rch som 01 introduction

7/23/2019 Rch Som 01 Introduction

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Mechanics of Solids

Introduction

Prof. TVK Bhanuprakash


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External Loads.

A body is subjected to only two types of external loads; namely, surfaceforces and body forces

Equilibrium of a Deformable Body


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Support Reactions


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Internal Resultant Loadings


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Normal force, N

This force acts perpendicular to the area. It is developed whenever the external

loads tend to push or pull on the two segments of the body.

Shear force, V

The shear force lies in the plane of the area and it is developed when the

external loads tend to cause the two segments of the body to slide over one

another.

Torsional moment or torque, T

This effect is developed when the external loads tend to twist one segment of

the body with respect to the other about an axis perpendicular to the area.

Bending moment, M

The bending moment is caused by the external loads that tend to bend the

body about an axis lying within the plane of the area.


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Coplanar Loadings


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Example 1

Determine the resultant internal loadings acting on the cross

section at C of the cantilevered beam shown


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Example 2

Determine the resultant internal loadings acting on the cross

section at C of the machine shaft shown. The shaft is supported

by journal bearings at A and B, which only exert vertical forceson the shaft.


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F1-1

Determine the internal normal force, shear force, and bending

moment at point C in the beam.


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F1-6

Determine the internal normal force, shear force, and bending

moment at point C in the beam.


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Stress

Units.Since stress represents a force per unit area, SI units for both normal

and shear stress are specified in the basic units of Newton per square

meter

This unit, called a pascal is rather small, and in engineering work

prefixes such as kilo- symbolized by k, mega- symbolized by M, or

giga- symbolized by G, are used to represent larger, more realistic

values of stress.


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Homogeneous and Isotropic


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Uniaxial State of Stress


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Example 6

A bar has a constant width of 35 mm and a thickness of 10 mm.

Determine the maximum average normal stress in the bar when it

is subjected to the loading shown.


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The 80-kg lamp is supported by two rods AB and BC as shown. If AB has a

diameter of 10 mm and BC has a diameter of 8 mm, determine the average

normal stress in each rod.


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Stresses on Inclined Planes

13-09-2013 20Strength of Materials - I (Introduction)


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1. Maximum normal stress is P/A, and it acts on the cross

section of the bar (that is, on the plane q = 0).2. The shear stress is zero when q = 0, as would be expected.3. The maximum shear stress is P/2A, which acts on the

planes inclined atq = 45o to the cross section.In summary, an axial load causes not only normal stress but also shear stress. The

magnitudes of both stresses depend on the orientation of the plane on which they act.


Repeated for convenience


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Stresses acting on two mutually

perpendicular inclined sections of a bar.

By Substituting q=q+90o, we get stresses ona plane perpendicular to q plane

Stresses acting on mutually perpendicular, or complementary planes, they

are called complementary stresses.



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The shear stresses that act on complementary planes

have the same magnitude but opposite sense.

1. The design of axially loaded bars is usually based on the maximum normal

stress in the bar.

2. This stress is commonly called simply the normal stress and denoted bys.

3. The design criterion thus is thats = P/A must not exceed the working stress

of the material from which the bar is to be fabricated.

4. The working stress, also called the allowable stress, is the largest value of

stress that can be safely carried by the material.

5. Working stress, denoted bysw, will be discussed more fully later



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Shear Stress

1. Shear stress is tangent to the plane on which it acts.

2. Shear stress arises whenever the applied loads cause one

section of a body to slide past its adjacent section.

3. The two plates that are joined by a rivet.

4. From FBD, the rivet must carry the shear force V.

5. Because only one cross section of the rivet resists the shear, the

rivet is said to be in single shear.


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Shear Stress (Contd)

1. The bolt of the clevis carries the load P across two cross-sectional areas,

the shear force being V = P/2 on each cross section.

2. Therefore, the bolt is said to be in a state of double shear.


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Shear Force (contd)

1. A circular slug is being punched out of a metal sheet.

2. Here the shear force is P and the shear area is similar to the

milled edge of a coin.

3. The loads are sometimes referred to as direct shear to

distinguish them from the induced shear.

Induced Shear

(Fig Redrawn for clarity)

Average Shear Stress


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Pure Shear

All four shear stresses must have equal magnitude and be directed either toward or

away from each other at opposite edges of the element

This is referred to as the complementary property of shear


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Example 2

If the wood joint has a width of 150 mm, determine the average

shear stress developed along shear planes aa and bb.


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Example 1.12

The inclined member is subjected to a compressive force of 600 lb.

