chapter objectives to determine the deformation of axially loaded members. to determine the support...
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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.
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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
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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
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READING QUIZ (cont.)
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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
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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
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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
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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
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APPLICATIONS
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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)
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Thermal Stress Stress Concentration
Inelastic Axial Deformation
ELASTIC DEFORMATION OF AN AXIALLY LOADED MEMBER
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• 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
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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)
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• 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
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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)
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• 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 ;
32
22
33y
L
ryxV o
EXAMPLE 2 (cont)
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• 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
022
22
0 E
L
ELr
dyLr
EyA
dyyP L
o
oL
32
2
3 ;0 y
L
ryPF o
y
VW
PRINCIPLE OF SUPERPOSITION
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• 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
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• 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
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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)
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• 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)
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• 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 9.133510
31556
(Ans) MPa 8.401N/mm 8.4015
31556
222
2
t
ts
b
bb
A
F
A
F
FORCE METHOD OF ANALYSIS
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• 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
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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)
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• Using the principle of superposition,
• From Eq. 4-2,
• Substituting into Eq. 1, we get
Solution
B
BABBB
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
103944.76105093.00002.03
93
B
B
F
F
EXAMPLE 4 (cont)
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• From the free-body diagram,
Solution
(Ans) kN 0.16
005.420
0
A
A
x
F
F
F
THERMAL STRESS
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• Ordinarily, the expansion or contraction δT is linearly related to the temperature increase or decrease ΔT that occurs.
• If the change in temperature varies throughout the length of the member, i.e. ΔT = ΔT (x), or if α 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
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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 = 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.
EXAMPLE 5 (cont)
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• 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)
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• 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
926
926
alst
alst
FF
FF
(Ans) kN 123 and kN 4.16 alst FF
STRESS CONCENTRATION
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• 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)
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• K is independent of the material properties• K depends only on the specimen’s geometry and the type
of discontinuity
INELASTIC AXIAL DEFORMATION
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• 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
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The bar in Fig. 4–29a 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.
EXAMPLE 6 (cont)
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• (a) Finding the stress concentration factor,
• Using the table and geometry ratios, we get K = 1.7. We have
Solution
25.1840
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)
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• 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
pY
P
PA
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.
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CONCEPT QUIZ (cont)
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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
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CONCEPT QUIZ (cont)
3) The greatest load that the bar can sustain is
a) σ1.Afull
b) σY.Afull
c) σY.Afull
d) σY.Aa-a
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