equal to a quarter of the extension of the whole wire.

In general we find that, at a given load, the ratio of the extension of any length to that length is constant for all parts of the wire; this ratio is known as the tensile strain.

Suppose the initial unstrained length of the wire is Lo, and e is the extension due to straining; the tensile strain ε is defined as

This definition of strain is useful only for small distortions, in which the extension e is small compared with the original length Lo; this definition is adequate for the study of most engineering problems, where we are concerned with values of ε of the order 0.001, or so.

If a material is compressed the resulting strain is defined in a similar way, except that e is the contraction of a length.

We note that strain is a non-dimensional quantity, being the ratio of the extension, or contraction, of a bar to its original length.

A cylindrical block is 30 cm long and has a circular cross-section 10 cm in diameter. It carries a total compressive load of 70 kN, and under this load it contracts by 0.02 cm. Estimate the average compressive stress over a normal cross-section and the compressive strain.

Stress-strain curves for brittle materials

Many of the characteristics of a material can be deduced from the tensile test. In the experiment we measured the extensions of the wire for increasing loads; it is more convenient to compare materials in terms of stresses and strains, rather than loads and extensions of a particular specimen of a material.

The tensile stress-strain curve for a high-strength steel has the form shown. The stress at any stage is the ratio of the load of the original cross-sectional area of the test specimen; the strain is the elongation of a unit length of the test specimen.

For stresses up to about 750 MN/m2 the stress-strain curve is linear, showing that the material obeys Hooke’s law in this range; the material is also elastic in this range, and no permanent extensions remain after removal of the stresses.

The ratio of stress to strain for this linear region is usually about 200 GN/m2 for steels;this ratio is known as Young’s modulus and is denoted by E.

Similarly, the strain at the limit of proportionality is of the order 0.003, and is small compared with strains of the order 0.100 at fracture.

We note that Young’s modulus has the units of a stress; the value of E defines the constant in the linear relation between stress and strain in the elastic range of the material. We have, for the linear-elastic range.

If P is the total tensile load in a bar, A its cross-sectional area, and Lo its length, e is the extension of the length Lo, then

Thus the expansion is given by

If the material is stressed beyond the linear-elastic range, the limit of proportionality isexceeded, and the strains increase non-linearly with the stresses. Moreover, removal of the stress leaves the material with some permanent extension; this range is then both non-linear and inelastic. The maximum stress attained may be of the order of 1500 MN/m2, and the total extension, or elongation, at this stage may be of the order of 10%. When a material is stressed beyond the limit of proportionality and is then unloaded,permanent deformations of the material take place.

Unloading and reloading of a material in the inelastic range; the paths bc and cd are approximately parallel to the linear-elastic line oa.

Suppose the tensile test-specimen is stressed beyond the limit of proportionality, in figure, point a, to a point b on the stress-strain diagram. If the stress is now removed, the stress-strain relation follows the curve bc ; when the stress is completely removed there is a residual strain given by the intercept Oc on the ε-axis.

If the stress is applied again, the stress-strain relation follows the curve cd initially, and finally the curve df to the breaking point. Both the unloading curve bc and the reloading curve cd are approximately parallel to the elastic line Oa; they are curved slightly in opposite directions. The process of unloading and reloading, bcd, had little or no effect on the stress at the breaking point, the stress-strain curve being interrupted by only a small amount bd. The stress-strain curves of brittle materials for tension and compression are usually similar in form, although the stresses at the limit of proportionality and at fracture may be very different for the two loading conditions.

Typical tensile and compressive stress-strain curves for concrete are shown in Figure; the maximum stress attainable in tension is only about one-tenth of that in compression, although the slopes of the stress-strain curves in the region of zero stress are nearly equal.

Ductile materials

A brittle material is one showing relatively little elongation at fracture in the tensile test; but some materials, such as mild steel, copper, and synthetic polymers, may be stretched appreciably before breaking. These latter materials are ductile in character.

If tensile and compressive tests are made on a mild steel, the resulting stress-strain curves are different in form from those of a brittle material, such as a high-strength steel.

If a tensile test specimen of mild steel is loaded axially, the stress-strain curve is linear and elastic up to a point a, refer figure 1. (The small strain region of Figure 1, is reproduced to a larger scale in Figure 2).

The ratio of stress to strain, or Young’s modulus, for the linear portion Oa is usually about 200 GN/m2, ie, 200 x109 N/m2.

The tensile stress at the point a is of order 300 MN/m2, i.e. 300 x 106 N/m2.

The stress for a tensile specimen attains a maximum value at d if the stress is evaluated on the basis of the original cross-sectional area of the bar; the stress corresponding tothe point d is known as the ultimate stress, (σult) of the material.

From d to f there is a reduction in the nominal stress until fracture occurs at f. The ultimate stress in tension is attained at a stage when necking begins; this is a reduction of area at a relatively weak cross-section of the test specimen.

Tensile stress-strain curve for an annealed mild steel, showing the drop in stress at yielding from the upper yield point a to the lower yield point b.

Upper and lower yield points of a mild steel.

If the test specimen is strained beyond the point a, figures 1 and 2, the stress must be reduced almost immediately to maintain equilibrium.

The reduction of stress, ab, takes place rapidly, and the form of the curve ab is difficult to define precisely.

Continued straining proceeds at a roughly constant stress along bc.

In the range of strains from a to c the material is said to yield;

Please note: a is the upper yield point, and b the lower yield point.

Yielding at constant stress along bc proceeds usually to a strain about 40 times greater than that at a.

Beyond the point c the material strain-hardens, and stress again increases with strain where the slope from c to d is about 1/50th that from 0 to a.

Compressive tests of mild steel give stress-strain curves similar to those for tension. If we consider tensile stresses and strains as positive, and compressive stresses and strains as negative, we can plot the tensile and compressive stress-strain curves on the same diagram.

In determining the stress-strain curves experimentally, it is important to ensure that the bar is loaded axially;

The figure shows the stress-strain curves for an annealed mild steel. In the annealed condition the yield stresses in tension andCompression are approximately equal.

The lower yield point stress is taken usually as a more realistic definition of yielding of the material. Some ductile materials show no clearly defined upper yield stress; for these materials the limit of proportionality may be lower than the stress for continuous yielding.

The term yield stress refers to the stress for continuous yielding of a material; this implies the lower yield stress for a material in which an upper yield point exists; the yield stress is denoted by σγ.

Tensile failures in steel specimens showing necking in mild steel, (i) and (iii), and brittle fracture in high-strength steel, (ii).

Definition of Stress The concept of stress originated from the study of strength and failure of solids. The stress field is the distribution of internal "tractions" that balance a given set of external tractions and body forces. First, we look at the external traction T that represents the force per unit area acting at a given location on the body's surface. Traction T is a bound vector, which means T cannot slide along its line of action or translate to another location and keep the same meaning.

In other words, a traction vector cannot be fully described unless both the force and the surface where the force acts on has been specified. Given both ΔF and Δs, the traction T can be defined as

The internal traction within a solid, or stress, can be defined in a similar manner. Suppose an arbitrary slice is made across the solid shown in the above figure, leading to the free body diagram shown at right. Surface tractions would appear on the exposed surface, similar in form to the external tractions applied to the body's exterior surface. The stress at point P can be defined using the same equation as was used for T. Stress therefore can be interpreted as internal tractions that act on a defined internal datum plane. One cannot measure the stress without first specifying the datum plane.

Surface tractions, or stresses acting on an internal datum plane, are typically decomposed into three mutually orthogonal components.

One component is normal to the surface and represents direct stress. The other two components are tangential to the surface and represent shear stresses.

What is the distinction between normal and tangential tractions, or equivalently, direct and shear stresses?

The Stress Tensor (or Stress Matrix)

Direct stresses tend to change the volume of the material (e.g. hydrostatic pressure) and are resisted by the body's bulk modulus (which depends on the Young's modulus and Poisson ratio). Shear stresses tend to deform the material without changing its volume, and are resisted by the body's shear modulus.

For example, the stress state at point P can be represented by an infinitesimal cube with three stress components on each of its six sides (one direct and two shear components). Since each point in the body is under static equilibrium (no net force in the absence of any body forces), only nine stress components from three planes are needed to describe the stress state at a point P. These nine components can be Organised into the matrix:

where shear stresses across the diagonal are identical (i.e. σxy = σyx, σyz = σzy, and σzx = σxz) as a result of static equilibrium (no net moment). This grouping of the nine stress components is known as the stress tensor (or stress matrix). The subscript notation used for the nine stress components have the

following meaning:     

Note: The stress state is a second order tensor since it is a quantity associated with two directions. As a result, stress components have 2 subscripts.

A surface traction is a first order tensor (i.e. vector) since it a quantity associated with only one direction. Vector components therefore require only 1 subscript.

Mass would be an example of a zero-order tensor (i.e. scalars), which have no relationships with directions (and no subscripts).

Consider the static equilibrium of a solid subjected to the body force vector field b. Applying Newton's first law of motion results in the following set of differential equations which govern the stress distribution within the solid,

In the case of two dimensional stress, the above equations reduce to,

Consider a rod with initial length L which is stretched to a length L'. The strain measure ε , a dimensionless ratio, is defined as the ratio of elongation with respect to the original length,

Global 1D Strain

Consider an arbitrary point in the bar P, which has a position vector x, and its infinitesimal neighbour dx. Point P shifts to P', which has a position vector x', after the stretch. In the meantime, the small "step" dx is stretched to dx'.

The strain at point p can be defined

Since the displacement , the strain can hence be rewritten as,

Infinitesimal 1D Strain

The above strain measure is defined in a global sense. The strain at each point may vary dramatically if the bar's elastic modulus or cross-sectional area changes.

There are a total of 6 strain measures which can be organised into a matrix.

3D Strain Matrix

Engineering Shear Strain

Focus on the strain εxy for a moment. The expression inside the parentheses can be rewritten as, where called the

engineering shear strain, γxy is a total measure of shear strain in the x-y plane.

In contrast, the shear strain εxy is the average of the shear strain on the x face along the y direction, and on the y face along the x direction. Engineering shear strain is commonly used in engineering reference books. However, please beware of the difference between shear strain and engineering shear strain, so as to avoid errors in mathematical manipulations.

In the strain-displacement relationships, there are six strain measures but only three independent displacements. That is, there are 6 unknowns for only 3 independent variables. As a result there exist 3 constraint, or compatibility, equations. These compatibility conditions for infinitesimal strain referred to rectangular Cartesian coordinates are,

In two dimensional problems (e.g. plane strain), all z terms are set to zero. The compatibility equations reduce to,

Note that some references use engineering shear strain (         ) when referencing compatibility equations.

Compatibility Conditions

General Definition of 3D Strain

As in the one dimensional strain derivation, suppose that point P in a body shifts to point P’ after deformation. The infinitesimal strain-displacement relationships can be summarised as,

where u is the displacement vector, x is coordinate, and the two indices i and j can range over the three coordinates {1, 2, 3} in three dimensional space. Expanding the above equation for each coordinate direction gives,

where u, v, and w are the displacements in the x, y, and z directions respectively (i.e. they are the components of u).

Bulk Elastic PropertiesThe bulk elastic properties of a material determine how much it will compress under a given amount of external pressure. The ratio of the change in pressure to the fractional volume compression is called the bulk modulus of the material.

A representative value for the bulk modulus for steel is and that for water is

The reciprocal of the bulk modulus is called the compressibility of the substance. The amount of compression of solids and liquids is seen to be very small.

The bulk modulus of a solid influences the speed of sound and other mechanical waves in the material. It also is a factor in the amount of energy stored in solid material.

A common statement is that water is an incompressible fluid. This is not strictly true, as indicated by its finite bulk modulus, but the amount of compression is very small.

At the bottom of the Pacific Ocean at a depth of about 4000 meters, the pressure is about 4 x 107 N/m2. Even under this enormous pressure, the fractional volume compression is only about 1.8% and that for steel would be only about 0.025%.

So it is fair to say that water is nearly incompressible.

Young's ModulusIn solid mechanics, Young's modulus (E) is a measure of the stiffness of an isotropic elastic material. It is also known as the Young modulus, modulus of elasticity, elastic modulus (though Young's modulus is actually one of several elastic

moduli such as the bulk modulus and the shear modulus) or tensile modulus. It is defined as the ratio of the uni-axial stress over the uni-axial strain in the range of stress in which Hooke's Law holds.

Young's modulus is named after Thomas Young, the 18th century British scientist. However, the concept was developed in 1727 by Leonhard Euler.

For the description of the elastic properties of linear objects like wires, rods, columns which are either stretched or compressed, a convenient parameter is the ratio of the stress to the strain, or Young's modulus of the material. Young's modulus can be used to predict the elongation or compression of an object as long as the stress is less than the yield strength of the material. This can be experimentally determined from the slope of a stress-strain curve created during tensile tests conducted on a sample of the material.

Material Density(kg/m3)

Young's Modulus109 N/m2

Ultimate Strength

Su

106 N/m2

Yield Strength

Sy

106 N/m2

Steela 7860 200 400 250

Aluminum 2710 70 110 95a Structural steel

shear modulus or modulus of rigidity, denoted by G, or sometimes S or μ, is defined as the ratio of shear stress to the shear strain:

Where = shear stress;

F is the force which acts and A is the area on which the force acts = shear strain; Δx is the transverse displacement , I is the initial length

Shear modulus is usually measured in GPa (giga pascal)

The shear modulus is concerned with the deformation of a solid when it experiences a force parallel to one of its surfaces while its opposite face experiences an opposing force (such as friction). In the case of an object that's shaped like a rectangular prism, it will deform into a parallelepiped.

Anisotropic materials such as wood and paper exhibit differing material response to stress or strain when tested in different directions. In this case, when the deformation is small enough so that the deformation is linear, the elastic moduli, including the shear modulus, will then be a tensor, rather than a single scalar value. In homogeneous and isotropic solids, there are two kinds of waves, pressure waves and shear waves. The velocity of a shear wave, (vs) is controlled by the shear modulus,

where

G is the shear modulus ρ is the solid's density.

shear modulus describes the material's response to shearing strains.

Hooke's law

Hooke's law is named after the 17th century British physicist Robert Hooke. He first stated this law in 1676 as a Latin anagram, whose solution he published in 1678 as Ut tensio, sic vis, meaning: “As the extension, so the force”.

Objects that quickly regain their original shape after being deformed by a force, with the molecules or atoms of their material returning to the initial state of stable equilibrium, often obey Hooke's law. We may view a rod of any elastic material as a linear spring. The rod has length L and cross-sectional area A. Its extension (strain) is linearly proportional to its tensile stress σ by a constant factor, the inverse of its modulus of elasticity E, hence, or

Hooke's law only holds for some materials under certain loading conditions. Steel exhibits linear-elastic behavior in most engineering applications; Hooke's law is valid for it throughout its elastic range (i.e., for stresses below the yield strength). For some other materials, such as aluminium, Hooke's law is only valid for a portion of the elastic range. For these materials a proportional limit

stress is defined, below which the errors associated with the linear approximation are negligible. For systems that obey Hooke's law, the extension produced is directly proportional to the load: where:

is the distance that the spring has been stretched or compressed away from the equilibrium position, (meters), is the restoring force exerted by the material (Newtons), and is the force constant (or spring constant). The constant has units of force per unit

length (newtons per meter). When this holds, we say that the behavior is linear.negative sign on the right hand side of the equation because the restoring force always acts in the opposite direction of the x displacement.

Poisson's ratio ( ), named after Simeon Poisson, is the ratio of the relative contraction strain, or transverse strain (normal to the applied load), to the relative extension strain, or axial strain (in the direction of the applied load).When a sample of material is stretched in one direction, it tends to contract (or rarely, expand) in the other two directions.Conversely, when a sample of material is compressed in one direction, it tends to expand (or rarely, contract) in the other two directions. Poisson's ratio ( ) is a measure of this tendency.The Poisson's ratio of a stable material cannot be less than −1.0 nor greater than 0.5 due to the requirement that the elastic modulus,the shear modulus and bulk modulus have positive values. Assuming that the material is compressed along the axial direction:

where, is the resulting Poisson's ratio, is transverse strain (negative for axial tension, positive for axial compression) is axial strain (positive for axial tension, negative for axial compression).

On the molecular level, Poisson’s effect is caused by slight movements between molecules and the stretching of molecular bonds within the material lattice to accommodate the stress. When the bonds elongate in the stress direction, they shorten in the other directions. This behavior multiplied millions of times throughout the material lattice is what drives the phenomenon.

Most materials have between 0.0 and 0.5. Cork is close to 0.0, showing almost no Poisson contraction, most steels are around 0.3, and rubber is nearly incompressible and so has a Poisson ratio of nearly 0.5. A perfectly incompressible material deformed elastically at small strains would have a Poisson's ratio of exactly 0.5. Some materials, mostly polymer foams, have a negative Poisson's ratio; if these auxetic materials are stretched in one direction, they become thicker in perpendicular directions.

Generalised Hooke's law : For an isotropic material, the deformation of a material in the direction of one axis will produce a deformation of the material along the other axes in three dimensions. Thus it is possible to generalise Hooke's Law into three dimensions:

Where , and are strain in the direction of x, y and z axis σx , σy and σz are stress in the direction of x, y and z axis E is Young's modulus (the same in all directions: x, y and z for isotropic materials) is Poisson's ratio (the same in all directions: x, y and z for isotropic materials)

Volumetric change: The relative change of volume ΔV/V due to the stretch of the material can be calculated using a simplified formula (only for small deformations):Where V is material volume

ΔV is material volume change L is original length, before stretch ΔL is the change of length: ΔL = Lnew − Lold

If a rod with diameter (or width, or thickness) d and length L is subject to tension so that its length will change by ΔL then its diameter d will change by (the value is negative, because the diameter will decrease with increasing length):

The above formula is true only in the case of small deformations; if deformations are large then the following (more precise) formula can be used

For Orthotropic material, such as wood in which Poisson's ratio is different in each direction (x, y and z axis) the relation between Young's modulus and Poisson's ratio is described as

Ei is a Young's modulus along axis i and νjk is a Poisson's ratio in plane jk

A 2.5 cm diameter steel bolt passes through a steel tube 5 cm internal diameter, 6.25cm external diameter, and 40 cm long. The bolt is then tightened up onto the tube through rigid end blocks until the tensile force in the bolts is 40 kN. The distance between the head of the bolt and the nut is 50 cm. If an external force of 30 kN is applied to the end blocks, tending to pull them apart, estimate the resulting tensile force in the bolt.

A cylindrical block is 30 cm long and has a circular cross-section 10 cm in diameter. It carries a total compressive load of 70 kN, and under this load it contracts by 0.02 cm. Estimate the average compressive stress over a normal cross-section and the compressive strain.

The pressure on the back of the piston acts on a net area

The load on the piston is then

Area of the piston rod is

The average tensile stress in the rod is then

From equation , the elongation of a length of 1 m is,

The cross-sectional area is

The tensile stress is then

The measured tensile strain is

Then Young’s modulus is defined by

1.4

a)

Composite bars in tension or compression

A composite bar is one made of two materials, such as steel rods embedded in concrete. Theconstruction of the bar is such that constituent components extend or contract equally under load. To illustrate the behaviour of such bars consider a rod made of two materials, 1 and 2, Figure ;A1, A2, are the cross-sectional areas of the bars, and E1, E2 are the values of Young's modulus. We imagine the bars to be rigidly connected together at the ends; then for compatibility, thelongitudinal strains to be the same when the composite bar is stretched we must have

Statically Determinate Structures:

Figure shows a rigid beam BD supported by two vertical wires BF and DG; the beam carries a force of 4W at C. We suppose the wires extend by negligibly small amounts, so that the geometrical configuration of the structure is practically unaffected; then for equilibrium the forces in the wires must be 3 W in BF and W in DG. As the forces in the wires are known, it is a simple matter to calculate their extensions and hence to determine the displacement of any point of the beam.

The calculation of the forces in the wires and structure of Figure is said to be statically determinate.

If, however, the rigid beam be supported by three wires, with an additional wire, say, between H and J, then the forces in the three wires cannot be solved by considering statical equilibrium alone; such a structure is statically indeterminate.

If the frame has just sufficient bars or rods to prevent collapse without the application of external forces, it is said to be simply-stiff, when there are more bars or rods than this, the frame is said to be redundant.

If m be the total number of members and j is the total number of joints, we must have m = 2j – 3, if the frame is to be simply-stiff or statically determinate.

A type of stress analysis problem in which internal stresses are not calculable on considering statical equilibrium alone; such problems are statically indeterminate. Consider the rigid beam BD of Figure which is supported on three wires; suppose the tensions in the wires are T1 , T2 and T3. Then by resolving forces vertically, we have T 1 + T 2 + T 3 = 4W …….. (1)and by taking moments about the point C, T 1 – T 2 – 3 T 3 = 0 …….. (2)From these equilibrium equations alone we cannot derive the values of the three tensile forces; a third equation is found by discussing the extensions of the wires or considering compatibility. If the wires extend by amounts e 1 , e 2 , e 3 we must have from Figure (ii) thate 1 + e 3 = 2e 2 ……. (3) because the beam BD is rigid.

