chapter objectives - gtu.edu.tr

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) Poisson’s effect

d) Saint Venant’s 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 Poisson’s ratio v ≤ 0.45

4. Young’s 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 Hooke’s Law, i.e. σ = E ε

dx

dδε

xA

xP and

L

ExA

dxxP

ExA

dxxPd

dx

dE

xA

xP

0

EXAMPLE 1


The assembly shown in Fig. 4–7a 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 001143.0001143.0107010400

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. 4–9a, 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

2

2

22 y

L

rxyA o

(Ans) 6

3 2

0

22

22

0E

L

ELr

dyLr

EyA

dyyPL

o

o

L

3

2

2

3 ;0 y

L

ryPF o

y

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 Hooke’s

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, δx = δx1 + δx2

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.

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. 4–17a 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

BBABB

B

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

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

chapter objectives - gtu.edu.tr

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