4.5 FORCE METHOD OF ANALYSIS (FLEXIBILITY METHOD) It is...
Transcript of 4.5 FORCE METHOD OF ANALYSIS (FLEXIBILITY METHOD) It is...
4.5 FORCE METHOD OF ANALYSIS (FLEXIBILITY METHOD)
• 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.
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0
p B
AC ACBB
BCA B A
PL LF L F PAE AE L
LF F P F PL
EXAMPLE 4.8
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 = 200GPa.
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EXAMPLE 4.8 (cont.)
• Using the principle of superposition,
• From Eq. 4-2,
• Substituting into Eq. (1), we get
Solutions
B
BABBB
ACP
FFAELF
AEPL
992
392
3
103944.7610200005.0
2.1
105093.010200005.04.01020
1 0002.0 BP
(Ans) kN 05.41005.4
103944.76105093.00002.03
93
B
B
F
F
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EXAMPLE 4.8 (cont.)
• From the free-body diagram,
Solutions
(Ans) kN 0.16 005.4200
A
A
x
FF
F
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APPLICATIONS
Thermal Stress Stress Concentration
Inelastic Axial Deformation
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4.6 THERMAL STRESS
• 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
TTT
dxTT
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T TL
TFLTLAE
F TAETE
EXAMPLE 4.10 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 = 80°C.
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EXAMPLE 4.10 (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.
Solutions
(2) alst
(1) 010902 ;0 3 alsty FFF
FalTalal
FstTstst
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EXAMPLE 4.10 (cont.)
• Applying Eq. 2 gives
• With reference from the material properties, we have
• Solving Eqs. (1) and (3) simultaneously yields
Solutions
FalTstFstTst
(3) 109.165216.1
101.7303.025.025.020801023
1020002.025.025.020801012
3
926
926
alst
alst
FF
FF
(Ans) kN 123 and kN 4.16 alst FF
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4.7 STRESS CONCENTRATION
• When a member’s cross section area is suddenly changed, the maximum normal stress occurs at the smallest cross section area.
• 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
Kmax
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• K is independent of the material properties.• K depends only on the specimen’s geometry and the type
of discontinuity.
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max . .
ys uall
orF S