Suppose the wires are all of the same material and cross-sectional area, and that they remain elastic. Then we may write , , .Where λ is a constant common to the three wires.

Then equation (3) becomes T 1 + T 3 = 2 T 2 …….. (4)The three equations (1), (2) and (4) then give

Equation (3) is a condition which the extensions of the wires must satisfy. It is called a strain compatibility condition.

Statically indeterminate problems are soluble if strain compatibilities are considered as well as statical equilibrium.

Statically Indeterminate Structures:

Temperature stresses

Temperature stresses in composite bars:Suppose the bar and tube are quite free of each other; if Lo is the original length of each bar, the extensions due to a temperature increase

Difference in lengths of the two members isRefer Figure (iii),this is now eliminated by compressing the inner bar with a force P, and pulling the outer tube with an equal force P.

Direct stresses tend to change the volume of the material (e.g. hydrostatic pressure) and are resisted by the body's bulk modulus (which depends on the Young's modulus and Poisson ratio).

Shear stresses tend to deform the material without changing its volume, and are resisted by the body's shear modulus. Defining a set of internal datum planes aligned with a Cartesian coordinate system allows the stress state at an internal point P to be described relative to x, y, and z coordinate directions.

The subscript notation used for the nine stress components have the following meaning:

For example, the stress state at point P can be represented by an infinitesimal cube with three stress components on each of its six sides (one direct and two shear components). Since each point in the body is under static equilibrium (no net force in the absence of any body forces), only nine stress components from three planes are needed to describe the stress state at a point P. These nine components can be organised into the matrix:

where shear stresses across the diagonal are identical (i.e. σxy = σyx , σyz = σzy and σzx = σxz) as a result of static equilibrium (no net moment).

This grouping of the nine stress components is known as the stress tensor (or stress matrix).

Consider the static equilibrium of a solid subjected to the body force vector field b. Applying Newton's first law of motion results in the following set of differential equations which govern the stress distribution within the solid,

In the case of two dimensional stress, the above equations reduce to,

Equations of Equilibrium

Note: The stress state is a second order tensor since it is a quantity associated with two directions. As a result, stress components have 2 subscripts.

A surface traction is a first order tensor (i.e. vector) since it is a quantity associated with only one direction. Vector components therefore require only 1 subscript.

Mass would be an example of a zero-order tensor (i.e. scalars), which have no relationships with directions (and no subscripts).

PLANE STATE of STRESS

Common engineering problems involving stresses in a thin plate or on the free surface of a structural element, such as the surfaces of thin-walled pressure vessels under external or internal pressure, the free surfaces of shafts in torsion and beams under transverse load, have one principal stress that is much smaller than the other two.

By assuming that this small principal stress is zero, the three-dimensional stress state can be reduced to two dimensions. Since the remaining two principal stresses lie in a plane, these simplified 2D problems are called plane stress problems.

Assume that the negligible principal stress is oriented in the z-direction. To reduce the 3D stress matrix to the 2D plane stress matrix, remove all components with z subscripts to get,

Where xy = yx for static equilibrium. The sign convention for positive stress components in plane stress is illustrated in the above figure on the 2D element.

The normal stresses (σx' and σy') and the shear stress (τx'y') vary smoothly with respect to the rotation angle θ, in accordance with the coordinate transformation equations. There exist a couple of particular angles where the stresses take on special values. First, there exists an angle θp where the shear stress τx'y' becomes zero. That angle is found by setting τx'y' to zero in the above shear transformation equation and solving for θ (set equal to θp). The result is,

The angle θp defines the principal directions where the only stresses are normal stresses. These stresses are called principal stresses and are found from the original stresses (expressed in the x,y,z directions) via,

Principal Directions, Principal Stress

Coordinate Transformations

The coordinate directions chosen to analyse a structure are usually based on the shape of the structure. As a result, the direct and shear stress components are associated with these directions. For example, to analyse a bar one almost always directs one of the coordinate directions along the bar's axis.

Stresses in directions that do not line up with the original coordinate set are also important. For example, the failure plane of a brittle shaft under torsion is often at a 45°

angle with respect to the shaft's axis. Stress transformation formulas are required to analyse these stresses.

The transformation of stresses with respect to the {x,y,z} coordinates to the stresses with respect to {x',y',z'} is performed via the equations, where q is the rotation angle between the two coordinate sets (positive in the counterclockwise direction).

The transformation to the principal directions can be illustrated as:

Maximum Shear Stress DirectionAnother important angle θs is where the maximum shear stress occurs. This is found by finding the maximum of the shear stress transformation equation, and solving for θ. The result is,

The maximum shear stress is equal to one-half the difference between the two principal stresses,

The transformation to the maximum shear stress direction can be illustrated as:

Mohr's Circle

Introduced by Otto Mohr in 1882, Mohr's Circle illustrates principal stresses and stress transformations via a graphical format,

The two principal stresses are shown in red, and the maximum shear stress is shown in orange. Recall that the normal stresses equal the principal stresses when the stress element is aligned with the principal directions, and the shear stress equals the maximum shear stress when the stress element is rotated 45° away from the principal directions.

As the stress element is rotated away from the principal (or maximum shear) directions, the normal and shear stress components will always lie on Mohr's Circle.

Derivation of Mohr's Circle

To establish Mohr's Circle, we first recall the stress transformation formulas for plane stress at a given location,

Using a basic trigonometric relation (cos2 2θ + sin2 2θ = 1) to combine the two above equations we have,

This is the equation of a circle, plotted on a graph where the abscissa is the normal stress and the ordinate is the shear stress. This is easier to see if we interpret σx and σy as being the two principal stresses, and τxy as being the maximum shear stress. Then we can define the average stress, σavg and a "radius" R (which is just equal to the maximum shear stress),      

The circle equation above now takes on a more familiar form,

The circle is centered at the average stress value, and has a radius R equal to the maximum shear stress, as shown in the figure below,

Principal Stresses from Mohr's Circle

A chief benefit of Mohr's circle is that the principal stresses σ1 and σ 2 and the maximum shear stress τmax are obtained immediately after drawing the circle,

where,

Principal Directions from Mohr's CircleMohr's Circle can be used to find the directions of the principal axes. To show this, first suppose that the normal and shear stresses, σx , σy and τxy are obtained at a given point O in the body. They are expressed relative to the coordinates XY, as shown in the stress element at right below.

The Mohr's Circle for this general stress state is shown at left above. Note that it's centered at σavg and has a radius R, and that the two points {σx , txy} and {σy , -txy} lie on opposites sides of the circle. The line connecting σx and σy will be defined as Lxy.

The angle between the current axes (X and Y) and the principal axes is defined as θp and is equal to one half the angle between the line Lxy and the σ -axis as shown in the schematic below,

Rotation Angle on Mohr's CircleNote that the coordinate rotation angle θp is defined positive when starting at the XY coordinates and proceeding to the XpYp coordinates. In contrast, on the Mohr's Circle θp is defined positive starting on the principal stress line (i.e. the σ-axis) and proceeding to the XY stress line (i.e. line Lxy). The angle θp has the opposite sense between the two figures, because on one it starts on the XY coordinates, and on the other it starts on the principal coordinates.

The principal axes are counterclockwise to the current axes (because τxy > 0) and no more than 45º away (because σx > σy).

Case 1:   τxy > 0 and σx > σy Case 2:   τxy < 0 and σx > σy

The principal axes are clockwise to the current axes (because τxy < 0) and no more than 45º away (because σx > σy)

A set of six Mohr's Circles representing most stress state possibilities are presented

Case 3:   τxy > 0 and σx < σy

The principal axes are counterclockwise to the current axes (because τxy > 0) and between 45º and 90º away. (because σx < σy)

Case 4:   τxy < 0 and σx < σy

The principal axes are clockwise to the current axes (because τxy < 0) and between 45º and 90º away. (because σx < σy)

Case 5:   τxy = 0 and σx > σy

The principal axes are aligned with the current axes. (because σx > σy and τxy = 0).

Case 6:   τxy = 0 and σx < σy

The principal axes are exactly 90° from the current axes (because σx < σy and τxy = 0).

Stress Transform by Mohr's Circle

Mohr's Circle can be used to transform stresses from one coordinate set to another, similar to that described on the plane stress . Suppose that the normal and shear stresses, σx , σy and τxy, are obtained at a point O in the body, expressed with respect to the coordinates XY. We wish to find the stresses expressed in the new coordinate set X'Y', rotated an angle q from XY, as shown below:

Plane Stress

To do this we proceed as follows:

Step 1: Draw Mohr's circle for the given stress state (σ x , σy and τxy).

Step 2: Draw the line Lxy across the circle from (σx , τxy) to (σy , - τxy).

Step 3: Rotate the line Lxy by 2*θ (twice as much as the angle between XY and X'Y') and in the opposite direction of θ.

Step 4: The stresses in the new coordinates (σx' , σy' and τx'y‘ ) are then read off the circle.

Mohr's Circle For PLANE STRAIN

Strains at a point in the body can be illustrated by Mohr's Circle. The idea and procedures are exactly the same as for Mohr's Circle for plane stress.

The two principal strains are shown in red, and the maximum shear strain is shown in orange. Recall that the normal strains are equal to the principal strains when the element is aligned with the principal directions, and the shear strain is equal to the maximum shear strain when the element is rotated 45° away from the principal directions. As the element is rotated away from the principal (or maximum strain) directions, the normal and shear strain components will always lie on Mohr's Circle.

To establish the Mohr's circle, we first recall the strain transformation formulas for plane strain,

Derivation of Mohr's Circle

Using a basic trigonometric relation (cos22θ+ sin22θ = 1) to combine the above two formulas we have,

This equation is an equation for a circle. To make this more apparent, we can rewrite it as,

where,

The circle is centered at the average strain value εAvg and has a radius R equal to the maximum shear strain, as shown in the figure.

Principal Strains from Mohr's CircleA chief benefit of Mohr's circle is that the principal strains ε1 and ε2 and the maximum shear strain εxyMax are obtained immediately after drawing the circle,

where,

Mohr's Circle can be used to find the directions of the principal axes. To show this, first suppose that the normal and shear strains, εx , εy and εxy are obtained at a given point O in the body. They are expressed relative to the coordinates XY, as shown in the strain element at right below.

Principal Directions from Mohr's Circle

The Mohr's Circle for this general strain state is shown at left above. Note that it's centered at εAvg and has a radius R, and that the two points (εx , εxy) and (εy , - εxy) lie on opposites sides of the circle. The line connecting εx and εy will be defined as Lxy.

The angle between the current axes (X and Y) and the principal axes is defined as θp and is equal to one half the angle between the line Lxy and the ε-axis as shown

Note that the coordinate rotation angle θp is defined positive when starting at the XY coordinates and proceeding to the XpYp coordinates. In contrast, on the Mohr's Circle θp is defined positive starting on the principal strain line (i.e. the ε-axis) and proceeding to the XY strain line (i.e. line Lxy). The angle θp has the opposite sense between the two figures, because on one it starts on the XY coordinates, and on the other it starts on the principal coordinates.

Rotation Angle on Mohr's Circle

Strain Transform by Mohr's Circle

Mohr's Circle can be used to transform strains from one coordinate set to another, similar that that described on the plane strain page. Suppose that the normal and shear strains, εx , εy and εxy are obtained at a point O in the body, expressed with respect to the coordinates XY. We wish to find the strains expressed in the new coordinate set X'Y', rotated an angle θ from XY, as shown below:

To do this we proceed as follows:

Step 1: Draw Mohr's circle for the given strain state (εx , εy and εxy shown below). Step 2: Draw the line Lxy across the circle from (εx, εxy) to (εy, - εxy).

Step 3: Rotate the line Lxy by 2*θ (twice as much as the angle between XY and X'Y') and in the opposite direction of θ.

Step 4: The strains in the new coordinates (εx' , εy' and εx'y‘ ) are then read.

Case 1:   εxy > 0 and εx > εy

The principal axes are counterclockwise to the current axes (because εxy > 0)

and no more than 45º away

(because εx > εy).

Case 2:   εxy < 0 and εx > εy

The principal axes are clockwise to the current axes (because εxy < 0)

and no more than 45º away (because εx > εy).

Case 3:   εxy > 0 and εx < εy

The principal axes are counterclockwise to the current axes (because εxy > 0)

and between 45º and 90º away (because εx < εy).

Case 4:   εxy < 0 and εx < εy

The principal axes are clockwise to the current axes (because εxy < 0)

and between 45º and 90º away (because εx < εy).

Case 5:   εxy = 0 and εx > εy

The principal axes are aligned with

the current axes (because εx > εy and εxy = 0).

Case 6:   εxy = 0 and εx < εy

The principal axes are exactly 90°

from the current axes (because εx < εy and εxy = 0).

One-dimensional Hooke's Law : Robert Hooke, who in 1676 stated

Generalised Hooke's Law (Anisotropic Form)

Cauchy generalised Hooke's law to three dimensional elastic bodies and stated that the 6 components of stress are linearly related to the 6 components of strain. The stress-strain relationship written in matrix form, where the 6 components of stress and strain are organised into column vectors, is,

  ie.,   ε = S·σ

The power (sic.) of any springy body is in the same proportion with the extension.announced the birth of elasticity. Hooke's statement expressed mathematically , where F is the applied force (and not the power, as Hooke mistakenly suggested), u is the deformation of the elastic body subjected to the force F, and k is the spring constant (i.e. the ratio of previous two parameters).

or

 ie., s = C·e

where, C is the compliance matrix, S is the stiffness matrix and S = C-1.

Orthotropic Definition

Some engineering materials, including certain piezoelectric materials (e.g. Rochelle salt) and 2-ply fiber-reinforced composites, are orthotropic.

By definition, an orthotropic material has at least 2 orthogonal planes of symmetry, where material properties are independent of direction within each plane. Such materials require 9 independent variables (i.e. elastic constants) in their constitutive matrices.

In contrast, a material without any planes of symmetry is fully anisotropic and requires 21 elastic constants, whereas a material with an infinite number of symmetry planes (i.e. every plane is a plane of symmetry) is isotropic, and requires only 2 elastic constants.

Hooke's Law in Compliance Form

By convention, the 9 elastic constants in orthotropic constitutive equations are comprised of 3  Young's moduli Ex , Ey , Ez , the 3 Poisson's ratios yz ,

zx , xy and the 3 shear moduli Gyz , Gzx , Gxy

The shear modulus G is related to E and

A wire strain gage can effectively measure strain in only one direction. To determine the three independent components of plane strain, three linearly independent strain measures are needed, i.e., three strain gages positioned in a rosette-like layout.

Consider a strain rosette attached on the surface with an angle α from the x-axis. The rosette itself contains three strain gages with the internal angles β and γ , as illustrated. Suppose that the strain measured from thesethree strain gages are ea , eb and ec respectively. The following coordinate transformation equation is used to convert the longitudinal strain from each strain gage into strain expressed in the x-y coordinates,

Applying this equation to each of the three strain gages results in the following system of equations,

Strain Rosette for Strain Measurement

These equations are then used to solve for the three unknowns, ex, ey, and exy.

Note: 1. The above formulas use the strain measure εxy as opposed to the engineering shear strain γxy . .To use γxy, the above equations should be adjusted accordingly.2. The free surface on which the strain rosette is attached is actually in a state of plane stress, while the formulas used above are for plane strain. However, the normal direction of the free surface is indeed a principal axis for strain. Therefore, the strain transform in the free surface plane can be applied.

Special Cases of Strain Rosette Layouts

Case 1: 45º strain rosette aligned with the x-y axes, i.e., α = 0º, β= γ = 45º.

Case 2: 60º strain rosette, the middle of which is aligned with the y-axis, i.e., α = 30º, β = γ = 60º.

The strain gage is one of the most widely used strain measurement sensors. It is a resistive elastic unit whose change in resistance is a function of applied strain. where R is the resistance, ε is the strain, and S is the strain sensitivity factor of the gage material (also called as gage factor).

Among strain gages, an electric resistance wire strain gage has the advantages of lower cost and being an established product. Thus it is the most commonly used type of device.

Other types of strain gages are acoustic, capacitive, inductive, mechanical, optical, piezo-resistive, and semi-conductive.

A wire strain gage is made by a resistor, usually in metal foil form, bonded on an elastic backing.

Its principle is based on fact that the resistance of a wire increases with increasing strain and decreases with decreasing strain, as first reported by Lord Kelvin in 1856.

Consider a wire strain gage, as illustrated above. The wire is composed of a uniform conductor of electric resistivity r with length l and cross-section area A. Its resistance R is a function of the geometry given by

Electric Resistance Strain Gages

The resistance change rate is a combination effect of changes in length, cross-section area, and resistivity.

When the strain gage is attached and bonded well to the surface of an object, the two are considered to deform together.

The strain of the strain gage wire along the longitudinal direction is the same as the strain on the surface in the same direction.

However, its cross-sectional area will also change due to the Poisson's ratio. Suppose that the wire is cylindrical with initial radius r, the normal strain along the radial direction is

The change rate of cross-section area is twice as the radial strain, when the strain is small.

The resistance change rate becomes

For a given material, the sensitivity of resistance versus strain can be calibrated by the following equation.

When the sensitivity factor S is given, (usually provided by strain gage vendors) the average strain at the point of attachment of the strain gage can be obtained by measuring the change in electric resistance of the strain gage.

Since most metal materials have the Poisson's ratio around 0.25 to 0.35, the (1 + 2   ) term in the strain sensitivity factor S is expected to be 1.5 to 1.7.

However, the strain sensitivity factor S itself ranges from -12.1 in Nickel up to 6.1 in Platinum. This wide variation indicates that the change in electric resistivity r , the so called piezo-resistance effect, can be quite large in some materials.

UNIT 3

Distribution of Bending Stress and Shear Stress in a cross section of BeamsSFD and BMD for propped cantilever,

built-in beams, continuous beams.

Prof. Dr. R. Chandra Sekaran Ph.D (C.Engg)

Beams:

Consider first the simple case of a beam which is fixed rigidly at one end B and is quite free atits remote end D, Figure 1 ; such a beam is called a cantilever. Imagine that the cantilever is horizontal, with one end B embedded in a wall, and that a lateral force W is applied at the free end D. Suppose the cantilever is divided into two lengths by an imaginary section C; the lengths BC and CD must individually be in a state of statical equilibrium. If we neglect the mass of the cantilever itself, the loading actions over the section C of CD balance the actions of the force W at C. The length CD of the cantilever is in equilibrium if we apply an upwards vertical force F and an anti-clockwise couple M at C; F is equal in magnitude to W, and M is equal to W(L - z), where z is measured from B. The force F at C is called a shearing force, and the couple M is a bending moment. But at the imaginary section C of the cantilever, the actions F and M on CD are provided by the length BC of the cantilever. In fact, equal and opposite actions F and M are applied by CD to BC. For the length BC, the actions at C are a downwards shearing force F, and a clockwise couple M.

When the cantilever carries external loads which are not applied normally to the axis of the beam, Figure 2, axial forces are set up in the beam. If W is inclined at an angle θ to the axis of the beam the axial thrust in the beam at any section isThe bending moment and shearing force at a section a distance z from the built-in end are

Relation between the intensity of loading, SF and BM:Consider a straight beam under any system of lateral loads and external couples, Figure 3; an element length δz of the beam at a distance z from one end is acted upon by an external lateral load,and internal bending moments and shearing forces.

Suppose external lateral loads are distributed so that the intensity of loading on the elemental length δz is w.

Then the external vertical force on the element is w δz ; this is reacted by an internal bending moment M and shearing force F on one face of the element, and M + δM and F + δF onthe other face of the element. For vertical equilibrium of the element we have,

If δz is infinitesimally small,

Suppose this relation is integrated between the limits z1 and z2 then

If F1 and F2 are the shearing forces at z = z1 and z = z2 respectively, then

Furthermore, for rotational equilibrium of the elemental length δz,

neglecting higher orders Then, in the limit as δz approaches zero,

On integrating between the limits z = z1 and z2 we have Thus

The shearing force F at a section distance z from one end of the beam is

On substituting this value of F into equation,

From equations & we have that the bending moment M has a stationary value when the shearing force F is zero. Thus, we get

All the relations developed in this section are merely statements of statical equilibrium, and are therefore true independently of the state of the material of the beam.

Basic Calculus Concepts

This is a very basic review of introductory Calculus concepts. We first look at a quadratic function: y = 9 x2 - 50 x + 50, which is graphed in Diagram 2.

This function has a slope at every point. If we take the "derivative" of our quadratic function, we obtain a new function (y' = 18 x - 50), which is graphed in Diagram 1.

