Axial Load

Chapter Objectives

To determine the deformation of axially loaded members.

To determine the support reactions when these reactions cannot be determined solely from the

equations of equilibrium.

To analyze the effects of thermal stresses.

Copyright 2011 Pearson Education South Asia Pte Ltd

1. Reading Quiz

2. Applications

3. Elastic deformation in axially loaded member

4. Principle of superposition

5. Compatibility conditions

6. Force method of analysis

7. Thermal Stress

8. Stress Concentration

9. Concept Quiz

In-class Activities


READING QUIZ

1) The stress distributions at different cross

sections are different. However, at locations

far enough away from the support and the

applied load, the stress distribution becomes

uniform. This is due to

a) Principle of superposition

b) Inelastic property

c) Poissons effect

d) Saint Venants Principle


READING QUIZ (cont.)


READING QUIZ (cont)

2) The principle of superposition is valid

provided that

1. The loading is linearly related to the stress or displacement

2. The loading does not significantly change the original

geometry of the member

3. The Poissons ratio v 0.45

4. Youngs Modulus is small

a) a, b and c

b) a, b and d

c) a and b only

d) All


READING QUIZ (cont)

3) The units of linear coefficient of thermal

expansion are

a) per C

b) per F

c) per K (Kelvin)

d) all of them


READING QUIZ (cont)

4) Stress concentrations become important in

design if

a) the material is brittle

b) the material is ductile but subjected to fatigue loading

c) the material is subjected to fatigue loadings to dynamic

loading

d) All of them


READING QUIZ (cont)

5) The principle of superposition is applicable to

a) inelastic axial deformation

b) residual stress evaluation

c) large deformation

d) None of the above


APPLICATIONS


Most concrete columns are reinforced with steel rods; and

these two materials work together in supporting the applied

load. Are both subjected to axial stress?

APPLICATIONS (cont)


Thermal Stress Stress

Concentration Inelastic Axial

Deformation

ELASTIC DEFORMATION OF AN AXIALLY

LOADED MEMBER


Provided these quantities do not exceed the proportional

limit, we can relate them using Hookes Law, i.e. = E

dx

d

xA

xP and

L

ExA

dxxP

ExA

dxxPd

dx

dE

xA

xP

0

EXAMPLE 1


The assembly shown in Fig. 47a 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. Take Est = 200 GPa, Eal = 70 GPa.

EXAMPLE 1 (cont)


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

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

Since both displacements are to the right,

Solution

m 001143.0001143.0

107010400

4.0108096

3

AE

PLB

m 003056.010200005.0

6.010809

3

/

AE

PLBC

mm 20.4m 0042.0/BCCC

EXAMPLE 2


A member is made from a material that has a specific weight and modulus of elasticity E. If it is in the form of a cone having the dimensions shown in Fig. 49a, determine how far its end is displaced due to gravity when it is suspended in the vertical position.

EXAMPLE 2 (cont)


Radius x of the cone as a function of y is determined by proportion,

The volume of a cone having a base of radius x and height y is

Solution

yL

rx

L

r

y

x oo ;

3

2

22

33y

L

ryxV o

EXAMPLE 2 (cont)


Since , the internal force at the section becomes

The area of the cross section is also a function of position y,

Between the limits of y =0 and L yields

Solution

22

22 y

L

rxyA o

(Ans) 6

3 2

0

22

22

0E

L

ELr

dyLr

EyA

dyyPL

o

o

L

32

2

3 ;0 y

L

ryPF oy

VW

PRINCIPLE OF SUPERPOSITION


It can be used for simple problems having complicated loadings. This is done by dividing the loading into

components, then algebraically adding the results.

It is applicable provided the material obeys Hookes Law and the deformation is small.

If P = P1 + P2 and d d1 d2, then the deflection at location x is sum of two cases, x = x1 + x2

COMPATIBILITY CONDITIONS


When the force equilibrium condition alone cannot determine the solution, the structural member is called

statically indeterminate.

In this case, compatibility conditions at the constraint locations shall be used to obtain the solution. For

example, the stresses and elongations in the 3 steel

wires are different, but their displacement at the

common joint A must be the same.

EXAMPLE 3


The bolt is made of 2014-T6 aluminum alloy and is tightened

so it compresses a cylindrical tube made of Am 1004-T61

magnesium alloy. The tube has an outer radius of 10 mm,

and both the inner radius of the tube and the radius of the bolt

are 5 mm. The washers at the top and bottom of the tube are

considered to be rigid and have a negligible thickness. Initially

the nut is hand-tightened slightly; then, using a wrench, the

nut is further tightened one-half turn. If the bolt has 20

threads per inch, determine the stress in the bolt.

EXAMPLE 3 (cont)


Equilibrium requires

When the nut is tightened on the bolt, the tube will shorten.

