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
Copyright 2011 Pearson Education South Asia Pte Ltd
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
Copyright 2011 Pearson Education South Asia Pte Ltd
READING QUIZ (cont.)
Copyright 2011 Pearson Education South Asia Pte Ltd
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
Copyright 2011 Pearson Education South Asia Pte Ltd
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
Copyright 2011 Pearson Education South Asia Pte Ltd
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
Copyright 2011 Pearson Education South Asia Pte Ltd
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
Copyright 2011 Pearson Education South Asia Pte Ltd
APPLICATIONS
Copyright 2011 Pearson Education South Asia Pte Ltd
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)
Copyright 2011 Pearson Education South Asia Pte Ltd
Thermal Stress Stress
Concentration Inelastic Axial
Deformation
ELASTIC DEFORMATION OF AN AXIALLY
LOADED MEMBER
Copyright 2011 Pearson Education South Asia Pte Ltd
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
Copyright 2011 Pearson Education South Asia Pte Ltd
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)
Copyright 2011 Pearson Education South Asia Pte Ltd
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
Copyright 2011 Pearson Education South Asia Pte Ltd
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)
Copyright 2011 Pearson Education South Asia Pte Ltd
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)
Copyright 2011 Pearson Education South Asia Pte Ltd
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
Copyright 2011 Pearson Education South Asia Pte Ltd
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
Copyright 2011 Pearson Education South Asia Pte Ltd
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
Copyright 2011 Pearson Education South Asia Pte Ltd
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)
Copyright 2011 Pearson Education South Asia Pte Ltd
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)
Copyright 2011 Pearson Education South Asia Pte Ltd
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
Copyright 2011 Pearson Education South Asia Pte Ltd
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
Copyright 2011 Pearson Education South Asia Pte Ltd
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)
Copyright 2011 Pearson Education South Asia Pte Ltd
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)
Copyright 2011 Pearson Education South Asia Pte Ltd
From the free-body diagram,
Solution
(Ans) kN 0.16
005.420
0
A
A
x
F
F
F
THERMAL STRESS
Copyright 2011 Pearson Education South Asia Pte Ltd
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
Copyright 2011 Pearson Education South Asia Pte Ltd
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)
Copyright 2011 Pearson Education South Asia Pte Ltd
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)
Copyright 2011 Pearson Education South Asia Pte Ltd
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
Copyright 2011 Pearson Education South Asia Pte Ltd
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)
Copyright 2011 Pearson Education South Asia Pte Ltd
K is independent of the material properties
K depends only on the specimens geometry and the type of discontinuity
INELASTIC AXIAL DEFORMATION
Copyright 2011 Pearson Education South Asia Pte Ltd
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
Copyright 2011 Pearson Education South Asia Pte Ltd
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)
Copyright 2011 Pearson Education South Asia Pte Ltd
(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)
Copyright 2011 Pearson Education South Asia Pte Ltd
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.
Copyright 2011 Pearson Education South Asia Pte Ltd
CONCEPT QUIZ (cont)
Copyright 2011 Pearson Education South Asia Pte Ltd
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
Copyright 2011 Pearson Education South Asia Pte Ltd
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
Copyright 2011 Pearson Education South Asia Pte Ltd
Top Related