The 'derivative' function gives us the value of the slope of our quadratic function at every point. (Thus at x = 2, the slope of the quadratic function is 18 * 2 - 50 = -14)

If on the other hand we "integrate" our quadratic function, we obtain a new function (y* = 3 x3 - 25 x2 + 50 x), which is graphed in Diagram 3 directly below Diagram 1.

The 'integrated' function tells us the net area under the quadratic function curve (between the function curve and the x axis). It actually tells us the area between some beginning x-value and ending x-value (when we do what is called a definite integral).

For the integrated function above, the initial x value is zero and the ending x-value is what ever value we choose. That is for x = 2, the area under the quadratic function curve between zero and 2 is: A = 3 (2)3 -25 (2)2 + 50 (2) = 24.

Thus every y value on the curve in Diagram 3 is equal to the sum of the area under the curve in Diagram 2 up to that point.

We can also do an "indefinite" integral which results in 'integrated' function involving a constant - which is evaluated by applying a boundary condition.

Let us review some basic derivatives d/dx (any constant) = 0 , d/dx (x) = 1, d/dx (x2) = 2x , d/dx (x3) = 3x2 , d/dx (xn) = n xn-1

A second way to think of derivatives and integrals is as inverse functions of each other. That is, the integral asks the question - what function must we take the derivative of to obtain what is inside the integral sign.

Look at the indefinite integral . This integral asks the question - what function must we take the derivative of to obtain 'x' (what is inside the integral sign). Some basic integrals,

that is, the integral of a sum (or difference) is the sum (or difference) of the integrals.

, where A is any constant

Shearing Force The shearing force (SF) at any section of a beam represents the tendency for the portion of the beam on one side of the section to slide or shear laterally relative to the other portion.

The diagram shows a beam carrying loads of W1, W2 and W3 and is simply supported at two points where the reactions are  R1 and R2. Assume that the beam is divided into two parts by a section XX .

The resultant of the loads and reaction acting on the left of AA is F vertically upwards and since the whole beam is in equilibrium, the resultant force to the right of AA must be F downwards. F is called the Shearing Force at the section AA. It may be defined as follows:-The shearing force at any section of a beam is the algebraic sum of the lateral components of the forces acting on either side of the section.

Where forces are neither in the lateral or axial direction they must be resolved in the usual way and only the lateral components used to calculate the shear force.

In a similar manner if the Bending moments (BM) of the forces to the left of AA are clockwise then the bending moment of the forces to the right of AA must be anticlockwise.

Bending Moment at AA is defined as the algebraic sum of the moments about the section of all forces acting on either side of the section

Bending moments are considered positive when the moment on the left portion is clockwise and on the right anticlockwise.

This is referred to as a sagging bending moment as it tends to make the beam concave upwards at AA. A negative bending moment is termed hogging.

Bending Moments

Shear Force - Moment RelationshipConsider a short length of a beam under a distributed load separated by a distance δx.

The bending moment at section AD is M and the shear force is S.  The bending moment at BC = M + δM and The shear force is S + δS.

The equations for equilibrium in 2 dimensions results in the equations.. Forces. S - w.δx = S + δS

Therefore making δx infinitely small, dS /dx = - wMoments.. Taking moments about C M + Sδx - M - δM - w(δx)2 /2 = 0

Therefore making δx infinitely small then dM /dx = Sand putting the relationship into integral form.

The integral (Area) of the shear diagram between any limits results in the change of the shearing force between these limits and the integral of the Shear Force diagram between limits results in the change in bending moment...

To understand the shear forces and bending moments in a beam, we will look at a simple example. In Diagram 1, we have shown a simply supported 20 m beam with a load of 10,000 N . acting downward right at the center of the beam. Due to symmetry the two support forces will be equal, with a value of 5000 each. This is the static equilibrium condition for the whole beam.

Next examine a section of the beam.

We will cut the beam at an arbitrary distance (x) between 0 and 10 feet, and apply static equilibrium conditions to the left end section as shown in Diagram 2..

We can do this since as the entire beam is in static equilibrium, then a section of the beam must also be in equilibrium.

Note: M is a moment or torque - not a force. It does not appear in the sum of forces equation when we apply static equilibrium to the section

Equilibrium Conditions:Sum of Forces in y-direction: + 5000 - V = 0 , solving V = 5000.Sum of Toque about left end: -V * x + M = 0 , or - 5000 * x + M = 0 , M = 5000 x.

These are the equations for the shear force and bending moments for the section of the beam from 0 to 10 feet.

Notice that the internal shear force is a constant value of 5000 for the section, but the value of the internal torque (bending moment) varies from 0 at x = 0, to a value of 50,000 at x = 10.

We cut the beam at distance x from the left end, where x is now greater than 10 and less then 20 and then look at entire section to the left of where we cut the beam (See Diagram 3).

Where the beam was cut, we have an internal shear force and bending moment - which now become external. These are shown in Diagram 3 as V2 and M2.

(We add the '2', to indicate we are looking at section two of the beam.)

Equilibrium Conditions:Sum of Forces in y-direction: + 5000 - 10,000 - V = 0 , solving V2 = -5000Sum of Toque about left end: -10,000 * 10 -V * x + M = 0 , or -10,000 * 10 - (-5000 ) * x + M = 0 , then solving M2 = -[5000x - 100,000]

The two expressions above give the value of the internal shear force and bending moment in the beam, between the distances of the 10 and 20. A useful way to visualise this information is to make Shear Force and Bending Moment Diagrams - which are really the graphs of the shear force and bending moment expressions over the length of the beam. (See Diagram 4.)

SIGN CONVENTION

The signs associated with the shear force and bending moment are defined in a different manner than the signs associated with forces and moments in static equilibrium.

•The Shear Force is positive if it tends to rotate the beam section clockwise with respect to a point inside the beam section.

•The Bending Moment is positive if it tends to bend the beam section concave facing upward. (Or if it tends to put the top of the beam into compression and the bottom of the beam into tension.)

Diagram 1, Shear Force V and Bending Moment M acting in positive directions.

Diagram 2, V is not shown but the horizontal force Fx is shown.

A loaded, simply supported beam is shown. For this beam we would like to determine expressions for the internal shear forces and bending moments in each section of the beam, and to make shear force and bending moment diagrams for the beam.

Part A. We first find the support forces acting on the structure. We do this in the normal way, by applying static equilibrium conditions for the beam.

STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces, as well asany needed angles and dimensions.

STEP 2: Break any forces not already in x and y direction into their x and y components.

STEP 3: Apply the equilibrium conditions.Sum Fy = -4,000 - (1,000 )(8) - 6,000 + By + Dy = 0

Sum TB = (Dy)(8) - (6,000)(4) + (1,000)(8)(4) + (4,000)(8) = 0

Solving for the unknowns: By =23,000; Dy = -5,000

(The negative sign indicates that Dy acts the opposite

of the initial direction we chose.)

Part B: Now we will determine the Shear Force and Bending Moment expressions for each section of the loaded beam. For this process we will ‘cut’ the beam into sections, and then use the translational equilibrium condition for the beam section (Sum of Forces = zero) to determine the Shear Force expressions in each section. Determining the Bending Moment expression for each section of the beam may be done in two ways.

1) By applying the rotational equilibrium condition for the beam section (Sum of Torque = zero), and solving for the bending moment.

2) By Integration. The value of the bending moment in the beam may be found from . That is, the bending moment expression is the integral of the shear force expression for the beam section.

We begin by starting at the left end of the beam, and cutting the beam a distance "x" from the left end - where x is a distance greater than zero and less the position where the loading of the beam changes in some way. In this problem we see that from zero to eight feet there is a uniformly distributed load of 1000 lb./ft. However this ends at eight feet (the loading changes). Thus for section 1, we will cut the beam at distance x from the left end, where x is greater than zero and less then eight feet.

Section 1: Cut the beam at x, where 0 < x < 8 ft., and analyse left hand section.

1. Draw a FBD of the beam section shown and labeling all forces and toque acting - including the shear force and bending moment (which act as an external force and torque at the point where we cut the beam.) (See Diagram - Section 1) Notice we have drawn the shear force and bending moment in their positive directions according to the defined sign convention and have labeled them as V1 and M1, as this is section 1 of the beam.

2. We check that we have all forces in x & y components

3. Apply translational equilibrium conditions to determine the shear force expression.Sum Fx = 0 (no net external x- forces)Sum Fy = -4,000 - 1,000 *(x) - V1 = 0 ; and solving: V1 = [-4,000 - 1,000x]

This expression gives us the values of the internal shear force in the beam between 0 and 8. Notice as x nears zero, the shear force value in the beam goes to – 4000 , and as x approaches 8 the shear force value becomes -12,000 , and that is negative everywhere between 0 and 8

Referring to the free body diagram for beam section 1, Sum Torque left end = -1000 * (x) * (x/2) - V1 (x) + M1 = 0 Substitute the expression for V1 (V1 = [-4,000 - 1,000x]) Sum Torque left end = -1000 * (x) * (x/2) -[-4,000 - 1,000x] (x) + M1 = 0 ;and solving for M1 = [-500x2 - 4,000x].Obtain the expression for the bending moment by integration of the shear force expression.

- 1000(1/2 x2) - 4000 (x) + C1; so M1 = -500x2 - 4,000x + C1

To determine the correct value for C1 for our problem we must apply a boundary condition: That is, we must know the value of the bending moment at some point on our interval into, to find the constant. For simply supported beams (with no external torque applied to the beam) the value of the bending moment will be zero at the ends of the beam. So we have for our "boundary condition" that at x = 0, M1 = 0. for the bending moment (M1 = -500x2 - 4,000x + C1), that is:0 = -500(0)2 - 4,000(0) + C1, and solving: C1 = 0Therefore: M1 = [-500x2 - 4,000x]for 0 < x < 8 ft., is the expression for the bending

Now continue with the next section of the beam. Referring to the beam diagram, we see that at a location just greater than 8 ft., there is no loading, and that this continues until 12 ft. where there is a point load of 6,000 lb. So for our second section, we cut the beam at a location "x", where x is greater than 8 ft., and less than 12 ft - and then analyse the entire left hand section of the beam.

1. Draw a FBD of the beam section shown and labeling all forces and toque acting - including the shear force and bending moment (which act as an external force and torque at the point where we cut the beam.) (See Diagram - Section 2) Notice we have drawn the shear force and bending moment in their positive directions according to the defined sign convention discussed earlier, and have labeled them as V2 and M2, as this is section 2 of the beam.

2. We check that we have all forces in x & y components 3. Apply translational equilibrium conditions (forces only):Sum Fx = 0 (no net external x- forces)Sum Fy = -4,000 - 1,000 (8 ) + 23,000 - V2 = 0, Solving: V2 = 11,000.4. Determine the bending moment expression by applying rotational equilibrium conditions, or by integration.

Rotational Equilibrium:Sum of Toque left end = - (1000 * 8) * 4 + 23,000 * 8 - V2 * x + M2 = 0;

then we substitute the value for V2 (V2 = 11,000) from above and obtain:- (1000 * 8 ) * 4 + 23,000 * 8 - (11,000) * x + M2 = 0; and then solving for M2 we find: M2 = [11,000x - 152,000]

Similarly from integration of the shear force, we find:Integration:

We get our boundary condition from another characteristic of the bending moment expression - which is that the bending moment must be continuous. That is, the value of the bending moment at the end of the first beam section, and the value of the bending moment at the beginning of the second beam section must agree - they must be equal.

We determine the value of the bending moment from our M1 equation as x approaches 8 . (M1 = [-500 (8)2 - 4,000(8)] = -64,000)

Then our boundary condition to find C2 is: at x=8 , M=-64,000 . We apply our boundary condition to find C2.Apply BC: -64,000 = 11,000 (8) + C2, Solving: C2 = -152,000Therefore: M2 = [11,000x - 152,000] for 8 < x < 12

In like manner we proceed with section 3 of the beam, cutting the beam at a location greater than 12 and less 16 , and then analysing the entire section left of where we cut the beam.

, or M2 = 11,000 x + C2

Section 3: Cut the beam at x, where 12 < x < 16 . Analyse left hand section.1. FBD. (See Diagram Section 3)

2. All forces in x & y components 3. Apply translational equilibrium conditions (forces only): Sum Fx = 0 (no net external x- forces), Sum Fy = -4,000 - 1,000 (8 ) + 23,000 - 6,000 - V3 = 0, and Solving: V3 = 5,000.4. We may determine the bending moment expression by applying rotational equilibrium conditions, or by integration. Once more we will do it both ways for this section.

Sum of Toque left end = - (1000 * 8) * 4 + 23,000 * 8 - 6,000 * 12 -V3 * x + M3 = 0;

then we substitute the value for V3 (V3 = 5,000 ) from above and obtain:- (1000 * 8) * 4 + 23,000 * 8 -  6,000 * 12 - (5,000) * x + M3 = 0; and then solving for M3 we find: M3 = [5,000x - 80,000]. We will find the bending moment expression for this section using integration only. Integration , or M3 = 5,000x + C3. We obtain a boundary condition for section 3 by remembering that at a free end or simply supported (no external torque) end, the bending moment must go to zero, thus we have the boundary condition to find C3: at x = 16 , M = 0.Apply BC: 0 = 5,000(16) + C3, and Solving: C3 = -80,000. Therefore: M3 = [5,000x - 80,000] for 12 < x < 16. Additional, we have shown the shear force and bending moment diagrams for the entire beam - which is a visual representation of the internal shear forces and internal torque in the beam due to the loading.

Part C: Shear Force and Bending Moment Diagrams: Using the expressions found , we can draw the shear force and bending moment diagrams for our loaded beam.

V1 = -1,000x+4,000; V2 = 11,000; V3 = 5,000 M1 =-500x2+4,000x .; M2 = 11,000x -152,000 ; M3 = 5,000x-80,000

The sketches below show simply supported beams with one concentrated force.

The sketches below show Cantilever beams with three different load combinations

A cantilever 5 m long carries a uniformly distributed vertical load 480 N per metre from C from H, and a concentrated vertical load of 1000 N at its mid-length,D. Construct the SFD and BMD. The SF due to the distributed load increases

uniformly from zero at H to + 1920 N at C, and remains constant at +1920 N from C to B; this is shown by the lines (i). Due to the concentrated load at D, the shearing force is zero from H to D, and equal to +1000 N from D to B, as shown by lines (ii). Adding the two together we get the total shearing force shown by lines (iii).The bending moment due to the distributed load increases parabolically from zero at H to

at C. The total load on CH is 1920 N with its centre of gravity 3 m from B; thus the bending moment at B due to this load is

From C to B the bending moment increases uniformly, giving lines (i). The bending moment due to the concentrated load increases uniformly from zero at D to

at B, as shown by lines (ii). Combining (i) and (ii), the total bending moment is given by (iii).

The method used here for determining shearing-force and bending-moment diagrams is known as the principle of superposition.

Cantilever with non-uniformly distributed load:Where a cantilever carries a distributed lateral load of variable intensity, we can find the bending moments and shearing forces from equations (4) and (6). When the loading intensity w cannot be expressed as a simple analytic function of z, these equations can be integrated numerically.

A cantilever of length 10 m, built in at its left end, carries a distributed lateral load of varying intensity w N per metre length. Construct curves of SF and BM in the cantilever. If z is the distance from the free end of

cantilever, the shearing force at a distance z from the free end is

We find first the shearing force F by numerical integration of the w-curve. The greatest force occurs at the built-in end, and has the value

The bending moment at a section a distance z from the free end is

and is found therefore by numerical integration of the F-curve. The greatest bending moment occurs at the built-in end, and has the value

It should be noted that by inspection the bending moment and the shearing force at the free end of the cantilever are zero; these are boundary conditions.

m

L x

m

x

( )zM x

x’=L-x

( )yV xcut beam and find internal forces and moment

( )P x m

x

( ) 0yV x

( )zM x m

xL

zM

1. Cantilever beam with end moment, m

Sum forces in x and y direction to obtain P(x)=0 and

Sum moments at any point x to obtain ( )zM x m

L x

x

( )zM x

x’=L-x

( )yV xcut beam and find internal forces and moment

( )P x

x

F

F

F

2. Cantilever beam with end load, F.

0 ( ) ( )zM M x F L x

( ) ( )zM x F L x or

() ( )zMx FLx

L

FL()yVx F

zMyV

F

xL

x

The shear and moment diagrams:

Sum forces in y direction to find

( )yV x F

x

()zMx

x’=L-x

()yVxcut beam and find internal forces and moment

()Px

x

L

x

0p

0p 0p

3. Cantilever beam with uniform distributed load, po

Sum forces in y direction to obtain

0( ) ( )yV x p L x

L

zMyV

xL

x

0( ) ( )yV x p x L 0pL2

0( ) ( ) /2zM x p L x 2

0 /2pL

Sum moments at any point x to obtain

00 ( ) [ ( )]( ) / 2zM M x p L x L x

20( ) ( ) / 2zM x p L x

or

Shear and bending moment diagrams :

3a. Cantilever beam with uniform distributed load, po. Use the integration method to obtain V&M diagrams.

L

x

0p0

yy

Vp p

x

We know the shear at x=L is 0, so lets integratefrom x=L to any point x:

0x x

yL LdV p dx

( ) ( )y yV x V L0

0 0( )xL

p p x L 0( ) ( )yV x p L x

Next, integrate the shear equation to obtain the moment equation.

0( )zy

MV p L x

x

We know the moment is zero at the free end,x=L, so integrate from x=L to any point x:

0( )x x

zL LdM p L x dx ( ) ( )z zM x M L

0 20

2 2 20 0

( / 2)

( / 2 / 2) ( ) / 2

x

Lp Lx x

p Lx x L p x L

4. Simply Supported Beam with Point Load

From equilibrium: R1 + R2 = P

5. Simply Supported Beam with Distributed Normal Load

V(x)

M(x)

x'=L-x

x

R 1 R 2R 2

NO MOMENTS AT ENDS

WITH SIMPLE SUPPORTS

po

po po

L

From equilibrium: R1 = R2 = poL/2.

M(x)

x L

V(x)

x L

L/2 p o

M(x)=(p o /2) x (x-L)

dsd1

Small deflection and

slope dxd

dsd1

We assume that the predominate deflection is normal to the x axis. Predominate strain is in the axial (x) direction. Assume small strain and rotations of beam so that “axial strains vary linearly over cross-section” or “plane sections remain plane”.In terms of the general elasticity problem, the above can be stated as1. If the cross-sectional dimensions are small compared to the beam length, then applied transverse tractions (in y and z directions) will be small compared to the resultant internal stress in the x direction. --> small transverse loads produce large axial stresses. Thus we assume that the only major stress is Txx (all other stresses are zero or negligible). 0 0

[ ] 0 0 0

0 0 0

Txx

T

The stress tensor reduces to

.

Equilibrium Conservation of Linear Momentum reduces to

This implies that Txx=Txx(y,z) for any point x.0

Txxx

2. Stress-Strain. We assume a linear isotropic material so that stress and strain are linearly related to each other: Txx = E xx and yy = zz = - xx = - (/E) Txx.

If Txx=Txx(y,z), then xx=xx(y,z) also.3. Strain-Displacementu

xxx

x

4. Kinematic assumptions: All deformation is described by the displacement, ( )u xoy

and rotation of the centroidal axis. Vertical displacement is function of x only:

( , ) ( )

( , )

u x y u xy oy

duoyu x y yx dx

TYPES of BENDING:Pure or circular bending: This type of bending occurs when the only internal force in the bar is a constant bending moment, i.e., the axial (N) and shear (V ) forces and the torsional moment (T) are zero. The designation of circular bending is suggested by the fact that the deformed axis of theinitially prismatic bar is an arc of circumference, when the bending moment is constant (a constant moment implies a constant curvature). Non-uniform bending: This designation is normally used for a loading causing bending moment and shear force, that is, for a non-constant bending moment. The axial force and the torsional moment are zero.Composed bending: This designation is used for a loading causing bending moment and axial force. Constant (circular composed bending):

Variable (non-uniform composed bending):

Each of these three types of bending may be sub-divided into plane and inclined bending.

An elementary bending problem is that of a rectangular beam under end couples.

Consider a straight uniform beam having a rectangular cross-section of breadth b and depth h, figure a, the axes of symmetry of the cross-section are Cx , Cy. A long length of the beam is bent in the yz-plane, Figure b, in such a way that the longitudinal centroidal axis, Cz, remains unstretched and takes up a curve of uniform radius of curvature, R.

We consider an elemental length δz of the beam, remote from the ends; in the unloadedcondition, AB and FD are transverse sections at the ends of the elemental length, and these sections are initially parallel. In the bent form we assume that planes such as AB and FD remain flat planes; A’ B’ and F‘ D‘ in Figure b are therefore cross-sections of the bent beam, but are nolonger parallel to each other.

In the bent form, some of the longitudinal fibres, such as A ‘F’ ; are stretched, whereas others, such as B ‘D ’are compressed. The unstrained middle surface of the beam is known as the neutral axis.

