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Chapter 2

AXIAL LOADING

2.1 INTRODUCTION

2.2 PRELIMINARY DISCUSSION

2.3 MATERIAL BEHAVIOURS

2.4 AXIAL DEFORMATIONS

2.5 STATIC INDETERMINACY

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2.1 Introduction

There are many engineering structures of

interest that are composed of members having

one-dimensional stress distribution over much

of their volumes, e.g. the planar truss

structures. In the analysis of truss structures,

the truss is treated as rigid body, i.e. structural

member that does not undergo deformation.

Real structural members, however, do deform

when subjected to external loads.

Determination of both the stresses as well as

deflections in such systems under the action of

given loadings for a given geometry of the

members is called analysis. There is also the

possibility that we may wish to choose the

cross-sectional areas of the members in order

for them to withstand expected loads safely

and economically. This is what we call design

and consists of a series of analyses leading in a

convergent manner to the optimal system.

In this chapter, we are going to capitalize what

we have learnt in Chapter 1 – normal stress

and add an additional ingredient – normal

strain, to complement and complete our

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analysis ability of one-dimensional structural

member.

2.2 Preliminary discussion

In the study of mechanics of solid, there are

three major components:

1. Force and stress relationship;

2. Strain and displacement relationship;

3. Stress and strain relationship or the

constitutive hypothesis.

These three items need to be distinguished and

each of them can be separately relevant, if not

directly.

In HES1125, you have covered item 1, in

which you learnt about statics and the

conditions for equilibrium. You were exposed

to the analysis and design of simple

connection using the various force and stress

relationships. The problems that you have

been solving up to now are termed statically

determinate system. By statically determinate

system, we mean that the system possesses a

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determinate solution provided that item 1 is

satisfied.

When item 1 along is insufficient to obtain a

solution for a given system, we say the system

is statically indeterminate. Under this

situation, item 2 and 3 need to come in. Let us

put item 1-3 in mathematical form that you can

recognize

1. Force and stress relationship:

0

limA

F

Aσ

∆ →

∆=

∆ (2.1)

2. Strain and displacement relationship

0

limx

u

xε

∆ →

∆=

∆ (2.2)

3. Stress and strain relationship

( )σ σ ε= (2.3)

You are familiar with (2.1) already. Let us see

what we can say about (2.2) and (2.3).

• Eq. (2.2) states that strain is defined as

displacement per unit length of material.

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• One of the common form of strain

definition we use frequently is the

engineering strain:

o

o o

l ll

l lε

−∆= = (2.4)

• Apart from (2.4) we can also define the so

called true strain:

lno

L

Lo

dl L

l Lε = =∫ (2.5)

• Eq. (2.3) hypothesized that stress is a

unique function of strain. One of the most

familiar form of (2.3) is the Hooke’s linear

elastic relationship

Eσ ε= (2.6)

• Eq. (2.6) is not unique. We can also have

non-linear elastic relationship, for example

n

o Eσ σ ε= + (2.7)

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2.3 Material Behaviors

Equation (2.6) and (2.7) are called constitutive

equations or material models. Material model

is used to describe the relationship between

stress and strain. This stress-strain relationship

is determined via tensile test experiment as

shown in Fig. 2.1

There are many types of material models:

linear and nonlinear elastic, elastoplastic,

viscoelastic, viscoplastic, thermoplastic etc.

These models are used to describe different

material behaviors.

Fig. 2.1

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There are many types of materials in used in

engineering. Steel, brass, iron, copper and

various types of alloys are conventional

materials used in industries. Composite and

reinforced-polymer etc are non-conventional

type of expensive materials used in high-tech

industries, e.g. aerospace.

Among all these, two types of material

behavior are common – ductile and brittle

materials. Materials exhibiting little or no

plastic deformation up to fracture, such as cast

iron and glass, are called brittle materials.

Materials exhibiting substantial plastic

deformation up to the point of fracture, such as

low-carbon steel, are called ductile materials.

There is a wide range of behavior between the

brittle and ductile designation.

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Fig. 2.3 Ductile material

Fig. 2.4 Brittle material

Fig. 2.2 Ductile Fig. 2.3 Brittle

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Below the elastic limit, Hooke’s law usually

applies.

Fig. 2.5

When stress exceeds the elastic limit, plastic

deformation takes place. Plastic deformation is

characterized by permanent irreversible

straining of the material.

Fig. 2.6

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When the stress is repeatedly cycled in time,

we find that the stress required for rupture is

reduced. This important phenomenon is called

fatigue failure. This is a highly specialized

subject – fracture mechanics and damage

theory.

Fig. 2.7

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2.4 Axial deformation

Fig. 2.8

From Hooke’s law

, P

EE AE

σσ ε ε= = = (2.8)

From definition of strain

L

δε = (2.9)

Equating and solving for displacement,

PL

AEδ = (2.10)

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Eq. (2.10) can be generalized as

1

ni i

i i i

PL

A Eδ

=

=∑ (2.11)

Notice that in this case, the bars are aligned in

parallel. If the bars are aligned in series, the

force will be constant.

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Example: Determine the deformation of the

steel rod shown under the given loads given

that 629(10 )psi, 1.07 '', 0.618''E D d= = =

Solution:

Using the method of section, divide the rod

into three sections:

Obtaining force for each section

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3 3 3

1 1 160(10 )lb, 15(10 )lb, 30(10 )lbP P P= = − =

Evaluate the total deflection

375.9(10 ) ''i i

i i

PL

A Eδ

−= =∑

2.5 Static indeterminacy

Structure for which internal forces and

reactions cannot be determined from statics

alone are said to be statically indeterminate. A

structure will be statically indeterminate

whenever it is held by more supports than are

required to maintain equilibrium.

In order to solve indeterminate problems,

redundant reactions are replaced with

unknown loads, which along with the other

loads must produce compatible deformations.

Deformations due to actual loads and

redundant reactions are then determined

separately and added or superposed:

0L Rδ δ δ= + = (2.12)

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Fig. 2.9

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Example: Determine the reactions at A and B

for the steel bar and loading shown in Fig. 2.9,

assuming a close fit at both supports before the

loads are applied.

Solution:

Step 1: Remove the constraint

Solve for the displacement at

B due to the loads with the

constraint released. 3

1 2 3

3

4

0, 600(10 ) ,

900(10 )

P P P N

P N

= = =

=

91.125 10i iL

i i

PL

A E Eδ

×= =∑ .

Step 2: Solve for the constraint

Solve for the displacement at

B due to the redundant

constraint.

1 2 BP P R= = − 31.95 10i i B

R

i i

PL R

A E Eδ

×= = −∑

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Step 3: Compatibility of strain

Require that the displacements

due to the loads and due to the

redundant reaction be

compatible

9 31.125 10 1.95 10 = 0

577

L R

B

B

R

E E

R kN

δ δ δ= +

⋅ ⋅− =

∴ =

Find the reaction at A due to the loads and the

reaction at B

0 300 600 577

323kN

y A

A

F R

R

= = − − +

∴ =

∑