Strength of Materials PPB 25403

Lecture 4: Axial Load

Saint-Venant’s Principle

Saint-Venant’s principle states that both localized

deformation and stress tend to “even out” at a

distance sufficiently removed from these regions.

Elastic Deformation of an Axially Loaded Member

Using Hooke’s law and the definitions of stress and

strain, we are able to develop the elastic deformation

of a member subjected to axial loads.

Suppose an element subjected to loads,

dx

dδε

xA

xP and

L

ExA

dxxP

0

= small displacement

L = original length

P(x) = internal axial force

A(x) = cross-sectional area

E = modulus of elasticity

Elastic Deformation of an Axially Loaded Member

Constant Load and Cross-Sectional Area

When a constant external force is applied at each

end of the member,

Sign Convention

Force and displacement is positive when tension and

elongation and negative will be compression and

contraction.

AE

PL

2- 5

Deformations Under Axial Loading

AE

P

EE

From Hooke’s Law:

From the definition of strain:

L

Equating and solving for the deformation,

AE

PL

With variations in loading, cross-section or

material properties,

i ii

ii

EA

LP

Example 4.1The assembly consists of an aluminum tube AB having a cross-sectional area of

400 mm2. A steel rod having a diameter of 10 mm is attached to a rigid collar and

passes through the tube. If a tensile load of 80 kN is applied to the rod, determine

the displacement of the end C of the rod. (Est = 200 GPa, Eal = 70 GPa )

Solution:Find the displacement of end C with respect to end B.

m 001143.0001143.0107010400

4.0108096

3

AE

PLB

m 003056.010200005.0

6.010809

3

/AE

PLBC

Displacement of end B with respect to the fixed end A,

Since both displacements are to the right, mm 20.4m 0042.0/ BCCC

(Ans)

Principle of Superposition

Principle of superposition is to simplify stress and

displacement problems by subdividing the loading

into components and adding the results.

A member is statically indeterminate when equations

of equilibrium are not sufficient to determine the

reactions on a member.

Statically Indeterminate Axially Loaded Member

Example 4.2The steel rod has a diameter of 5 mm. It is attached to the fixed wall at A, and

before it is loaded, there is a gap between the wall at and B’ and the rod of 1 mm.

Find the reactions at A and B’ if the rod is subjected to an axial force of P = 20 kN.

Neglect the size of the collar at C. (Est = 200 GPa)

Solution:Equilibrium of the rod requires

(1) 01020 ;0 3

BAx FFF

(2) mN 0.39278.04.0

001.0/

BA

CBBACAAB

FF

AE

LF

AE

LF

The compatibility condition for the rod is .m 001.0/ AB

By using the load–displacement relationship,

Solving Eqs. 1 and 2 yields FA = 16.6 kN and FB = 3.39 kN. (Ans)

Example 4.3The A-36 steel rod shown has a diameter of 5 mm. It is attached to the fixed wall at

A, and before it is loaded there is a gap between the wall at and the rod of 1 mm.

Determine the reactions at A and B’.

Solution:

(1) 0.001 Bp

BBABB

B

ACP

FF

AE

LF

AE

PL

6

9

9

3

103056.0102000025.0

2.1

m 002037.0102000025.0

4.01020

Consider the support at B’ as redundant and using principle of superposition,

Thus,

By substituting into Eq. 1,

Solution:

(Ans) kN 39.31039.3

103056.0002037.0001.0

3

6

B

B

F

F

(Ans) kN 6.16

039.320 ;0

A

Ax

F

FF

From the free-body diagram,

Thermal Stress

Change in temperature cause a material to change

its dimensions.

Since the material is homogeneous and isotropic,

TLT

= linear coefficient of thermal expansion, property of the material

= algebraic change in temperature of the member

= original length of the member

= algebraic change in length of the member

TT

T

Example 4.12The rigid bar is fixed to the top of the three posts made of A-36 steel and 2014-T6

aluminum. The posts each have a length of 250 mm when no load is applied to the

bar, and the temperature is T1 = 20°C. Determine the force supported by each post

if the bar is subjected to a uniform distributed load of 150 kN/m and the temperature

is raised to T2 = 20°C.

Solution:

(2) alst

(1) 010902 ;0 3

alsty FFF

From free-body diagram we have

The top of each post is displaced by an equal amount and hence,

The final position of the top of each post is equal to its displacement caused by the

temperature increase and internal axial compressive force.

Solution:

FalTstFstTst

FalTalal

FstTstst

Applying Eq. 2 gives

With reference from the material properties, we have

(3) 109.165216.1

101.7303.0

25.025.020801023

1020002.0

25.025.020801012

3

92

6

92

6

alst

alst

FF

FF

Solving Eqs. 1 and 3 simultaneously yields (Ans) kN 123 and kN 4.16 alst FF

© 2008 Pearson Education South Asia Pte Ltd

Chapter 4: Axial Load

Mechanics of Material 7th Edition

Exercise 1

allow

fail

allow

fail

allow

fail

SF

SF

F

FSF

.

.

.

Ffail=406kN

Answer: 12.5mm, 5.8

© 2008 Pearson Education South Asia Pte Ltd

Chapter 4: Axial Load

Mechanics of Material 7th Edition

Solution

2- 16

Exercise 2

Determine the deformation

of the steel rod shown under

the given loads.

Steps:

Divide the rod into components

at the load application points.

Apply a free-body analysis on

each component to determine

the internal force

Evaluate the total of the

component deflections.

mm 73.1Answer

2- 17

SOLUTION:

Divide the rod into three

components:

26-

21

21

m10581

m3.0

AA

LL

26-

3

3

m10194

m4.0

A

L

Apply free-body analysis to each

component to determine internal forces,

N10120

N1060

N10240

3

3

3

2

3

1

P

P

P

Evaluate total deflection,

m1073.1

10194

4.010120

10581

3.01060

10581

3.010240

10200

1

1

3

6

3

6

3

6

3

9

3

33

2

22

1

11

A

LP

A

LP

A

LP

EEA

LP

i ii

ii

mm 73.1

Exercise 3

Answer Pmax= 186N

Answer