Now consider an elemental fibre HJ of the beam, parallel to the longitudinal axis Cz, Figure c; this fibre is at a distance y from the neutral surface and on the tension side of the beam. The original length of the fibre HJ in the unstrained beam is δz; the strained length is

because the angle between A’ B’ and F‘ D‘ in Figure b and c is (δzR). Then during bending HJ stretches an amount

The longitudinal strain of the fibre HJ is therefore

Then the longitudinal strain at any fibre is proportional to the distance of that fibre from the neutral surface; over the compressed fibres, on the lower side of the beam, the strains are of course negative.If the material of the beam remains elastic during bending then the longitudinal stress on the fibre HJ is

….. 1

The distribution of longitudinal stresses over the cross-section takes the form shown in Figure d; because of the symmetrical distribution of these stresses about Cx, there is no resultant longitudinal thrust on the cross-section of the beam. The resultant hogging moment is

….. 2

On substituting for σ from equation (1), we have

….. 3

where Ix is the second moment of area of the cross-section about Cx. From equations (1) and (3), we have

….. 4

Equation (3) implies a linear relationship between M, the applied moment, and (l/R), the curvature of the beam. The constant EI, in this linear relationship is called the bending stiffness or flexural stiffness of the beam;

E I is also known as Flexural Rigidity.

Bending of a beam about a principal axisWe considered the bending of a straight beam of rectangular cross-section; this formof cross-section has two axes of symmetry. More generally we are concerned with sections having only one, or no, axis of symmetry.Consider a long straight uniform beam having any cross-sectional form; the axes Cx and Cy are principal axes of the cross-section. The principal axes of a cross-section are thosecentroid axes for which the product second moments of area are zero. In Figure a, C is thecentroidal of the cross-section; Cz is the longitudinal centroidal axis. When end couples M are applied to

the beam, we assume as before that transverse sections of the beam remain plane during bending. Suppose further that, if the beam is bent in the yz-plane only, there is a neutral axis C ' x ; Figure a, which is parallel to Cx and is unstrained; radius of curvature of this neutral surface is R, Figure b. As before, the strain in a longitudinal fibre at a distance y’

from C ‘x‘ isIf the material of the beam remains elastic during bending the longitudinal stress on this fibre is

If there is to be no resultant longitudinal thrust on the beam at any transverse section we must haveWhere b is the breadth of an elemental strip of the cross-section parallel to Cx, and the integration is performed over the whole cross-sectional area, A. But

This can be zero only if C ' x‘ is a centroidal axis; now, Cx is a principal axis, and is therefore a centroidal axis, so that C ‘x‘ and Cx are coincident, and the neutral axis is Cx in any cross-section of the beam. The total moment about Cx of the internal stresses is

….. 5

The stress in any fibre a distance y from Cx is….. 6

No moment about Cy is implied by this stress system, for

because Cx and Cy are principal axes for which , or the product second moment of area,is zero; δA is an element of area of the cross-section.Beams having two axes of symmetry in the cross-section:

I-section beam.

Solid circular cross-section

Hollow circular cross-section.

For bending about the axis Cx

Similarly for bending by a couple My about Cy,

From above equations (7) and (8) we see that the greatest bending stresses occur in the extreme longitudinal fibres of the

beams.

….. 7

….. 8

A steel bar of rectangular cross-section, 10 cm deep and 5 cm wide, is bent in the planes of the longer sides. Estimate the greatest allowable bending moment if the bending stresses are not to exceed 150 MN/m2 in tension and compression.

The bending moment is applied about Cx. The second moment of area about this axis is

The bending stress σ at a fibre a distance y from Cx is,

where M is the applied moment. If the greatest stresses are not to exceed 150 MN/m2, we must have

The greatest bending stresses occur in the extreme fibres where y = 5 cm. Then

The greatest allowable bending moment is therefore 12 500 Nm.

The second moment of area about Cy is

The greatest allowable bending moment about Cy is

which is only half that about Cx.

A light-alloy I-beam of 10 cm overall depth has flanges of overall breadth 5 cm and thickness 0.625 cm, the thickness of the web is 0.475 cm. If the bending stresses are not to exceed 150 MN/m2 in tension and compression estimate the greatest moments which may be applied about the principal axes of the cross section.

Consider, first, bending about Cx. the second moment of area about Cx is

The allowable moment M, is

Second, for bending about Cy.

The allowable moment about Cy is

A steel scaffold tube has an external diameter of 5 cm, and a thickness of 0.5 cm. Estimate the allowable bending moment on the tube if the bending stresses are limited to 100 MN/m2.

A steel scaffold tube has an external diameter of 5 cm, and a thickness of 0.5 cm. Estimate the allowable bending moment on the tube if the bending stresses are limited to 100 MN/m2.The second moment of area about a centroid axis Cx is

The allowable bending moment about Cx is

Beams having only one axis of symmetry:

(i) Channel section. (ii) Equal angle section. (iii) T-section.

A T-section of uniform thickness 1 cm has a flange breadth of 10 cm and an overall depth of 10 cm. Estimate the allowable bending moments about the principal axes if the bending stresses are limited to 150 MN/m2.

Suppose is the distance of the principal axis Cx from the remote edge of the flange. The totalarea of the section isOn taking first moments of areas about the upper edge of the flange,

The second moment of area of the flange about Cx is

The second moment of area of the web about Cx is

For bending about Cx, the greatest bending stress occurs at the toe of the web.The maximum allowable moment is

The second moment of area about Cy is

The T-section is symmetrical about Cy, and for bending about this axis equal tensile and compressive stresses are induced in the extreme fibres of the flange; The bending stress in the extreme fibres of the flange is only 60.4 MN/m2 at this bending moment.

The greatest allowable moment is

A light-alloy I-beam of 10 cm overall depth has flanges of overall breadth 5 cm and thickness 0.625 cm, the thickness of the web is 0.475 cm. The I-section is bent by couples of 2500 Nm about Cx and 500 Nm about Cy. Estimate the maximum bending stress in the cross-section, andfind the equation of the neutral axis of the beam.

For bending about Cx the bending stresses in the extreme fibres of the flanges are

For bending about Cy the bending stresses at the extreme ends of the flanges are

On superposing the stresses due to the separate moments, the stress at the comer a is tensile, and of magnitude

The total stress at the comer a 'is also 172.2 MN/m2, but compressive. The total stress at the comer b is compressive, and of magnitude

The total stress at the comer b 'is also 20.0 MN/m2, but tensile. The equation of the neutral axis is given by

and

The greatest bending stresses occur at points most remote from the neutral axis; these are thepoints a and a’ the greatest bending stresses are therefore + 172.2 MN/m'.

 The shear load and bending moment diagrams are constructed by integrating the distributed load to get the shear diagram(adding jumps at all point loads), andintegrating the shear diagram to get the bending moment (adding jumps at all point couples).

Drawing shear force and bending moment diagrams

Example of one shear load and bending moment diagram.

1.  First draw the free-body-diagram of the beam with sufficient room under it for the shear and moment diagrams (if needed, solve for support reactions first).

2.  Draw the shear diagram under the free-body-diagram. The distributed load is the slope of the shear diagram and each point load represents a jump in the shear diagram. Label all the loads on the shear diagram

3.      Draw the moment diagram below the shear diagram. The shear load is the slope of the moment and point moments result in jumps in the moment diagram. The area under the shear diagram equals the change in moment over the segment considered (up to any jumps due to point moments). Label the value of the moment at all important points on the moment diagram.

For the beam and loading shown, (a) draw the shear and bending moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.

(a) FBD Beam:

Along AB:

(b) From diagrams:

Built -in (Encastre) beams are fixed at both ends. Continuous beams, which are beams with more than two supports and covering more than one span, are not statically determinate using the static equilibrium laws.

Nomenclatureε = strain, σ = stress (N/m2), E = Young's Modulus = σ /e (N/m2), y = distance of surface from neutral surface (m), R = Radius of neutral axis (m).I = Moment of Inertia (m4 - more normally cm4) Z = section modulus = I/y max(m3 - more normally cm3)M = Moment (Nm), w = Distributed load on beam (kg/m) or (N/m as force units) W = total load on beam (kg ) or (N as force units), F= Concentrated force on beam (N)L = length of beam (m), x = distance along beam (m)

Built in beamsA built in beam is normally considered to be horizontal with both ends built-in at the same level and with zero slope at both ends. A loaded built in beam has a moment at both ends and normally the maximum moments at one or both of the two end joints.

A built in beam is generally much stronger than a simply supported beam of the same geometry. The bending moment reduces along the beam and changes sign at points of contra flexure between the supports and the load.

A typical built-in beam is shown below: It is not normally possible to determine the bending moments and the resulting stress using static equilibrium. Deflection calculations are often used to enable the moments to be determined. It has been proved that dM/dx = S and dS/dx = -w = d2M /dxWhere S = the shear force, M is the moment and w is the distributed load /unit length of beam. Therefore

Using the above equations the bending moment, shear force, deflection, slope can be determined at any point along the beam.

M = EI d 2y/dx 2 = w(- 6x2+6lx -l2)/12 at x = 0 & l then M = -wl2 /12 and at x = l/2 then M = wl2 /24

S = EI d 3y/dx 3 = w(l/2 - x) at x = 0 then S = w.l/2 at x = l then S = -w.l/2

Built-in beam carrying a single lateral load:

Consider a uniform beam, of flexural stiffness EI, and length L, which is built-in to end supports C and G, Fig. Suppose a concentrated vertical load W is applied to the beam at a distance a from C. If MC and MG are the restraining moments at the supports, then the vertical reaction is at C is

First boundary conditions are: when z = 0, v = dv/dz = 0As the Macaulay brackets will be negative when these boundary conditions are substituted, the terms on the right of equations can be ignored, hence A = B = OOther boundary conditions are: at z = L, v = dv/dz = 0which on substituting into equations give the following two simultaneous equations:

MC and MG are referred to as the fixed-end moments of the beam; MC is measured anticlockwise, and MG

clockwise.In the particular case when the load W is applied at the mid-length, a = L/2., and

The bending moment in the beam vary linearly from hogging moments of WL/8 at each end to a sagging moment of WL/8 at the mid-length, Figure above. There are points of contra-flexure, or zero bending moment, at distances L/4 from each end.

Fixed-end moments for other loading conditionsThe built-in beam of Fig. carries a uniformly distributed load of w per unit length over the section of the beam from z = a to z = b.

Consider the loading on an elemental length δz of the beam; the vertical load on the element is wδz, and this induces a retraining moment at C of amount

The total moment at C due to all loads is

which gives

MG may be found similarly. When the load covers the whole of the span, a = 0 and b = L, and equation reduces to

In this particular case, MC = MG; the variation of bending moment is parabolic, and of the form shown in Fig. the bending moment at the mid-length is so the fixed-end moments are also the greatest bending moments in the beam. The points of contra-flexure: from each end of the beam.

When a built-in beam carries a number of concentrated lateral loads, W1, W2, and W3, the fixed-end moments are found by adding together the fixed-end moments due to the loads acting separately. For example,

The case of a concentrated couple M0 applied a distance a from the end C, Fig, as a limiting case of two equal and opposite loads W, small distance δa apart. The fixed-endmoment at C is

Disadvantages of built-in beamsThe results we have obtained above show that a beam which has its ends firmly fixed in direction is both stronger and stiffer than the same beam with its ends simply-supported. On this account it might be supposed that beams would always have their ends built-in whenever possible; in practice it is not often done.

There are several objections to built-in beams: in the first place a small subsidence of one of the supports will tend to set up large stresses, and, in erection, the supports must be aligned with the utmost accuracy; changes of temperature also tend to set up large stresses. Again, in the case of live loads passing over bridges, the frequent fluctuations of bending moment, and vibrations, would quickly tend to make the degree of fixing at the ends extremely uncertain.

Most of these objections can be obviated by employing the double cantilever construction. As the bending moments at the ends of a built-in beam are of opposite sign to those in the central part of the beam, there must be points of inflexion, i.e. points where the bending moment is zero. At these points a hinged joint might be made in the beam, the axis of the hinge being parallel to the bending axis, because there is no bending moment to resist. If this is done at each point of inflexion, the beam will appear as a central girder freely supported by two end cantilevers; the bending moment curve and deflection curve will be exactly the same as if the beam were solid and built in. With this construction the beam is able to adjust itself to changes of temperature or subsistence of the supports.

A horizontal beam 6 m long is built-in at each end. The elastic section modulusis . Estimate the uniformly-distributed load over the whole span causing an elastic bending stress of 150 MN/m2.

The maximum bending moments occur at built-in ends, and have value

The bending stress is 150 MN/m2,

Continuous BeamsThis type of beam is normally considered using the Clapeyron's Theorem ( Three Moments theorem) . The three moments theorem identifies the relationship between the bending moments found at three consecutive supports in a continuous beam. This is achieved by evaluating the slope of the beam at the end where the two spans join. The slopes are expressed in terms of the three moments and the supported loads which are then equated and the resulting equations solved. This relationship for spans with supports at the same height and with spans of constant section results in the following expression.

If the beams has a different section for each span then the more general expression applies as shown below.

Theorem of Three MomentsA beam that is continuous across a support can have a bending moment there, unlike a beam that is merely supported. As an example, consider a prop placed at the centre of a beam that is simply supported at both ends. The amount of weight taken by the prop depends on how long it is, and the elasticity of the beam, and cannot be found by the application of the equations of statics. Such a system is called statically indeterminate.

A remarkable relation between the bending moments at three supports in a row, that depends only on the loads between the three supports, was discovered by Clapeyron, and is called the Theorem of Three Moments.

It is derived by integrating the equation for the elastic curve, EI(d4y/dx4) = -w(x), where y is the deflection, and w(x) is the load per unit length. There are enough conditions to eliminate the constants of integration, and the result is the desired theorem.

The functions f(x) and F(x) are found by integrating the load density with respect to x from x = 0 four times, setting the constants of integration to zero.

For a uniform load w, f(x) = wx4/24. The quantities y are the upward displacements of the supports 1 and 2.

Areas and x 1 value calculations.

This is a two span continuous beam with the ends simple supported, therefore with no moments at the end support points..

1) Bending Moments.

This is a three span continuous beam with the ends simple supported, therefore with no moments at the end support points. The values of A1,x1,A2, and x2 are calculated using the methods above.

UNIT 4Deflection of Beams

Differential Equation of elastic curve- deflection of beam by double integration method – area moment theorems

– application to simply supported beams, cantilever beams, overhanging beams. Deflection due to shear, SFD and BMD for propped cantilever, built-in beams,

continuous beams.

Prof. Dr. R. Chandra Sekaran Ph.D (C.Engg)

DEFLECTION of BEAMS:The loading actions at any section of a simply-supported beam or cantilever can be resolved into a bending moment and a shearing force. There are ways of estimating the stresses due to these bending moments and shearing forces. There is, however, another aspect of the problem of bending which remains to be treated, namely, the calculation of the stiffness of a beam. In most practical cases, it is necessary that a beam should be not only strong enough for its purpose, but also that it should have the requisite stiffness, that is, it should not deflect from its original position by more than a certain amount.

Again, there are certain types of beams, such as those carried by more than two supports and beams with their ends held in such a way that they must keep their original directions, for which we cannot calculate bending moments and shearing forces without studying the deformations of the axis of the beam; these problems are in fact statically indeterminate. In this module we consider methods of finding the deflected form of a beam under a givensystem of external loads and having known conditions of support.

Elastic bending of straight beams:From Beam Equation a straight beam of uniform cross-section, when subjected to end couples M applied about a principal axis, bends into a circular arc of radius R, given by where EI, the product of Young's modulus E and the second moment (1)of area I about the relevant principal axis, is the flexural stiffness of the beam; equation (1) holds only for elastic bending.

Where a beam is subjected to shearing forces, as well as bending moments, the axis of the beamis no longer bent to a circular arc. To deal with this type of problem, we assume that equation still defines the radius of curvature at any point of the beam where the bending moment isM. This implies that where the bending moment varies from one section of the beam to another, the radius of curvature also varies from section to section, in accordance with eqn.(1)

Differential Equation of elastic curve:In the unstrained condition of the beam, Cz is the longitudinal centroidal axis, Figure.1, and Cx, Cy are the principal axes in the cross-section. The co-ordinate axes Cx, Cy are so arranged that the y-axis is vertically downwards. (This is convenient as most practical loading conditions give rise to vertically downwards deflections.) Suppose bending moments are applied about axes parallel to Cx, so that bending is restricted to the yz-plane, because Cx and Cy are principal axes.

Fig.2 Displacements of the longitudinal axis of the beam.

Fig.1 Longitudinal and principal centroidal axes for a straight beam.

Consider a short length of the unstrained beam, corresponding with DF on the axis Cz, Figure 2.In the strained condition D and F are displaced to D' and F', respectively, which lies in the yz plane.Any point such as D on the axis Cz is displaced by an amount v parallel to Cy; it is also displaced a small, but negligible, amount parallel to Cz.The radius of curvature R at any section of the beam is then given by

We are concerned generally with only small deflections, in which v is small; this implies that(dv/dz) is small, and that (dv/dz)2 is negligible compared with unity.

(2)

(3)

Sign Convention: we have already adopted the convention that sagging bending moments are positive. When a length of the beam is subjected to sagging bending moments, as in Figure 3, the value of (dv/dz) along the length diminishes as z increases; hence a sagging moment implies that the curvature is negative. Then (5) where M is the sagging bending moment.

Where the beam is loaded on its axis of shear centres, so that no twisting occurs, M may be writtenin terms of shearing force F and intensity w of vertical loading at any section.

Hence from 1 and 3 (4)

On substituting for M (6)

This relation is true if EI varies from one section of a beamto another. Where El is constant along the length of a beam, (7)

Fig 3.Deflected form of a beam in pure bending.

As an example of the use of equation (4), consider the case of a uniform beam carrying couples M at its ends, Fig.3. The bending moment at any section is M, so the beam is under aconstant bending moment. Equation (5) givesOn integrating once, we have

(8) where A is a constant.

On integrating once more (9)

where B is another constant. If we measure v relative to a line CD joining the ends of the beam,v is zero at each end. Then v = 0, for z = 0 and z = L.On substituting these two conditions into equation (9), we have and

Therefore

At the mid-length, z = , and

(10)

(11) which is the greatest deflection.

(12)

It is important to appreciate that equation (3), expressing the radius of curvature R in terms of v, is only true if the displacement v is small.

Fig.4 Distortion of a beam in pure bending.

We can study more accurately the pure bending of a beam by considering it to be deformed into the arc of a circle, Figure 4; as the bending moment M is constant at all sections of the beam,the radius of curvature R is the same for all sections. If L is the length between the ends, and D is the mid-point,

Thus the central deflection v is

Suppose L/R is considerably less than unity; then

which can be written

But

Clearly, if is negligible compared with unity we have, approximately,

which agrees with equation (11). The more accurate equation (13) shows that, when is not negligible, the relationship between v and M is non-linear; for all practical purposes thisrefinement is unimportant, and we find simple linear relationships of the type of equation (11)are sufficiently accurate for engineering purposes.

(13)

Simply-supported beam carrying a uniformly distributed load

A beam of uniform flexural stiffness EI and span L is simply-supported at its ends, Figure 5;it carries a uniformly distributed lateral load of w per unit length, which induces bending in the yzplane only. Then the reactions at the ends are each equal to ½ wL; if z is measured from the endC, the bending moment at a distance z from C is

Fig. 5 Simply-supported beam carrying a uniformly supported load.

On integrating twice,

From equation (5)

Then equation (14) becomes

The deflection at the mid-length, z = ½ L , is

(14)

(15)

(16)

Cantilever with a concentrated load

A uniform cantilever of flexural stiffness El and length L carries a vertical concentrated load W atthe free end, Figure 6. The bending moment a distance z from the built-in end is

Fig. 6 Cantilever carrying a vertical load at the remote end.

At the end z = 0, there is zero slope in the deflected form, so that dv/dz = 0; then equation (17) gives A = 0. Furthermore, at z = 0 there is also no deflection, so that B = 0. Then

(17)

From equation (5)

On integrating twice,

At the free end, z = L (18)

The slope of the beam at the free end is(19)

When the cantilever is loaded at some point between the ends, at a distance a , say, from the built-in support, Figure 7, the beam between G and D carries no bending moments and thereforeremains straight. The deflection at G can be deduced from equation (18); for z = a,

(21)

(20)

and the slope at z = a is

Then the deflection at the free end D of the cantilever isFig. 7 Cantilever with a load

applied between the ends.

(22)

Cantilever with a uniformly distributed load

A uniform cantilever, Figure 8, carries a uniformly distributed load of w per unit length over the whole of its length. The bending moment at a distance z from C is

Fig. 8 Cantilever carrying a uniformly distributed load.

