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CN4119 Design I

Part 2 Mechnical Design

References

Beer, F. P. and E. R. Johnston, Jr.Mechanics of Materials, Second Edition in SI Units,McGraw Hill, Book Company.Singapore, 1992.

Chapter 1. Tension

1. Stress

Mechanics of Materials. Strength and Rigidity. Determination of stresses and deformations of members in a structure.

Mechanical design:- Strength requirement: [].- Rigidity requirement: deformations an allowable value.

Analysis of stresses and strains.

1.1. Stress Normal stress

Consider a rod subjected to a tensile force P at its end. To determine the internal force at any point, we pass through a cross section, considerthe equilibrium of either portion of the rod, we have a force equilibrium equation in theaxial direction. The internal force at any cross section within the rod is thus equal to theload P. Will the rod break under the action of P?

The ability of the material to withstand the intensity of the internal force, but not theinternal force itself. Assume that the internal force P is uniformly distributed over the entire cross sectionA, we call = P/A the normal (tensile) stress on any point of area A.

In general, consider a small area A around a point Q on a cross section A, on whichthe internal force is F, we define

= limA0

F

A

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as the normal stress at point Q. Therefore = (Q) represents the stress distributionover cross section A. Accordingly,

P = A

(Q) dA.

SI units of stress: N/m2 = P a (Pascal).

Multiples of P a:1 kP a = 103 P a = 103 N/m2

1 M P a = 106 P a = 106 N/m2

1 GP a = 109 P a = 109 N/m2

The design of a rod requires [],

where

[] = y/ns for ductile materials,

[] = c/ns for brittle materials.

y is the yielding stress of the ductile material specimen and c is the stress of thebrittle specimen at break in tensile test, which is coducted in a tensile machine. ns iscalled the factor of safety, which should be determined by considering the imporatnce ofthe structure componenet and the price of the material.

1.2. Shear Stress

Consider a member AB subjected to

transverse forces P and P in a close distance. To determine the internal force, we passthrough a section at position C between the two forces and consider the equilibrium foreither portion of the member. We conclude the internal force must be in the section andits magnitude is P. We call P in the section shearing force, or simply, shear.

Whether the member breaks depends

upon both the magnitude of P and the value of A. Assume a uniform distribution of Pover A, we define

= P/A

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the shear stress at any point on A.

General definition:

= limA0

F

A

= (Q) is a function of the position Q on section A and represents the intensity of theshearing forces at point Q. Accordingly

P =

A

(Q) dA.

Design of bolts, pins and rivets: [],

where [] is the allowable shear stress of the material used.

1.3. Stress State at a Point and Stress Components

Consider a two force member subjected to axial forces P and P

. We pass though asection at an angle with the normal plane. What would be the stresses on this obliqueplane? Resolving P into components F and V, which are normal and tangent to the sectionrespectively

F = Pcos, V = P sin.

The area A0 of this section is A0 = A/ cos , the normal and shear stress on this section:

=F

A0=

P cos

A/ cos =

P

Acos2

=V

A0 =P sin

A/ cos =P

A sin cos

We can see

= 0, =P

A= max, = 0;

= 45, =P

2A, =

P

2A= max

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i.e., and depend on the orientation of the section.

Scholar, vector and tensor. Stress is a (second order) tensor. It has 32 = 9 components.

Consider a body subjected to loads P1, P2, P3, . Consider point Q within this body.To know the stresses at Q, we should calculate the stresses on the three co-ordinate planesat Q.

First, pass through point Q a section perpendicular to x-axis. Consider an area elementA around Q.

Resolve the internal force on A to a normal force Fx

and a shearing force Vx

.Further resolve Vx to Vxy and Vxz in y and z direction, respectively.

The definition of the normal & shear stress at Q on this section:

x = limA0

Fx

A, xy = lim

A0

VxyA

, xz = limA0

VxzA

.

Note

1) x = xx is the normal stress.

2) xy and xz have two subscripts: the first indicates the direction of the section, thesecond indicates the direction of the shear force.

3) If the section is in +x direction, x, xy and xz are + if the force is in + co-orddirection. They are - if they are in - co-ord direction. However, If the section is in -xdirection, x, xy and xz are + if the force is in - co-ord direction. They are - if they arein + co-ord direction.

Similarly, at point Q, we can define

y, yx and yz on the section of outnormal in y direction and

z, zx and zy on the section of outnormal in z direction.

The stress state at Q,

x xy xz

yx y yz

zx zy z

which is called the stress tensor, or stress matrix.

The stress state at Q can also be represented by a small cube centred at Q with variousstresses exerted on each of its six faces.

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The stress tensor is symmetric. The two shear stresses of the same indices is equal.

2. Hookes Law2.1. 1-D Hookess Law

For small deformation, i.e., for < y, from the - diagram, we have

= E

which is known as the 1D Hookes law after the English mathematician Robert Hooke(1635-1703). E is called modulus of elasticity, or Youngs modulus after the Englishscientist Thomas Young (1773-1829).

Note since is dimensionless, E has the same dimension as the stress.2.2. Poissons Ratio

We assume that the materials considered are 1) homogeneous; and 2) isotropic. Letsconsider a slender bar under an axis loading P in x direction, the stress within the bar isx = P/A. According to the Hookes law, the strain in x direction is x = x/E. Fromequilibrium consideration, the stress is zero in any lateral direction, i.e., y = z = 0.However, y = 0 and z = 0. The elongation in the axial direction is always accompaniedby a contraction in any transverse direction. We call

= y

x=

z

x

Poissons Ratio after the French mathematician Simeon Denis Poisson (1781-1840). In the case of 1D axial loading

x =x

E, y = z =

x

E

2.3. Hookes Law for Multiaxial LoadingsWe consider a cube and assume it is subjected axial loading x, y, and z, in x, y, and

z direction, respectively.If the cube is subjected to x alone,

x =x

E, y = z =

x

E

Similarly, we have

y =y

E, z = x =

y

E

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for the single axial load y, and

z =z

E, x = y =

z

E

for the single axial load z.For small deformation, superposition can be applied. That is, we assume that the

deformation produced by each load is not affected by other loads. The Hookes law formultiaxial loading:

x =1

E[x (y + z)]

y =1

E[y (z + x)]

z =1

E[z (x + y)]

For < y, there is a proportional segment of the - diagram, which can be determinedby test of a specimen in a torsion machine..

Hookes law in shear:xy = Gxy

G is called the modulus of rigidity, or the shear modulus.Similarly,

yz = Gyz zx = Gzx

The six equations are called the 3D Hookes law. They are also called the 3D constitutiveequations for linearly elastic materials.

1-D Hookes Law: = EFor members under axial loading, both the stress and the strain are uniformly distribu-

ted over the member far from the two ends, i.e.

=P

A =

L

Therefore

=P L

EA

i.e., the deformation is proportional to the load and the length, inversely proportional tothe cross section area.

this tension formula is valid for1) homogeneous rod,2) uniform cross section area A, and3) P is loaded at the two end of the rod.

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