Determine the average compressive stress along the smooth areas of

contact defined by AB and BC, and the average shear stress along the

horizontal plane defined by DB


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Exercise 1-50

The block is subjected to a compressive force of 2 kN.

Determine the average normal and average shear stress

developed in the wood fibers that are oriented along section a

a at 30 with the axis of the block.


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174. The lever is attached to the shaft A using a key that has a width d

and length of 25 mm. If the shaft is fixed and a vertical force of 200 N is

applied perpendicular to the handle, determine the dimension d if the

allowable shear stress for the key is tallow= 35 MPa.


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175. The joint is fastened together using two bolts. Determine the

required diameter of the bolts if the failure shear stress for the bolts is

350 Mpa. Use a factor of safety for shear of F.S. = 2.5.


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Bearing Stress

1. If two bodies are pressed against each other, compressive forces are

developed on the area of contact.

2. The pressure caused by these surface loads is called bearing stress.

3. Examples of bearing stress are the soil pressure beneath a pier and

the contact pressure between a rivet and the side of its hole.

4. If the bearing stress is large enough, it can locally crush the

material, which in turn can lead to more serious problems.

5. To reduce bearing stresses, engineers sometimes employ bearing

plates, so that the contact forces are distributed over a larger area.


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Bearing Stress (Contd)

1. Consider the lap joint formed by the two plates that are riveted

together as shown.2. The bearing stress caused by the rivet is not constant; it actually

varies from zero at the sides of the hole to a maximum behind

the rivet.


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Bearing Stress (Contd)

1. The difficulty is avoided by assuming that the bearing stress

sb

is uniformly distributed over a reduced area.

2. The reduced area Ab is taken to be the projected area of rivet

3. From FBD bearing force Pb = P

4. The bearing stress becomes


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Example 1 (Shear and Bearing Stress)

The lap joint is fastened by four rivets of 3/4-in. diameter. Find the

maximum load P that can be applied if the working stresses are 14 ksi for

shear in the rivet and 18 ksi for bearing in the plate. Assume that the

applied load is distributed evenly among the four rivets, and neglect friction

between the plates.

We see that the equilibrium


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Solution:

(FBD of Lower plate is drawn)

We see that the equilibrium

condition is V =P/4.

Given problem


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Design for Shear Stress in Rivets

The value of P that would cause the shear stress in the rivets to reach its

working value is found as follows:


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Design for Bearing Stress in Plate

1. The shear force V=P/4 that acts on the cross section of one rivet is

equal to the bearing force Pb due to the contact between the rivet and

the plate.

2. The value of P that would cause the bearing stress to equal its working

value is computed from

Comparing, the maximum safe load P that can be applied to the lap joint is

P = 24 700 lb, with the shear stress in the rivets being the governing design

criterion.


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Example 2: (Normal and Bearing Stresses)

The shaft is subjected to the axial force of 40 kN. Determine the average

bearing stress acting on the collar C and the normal stress in the shaft


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Solution:

( )

( ) 2

2

1044

102534m..A

m..A

b

s

==

==


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P = 82.5 kN P = 72.5 kN (Controls!)


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Example 3: (Bearing Stress)

The assembly consists of three disks A, B, and C that are used to support the

load of 140 kN. Determine the smallest diameter of the top disk, the diameter

within the support space, and the diameter of the hole in the bottom disk. The

allowable bearing stress for the material is (sallow)b = 350 Mpa, and allowable

shear stress is tallow= 125 MPa.


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Solution


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Allowable Stress

In some cases, such as columns, the applied load is not linearly related to

stress and therefore only the first equation can be used to determine thefactor of safety.


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Simple Connections

Ex 1.14

The suspender rod is supported at its end by a fixed-connected circular

disk. If the rod passes through a 40-mm-diameter hole, determine theminimum required diameter of the rod and the minimum thickness of the

disk needed to support the 20-kN load. The allowable normal stress for

the rod is 35 Mpa and the allowable shear stress for the disk is 60 MPa,

Di t f th d


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Diameter of the rod

Thickness of the disk


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F118. Determine the maximum average shear stress

developed in the 30-mm-diameter pin.

Total force = 50 kN. Double shear .

So Stress = 50/(2*pi*r2)


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F122. The pin is made of a material having a failure shear

stress of 100 MPa . Determine the minimum required

diameter of the pin to the nearest mm. Apply a factor of safety

of F.S. = 2.5 against shear failure.


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Exercise 1-101The 200-mm-diameter aluminum cylinder supports a

compressive load of 300 kN. Determine the average normal

and shear stress acting on section a

a.

rch som 01 introduction

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