At the built end, z = 0, and we have

From equation (5)

On integrating twice,

Thus A = B = 0. Then

At the free end, D, the vertical deflection is(23)

Propped cantilever with distributed load

The uniform cantilever of Figure 9 (i) carries a uniformly distributed load w and is supportedon a rigid knife edge at the end D. Suppose P is the force on the support at D. Then we regardFigure (i) as the superposition of the effects of P and w acting separately.

Fig 9 (i) Uniformly loaded cantilever propped at one end. (ii) Deflections due to w alone. (iii) Deflections due to P alone.

If w acts alone, the deflection at D is given by equation (23), and has the value

If the reaction P acted alone, there would be an upward deflection

at D. If the support maintains zero deflection at D,

This gives

or (24)

Simply-supported beam carrying a concentrated lateral loadConsider a beam of uniform flexural stiffness EI and length L, which is simply-supported at itsends C and G, Figure 13.1 1. The beam carries a concentrated lateral load W at a distance a fromC. Then the reactions at C and G are

Fig 10. Deflections of a simply-supported Beam carrying a concentrated lateral load.

Now consider a section of the beam a distance z from C;if z < a, the bending moment at the section is M = V c z and if z > a, M = V c z – W (z - a)

In these equations A, B, A' and B' are arbitrary constants. Now for z = a the values of v given byequations (27) and(28) are equal, and the slopes given by equations (25) and (26) are also equal, as there is continuity of the deflected form of the beam through the point D. Then

At the extreme ends of the beam v = 0, so that when z = 0 equation (27) gives B = 0, and when z =L, equation (28) gives

Then equations (27) and (28) may be written

The second relation, for z > a, may be written

Then equations (32) and (33) differ only by the last term of equation (34); if the last term of equation (34) is discarded when z < a, then equation (34) may be used to define the deflected form in all parts of the beam. On putting z = a, the deflection at the loaded point D is

When W is at the centre of the beam, a = L/2 , and

This is the maximum deflection of the beam only when a = ½ L.

A steel rod 5 cm diameter protrudes 2 m horizontally from a wall. (i) Calculate the deflection due to a load of 1 kN hung on the end of the rod. The weight of the rod may be neglected. (ii) If a vertical steel wire 3 m long, 0.25 cm diameter, supports the end of the cantilever, being taut but unstressedbefore the load is applied, calculate the end deflection on application of the load. Take E = 200GN/m2.

The deflection at the end is then

If this equals the stretching of the wire, then

This gives T = 934 N, and the deflection of the cantilever becomes

A platform carrying a uniformly distributed load rests on two cantilevers projecting a distance 1 m from a wall. The distance between the two cantilevers is ½ l. In what ratio might the load on the platform be increased if the ends were supported by a cross girder of the same section as the cantilevers, resting on a rigid column in the centre, as shown. It may be assumed that when thereis no load on the platform the cantilevers just touch the cross girder without pressure.

“I” having the same value for the cantilevers and cross girder. Substituting this value of δ

Shear Stress in a cross section of Beams

Prof. Dr. R. Chandra Sekaran Ph.D (C. Engg)

Shearing stress on any surface is defined as the intensity of shearing force tangential to the surface. If the block of material of Figure (a) has an area A over any section such as ab, the average shearing stress τ over the section ab is

In many cases the shearing force is not distributed uniformly over any section; if δF is the shearing force on any elemental area δA of a section, the shearing stress on that elemental area is

Three steel plates are held together by a 1.5 cm diameter rivet. If the load transmitted is 50 kN, estimate the shearing stress in the rivet.

A lever is keyed to a shaft 4 cm in diameter, the width of the key being 1.25 cm and its length 5 cm. What load P can be applied at an arm of a = 1 m if the average shearing stress in the key is not to exceed 60 MN/m2?The torque applied to the shaft is Pa. If this is resisted by a shearing force F on the plane ab of the key, then

where r is the radius of the shaft. Then

The area resisting shear in the key is

The permissible shearing force on the plane ab of the key is then

The permissible value of P is then

Two steel rods are connected by a cotter joint. If the shearing strength of the steel used in the rods and the cotter is 150 MN/m2, estimate which part of the joint is more prone to shearing failure.Shearing failure may occur in the following ways:(i) Shearing of the cotter in the planes ab and cd.The area resisting shear is

For a shearing failure on these planes, the tensile force is

(ii) By the cotter tearing through the ends of the socket q, i.e. by shearing the planes ef and gh. The total area resisting shear is

For a shearing failure on these planes

(iii) By the cotter tearing through the ends of the rod p, i.e. by shearing in the planes kl and mn.The total area resisting shear is

For a shearing failure on these planes

Thus, the connection is most vulnerable to shearing failure in the cotter itself, as discussed in (i); the tensile load for shearing failure is 338 kN.

Relative sliding of two separate beams due to shear

Small Slice of Beam

Equivalent Systems

Shear Stress on a Horizontal Plane

The first step in determining the shear stress at any location is to look at a section in a small slice from the beam. Summing the forces due to the normal bending stresses in the horizontal direction gives      ΣFx = 0  - P + (P + dP) + τ b dx = 0   dP/dx = τ b                   (1) where b is the beam depth at the location of the shear stress being calculated.

Horizontal Shear Stress:

P can be found by integrating the normal stress over a section A (shown in diagram), giving     

But the bending stress is σb = My/I. Substituting and simplifying gives Since M and I do not change, they can be moved outside the integral.     

The integral now is just the first moment of the area that is commonly used to find the centroid of an area, and is called "Q". Substituting P into the equation (1) and using Q, gives

Recall, the derivative, dM/dx is equal to the vertical shear load V. This gives,      

The final horizontal shear stress equation is

Calculating Q:

One of the confusing aspects of determining the shear stress τ, is calculating Q, the first moment of the area about the neutral axis. It is rare when the full integral,      is needed. Generally, Q can be determined using parts which is can be written as

Determining Q by Parts at Section a-a:

In determining Q, there are a number of steps that should be considered:1. Locate the neutral axis (NA) for the full cross section. 2. Select the location where the stress will be calculated (generally at the mid-section) and determine b (the cross section width at that location). 3. Split the area above or below the stress location into common geometric shapes (rectangles, circles, triangles, etc.). 4. Sum the product of each area with their respective centroid location from the NA (Important: from the NA, not from the stress location). Note, Q can be negative, but the shear stress equation assume all positive values.

Examples of Q for various shapes and stress locations

We consider first the simple problem of a cantilever of narrow rectangular cross-section, carryinga concentrated lateral load F at the free end, Figure a.; h is the depth of the cross-section, andc is the thickness, Figure b; the depth is assumed to be large compared with the thickness. Theload is applied in a direction parallel to the longer side h.

Shearing actions on an elemental length of a beam of narrow rectangular cross-section

Consider an element a1 length δz of the beam at a distance z from the loaded end. On the face BC of the element the hogging bending moment is M = F zWe suppose the longitudinal stress σ at a distance y from the centroidal axis Cx is the same as that for uniform bending of the element. Then

where I, is the second moment of area about the centroidal axis of bending, Cx, which is also a neutral axis . On the face DE of the element the bending moment has increased to M + δ M = F (z + δ z)The longitudinal bending stress at a distance y from the neutral axis has increased correspondinglyto

Now consider a depth of the beam contained between the upper extreme fibre BD, given by and the fibre GH, given by y = y, Figure (ii). The total longitudinal force on the face BG due to bending stresses σ is

By a similar argument we have that the total force on the face DH due to bending stresses is

These longitudinal force, which act in opposite directions, are not quite in balance; they differ bya small amount

Now the upper surface BD is completely free of shearing stress, and this out-of-balance force canonly be equilibrated by a shearing force on the face GH. We suppose this shearing force isdistributed uniformly over the face GH; the shearing stress on this face is thenThis shearing stress acts on a plane parallel to the neutral surface of the beam; it gives rise therefore to a complementary shearing stress at a point of the cross-section a distance y, from the neutralaxis, and acting tangentially to the cross-section.

…… 1

For this simple type of cross-sectionand so

…… 2

We note firstly that is independent of z; this is so because the resultant shearing force is the same for all cross-sections, and is equal to F. The resultant shearing force implied by the variation of is

The shearing stresses are sufficient then to balance the force F applied to every cross-section ofthe beam.The variation of over the cross-section of the beam is parabolic, attains a maximum value on the neutral axis of the beam, where y1 = 0, and

Shear stress acts on two different parallel surfaces of any element as shown in the diagram. One side cannot be under a different shear stress magnitude than the other. If a small element is taken from a structure under shear, parallel sides will have shear stress loading in the opposite direction, causing it to shear as shown in the diagram. Notice, the other two sides try to resist the sliding motion, and the stress element stays in equilibrium.

Vertical Shear Stress:

Shear Stress on Element

Horizontal and Vertical Shear Stressat the Same Location in a Beam

Similarly, a small element taken from a beam under a shear loading will have equal shear stresses in the vertical and horizontal directions as shown in the diagram at the left. The magnitude of the shear stress will depend on the location of the stress element.

There are three possible shear stresses on a three dimensional cube. This section has only examined one dimension since shear loading in beams is generally only in one direction. But just like uni-axial loading, shear loading can be in three directions.

Rectangular beams are so common, it is helpful to plot the shear stress from top to bottom. The resulting equation is

Rectangular Beams

Shear Stress Distribution in Rectangular Beam

It is a parabolic shape with the maximum at the center. The center shear stress is

τmax = 1.5 V/A

Distribution of shear stresses in a beam of rectangular cross section: (a) cross section of beam, and (b) diagram showing the parabolic distribution of shear stresses over the height of the beam.

Consider the shearing stresses induced in a thin-walled I-beam carrying a concentrated load Fat the free end, acting parallel to Cy, Figure 1. The cross-section has two axes of symmetry Cx and Cy; the flanges are of breadth b, and the distance between the centres of the flanges is h, flanges and web are assumed to be of uniform thickness t.Equation gives the shearing force q per unit length of beam at any region

of the cross section. Consider firstly a point I of the flange at a distance s, from a free edge, Figure (iii); the area of flange cut off by a section through the point I is

The distance of the centroid of this area from the neutral axis Cx is

The shearing force at point I of the cross-section is

If the wall thickness t is small compared with the other linear dimensions of the cross-section, we may assume that q is distributed uniformly over the wall thickness t; the shearing stress is then

at point I. At the free edge, given by s1 = 0, we have = 0, since there can be no longitudinal shearing stress on a free edge of the cross-section. The shearing stress increases linearly in intensity as s, increases from zero to ½ b; at the junction of web and flanges s, = ½ b, and

As the cross-section is symmetrical about Cy, the shearing stress in the adjacent flange also increases linearly from zero at the free edge.

Consider secondly a section through the web at the point 2 at a distance s2 from the junctions of the flanges and web. In evaluating for this section we must consider the total area cut off by the section through the point 2. However, we can evaluate for the component areas cut off bythe section through the point 2; we have

If this shearing force is assumed to be uniformly distributed as a shearing stress, then

At the junction of web and flanges s2 = 0, and

We notice that varies parabolically throughout the depth of the web, attaining a maximum value at s2 = ½ h, the neutral axis, (Figure). In any cross-section of the beam the shearing stresses vary in the form shown; in the flanges the stresses are parallel to Cx, and contribute nothing to the total force on the section parallel to Cy.

At the junctions of the web and flanges the shearing stress in the web is twice the shearing stresses in the flanges. The reason for this is easily seen by considering the equilibrium conditions at the junction. Consider a unit length of the beam along the line of the junction, (Figure); the shearing stresses in the flanges are

the shearing stress in the web

For longitudinal equilibrium of a unit length of the junction of web and flanges, we have

which gives

If this is true, in fact, for the relations we have derived above; longitudinal equilibrium is ensured at any section of the cross-section in our treatment of the problem. If the flanges and web were of different thicknesses, and tw, respectively, the equilibrium condition at the junction would be Then

The implication of this equilibrium condition is that at a junction, such as that of the flanges and web of an I-section, the sum of the shearing forces per unit length for the components meeting atthat junction is zero when account is taken of the relevant directions of these shearing forces. For a junction Where is the shearing stress in an element at the junction,

and t is the thickness of the element; the summation is carried out for all elements meeting at the junction. For an I-section carrying a shearing force acting parallel to the web we see that the maximum shearing stress occurs at the middle of the web, and is given by equation. Now, Ix, for the section is given approximately Then

The total shearing force in the web of the beam parallel to Cy is F, if this were Distributed uniformly over the depth of the web the average shearing stress would be

Then for the particular case when h = 3b, we have

, is only one-sixth greater than the mean shearing stress over the web.

The web of a girder of I-section is 45 cm deep and 1 cm thick; the flanges are each 22.5 cm wide by 1.25 cm thick. The girder at some particular section has to withstand a total shearing force of 200 kN. Calculate the shearing stresses at the top and middle of the web.

Within the limits of our present theory we can employ certain formulae to find the principal stresses and the maximum shearing stress. On a side elevation of the beam we can draw, lines showing the direction of the principal stresses. Such lines are called the lines of principal stress; they are such that the tangent at any point gives the direction of principal stress. As an example, the lines of principal stress have been drawn in Figure for a simply-supported beam of uniform rectangular cross-section, carrying a uniformly distributed load. The stresses are a maximum where the tangents to the curves are parallel to the axis of the beam, and diminish to zero when the curves cut the faces of the beam at right angles. On the neutral axis, where the stress is one of shear, the principal stress curves cut the axis at 45".

Principal stress lines in a simply-supported rectangular beam carrying a uniformly distributed load.

The flanges of an I-girder are 30 cm wide by 2.5 cm thick and the web is 60 cmdeep by 1.25 cm thick. At a particular section the sagging bending moment is500 kNm and the shearing force is 500 kN. Consider a point in the section atthe top of the web and calculate for h s point; (i) the longitudinal stress, (ii) theshearing stress, (iii) the principal stresses.

First calculate the second moment of area about the neutral axis; the second moment of area of the web is

of each flange is

It should be noticed that the greater principal stress is about 30% greater than the longitudinal stress. At the top of the flange the longitudinal stress is -96 MN/m2, so the greatest principal stress at the top of the web is 20% greater than the maximum longitudinal stress.

Shearing stresses in a channel section; shear centreWe have discussed the general case of shearing stresses in the bending of a beam having an axis of symmetry in the cross-section; we assumed that the shearing forces were applied parallel to this axis of symmetry. This is a relatively simple problem to treat because there can be no twisting of the beam when a shearing force is applied parallel to the axis of symmetry. We consider now the case when the shearing force is applied at right angles to an axis of symmetry of the cross-section.Consider for example a channel section having an axis Cx of symmetry in the cross-section, Figure ; the section is of uniform wall-thickness t, b is the total breadth of each flange, and h is the distance between the flanges; C is the centroid of the cross-section. Suppose the beam is supported at one end, and that a shearing force F is applied at the free end in a direction parallel to Cy. We apply this shearing force at a point O on Cx such that no torsion of the channel occurs, Figure 2; if F is applied considerably to the left of C, twisting obviously will occur in a counterclockwise direction; if F is applied considerably to the right then twisting occurs in a clockwise direction. There is some intermediate position of O for which no twisting occurs; as we shall see this position is not coincident with the centroid C.

The problem is greatly simplified if we assume that F is applied at a point O on Cx to give no torsion of the channel; suppose O is a distance e from the centre of the web, at any section of the beam there are only bending actions present; therefore, we can again use the relation

acting parallel to the centre lines of the flanges; the total shearing forces in the two flanges are in opposite directions. If the distribution of shearing stresses T, and T* is statically equivalent to the applied shearing force F, we have, on taking moments about B- the centre of the web-- that

which, as we should expect, is independent of F. We note that O is remote from the centroid C of the cross-section; the point O is usually called the shear centre; it is the point of the cross-section through which the resultant shearing force must pass if bending is to occur without torsion of the beam.

2/hy

yy11

1

dAF

2/hy

yy22

1

dAF

zz1 I

yMzz

2 I

ydxx

MM

1F 2F

2/hy

yy 1

ydAQ

dAFFdxb2/hy

yy1212

1

zzbI

QV

0

Shear Stress in Webs

b=10t h=20t

t

t

t

b

h

y

z

zzbI

)s(QV)s(

)s(QQ)s(Q wF s

3F t95

2t

2h

btQ

2

st

2

hts)s(Qw

0t2hsQ0sQ ww

3w t

2

81t

2

ht

2

h

2

tt

2

hsQ

y

z

zz

3

zz

F

tI

t95V

tI

QV

zz

3

tI2

t81V

zz

4t2hs

0s

3

zzw I

Vt2196ds

2

st

2

htst95

tI

VtV

y

z

b

h

b-t

h-2t

3

t6878

12

t2htb

12

bhI

433

zz

3

t6878

12

t2htb

12

bhI

433

zz

zz

4t2hs

0s

3

zzw I

Vt2196ds

2

st

2

htst95

tI

VtV

V96.0Vw

y

z

s1

resultpreviousthecheckswhich

t952t

2h

bttsQ

2s

2h

bssQ

31F

111F

zzzz

FF bI

tV

bI

tsQV 31 95

y

z

zz

4

1

ts

0s

11

zzF I3

Vt145ds

2s

2h

bsbI

VbV

1

1

zzzz

FF bI

tV

bI

tQV 395

zz

3

zz

FF tI

t95V

tI

tQV

zz

4

F I3

Vt145V

zzw I

VtV

42196

V3

6878

I

Vt2196

3

290

I

VtVV2

zz

4

zz

4

wF

VF

VF

Vw

Limitations• The elementary shear theory is suitable for determining

vertical shear stresses in the web portion only of a wide-flange beam– we can no longer assume constant shear stresses across the width b

• The shear formula cannot be used to determine vertical shear stresses in the flanges– The distribution used in the I beam problem was provided to show

that the resultants of vertical shear stresses in the flanges were 4% of the applied shear force V.

• Shear formula does give good results for the shear stresses acting horizontally in the flanges.

Torsional Loads on Circular Shafts

• Interested in stresses and strains of circular shafts subjected to twisting couples or torques

• Turbine exerts torque T on the shaft

• Shaft transmits the torque to the generator

• Generator creates an equal and opposite torque T’

dAdFT

• Net of the internal shearing stresses is an internal torque, equal and opposite to the applied torque,

• Although the net torque due to the shearing stresses is known, the distribution of the stresses is not

• Unlike the normal stress due to axial loads, the distribution of shearing stresses due to Torsional loads can not be assumed uniform.

• Distribution of shearing stresses is statically indeterminate –

• must consider shaft deformations

• The existence of the axial shear components is demonstrated by considering a shaft made up of axial slats.

• The slats slide with respect to each other when equal and opposite torques are applied to the ends of the shaft.

• Torque applied to shaft produces shearing stresses on the faces perpendicular to the axis.

• Conditions of equilibrium require the existence of equal stresses on the faces of the two planes containing the axis of the shaft

• From observation, the angle of twist (φ) of the shaft is proportional to the applied torque and to the shaft length.

L

T

• When subjected to torsion, every cross-section of a circular shaft remains plane and undistorted.

• Cross-sections of noncircular (non axisymmetric) shafts are distorted when subjected to torsion.

UNIT 2 TRANSLATIONAL EQUILIBRIUM

Prof. Dr. R. Chandra Sekaran Ph.D (C.Engg)

Rotational Equilibrium : TRANSLATIONAL EQUILIBRIUM

The second condition for equilibrium is rotational equilibrium. We can see the need for this second condition if we look at the diagram 1.3a. In this diagram, if we apply the 1st condition of equilibrium and sum the forces in the y-direction, we obtain zero. (+100 lb. - 100 lb. = 0). This would indicate that the object is in translational equilibrium. However, we almost instinctively recognise that the object certainly will not remain at rest, and will experience rotational motion (and rotational acceleration).

Please notice that the object actually is in translational equilibrium. That is, even though it rotates, it rotates about the center of mass of the bar, and the center of mass of the bar will not move.

The second condition for equilibrium states that if we are to have rotational equilibrium, the sum of the Torque acting on the structure must be zero.

The 2nd condition for equilibrium may be written:

or in three dimensions:

,

,

Two painters are standing on a 300 Kg. scaffolding (beam) which is 12 m. long. One painter weighs 160 Kg. and the second painter weighs 140 Kg. The scaffolding is supported by two cables, one at each end. As they paint, the painters begin wondering what force (tension) is in each cable. The question is, what is the force (tension) in each cable when the painters are standing in the positions shown. (Notice that for a uniform beam or bar, as far as equilibrium conditions are concerned, the beam weight may be considered to act at the center (of mass) of the bar.)

1. Draw a Free Body Diagram (FBD) showing and labeling all external forces acting on the structure, and including a coordinate system. Notice in the diagram to the right, we have shown the forces in the cables supporting the beam as arrows upward, and labeled these forces TA and TB.

2. Resolve all forces into x and y-components. In this example, all the forces are already acting in the y-direction only, so nothing more needs to be done.

