Chapter 04 - Axial Load

Russell C. Hibbeler

Chapter 4: Axial Load

© 2008 Pearson Education South Asia Pte Ltd© 2008 Pearson Education South Asia Pte Ltd

Chapter 4: Axial LoadChapter 4: Axial LoadMechanics of Material 7Mechanics of Material 7thth Edition Edition

Saint-Venant’s PrincipleSaint-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 lengthP(x) = internal axial forceA(x) = cross-sectional area E = modulus of elasticity



Elastic Deformation of an Axially Loaded Member

Constant Load and Cross-Sectional AreaConstant Load and Cross-Sectional AreaWhen a constant external force is applied at each

end of the member,

Sign ConventionSign ConventionForce and displacement is positive when tension

and elongation and negative will be compression and contraction.

AE

PL



Example 4.2The 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 001143.0001143.0

107010400

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)



Example 4.4A member is made from a material that has a specific weight and modulus of and elasticity E. If it is formed into a cone, find how far its end is displaced due to gravity when it is suspended in the vertical position.

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

yL

rx

L

r

y

x oo ;

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

32

22

33y

L

ryxV o



Since , the internal force at the section becomes

Solution:

32

2

3 ;0 y

L

ryPF o

y

VW

22

22 y

L

rxyA o

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

Between the limits of y =0 and L yields

(Ans)

6

3 2

022

22

0 E

L

ELr

dyLr

EyA

dyyP L

o

oL



Principle of SuperpositionPrinciple 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 4.5The 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

FFAE

LF

AE

LF

The compatibility condition for the rod is .m 001.0/ ABBy using the load–displacement relationship,

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



Example 4.8The 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.

Solution:

(1) 0 ;0 tby FFF

bt 5.0

Equilibrium requires

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



Taking the 2 modulus of elasticity,

Solution:

(2) 911251255

10755

605.0

1045510

6032322

bt

bt

FF

FF

kN 56.3131556 tb FF

The stresses in the bolt and tube are therefore

Solving Eqs. 1 and 2 simultaneously, we get

(Ans) MPa 9.133N/mm 9.133510

31556

(Ans) MPa 8.401N/mm 8.4015

31556

222

2

t

ts

b

bb

A

F

A

F



Example 4.9The 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

B

BABBB

ACP

FF

AE

LF

AE

PL

69

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

103056.0002037.0001.03

6

B

B

F

F

(Ans) kN 6.16

039.320 ;0

A

Ax

F

FF

From the free-body diagram,



Thermal StressChange 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

926

926

alst

alst

FF

FF

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



Stress ConcentrationsStress concentrations occur when cross-sectional area

changes.Maximum stress is determined using a stress

concentration factor, K, which is a function of geometry.

avg

Kmax



Example 4.14The steel strap is subjected to an axial load of 80 kN. Find the maximum normal stress developed in the strap and the displacement of one end of the strap with respect to the other end. The steel has a yield stress of σY = 700 MPa, and Est = 200 GPa.

Solution:

(Ans) MPa 640

01.002.0

10806.1

3

max

A

PK

220

40 ,3.0

20

6

h

w

h

r

Maximum normal stress occurs at the smaller cross section (B-C),

Using the table and geometry ratios, we get K = 1.6. Thus the maximum stress is



Neglecting the localized deformations surrounding the applied load and at the sudden change in cross section of the shoulder fillet (Saint-Venant’s principle), we have

Solution:

(Ans) mm 20.2

1020001.002.0

8.01080

1020001.004.0

3.010802

9

3

9

3

/

AE

PLDA



*Inelastic Axial DeformationMember may be designed to yield and permanently

deform; this is referred to as being elastic perfectly plastic or elastoplastic.

The yield stress and A is the bar’s cross-sectional area at section a–a.

AdAP Y

A

Yp



Example 4.16The bar is made of steel that is assumed to be elastic perfectly plastic, with σY = 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.

Solution:

(Ans) kN 14.9032.0002.0

75.110250

;

6

Ymax

YY

Yavg

PP

A

PKK

25.1840

40 ,125.0

840

4

h

w

h

r

a) Finding the stress concentration factor,

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



Solution:

(Ans) kN 0.16

032.0002.010250

6

p

p

pY

P

PA

P

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

Chapter 04 - Axial Load

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