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CHAPTER ONE

Introduction—Concept of Stress

1.1 Introduction

1.2 Forces And Stresses

1.3 Axial Loading: Normal Stress

1.4 Shearing Stress

1.5 Bearing Stress In Connections

1.6 Application To The Analysis Of Simple Structures

1.7 Stress On An Oblique Plane Under Axial Loading1.8 Stress Under General Loading Conditions: Components Of Stress

1.9 Ultimate And Allowable Stress: Factor Of Safety

Review And Summary

1.1 INTRODUCTION

The main objective of the study of the mechanics of materials is to

provide the engineer with the means of analyzing and designing various

machines and load-bearing structures. Both the analysis and the design

of a given structure involve the determination of stresses and

deformations. This first chapter is devoted to the concept of stress.

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1.2 FORCES AND STRESSES

Forces ( external and internal ): Consider the structure in

Fig. 1.1, from the knowledge of statics we know that theboom AB and rod BC are two-force members. The forces

in the two members can be determined by “Equilibrium

Conditions” using the “Free-body Diagram” as the

following.

Fig. 1.1

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Fig. 1.2 Fig. 1.3

9

Fig. 1.4

From the free-body diagram of pin B and the force triangle (Fig. 1.3),

we obtain:

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Stresses: The force per unit area, or the intensity of the forces

distributed over a given section, is called the stress on that section and

is denoted by Greek letter σ . The stress formula:

=

(1.1)

Fig. 1.5 Fig. 1.6

The sign of stress: + tensile stress (member in tension)

− compressive stress (member in compression)

The unit of stress: =

, a Pascal. SI metric unites are used.

1 kPa = 103 Pa = 103 /2

1 MPa = 106 Pa = 106 /2

1 GPa = 109 Pa = 109 /2

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1.3 AXIAL LOADING: NORMAL STRESS

• Axial loading: Forces directed along the axis of

the rod

• Normal stress: Average of the force per unit area

in a member under axial loading

=

(1.1)

• Stress at point Q:

• The relation between the axial load P

And stress (by equilibrium)

• The actual distribution of stresses in any

given section is statically indeterminate.

Fig. 1.7

Fig. 1.8

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• A uniform distribution of stress is possible only if the line of action

of the concentrated load P pass through the centroid of the section

considered.

Fig. 1.9

• If a two-force member is loaded axially , but eccentrically as

shown in Fig. 1.11 , then the internal forces in a given section

must be equivalent to a force P applied at the centroid of the

section and a couple M of moment M = Pd , the distribution of

stresses cannot be uniform.

Fig. 1.10 Fig. 1.11

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1.4 SHEARING STRESS

• Shearing Force or Shear

• Average Shearing Stress

• Bolt, Pins and Rivets

• Single shear and Double shear

Fig. 1.12 Fig. 1.13

• Average shearing stress:

• The distribution of shearing stresses across the section cannot be

assumed uniform

• Rivet – single shear (free body diagram analysis)

Fig. 1.14 Fig. 1.15

A

P=aveτ

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“Equilibrium Condition” “Free-body diagram”

Fig. 1.16

Fig. 1.17

• Double shear (free-body diagram)

AF

AP

2ave

==τ

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1.5 BEARING STRESS IN CONNECTIONS

Fig. 1.18

Fig. 1.19

• Bearing stress:

d t P

AP == b

σ

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1.6 APPLICATION TO THE ANALYSIS OF

SIMPLESTRUCTURES

Fig. 1.20

• Determination of the normal stress in two-force members

First to calculate the force in each of the members by free-body

diagram analysis and equilibrium equation for a rigid body:

Σ = 0, Σ = 0, Σ M = 0

• Calculate the tensile stress in each cross-section of rod BC

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• Determination of the shearing stress

in various connections

Single shear (Pin at C)

Double shear (Pin at A)

For example:

MPa102m10491

N105026

3

=×

×==

− A

PPinC aveτ

Fig. 1.22

For example: Pin at B, the largest shearing stress

occur in section G and H, where

MPa9.50m10491

kN2526, =

×==

− A

PGave Bτ

• Determination of bearing stresses in members, pins and

brackets.

Fig. 1.21

Fig. 1.23

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1.7 STRESS ON AN OBLIQUE PLANE UNDER AXIAL

LOADING

• Axial force cause both normal and

shearing stresses on planes not

perpendicular to the axis of the member.

Considering the equilibrium of

Fig. 1.26 (b) and resolving P into normal

component F and shearing component V ,

we have:

θ θ sincos PV PF ==

The corresponding average normal and

shearing stresses are

θ θ

θ

θ

τ

θ

θ

θ σ

θ

θ

cossin

cos

sin

cos

cos

cos

00

2

00

A

P

A

P

A

V

A

P

A

P

A

F

===

===

Fig. 1.24

Fig. 1.26

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Or:

σ =

02 = 0

2

=

0 = 0

We can see that:

At = , σ = =

= 0

and 0 when 90°

At = and = °， = 0

and at = 45°,

= =

2=

1

20

At = °, = σ =

2

Fig. 1.27

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1.8 STRESS UNDER GENERAL LOADING

CONDITIONS: COMPONENTS OF STRESS

Fig. 1.28

• Consider a body subjected to several

loads 1, 2, etc. Now let’s define the stress at a

point Q inside the body. Define the stress on a

small area Δ normal to the x axis:

A

F x

∆

∆,

A

V x

y

∆

∆,

A

V x

z

∆

∆

Letting Δ 0, we define the stress

components at point Q as

A

V

A

V

A

F

x

z

A

xz

x

y

A

xy

x

A x

∆

∆=

∆

∆=

∆

∆=

→∆→∆

→∆

limlim

lim

00

0

τ τ

σ

• The sign of stresses

Fig. 1.30

Fig. 1.31

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Similarly we can define the stress

components on the small area (or

point Q) normal to the y and z axes:

, ,

, ,

• Important relations betweenshearing stress components:

Considering the six equilibrium

conditions of small cube of side a

0

0

===

===

∑∑∑

∑∑∑

z y x

z y x

M M M

F F F

From

( ) ( ) 0=∆−∆=∑ a Aa A M yx xy z τ τ

We have yx xy τ τ =

Similarly, zy yz zy yz τ τ τ τ ==

Fig. 1.32

Fig. 1.33

Fig. 1.34

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• Only 6 stress components are required to define the

condition of stress at a given point Q , instead of 9 as originally

assumed

3 Normal Stresses , ,

3 Shear Stresses , ,

• At a given point, shear cannot take place in one plane only; an equal

shear must be exerted on another plane perpendicular to the first

one

Fig. 1.35

• Same loading condition may

lead to different

interpretations of stress

situation at a given point, depending upon the orientation of the

element considered. More detail will be

discussed in Chapter 6.

Fig. 1.36

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1.9 ULTIMATE AND ALLOWABLE STRESS: FACTOR

OF SAFETY

• Concept of ultimate load and its determination

Ultimate normal stress and ultimate shear stress

• Concept of allowable load —working load or design load

• Factor of safety = F.S.= ultimate loadallowable load

Factor of safety =ultimae stress

allowable stress

• The determination of the factor of safety should be based upon

many considerations.

Fig. 1.37 Fig. 1.38

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REVIEW AND SUMMARY

(On PP.44-46 of the text book)

xial loading. Normal stress

Transverse forces. Shearing stress

Single and double shear

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Bearing stress

Stresses on an oblique section

Stress under general loading

Ultimate strength

Ultimate load and allowable load

Factor of safety

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