3. Apply the (2-dimensional) equilibrium conditions:

(No external x-forces acting on the structure, so this equation gives no information.)

TA - 300 lb - 160 lb - 140 lb + TE = 0

We can't solve this equation, as there are two unknowns (TA and TE) and only one equation. We obtain our second equation from the 2nd condition of equilibrium - sum of torque must equal zero.

Before we can sum torque we must first PICK A POINT, as we always calculate torque with respect to a point (or axis). Any point on (or off) the structure will work, however some points result in an easier equation(s) to solve. As an example, if we sum torque with respect to point E, we notice that unknown force TE acts through point E, and, if a force acts through a point, that force does not produce a torque with respect to the point (since the perpendicular distance is zero). Thus summing torque about point E will result in an equation with only one unknown, as shown

or -TA(12) +300 lb(6ft) + 160 lb(3 ft) + 140 lb(1 ft) = 0

Where we determined each term by looking at the forces acting on the beam one by one, and calculating the torque produced by each force with respect to the chosen point E from: Torque = Force x perpendicular distance

(from the point E to the line of action of the force).

The sign of the torque is determined by considering which way the torque would cause the beam to rotate, if the beam were actually pinned at the chosen point. That is, if we look at the 160 Kg. weight of painter one (and ignore the other forces), the 160 Kg. weight would cause the beam to start rotating counter clockwise (+), if the beam were pinned at point E.

It is important to note that the sign of the torque depends on the direction of the rotation it would produce with respect to the chosen pivot point, not on the direction of the force. That is, the 160 Kg. force is a negative force (downward) in the sum of forces equation, but it produces a positive torque with respect to point E in the sum of torque equation.

or, TA – 300 Kg – 160 Kg -140 Kg + TE = 0

or, -TA(12) +300 Kg(6 m) + 160 kg (3 m) + 140 Kg(1 m) = 0

TA = 201.67 Kg TE = 398 Kg.

What is the minimum cable strength required so that the painters could move anywhere on the scaffolding safely? (Answer = 450 Kg.)

Moment of a force: The magnitude of the moment of a force:

MO: Magnitude of the moment of F around point O

d: Perpendicular distance from O to the line of action of F

M FdO

F

Line of Action of F

Fr

θθ

O

F1

d

Note: moving a force along its line of action does not change its moment

rFFrFdM )(Sin0

Direction of the moment in 2-D: The direction of the moment is given by the right hand rule:Counter Clockwise (CCW) is out of the page, Clockwise (CW) is into the page.

Calculating the moment in 2-D using components: Moments add together as vectors. Select a positive direction (CCW or CW), then calculate each moment and add them using the proper sign for each term. For example:

M F d F d CCW positiveO x y 1 2

d1

F

Fx

Fy

d2

Moment of a force:

Moment of Force F around point O: MO

Calculating the moment using rectangular components:

Resultant moment: Mro

Moment of a force about a specified axis a-a: Ma

O: any point on a-a

Couple: C

Note: The moment of a couple does not depend on the point one takes the moment about. In other words, a moment of a couple is the same about all points in space.

Equivalent force systems: The basic idea: Two force systems are equivalent if they result in the same resultant force and the same resultant moment.

Moving a force along its line of action: Moving a force along its line of action results in a new force system which is equivalent to the original force system.

Moving a force off its line of action: If a force is moved off its line of action, a couple must be added to the force system so that the new system generates the same moment as the old system.

The resultant of a force and couple system: For any point O, every force and couple system can be made equivalent to a single force passing through O and a single couple. The single force passing through O is equal to the resultant force of the original system, and the couple is equal to the resultant moment of the original system around point O.

Note: All 2-D force systems can be reduced to a single force. To find the line of action of the force, the moment of the original system must be forced to be the same as the system with the single force.

Equivalent force systems: The wrenchA wrench is a force and couple system in which the force and couple are parallel.

Every force and couple system can be reduced to a wrench: The perpendicular part of the resultant moment can be replaced by sliding the resultant force along the line perpendicular to the plane of the resultant force and resultant couple. What remains after this operation is a wrench which has a moment equal to the component of the resultant couple parallel to the force.

When can one reduce a force and couple system to a single force?: For a force and couple system if the resultant force and the resultant couple are perpendicular, then one can find an equivalent system with a single force and no couple. To obtain this system, move the resultant force a distance d along the line perpendicular to the plane of the resultant force and resultant couple until the resultant force creates a moment equivalent to the resultant couple.

Note: All 2-D force systems can be reduced to a single force. To find the line of action of the force, the moment of the original system must be forced to be the same as the system with the single force.

Determine the un stretched length δ of spring AC if a force P causes the angle θ for equilibrium. Cord AB has length a. Given: P = 80 lb, θ = 60 deg, k =50 lb /ft, a = 2 ft, b= 2 ft.

A vertical force P is applied to the ends of cord AB of length a and spring AC. If the spring has an un stretched length δ, determine the angle θ for equilibrium.Given: P = 10 lb, δ = 2 ft, k=15 lb/ ft, a = 2 ft, b= 2 ft

Determine the mass of each of the two cylinders if they cause a sag of distance d when suspended from the rings at A and B. Note that s = 0 when the cylinders are removed.Given: d = 0.5 m, l1 = 1.5 m, l2 = 2 m, l3 = 1 m, k =100 N / m, g =9.81 m / s2.

Prove Lami's theorem, which states that if three concurrent forces are in equilibrium, each is proportional to the sine of the angle of the other two; that is, P/sin α = Q/sin β = R/sin γ.

The mast OA is supported by three cables. If cable AB is subjected to tension T, determine the tension in cables AC and AD and the vertical force F which the mast exerts along its axis on the collar at A. Given: T = 500 N, a = 6 m, b = 3 m, c = 6 m, d = 3 m, e = 2 m, f = 1.5 m, g = 2 m

Replace the force system by a single force resultant and specify its point of application, measured along the x axis from point O.Given: F1 = 125 kg, F2 = 350 kg, F3 = 850 kg, a = 2 m, b = 6 m, c = 3 m, d = 4 m.

m

kg

Given the three non zero vectors A, B, and C, show that if A (B×C) = 0, the three vectors ⋅ must lie in the same plane.

Consider,A (B×C) = A B×C cos(θ )⋅

= ( A cos(θ )) B×C

= h B×C

= BC h sin(φ )

= volume of parallelepiped.

If A (B×C) = 0, then the volume equals zero, so that A, B, and C are coplanar.⋅

Determine the magnitude and directional sense of the resultant moment of the forces at A and B about point O. Given: F1 = 40 lb, F2 = 60 lb, θ 1 = 30 deg, θ 2 = 45 deg, a = 5 in, b = 13 in, c = 3 in, d = 6 in, e = 3 in and f = 6 in.

Determine the magnitude and directional sense of the resultant moment of the forces at A and B about point P. Units Used: kip = 1000 lbGiven: F1 = 40 lb, b = 13 in, F2 = 60 lb, c = 3 in, θ 1 = 30 deg, d = 6 in, θ 2 = 45 deg, e = 3 in, a = 5 in and f = 6 in

Determine the magnitude and sense of the couple moment.Units Used: kN = 1000 N. Given: F = 5 kN, θ = 30 deg, a = 0.5 m, b = 4 m, c = 2 m, d = 1 m

Determine the magnitude and sense of the couple moment. Each force has a magnitude F.Given: F = 65 kg, a = 2 m, b = 1.5 m, c = 4 m, d = 6 m, e = 3 m.

Kg- m Counter Clockwise

A clockwise couple M is resisted by the shaft of the electric motor. Determine the magnitude of the reactive forces −R and R which act at supports A and B so that the resultant of the two couples is zero.Given: a = 150 mm, θ = 60 deg, M = 5 N m⋅

The ends of the triangular plate are subjected to three couples. Determine the magnitude of the force F so that the resultant couple moment is M clockwise.Given: F1 = 600 N, F2 = 250 N, a = 1 m, θ = 40 deg, M = 400 N m. ⋅

Initial Guess F = 1 N

Draw the free-body diagram of the sphere of weight W resting between the smooth inclined planes. Explain the significance of each force on the diagram.Given: W = 10 lb, θ 1 = 105 deg, θ 2 = 45 deg

NA, NB force of plane on sphere.W force of gravity on sphere.

Draw the free-body diagram of the hand punch, which is pinned at A and bears down on the smooth surface at B.Given: F = 8 lb, a = 1.5 ft, b = 0.2 ft, c = 2 ft

Ax, Ay, MA effect of wall on beam. NB force of roller on beam.Wa / 2 resultant force of distributed load on beam.

Draw the free-body diagram of the beam supported at A by a fixed support and at B by a roller. Explain the significance of each force on the diagram.Given: w =40 lb / ft, a = 3 ft, b = 4 ft, θ = 30 deg

Draw the free-body diagram of the jib crane AB, which is pin-connected at A and supported by member (link) BC. Units Used: kN = 1000 NGiven: F = 8 kN, a = 3 m, b = 4 m, c = 0.4 m, d = 3, e = 4.

Draw the free-body diagram of the uniform bar, which has mass M and center of mass at G. The supports A, B, and C are smooth.

Draw the free-body diagram of the beam, which is pin-connected at A and rocker-supported at B.

The drainpipe of mass M is held in the tines of the fork lift. Determine the normal forces at A and B as functions of the blade angle θ and plot the results of force (ordinate) versus θ (abscissa) for 0 ≤ θ ≤ 90 deg.

The articulated crane boom has a weight W and mass center at G. If it supports a load L,determine the force acting at the pin A and the compression in the hydraulic cylinder BC whenthe boom is in the position shown

The platform assembly has weight W1 and center of gravity at G1. If it is intended to support a maximum load W2 placed at point G2,,determine the smallest counterweight W that should be placed at B in order to prevent the platform from tipping over.Given: W1 = 250 lb, a = 1 ft, c = 1 ft, e = 6 ft, W2 = 400 lb, b = 6 ft, d = 8 ft, f = 2 ft

The man has weight W and stands at the center of the plank. If the planes at A and B are smooth, determine the tension in the cord in terms of W and θ.

The telephone pole of negligible thickness is subjected to the force F directed as shown. It is supported by the cable BCD and can be assumed pinned at its base A. In order to provide clearance for a sidewalk right of way, where D is located, the strut CE is attached at C, as shown by the dashed lines (cable segment CD is removed). If the tension in CD' is to be twice the tension in BCD, determine the height h for placement of the strut CE. F = 80 , θ = 30, a = 30, b = 10

The mobile crane has weight W1 and center of gravity at G1; the boom has weight W2 and center of gravity at G2. Determine the smallest angle of tilt θ of the boom, without causing the crane to overturn if the suspended load has weight W. Neglect the thickness of the tracks at A and B.Given: W1 =120000 lb, W2 =30000 lb, W =40000 lb, a = 4 ,b = 6 , c = 3 , d = 12, e = 15

The man attempts to support the load of boards having a weight W and a center of gravity at G. If he is standing on a smooth floor, determine the smallest angle θ at which he can hold them up in the position shown. Neglect his weight. Given: a = 0.5ft, b = 3ft, c = 4ft, d = 4 ft.

b)

Mo = 1200 lb.in

F = 57.7 lb F = 50 lb OB = 10 in.

For the bracket and loading of Previous Problem , determine the smallest distance “a” if the bracket is not to move.

Determine the magnitude of the force F that should be applied at the end of the lever suchthat this force creates a clockwise moment M about point O.Given: M = 15 Nm, φ = 60 deg, θ = 30 deg, a = 50 mm, b = 300 mm

Determine the direction θ ( 0° ≤ θ ≤ 180°) of the force F so that it produces (a) the maximummoment about point A and (b) the minimum moment about point A. Compute the moment ineach case.Given: F = 40 Kg, a = 8 m, b = 2 m

The rod on the power control mechanism for a business jet is subjected to force F. Determine the moment of this force about the bearing at A.Given: F = 80 N, θ 1 = 20 deg, a = 150 mm, θ 2 = 60 deg.

The boom has length L, weight Wb, and mass center at G. If the maximum moment that can bedeveloped by the motor at A is M, determine the maximum load W, having a mass center at G',that can be lifted.

The tool at A is used to hold a power lawnmower blade stationary while the nut is being loosened with the wrench. If a force P is applied to the wrench at B in thedirection shown, (a) Determine the moment it creates about the nut at C?(b) What is the magnitude of force F at A so that it creates the opposite moment about C ?Given: P = 50 N, a = 400 mm, b = 300 mm, θ = 60 deg, c = 5 and d = 12.

The force F acts on the end of the pipe at B. Determine the angles θ ( 0° ≤ θ ≤ 180° ) ofthe force that will produce maximum and minimum moments about point A. What are the magnitudes of these moments?Given: F = 70 N, a = 0.9 m, b = 0.3 m, c = 0.7 m

The towline exerts force P at the end of the crane boom of length L. Determine the placement x of the hook at A so that this force creates a maximum moment about point O. What is this moment?Unit Used: kN = 1000 NGiven:P = 4 kN, L = 20 m, θ = 30 dega = 1.5 m

Determine the resultant moment of the forces about point A. Solve the problem first byconsidering each force as a whole, and then by using the principle of moments.Units Used: kN = 1000 NGiven: F1 = 250 N, a = 2 m, F2 = 300 N, b = 3 m, d= 3, F3 = 500 N, c = 4 m, θ 1 = 60 deg, θ 2 = 30 deg, e = 4

Determine the moment of the force F at A about point O. Express the result as a Cartesianvector. Units Used: kN = 1000 N, Given: F = 13 kN, a = 6 m, b = 2.5 m, c = 3 m, d = 3 m, e = 8 m, f = 6 m, g = 4 m, h = 8 m

Determine the smallest force F that must be applied along the rope in order to cause the curvedrod, which has radius r, to fail at the support C. This requires a moment to be developed at C ofmagnitude M. Given: r = 5 ft, M = 80 lb ft, ⋅ θ = 60 deg, a = 7 ft and b = 6 ft

The man pulls on the rope with a force F. Determine the moment that this force exerts about the base of the pole at O. Solve the problem two ways, i.e., by using a position vector from O to A, then O to B. Given: F = 20 N, a = 3 m, b = 4 m, c = 1.5 m d = 10.5 m

The force F is applied to the handle of the box wrench. Determine the component of themoment of this force about the z axis which is effective in loosening the bolt.Given: a = 3 in, b = 8 in and c = 2 in.

The cutting tool on the lathe exerts a force F on the shaft in the direction shown. Determine the moment of this force about the y axis of the shaft. Units Used: kN = 1000 NGiven: a = 30 mm, θ = 40 deg

The bevel gear is subjected to the force F which is caused from contact with another gear. Determine the moment of this force about the y axis of the gear shaft.Given: a = 30 mm, b = 40 mm

The flex-headed ratchet wrench is subjected to force P, applied perpendicular to the handle as shown. Determine the moment or torque this imparts along the vertical axis of the bolt at A.Given: P = 16 Kg, a = 10 cm, θ = 60 deg, b = 0.75 cm.

Kg-m

UNIT 2FRAMES –

NON TRUSS RIGID BODIES

Prof. Dr. R. Chandra Sekaran Ph.D (C.Engg)

In Diagram 1 we have shown a horizontal beam supported at point A by a roller and at point C by a pinned support. Diagram 1a is the Free Body Diagram of the beam with the roller and the pinned joint now replaced by the support forces which they apply on the beam. The roller applies the vertical force A y and the pinned support applies the forces Cx and Cy.

Diagram 2 above shows examples of supports and the types of forces and/or torque which they may exert on a structure.

Axial Member is a member which is only in simple tension or compression. The internal force in the member is constant and acts only along the axis of the member. If forces (no matter how many) act at only Two Points on the member - it is an axial member. That is, the resultant of the forces must be two single equal forces acting in opposite directions along axis of the member. See Diagram 3.

Non-Axial Member is a member which is not simply in tension or compression. It may have shear forces acting perpendicular to the member and/or there may be different values of tension and compression forces in different parts of the member. A member with forces acting at More Than Two Points (locations) on the member is a non-axial member. See Diagram 4.

Frames (non-truss, rigid body)In this problem we wish to determine all the external support forces (reactions) acting on the structure shown in Diagram 1 below. Once again our procedure consists basically of three steps.

1.Draw a Free Body Diagram of the entire structure showing and labeling all external load forces and support forces, include any needed dimensions and angles. 2. Resolve (break) all forces into their x and y-components.

3. Apply the Equilibrium Equations ( ) and solve for the unknown forces.

In our example, the load forces are already shown by the downward arrows. We next look at the forces exerted on the structure by the supports. Since each support is a pinned joint, the worst case we could have is an unknown x and y-force acting on the structure at each support point. We also must choose directions for the x and y support forces.In some problems the directions of the support forces are clear from the nature of the problem. In other problems the directions the support forces act is not clear at all. However, this is not really a problem. We simple make our best guess for the directions of the support reactions. If our guess is wrong, when we solve for the value of the support forces, that value will be negative.

This is important. A negative value when solving for a force does not mean the force necessarily acts in the negative direction, rather it means that the force acts in the direction OPPOSITE to the one we initially chose.

We can draw a better FBD by reflecting on the concept of axial and non-axial members. Notice in our structure that member ABC is a non-axial member (since forces act on it at more than two points), while member CD is an axial member (since if we drew a FBD of member CD we would see forces act on it at only two points, D and C). This is important. Since CD is an axial member the force acting on it from the wall (and in it) must act along the direction of the member. This means that at point D, rather than having two unknown forces, we can draw one unknown force acting at a known angle (force D acting at angle of 37o, as shown in Diagram 3). This means we have only three unknowns, Ax, Ay, and D. In Diagram 3, we have also completed

STEP I: This is an accurate FBD, but it is not the best. The difficulty is that for our problem, we have three equilibrium conditions ( ), but we have four unknowns (Ax, Ay, Dx, Dy) in this FBD. And as we are well aware, we can not solve

for more unknowns than we have independent equations.

Step II, breaking any forces not in the x or y-direction into x and y-components. Thus, in Diagram 3, we have shown the two components of D (which act at 37o), D cos 37o being the x-component, and D sin 37o being the y-component.

[Please notice that there are not three forces at point D, there is either D acting at 37o or its two equivalent components, D cos 37o and D sin 37o. In Diagram 3 at this point we really should cross out the D force, which has been replaced by its components.]

Now before we proceed with the final step and determine the values of the support reactions, we should deal with several conceptual questions which often arise at this point. First, why can't we do at point A what we did at point D, that is put in one force acting at a known angle. Member ABC is a horizontal member, doesn't the wall just push horizontally on member ABC, can't we just drop the Ay force? The answer is NO,

because member ABC is not an axial member, it is not simply in compression or tension, and the wall does not just push horizontally on member ABC (as we will see in our solution). Thus the best we can do at point A is unknown forces Ax and Ay.

Now Step III. Apply the Equilibrium conditions.

Here we sum the x-forces, keeping track of their direction signs, forces to right, +, to left, -

Sum of y-forces, including load forces, keeping track of direction signs.

A second question is often, what about the wall, aren't there forces acting on the wall that we should consider? Well, yes and no. YES, there are forces acting on the wall (as a matter of fact they are exactly equal and opposite to the forces acting on the members, in compliance with Newton's Third Law). But NO we should not consider them, because we are making a FBD of the STRUCTURE, not of the wall, so we want to consider forces which act on the structure due to the wall, not forces on the wall due to the structure.

Sum of Torque about a point. We choose point A. Point D is also a good point to sum torque about since unknowns act through both points A and D, and if a force acts through a point, it does not produce a torque with respect to that point. Thus our torque equation will have less unknowns in it, and will be easier to solve. Notice that with respect to point A, forces Ax, Ay, and D sin 37o do not produce torque since their lines of action pass through point A. Thus in this problem the torque equation has only one unknown, D. We can solve for force D, and then use it in the two force equations to find the other unknowns, Ax and Ay. (Completing

the calculations, we arrive at the following answers.) D = +7500 Ax = +6000 Ay = +1500.Note that all the support forces we solved for are positive, which means the directions we choose for them initially are the actual directions they act. We have now solved our problem. The support force at point D is 7500 acting at 37o. The support forces at A can be left as the two components, Ax = 6000 and Ay = 1500 , or may be added (as vectors) obtaining one force at a known angle, as shown in Diagram 4.

Thus the force at point A is 6185 acting at 14o, as shown.

This information would help us purchase the correct size hinge (able to support 6185 at A, and able to support 7500 at D), or estimate if the wall is strong enough to support the structure.

Note: Both sides of the rope have 200 lb. force in them - one side due to the load, and the other side due to the pull of the person. Point B must support both the load and the pull of the person which results in a total force of 400 lb. acting on point B. At point A we have shown one unknown support force 'A' acting at a known angle (37o). We can do this at point A since we know member AB is an axial member. In an axial member the force is along the direction of the member, thus the floor must exert a force on the member also along the direction of the member (due to equal and opposite forces principle).