Solution

(1) 0 ;0 tby FFF

bt 5.0

EXAMPLE 3 (cont)


Taking the 2 modulus of elasticity,

Solving Eqs. 1 and 2 simultaneously, we get

The stresses in the bolt and tube are therefore

Solution

(2) 911251255

10755

605.0

1045510

6032322

bt

bt

FF

FF

kN 56.3131556 tb FF

(Ans) MPa 9.133N/mm 9.133

510

31556

(Ans) MPa 8.401N/mm 8.4015

31556

2

22

2

t

ts

b

bb

A

F

A

F

FORCE METHOD OF ANALYSIS


It is also possible to solve statically indeterminate problem by writing the compatibility equation using the superposition

of the forces acting on the free body diagram.

EXAMPLE 4


The A-36 steel rod shown in Fig. 417a has a diameter of 10 mm. It is fixed to the wall at A, and before it is loaded there is

a gap between the wall at B and the rod of 0.2 mm. Determine the reactions at A and Neglect the size of the

collar at C. Take Est = 200 GPa.

EXAMPLE 4 (cont)


Using the principle of superposition,

From Eq. 4-2,

Substituting into Eq. 1, we get

Solution

BBABBB

ACP

FF

AE

LF

AE

PL

9

92

3

92

3

103944.7610200005.0

2.1

105093.010200005.0

4.01020

1 0002.0 BP

(Ans) kN 05.41005.4

103944.76105093.00002.0

3

93

B

B

F

F

EXAMPLE 4 (cont)


From the free-body diagram,

Solution

(Ans) kN 0.16

005.420

0

A

A

x

F

F

F

THERMAL STRESS


Ordinarily, the expansion or contraction T is linearly related to the temperature increase or decrease T that occurs.

If the change in temperature varies throughout the length of the member, i.e. T = T (x), or if varies along the length, then

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

TL

T

dxTT

EXAMPLE 5


The 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 = 20C. 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 = 20C.

EXAMPLE 5 (cont)


From the free-body diagram we have

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

Final position of the top of each post is equal to its displacement caused by the temperature increase and internal axial compressive force.

Solution

(2) alst

(1) 010902 ;0 3 alsty FFF

FalTalal

FstTstst

EXAMPLE 5 (cont)


Applying Eq. 2 gives

With reference from the material properties, we have

Solving Eqs. 1 and 3 simultaneously yields

Solution

FalTstFstTst

(3) 109.165216.1 101.7303.0

25.025.020801023

1020002.0

25.025.020801012

3

92

6

92

6

alst

alst

FF

FF

(Ans) kN 123 and kN 4.16 alst FF

STRESS CONCENTRATION


The stress concentration factor K is a ratio of the maximum stress to the average stress acting at the

smallest cross section; i.e.

avg

K

max

STRESS CONCENTRATION (cont)


K is independent of the material properties

K depends only on the specimens geometry and the type of discontinuity

INELASTIC AXIAL DEFORMATION


When a material is stressed beyond the elastic range, it starts to yield and thereby causes permanent deformation.

Among various inelastic behavior, the common cases

exhibit elastoplastic or elastic-perfectly-plastic behavior.

EXAMPLE 6


The bar in Fig. 429a is made of steel that is assumed to be elastic perfectly plastic, with Y = 250 MPa. Determine (a) the maximum value of the applied load P that can be applied

without causing the steel to yield and (b) the maximum value

of P that the bar can support. Sketch the stress distribution at

the critical section for each case.

EXAMPLE 6 (cont)


(a) Finding the stress concentration factor,

Using the table and geometry ratios, we get K = 1.7. We have

Solution

25.1

840

40 125.0

840

4

h

w

h

r

(Ans) kN 14.9

032.0002.075.110250

;

6

max

Y

Y

YY

avg

P

P

A

PK

K

EXAMPLE 6 (cont)


b) As P is increased to the plastic load it gradually changes the stress distribution from the elastic state to the plastic state.

Solution

(Ans) kN 0.16

032.0002.010250 6

p

p

p

Y

P

P

A

P

CONCEPT QUIZ

1) The assembly consists of two posts made

from material 1 having modulus of elasticity

of E1 and a cross-sectional area A1 and a

material 2 having modulus of elasticity E2 and

cross-sectional area A2. If a central load P is

applied to the rigid cap, determine the force

in each post. The support is also rigid.


CONCEPT QUIZ (cont)


PrPPr

rP

PrrPPr

P

PrPPr

P

rPPPr

rP

AE

AEr

1 12

1 d) 12

1 b)

12 12

1

c) 12

a)

Let

22

11

22

11

22

11

CONCEPT QUIZ (cont)

2) The value of stress concentration factor

depends on the geometry. Which one of the

following is true?

a) Ka > Kb > Kc

b) Ka > Kb > Kd

c) A and B

d) None of the above


CONCEPT QUIZ (cont)

3) The greatest load that the bar can sustain is

a) 1.Afull

b) Y.Afull

c) Y.Afull

d) Y.Aa-a


Axial Load

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