However, at point D, since member D is a non-axial member, the best we can do is to show an unknown Dx and Dy support forces acting on the structure at point D. SolvingA = +343 lb. Dx = +274 lb. Dy = +354 lb. We can find Resultant D = Square Root [Dx2 + Dy2] = Square Root [(-274 lb)2 + (354 lb)2] = 447.7 lb. and Tangent (angle) = Dy/Dx = 354lb/-274 lb = -1.29, so Angle = Arc Tangent (-1.29) = 127.8o (from x -axis) so support force at Point D could also be expressed as: D = 447.7 lb. @ 127.8o .

Note: Force at D does not act along the direction of member BCD, which it would do if BCD were an axial member.

E = +10,800 lb. Ax = +8620 lb. Ay = +13500 lb.

To determine the force in axial member CD: We draw a FBD, not of the entire structure, but of a member of the structure, (choosing not the member we wish to find the force in, but a member it acts on). Thus, if we wish to find the force in member CD, we draw a FBD - not of member CD, but a member CD acts on, such as member ABC, or member AD. We will use member ABC to find the force in member CD.

Step 3: Apply the equilibrium conditions and solve for unknown forces.Fx: ACx + CD cos 53.8o = 0

Fy: ACy -12,000 lb. + CD sin 53.8o = 0 TA: -12,000 lb. (4 ft) + CD sin 53.8o (8 ft ) =0

CD = 7440 lb ACx = - 4390 lb. ACy = 6000 lb.(- sign shows force acts in opposite direction.),

The second good FBD of member ABC is shown in Diagram 4. What we have done in this diagram is to look more closely at the left end of member ABC and observe that the effect of the wall forces and the effect of member AD, is to give some net x and y-force acting on member ABC. Thus, rather than show both the wall forces and the force due to AD on ABC, we simple show an ACx and an ACy force which is

the net horizontal and vertical force acting on ABC at the left end. This is fine to do, as we are looking for force CD, and that is still present in our FBD. This second FBD is slightly easier than the first in that it will result in one less force (AD) in the equilibrium equations. We will use the second FBD in the rest of the problem.

Point A: As an aside, notice that the forces ACx and ACy (The horizontal and vertical forces acting on member ABC at end A.) are not the same as the forces Ax and Ay acting on the entire structure at joint A. This results since the forces of the wall at point A are not just acting on member ABC, but are distributed to both members ABC and AD, as shown in Diagram 6. Note in the diagram that if the forces ACx and ADx are summed (13010 lb. - 4390 lb. = 8620 lb.), and if forces ACy and ADy are summed (6000 lb. + 7500 lb. = 13500 lb.), that their vector sums equal the external forces (Ax and Ay) acting on point A, as we expect they should.

In the structure shown below members AD, DC, and ABC are assumed to be solid rigid members. Member ED is a cable. For this structure:A. Draw a Free Body Diagram showing all support forces and loads.B. Determine the values of all the support forces acting on the structure.C. Determine the force (tension or compression) in member DC. Unless otherwise indicated, all joints and support points are assumed to be pinned or hinged joints.

STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces, as well as any neededangles and dimensions.

STEP 2: Break any forces not alreadyin x and y direction into their x and y components.

STEP 3: Apply the equilibrium conditions. Sum Fx = -E cos (37o) + Ax = 0

Sum Fy = Ay + E sin (37o)

- 10,000 lbs - 8,000 lbs = 0

Sum TA = E cos (37o)(12ft)

- (10,000 lbs)(4 ft) - (8,000 lbs)(12ft) = 0

Solving for the unknowns: E = 14,200 lbs; Ay= 9,480 lbs; Ax = 11,400 lbs

STEP 1: Draw a free body diagram of a member that DC acts on - member ABC.STEP 2: Resolve all forces into x and y components (see diagram).STEP 3: Apply the equilibrium conditions.Fx = Acx - DC cos (56.3o) = 0

Fy = Acy - 10,000 - 8,000 + DC sin (56.3o) = 0

TA = (-10,000 )(4 ) - (8,000 )(12 ) +DC sin (56.3o)(12 ) = 0

Solving for the unknowns: DC = 13,600 ; Acx = 7,560 ; Acy = 6,670

These are external forces acting on member ABC. The force in DC is 13,600 (c).

PART C: Now find internal force in member DC

EXERCISES: In the structure shown below member ABC is assumed to be a solid rigid member. Member CD is a cable. For this structure: A. Draw a Free Body Diagram showing all support forces and loads.B. Determine the value of all the support forces acting on the structure.C. Determine the force (tension ) in member CD. Unless otherwise indicated, all joints and support points are assumed to be pinned or hinged joints.

STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces, as well as any needed angles and dimensions.STEP 2: Break any forces not already in x and y direction into their x and y components.STEP 3: Apply the equilibrium conditions. Sum Fx = -D cos (37o) + Ax = 0Sum Fy = D sin (37o) + Ay - 10,000 lbs - 8,000 lbs = 0Sum TA = (-10,000 lbs)(6.93 ft) - (8,000 lbs)(10.4 ft) + D sin (37o)(5.61 ft) + D cos (37o)(9.61 ft) = 0Solving for the unknowns:D = 13,800 lbs; Ax = 11,100 lbs; Ay = 9,710 lbs

PART C – Now find internal force in member DC.In this problem member DC is a single axial member connected to the wall at point D. Therefore, the force in member DC is equal and opposite the force exerted on DC by the wall.

From parts A and B, the force on DC due to the wall is 13,800 lbs. Therefore, force in DC is also 13,800 lbs (in tension since DC is a cable).

Fig 2

In the structure shown Fig 2 members ABC, ADE, and DB are assumed to be solid rigid members. Members ABC and ADE are pinned to the wall at point A. Member ADE is supported by a roller at point E. For this structure: A. Draw a Free Body Diagram showing all support forces and loads. B. Determine the value of all the support forces acting on the structure. C. Determine the force (tension or compression) in member DB. Unless otherwise indicated, all joints and support points are assumed to be pinned or hinged joints.

STEP 1: Draw a free body diagram showing and labelling all load forces and support (reaction) forces, as well as any neededangles and dimensions.

STEP 2: Break any forces not already in x and y direction into their x and y components.

STEP 3: Apply the equilibrium conditions.Sum Fx = Ex + Ax = 0Sum Fy = Ay - 12,000 lbs = 0Sum TA = Ex(8 ft) - (12,000 lbs)(12 ft) = 0Solving for the unknowns: Ex= 18,000 lbs; Ax= -18,000 lbs; Ay = 12,000 lbs After Drawing the FBD of a member that DB acts on - member ABC, and Resolving the Forces, apply the equilibrium conditions.Sum Fx = Acx + DB cos (33.7o) = 0Sum Fy = Acy + DB sin (33.7o) - 12,000 lbs = 0Sum TA = DB sin (33.7o)(6 ft) - (12,000 lbs)(12 ft) = 0Solving for the unknowns:DB = 43,300 lbs; Acx = -36,000 lbs; Acy = -12,000 lbsThese are external forces acting on member ABC.The force in DB is 43,300 lbs (c).

In the structure shown below members ABC , BDE and CD are assumed to be solid rigid members. The structure is pinned at A and supported by a roller at E. For this structure:A. Draw a Free Body Diagram showing all support forces and loads.B. Determine the value of all the support forces acting on the structure.C. Determine the force (tension or compression) in member CD.Unless otherwise indicated, all joints and support points are assumed to be pinned or hinged joints.

STEP 1: Draw a free body diagram showing And labelling all load forces and support(reaction) forces, as well as any needed angles and dimensions.STEP 2: Break any forces not already in x and y direction into x and y components.STEP 3: Apply the equilibrium conditions.Sum Fx = Ax = 0,

Sum Fy = Ay + Ey - 8,000 - 4,000 = 0 Sum TA = (-8,000 )(3 ) – (4,000 )(6.5 ) + Ey(8.5 ) = 0Solving for the unknowns: Ey = 5,880 ; Ay = 6,120 .

PART C: Draw a free body diagram of a member that CD acts on - member BDE.Resolve all forces into x and y components Apply the equilibrium conditions:Sum Fx = -Bx + CD cos (60o) = 0

Sum Fy = By - CD sin (60o) - 4,000 + 5,880 = 0

Sum TB = -CD sin (60o)(3 ) - (4,000 )(5 )

+ (5,880 )(7 ) = 0Solving for the unknowns:CD = 8,140 ; Bx = 4,070 ; By = 5,170

These are the external forces acting on member BDE.The force in CD is 8,140 (c).

Determine the reaction at supports A and B of the beam in the diagram shown.  Neglect the weight of the beam. Ans. (Ax = 938 lb., Ay = 728 lb., B = 814 lb.)

A brace is hinged at one end to a vertical wall and at the other end to a beam 14ft long.   The beam weighs 250 lb. and is also hinged to a vertical wall as shown.  The beam carries load of 500 lb. at the free end.  What will be the compressive force in the brace, and what will be the values of the vertical and horizontal components of the reaction at hinge A? Ans. (Ax = 596 lb., Ay = 45 lb., B = 994 lb.)

Problem 1

A pin-connected A-frame supports a load as shown.  Compute the pin reactions at all of the pins.  Neglect the weight of the members. (Ay=1500 lb., Ey=1000 lb., Bx=1250 lb., By=1750 lb., Cx=1250 lb., Cy=250 lb., Dx = 1250 lb., Dy=750 lb., directions not indicated)

The frame shown supports part of the roof of a small building. Knowing that the tension in the cable is 150 kN, determine the reaction at the fixed end E.

Free-Body Diagram. A free-body diagram of the frame and of the cable BDF is drawn. The reaction at the fixed end E is represented by the force components Ex and Ey and the couple ME.

The other forces acting on the free body are the four 20-kN loads and the 150-kN force exerted at end F of the cable.Equilibrium Equations.

Noting that DF

PROBLEM 4.2 Two children are standing on a diving board of mass 65 kg. Knowing that the masses of the children at C and D are 28 kg and 40 kg, respectively, determine (a) the reaction at A, (b) the reaction at B.

Four boxes are placed on a uniform 14-kg wooden plank which rests on two sawhorses. Knowing that the masses of boxes B and D are 4.5 kg and 45 kg, respectively, determine the range of values of the mass of box A so that the plank remains in equilibrium when box C is removed.

For the beam shown determine the range of values of P for which the beam will be safe knowing that the maximum allowable value of each of the reactions is 45 kips and that the reaction at A must be directed upward.

A check must be made to verify the assumption that the maximum value of P is based on the reaction force at A. This is done by making sure the corresponding value of B is < 45 kips.

Neglecting friction, determine the tension in cable ABD and the reaction at C , 1. when θ =60°., 2. when θ = 30°.

For θ = 30°

For θ = 60°

A 200-lb crate is attached to the trolley-beam system shown. Knowing that a = 1.5 ft, determine (a) the tension in cable CD, (b) the reaction at B. Similarly find if a = 3 ft.

UNIT 2TRUSS RIGID BODIES

Prof. Dr. R. Chandra Sekaran Ph.D (C.Engg)

Rigid Body Structures - TrussesA very common structure used in construction is a truss. An ideal truss is a structure which is composed completely of (weightless) Axial Members that lie in a plane, connected by pinned (hinged) joints, forming triangular substructures (within the main structure), and with the external loads applied only at the joints. See Diagram 1. In real trusses, of course, the members have weight, but it is often much less than the applied load and may be neglected with little error. Or the weight maybe included by dividing the weight in half and allowing half the weight to act at each end of the member. Also in actual trusses the joints may be welded, riveted, or bolted to a gusset plate at the joint. However as long as the centerline of the member coincide at the joint, the assumption of a pinned joint maybe used. In cases where there are distributed loads on a truss, these may be transmitted to a joint by use of a support system composed of stringers and cross beams, which is supported at the joints and transmits the load to the joints.

The procedure for determining the external support reactions acting on a truss is exactly the same as the procedure for determining the support forces in non-truss problems, however the method for determining the internal forces in members of a truss is not the same. The procedures for finding internal forces in truss members are Method of Sections and Method of Joints (either of which may be used), and in fact, one must be very careful not to use these methods with non-truss problems as they will not give correct results. In Diagram 1 we have a truss supported by a pinned joint at Point A and supported by a roller at point D. A vertical load of 500 acts at point F, and a horizontal load of 800 acts at Point C. For this structure we wish to determine the values of the support reactions, and the force (tension/compression) in members BE, BC, and EF. For the first part, determining the external support reactions, we apply the normal static equilibrium procedure:

I. Draw a Free Body Diagram of the entire structure showing and labeling all external load forces and support forces, include any needed dimensions and angles. Note that at the pinned support point A, the best we can do is to put both

an unknown x and y support force, however at point D we only need a unknown y support force since a roller can only be in compression and so must support vertically in this problem.

II. Resolve (break) all forces into their x and y-components.

III. Apply the Equilibrium Equations ( ) and solve for the unknown forces.

(Sum of x-forces)

(Sum of y-forces)

Part 2Once we have determined the values of the external support reactions, we may proceed to determining the values of the forces in the members themselves, the internal forces. In this first example, we will use Method of Joints to determine the force in the selected members. In Method of Joints, rather then analyse the entire structure, or even a member of the structure, we rather examine the joint (pin or hinge) where members come together. As the structure is in static equilibrium, so the pin or joint will be in static equilibrium, and we may apply the static equilibrium conditions (and procedure) to solve for the forces on the joint(s) due to the members, which will also equal the forces in the axial members - due to Newton's third law of equal and opposite reactions (forces).

There are several points to keep in mind as we use method of joints. One is that since we are analysing a point (joint) rather then an extended body, our sum of torque equation will be of no help. That is, since all the forces pass through the same point, they have no perpendicular distance to that point and so produce no torque. This means that to solve completely for the forces acting on a joints we must have a joint which has, at most, two unknown forces acting. In our example (Diagram 2), we notice that there are only two joints which initially have only two unknowns acting - Joint A and Joint D. Thus, we start our process at one of these. Let us begin with Joint A.

Step 1. FBD of the Joint A, showing and labeling all forces acting on the joint. Include needed angles. In Diagram 3 we have shown Joint A with the all the forces which act on the joint. The forces on Joint A, due to members AB and AE, act along the directions of the members (since the members are axial). We choose directions for AB and AE (into or out of the joint). When we solve for the forces AB and AE, if these values are negative it means that our chosen directions were incorrect and the forces act in the opposite direction. A force acting into the joint due to a member means that member is in compression. That is, if a member of a truss is in compression, it will push outward on it ends - pushing into the joint. And likewise, if a truss member is in tension, it will pull outward on the joint. In Diagram 3, we have assumed member AB is in compression, showing its direction into the joint, and that member AE is in tension, showing it acting out of the joint. [A little consideration will show that we have

actually chosen AB in an incorrect direction. We can see this if we consider the y-component of AB, which clearly will be in the - y direction. However, the y-support reaction, 33, is also in the -y direction. There are no other y-forces on the joint, so it can not be in equilibrium if both forces act in the same direction. We will leave AB as chosen to see, if indeed, that the solution will tell us that AB is in the wrong direction.]

Step 2: Resolve (Break) all forces into their x and y components.

Step 3: Apply the Equilibrium Conditions:

(Sum of x-forces)

(Sum of y- forces)

Solving: AE = 756 (Tension), AB = -55 . (The negative sign indicates we selected an incorrect initial direction for AB, AB is in Tension, not Compression.)

Remember, we are trying to find the forces in members BE, BC, and EF. Now that we have the forces in members AE (756 tension) and AB (55 tension), we can move unto a second joint (joint B) and find two of the unknowns we are looking for. We could not have solved for the forces acting at joint B initially, since there were three unknowns (initially) at joint B (AB, BC, and BE). However now that we have analysed joint A, we have the value of the force in member AB, and can proceed to joint B where we will now have only two unknowns to determine (BE & BC)

Joint B: Procedure - Method of Joints

Step 1. FBD of the Joint B, showing and labeling all forces acting on the joint. Include needed angles. Step 2: Resolve (Break) all forces into their x and y components.Step 3: Apply the Equilibrium Conditions:

In Diagram 5, we have on the left the FBD of joint B with all external forces acting on the joint shown, and our initial direction for the forces. If the directions we chose for the unknowns are correct, their values will be positive in the solution. If a value is negative it means the force acts in the opposite direction. On the right side of Diagram 5 is the FBD with all forces resolved into x and y-components. We now apply the Equilibrium Conditions (for joints).

(Sum of x-forces)

(Sum of y-forces)Solving : BC = 44 (Tension) BE = 33 (Compression) Finally, we can now proceed to analyse joint E and determine the force in member EF.

Joint E: Procedure - Method of Joints

Step 1. FBD of the Joint E, showing and labeling all forces acting on the joint. Include needed angles.Step 2: Resolve (Break) all forces into their x and y components.Step 3: Apply the Equilibrium Conditions:

In Diagram 6, we have on the left, the FBD of joint E with all external forces acting on the joint shown, and our initial direction for the forces. If the directions we chose for the unknowns are correct, their values will be positive in the solution. If a value is negative it means the force acts in the opposite direction. The right hand drawing in Diagram 6 is the FBD of joint E with all forces resolved into x and y-components. We now apply the Equilibrium Conditions (for joints).

(Sum of y-forces) Solving : EC = 55 (Tension) EF = 712 (Tension) Since both force values came out positive in our solution, this means that the initial directions selected for the forces were correct. We have now solved our problem, finding both the external forces and forces in members BE, BC, and EF (and along the way, the forces in several other members - See Diagram 7 )

(Sum of x-forces)

We will use the same truss and solve for several internal forces by Method of Sections. And since we previously solved for the external support reactions, we will not repeat that portion, but begin with the external support forces given and move to determine the internal forces in the selected members.

In Method of Sections, we will 'cut' the truss into two sections by drawing a line through the truss. This line may be vertical, horizontal, at some angle, or even curved depending on the problem. The criteria for this line is that we would like to cut through the unknown members (whose internal force value we wish to determine), but not to cut through more than three unknowns (since we will have three equilibrium conditions equations, we can only solve for three unknowns). In this example, cutting the truss once will enable us to find our selected unknowns, however, in some trusses, or for finding more internal forces, one may have to repeat Method of Sections several times to determine all the unknowns.

I. Draw a Free Body Diagram of the structure (section), showing and labeling all external forces, and indicating needed dimensions and angles. (Diagram 3)

II. Resolve (break) all forces into their x and y-components. III. Apply the Equilibrium Equations :

and solve for the unknown forces.

(Here we sum the x-forces)

(Sum of y-forces, including load forces.)

(Sum of Torque with respect to point E.)Solving we obtain: BC = 44 (T), EC = -50 (T), EF = 712 (T) (The negative sign for force EC means that we initially chose it in the incorrect direction. EC acts out of the section and so is in tension, not into the section as shown in the FBD). Thus, we have solved for the internal forces in the members BC, EC, and EF by method of sections.

The structure shown in Diagram 1 is a truss which is pinned to the floor at point A, and supported by a roller at point D. For this structure we wish to determine the value of all the support forces acting on the structure, and to determine the force in member FC by method of joints.

For the first part of the problem we proceed using our normal static's procedure.STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces as well as any needed angles and dimensions. (Note in Diagram 2, we have replaced the pinned support by an unknown x and y force (Ax , Ay), and replaced the roller support by the vertical unknown force Dy.

STEP 2: Break any forces not already in x and y direction into their x and y components.

STEP 3: Apply the equilibrium conditions

Ax = 0 Ay + Dy - 12,000 lbs - 20,000 lbs = 0

:

(-12,000 lbs)(4 ft) - (20,000 lbs)(12 ft) + Dy(24 ft) = 0

Solving for the unknowns: Dy = 12,000 lbs; Ay = 20,000 lbs. These are the external support reactions acting on the structure.

PART 2: Determine the internal force in member FC by method of joints. We begin at a joint with only two unknowns acting, joint D.JOINT D: STEP 1: Draw a free body diagram of the joint, showing and labeling all external forces and load, and including any needed angles.(Diagram 3) We select an initial direction for the unknowns, if their solution value is negative they act in a direction opposite to the direction initially selected.

STEP 2: Resolve all forces into x and y components. (Diagram 3).STEP 3: Apply equilibrium conditions:

-CD + ED cos (66.4o) = 0

12,000 lbs - ED sin (66.4o)= 0

Solving for the unknowns: ED = 13,100 lbs (C); CD = 5,240 lbs (T)

Now that we have calculated the values for ED and CD we can move to joint E. We could not solve joint E initially as it had too many unknowns forces acting on it.

JOINT E: STEP 1: Draw a free body diagram of the joint, showing and labeling all external forces and loads, and including any needed angles. (Diagram 4) We select an initial direction for the unknowns, if their solution value is negative they act in a direction opposite to the direction initially selected.STEP 2: Resolve all forces into x and y components. (Diagram 4).

STEP 3: Apply equilibrium conditions:

FE -(13,100 lbs) cos (66.4o) - CE cos (66.4o) = 0

(13,100 lbs) sin (66.4o) - CE sin (66.4o) = 0

Solving for the unknowns: FE = 10,500 lbs (C); CE = 13,100 lbs (T) (Since force values were positive, the initial direction chosen for the forces was correct.)Now that we have calculated the values for FE and CE we move to joint C. We could not solve joint C initially as it had too many unknowns forces acting on it.

JOINT C: STEP 1: Draw a free body diagram of the joint, showing and labeling all external forces and loads, and including any need angles. (Diagram 5) We select an initial direction for the unknowns, if their solution value is negative they act in a direction opposite to the direction initially selected.

STEP 2: Resolve all forces into x and y components. (Diagram 5).

STEP 3: Apply equilibrium conditions:

5,450 + (13,100 lbs) cos (66.4o) + FC cos (66.4o) - BC = 0

13,100 lbs sin (66.4o) - FC sin (66.4o) = 0

Solving for the unknowns: FC = 13,100 lbs (c); BC = 15,950 lbs (t) Thus, member FC is in compression with a force of 13, 100 lbs.

The structure shown in Diagram 1 is a truss which is pinned to the floor at point A, and supported by a roller at point H. For this structure we wish to determine the value of all the support forces acting on the structure, and to determine the force in member DG by method of sections.

We begin by determining the external support reactions acting on the structure.STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces, as well as any needed angles and dimensions. (Note in Diagram 1, we have replaced the pinned support by an unknown x and y force (Ax , Ay), and replaced the roller support by the vertical unknown force Hy )STEP 2: Break any forces not already in x and y direction into their x and y components. (All forces in x/y directions.)STEP 3: Apply the equilibrium conditions.

: Ax = 0 : Ay + Hy -12,000 lbs - 20,000 lbs - 10,000 lbs = 0

(-12,000 lbs)(20 ft)- (20,000 lbs)(40 ft)- (10,000 lbs)(60 ft)+ Hy(80 ft)=

0

Solving for the unknowns: Ay = 21,500 lbs; Hy = 20,500 lbs.

These are the external support reactions acting on the structure.

Part 2: Now we will find internal force in member DG by method of sections. Cut the truss vertically with a line passing through members DF, DG, and EG. We have shown the section of the truss to the right of the cut. We now treat this section of the truss as if it were a completely new structure. The internal forces in members DF, DG, and EG now become external forces with respect to this section. We have represented these forces with the arrows shown. The forces must act along the direction of the cut member (since all members in a truss are axial members), and we have selected an initial direction either into or away from the section for each of the forces. If we have selected an incorrect initial direction for a force, when we solve for the value of the force, the value will be negative indicating the force acts in the opposite direction of the one chosen initially. We may now proceed with the analysis of this structure using standard Static's techniques.

I. Draw a Free Body Diagram of the structure (section), showing and labeling all external forces, and indicating needed dimensions and angles. (Diagram 2) II. Resolve (break) all forces into their x and y-components. (Diagram 2) III. Apply the Equilibrium Equations ( )

EG + DG cos (51.3o) + DF cos (22.6o) = 0

-10,000 lbs + 20,500 lbs - DG sin (51.3o) - DF sin (26.6o) = 0

-DF cos (26.6o)(15 ft) + (20,500 lbs)(20 ft) = 0

Solving for the unknowns: DF = 30,600 lbs (C); DG = -4,090 (opposite direction)= 4,090 lbs (T); EG = -24,800(opposite direction)= 24,800 lbs (T)

The structure shown below is a truss which is pinned to the wall at point F, and supported by a roller at point A. For this structure:A. Draw a Free Body Diagram showing all support forces and loads.B. Determine the value of all the support forces acting on the structure.C. Determine the force (tension or compression) in member EB by method of joints.Unless otherwise indicated, all joints and support points are assumed to be pinned or hinged joints.

PARTS A & B:STEP 1 : Draw a free body diagram showing and labeling all load forces and support (reaction) forces, as well as any needed angles and dimensions.STEP 2: Break any forces not already in x and y direction into their x and y componentsSTEP 3: Apply the equilibrium conditions.Sum Fx = Ax + Fx = 0

Sum Fy = Fy - 8,000 lbs - 6,000 lbs = 0

Sum TA = (-8,000 lbs)(8 ft) - (6,000 lbs)(16 ft) - Fx(10 ft) = 0

Solving for the unknowns: Ax = 16,000 lbs; Fx = -16,000 lbs; Fy = 14,000 lbs

PART C - Now find internal force in member EB by method of joints.JOINT F:

STEP 1: Draw a free body diagram of the joint.STEP 2: Resolve all forces into x and y components (see diagram).STEP 3: Apply equilibrium conditions:Sum Fx = FE - 16,000 lbs = 0

Sum Fy = 14,000 lbs - FA = 0

Solving for the unknowns: FE = 16,000 lbs (t); FA = 14,000 lbs (t)

JOINT A:

STEP 1: Draw a free body diagram of the joint.STEP 2: Resolve all forces into x and y components (see diagram).STEP 3: Apply equilibrium conditions:Sum Fx = 16,000 lbs - AB -AE cos (68.2o) = 0

Sum Fy = 14,000 lbs - AE sin (68.2o) = 0

Solving for the unknowns: AE = 15,080 lbs (c); AB = 10,400 lbs (c)

JOINT E:

STEP 1: Draw a free body diagram of the joint.STEP 2: Resolve all forces into x and y components (see diagram).STEP 3: Apply equilibrium conditions:Sum Fx = -16,000 lbs + ED + 5,600 lbs + EB cos (68.2o) = 0

Sum Fy = 14,000 lbs - EB sin (68.2o) = 0

Solving for the unknowns: ED = 4,800 lbs (t); EB = 15,080 lbs (t)

The structure shown below is a truss which is pinned to the wall at point E, and supported by a roller at point A. For this structure:A. Draw a Free Body Diagram showing all support forces and loads.B. Determine the value of all the support forces acting on the structure.C. Determine the force in member GC by method of sections.Unless otherwise indicated, all joints and support points are assumed to be pinned or hinged joints.

PARTS A & B:STEP 1: Draw a free body diagram showing and labeling all load forces and support(reaction) forces, as well as any needed angles and dimensions.STEP 2: Break any forces not already in x and y direction into their x & y components STEP 3: Apply the equilibrium conditions. Sum Fx = Ex = 0Sum Fy = Ay + Ey - 12,000 lbs - 6,000 lbs =0, Sum TE = (12,000 lbs)(4 ft) - Ay(12 ft)= 0Solving for the unknowns: Ey = 14,000 lbs; Ay = 4,000 lbs.

PART C: - Now find the internal force in member GC by method of sectionSTEP 1: Cut the structure into "2 sections" with a vertical line which cuts through members GE, GC and BC, and is just to the right of point G (see diagram). The internal forces in members GE, GC and BC now become external forces acting on the left hand section as shown. (We chose directions for these forces which may or may not be correct, but which will become clear when we solve for their values.)

STEP 2: Now treat the section shown as a new structure and apply statics procedure- Draw a free body diagram of the left hand section. - Resolve all forces into x an y components (see diagram).- Apply equilibrium conditions:Sum Fx = - GE cos (37o) + GC cos (37o) + BC = 0Sum Fy = 4,000 lbs - GE sin (37o) - GC sin (37o) = 0Sum TG = (-4,000 lbs)(4 ft) + BC(3 ft) = 0Solving for the unknowns: BC = 5,330 lbs; GE = 6,670 lbs; GC = 0 lbsThe structure shown below is a truss which is pinned to the floor at point D, and supported by a roller at point A. For this structure:A. Draw a Free Body Diagram showing all support forces and loads.B. Determine the value of all the support forces acting on the structure.C. Determine the force in member GC by method of joints.Unless otherwise indicated, all joints and support points are assumed to be pinned or hinged joints.

Ans: GC = 730 lbs (t)

The structure shown below is a truss which is pinned to the floor at point A and also at point H. For this structure:A. Draw a Free Body Diagram showing all support forces and loads.B. Determine the value of all the support forces acting on the structure.C. Determine the force in member FB by any method.Unless otherwise indicated, all joints and support points are assumed to be pinned or hinged joints. Ans: FB = 12,8000 lbs (t)

The structure shown below is a truss which is pinned to the floor at point A and supported by a roller at point F. For this structure:A. Draw a Free Body Diagram showing all support forces and loads.B. Determine the value of all the support forces acting on the structure.C. Determine the force in member CD by any method.Unless otherwise indicated, all joints and support points are assumed to be pinned or hinged joints. Ans: CD = 43 lbs (c)

Determine the force in each member of the truss and state if the members are in tension or compression. Units Used: kN = 1000 N, Given: P1 = 7 kN, P2 = 7 kN, θ = 45 deg.

Joint A: FAB + FAD cos(θ ) = 0 −P1 − FAD sin(θ ) = 0Joint D: FDB cos(θ ) − FAD cos(θ ) + FDC cos(θ ) = 0 (FAD + FDB − FDC)sin(θ ) − P2 = 0Joint C: FCB + FDC sin(θ ) = 0

The truss, used to support a balcony, is subjected to the loading shown. Approximate each joint as a pin and determine the force in each member. State whether the members are in tension or compression. Given: P1 = 600 lb, P2 = 400 lb, a = 4 ft, θ = 45 deg

Joint A: FAB + FAD cos(θ ) = 0−P1 − FAD sin(θ ) = 0Joint B: FBC − FAB = 0−P2 − FBD = 0Joint D: (FDC − FAD)cos(θ ) + FDE = 0(FDC + FAD)sin(θ ) + FBD = 0

Determine the force in each member of the truss and state if the members are in tension or compression. Given:P1 = 20 kN, P2 = 10 kN, a = 1.5 m, e = 2 m

Determine the force in each member of the truss and state if the members are in tension or compression. Given: F1 = 3 kN, F2 = 8 kN, F3 = 4 kN, F4 = 10 kN , a = 2 m, b = 1.5 m

The maximum allowable tensile force in the members of the truss is Tmax, and the maximum allowable compressive force is Cmax. Determine the maximum magnitude P of the two loads that can be applied to the truss. Given: Tmax = 1500 lb, Cmax = 800 lb

Determine the force in each member of the truss and state if the members are in tension or compression. P1 = 10 kN, P2 = 15 kN, a = 2 m, b = 4 m, c = 4 m

Determine the force in each member of the truss in terms of the load P and state if the members are in tension or compression.

Determine the force in each member of the truss in terms of the external loading and state if the members are in tension or compression.

Determine the force in each member of the truss and state if the members are in tension or compression. P1 = 4 kN, P2 = 0 kN, a = 2 m, θ = 15 deg

The Fink truss supports the loads shown. Determine the force in each member and state if the members are in tension or compression. Approximate each joint as a pin.Given: F1 = 500 lb, a = 2.5 ft, F2 = 1ooo lb, θ = 30 deg, F3 = 1ooo lb

Determine the force in each member of the Pratt bridge truss shown. State whether each member is in tension or compression. Assume that the load at G has been removed.

ΣFx=0: Ax=0 , ΣFy =0:Ay−6 kN−6 kN+4.5 kN=0, Ay = 7.5 kN

ΣMA=0:(12 m)Hy−(6 m)(6 kN) − (3 m)(6 kN) = 0 , Hy = 4.5 kN

By inspection of joint G: FFG = 0 and FEG =FGH =3.38 kN (T)

By inspection of joint D:

A Warren bridge truss is loaded as shown. Determine the force in members CE, DE, and DF.

UNIT 5Torsion of circular shaft and power transmitted by shaft ,

Combined bending and torsion on circular shaft Equivalent BM and equivalent twisting moment

Prof. Dr. R. Chandra Sekaran Ph.D (C. Engg)

EXTRAStrain Energy for Axial load,

Bending and Torsion, Castigliono's theorem - Application

Prof. Dr. R. Chandra Sekaran Ph.D (C.Engg)

The principle of superposition is a central concept in the analysis of structures. This is applicable when there exists a linear relationship between external forces and corresponding structural displacements. The principle of superposition may be stated as “the deflection at a given point in a structure produced by several loads acting simultaneously on the structure can be found by superposing deflections at the same point produced by loads acting individually”.

Fig. 1 Cantilever loaded with two concentrated loads

Consider an elastic spring as shown in the Fig.2 . When the spring is slowly pulled, it deflects by a small amount . When the load is removed from the spring, it goes back to the original position. When the spring is pulled by a force, it does some work and this can be calculated once the load-displacement relationship is known. It may be noted that, the spring is a mathematical idealisation of the rod being pulled by a force P axially. It is assumed here that the force is applied gradually so that it slowly increases from zero to a maximum value P. Such a load is called static loading, as there are no inertial effects due to motion. Let the load-displacement relationship be as shown in Fig. 3. Now, work done by the external force may be calculated as,

The area enclosed by force-displacement curve gives the total work done by the externally applied load. Here it is assumed that the energy is conserved i.e. the work done by gradually applied loads is equal to energy stored in the structure. This internal energy is known as strain energy. Now strain energy stored in a spring is

Fig. 2 Linear Spring Fig. 3 Force Displacement relation

In SI system, the unit of work and energy is the joule (J), which is equal to one Newton metre (N.m). The strain energy may also be defined as the internal work done by the stress resultants in moving through the corresponding deformations. There are normal stresses ( ), and shear stresses ( ), acting on an element. Corresponding to normal and shear stresses we have normal and shear strains. Now strain energy may be written as,

in which σT is the transpose of the stress column vector ,

The strain energy may be further classified as elastic strain energy and inelastic strain energy as shown in Fig. 5

The shaded area BCD is known as the elastic strain energy.

Strain energy = Complementary strain energy Consider a member of constant cross sectional area A, subjected to axial force Pas shown in Fig. . Let E be the Young’s modulus of the material. Let the member be under equilibrium under the action of this force, which is applied through the centroid of the cross section. Now, the applied force P is resisted by uniformly distributed internal stresses given by average stress

Fig. 4 Infinitesimal element

Fig. 5 Elastic and Inelastic strain energy

The work done by external loads

The external work is stored as the internal strain energy. Hence, the strain energy stored in the bar in axial deformation is,

The incremental elongation du of small element of length dx of beam is given by,

The total elongation of the member of length L may be obtained by integration

(5)

(1)

(2)

(3)

(4)

Substituting equation (2) in (4) we get,

Strain energy under axial load

Fig. 6 Axially loaded beam and an element

Strain energy due to bending Consider a prismatic beam subjected to loads as shown in the Fig.7. The loads are assumed to act on the beam in a plane containing the axis of symmetry of the cross section and the beam axis. It is assumed that the transverse cross sections, which are perpendicular to centroidal axis, remain plane and perpendicular to the centroidal axis of beam.

Fig. 7 Bending Deformation

Consider a small segment of beam of length subjected to bending moment as shown in the Fig. 7. Now one cross section rotates about another cross section by a small amount dθ. From the figure,

where R is the radius of curvature of the bent beam and EI is the flexural rigidity of the beam. Now the work done by the moment M while rotating through angle dθ will be stored in the segment of beam as strain energy dU. Hence,

Substituting for dθ in equation (7), we get,

(6)

(7)

(8)

Now, the energy stored in the complete beam of span L may be obtained by integrating equation (8). Thus,

(9)

Strain energy due to transverse shear

The shearing stress on a cross section of beam of rectangular cross section may be found out by the relation

where Q is the first moment of the portion of the cross-sectional area above the point where shear stress is required about neutral axis, V is the transverse shear force, b is the width of the rectangular cross-section and Izz is the moment of inertia of the cross-sectional area about the neutral axis. Due to shear stress, the angle between the lines which are originally at right angle will change. The shear stress varies across the height in a parabolic manner in the case of a rectangular cross-section. Also, the shear stress distribution is different for different shape of the cross section. However, to simplify the computation shear stress is assumed to be uniform (which is strictly not correct) across the cross section. Consider a segment of length ds subjected to shear stress . The shear stress across the cross section may be taken as

(10)

(11)in which A is area of the cross-section and is k the form factor which is dependent on the shape of the cross section.

Fig. 8 Shearing Deformation

the deformation du as

Where is the shear strain and is given by

The total deformation of the beam due to the action of shear force is

Now the strain energy stored in the beam due to the action of transverse shear force is given by,

The strain energy due to transverse shear stress is very low compared to strain energy due to bending and hence is usually neglected. Thus the error induced in assuming a uniform shear stress across the cross section is very small.

(12)

(13)

(14)

(15)

(16)

Strain energy due to torsion Consider an elemental length ds of the shaft. Let the one end rotates by a small amount dφ with respect to another end. Now the strain energy stored in the elemental length is,

We know that

where, G is the shear modulus of the shaft material and J is the polar moment of area.

Substituting for dφ from (18) in equation (16), we obtain

(17)

(18)

(19)

The total strain energy stored in the beam may be obtained by integrating equation (19)

Hence the elastic strain energy stored in a member of length s (it may be curved or straight) due to axial force, bending moment, shear force and torsion is summarised below.

Fig. 9 Torsional Deformation

(20)

(20)

(16)(5)

(9)

Castigliano’s first theorem is being used in structural analysis for finding deflection of an elastic structure based on strain energy of the structure. The Castigliano’s theorem can be applied when the supports of the structure are unyielding and the temperature of the structure is constant. For linearly elastic structure, where external forces only cause deformations, the complementary energy is equal to the strain energy. For such structures, the Castigliano’s first theorem may be stated as “the first partial derivative of the strain energy of the structure with respect to any particular force gives the displacement of the point of application of that force in the direction of its line of action”.

Fig. 9 Simply supported beam with loadings

Let be the forces P1, …. acting at x1, , , ,…… from the left end on a simply supported beam of span L . Let u1 , ….. be the displacements at the loading points respectively as shown in Fig.9 . Now, assume that the material obeys Hooke’s law and invoking the principle of superposition, the work done by the external forces is given by

P2 P3 Pn x2 x3 xnu2 , u3 un

Work done by the external forces is stored in the structure as strain energy in a conservative system. Hence, the strain energy of the structure is,

Displacement u1 below point P1 is due to the action of P1 , P2 , P3 ........ Pn acting at distances respectively from left support. Hence, u1 may be expressed as,

In general,

where is the flexibility coefficient at i due to unit force applied at j. Substituting the values of in equation (22) from equation (24), we get,

(21)

(22)

(23)

(24)

From Maxwell- Betti’s reciprocal theorem Hence, equation (25) may be simplified as,

(25)

(26)

Equation (27) is nothing but displacement u1 at the loading point.

(27)

Now, differentiating the strain energy with any force P1 gives,

In general,

Hence, for determinate structure within linear elastic range the partial derivative of the total strain energy with respect to any external load is equal to the displacement of the point of application of load in the direction of the applied load, provided the supports are unyielding and temperature is maintained constant.

This theorem is advantageously used for calculating deflections in elastic structure.

(28)

Find the displacement and slope at the tip of a cantilever beam loaded as in Fig. Assume the flexural rigidity of the beam EI to be constant for the beam.

Moment at any section at a distance x away from the free end is given by

Strain energy stored in the beam due to bending is

Substituting

Now, according to Castigliano’s theorem, the first partial derivative of strain energy with respect to external force P gives the deflection at A in the direction of applied force. Thus,

To find the slope at the free end, we need to differentiate strain energy with respect to externally applied moment M at A. As there is no moment at A, apply a fictitious moment M 0 at A. Now moment at any section at a distance x away from the free end is given by

Now, strain energy stored in the beam may be calculated as,

Taking partial derivative of strain energy with respect to M 0 , we get slope at A.

But actually there is no moment applied at A. Hence substitute = 0 in equation we get the slope at A. M 0

A cantilever beam which is curved in the shape of a quadrant of a circle is loaded as shown in Fig. The radius of curvature of curved beam is R, Young’s modulus of the material is E and second moment of the area is I about an axis perpendicular to the plane of the paper through the centroid of the cross section. Find the vertical displacement of point A on the curved beam.

The bending moment at any section θ of the curved beam is given by

Strain energy U stored in the curved beam due to bending is,

Differentiating strain energy with respect to externally applied load, P we get

Castigliano’s Second Theorem In any elastic structure having n independent displacements corresponding to external forces along their lines of action, if strain energy is expressed in terms of displacements then n equilibrium equations may be written as follows.

The strain energy of an elastic body may be written as

We know

where is the stiffness coefficient and is defined as the force at i due to unit displacement applied at j. Hence, strain energy may be written as,

From Maxwell- Betti’s reciprocal theorem Hence, equation (31) may be simplified as,

(29)

(30)

(31)

(32)

Now, differentiating the strain energy with respect to any displacement u1 gives the applied force P1 at that point, Hence, (33)

Or, (34)