Approximate Analysis of Statically Indeterminate Structures.
Chapter 4-3 Statically Indeterminate Problemsqiw4/Academic/ENGR0135/Chapter4-3.pdfStatically...
Transcript of Chapter 4-3 Statically Indeterminate Problemsqiw4/Academic/ENGR0135/Chapter4-3.pdfStatically...
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Statics and Mechanics of Materials
Statically Indeterminate Problems
Chapter 4-3
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Example 1:
Given: – A is a tube (E=73 GPa, OD = 75 mm)
– B is a solid cylinder (E=200 GPa, d=25 mm)
– Load P is 35 kN
– Maximum deflection at the end of bar B is 0.4 mm
(Anyway, whose deflection is this?)
What is the thickness of A?
Notes:
Tube A experiences shortening
Cylinder B experiences extension
Total Displacement = dA + dB
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Example 2
EA
PLd
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Static Indeterminacy • Structures for which internal forces and reactions
cannot be determined from statics alone are said
to be statically indeterminate.
0 RL ddd
• Deformations due to actual loads and redundant
reactions are determined separately and then added
or superposed.
• Redundant reactions are replaced with
unknown loads which along with the other
loads must produce compatible deformations.
• A structure will be statically indeterminate
whenever it is held by more supports than are
required to maintain its equilibrium.
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Example Problem 2 Determine the reactions at A and B for the steel
bar and loading shown, assuming a close fit at
both supports before the loads are applied.
• Solve for the reaction at A due to applied loads
and the reaction found at B.
• Require that the displacements due to the loads
and due to the redundant reaction be compatible,
i.e., require that their sum be zero.
• Solve for the displacement at B due to the
redundant reaction at B.
SOLUTION:
• Consider the reaction at B as redundant, release
the bar from that support, and solve for the
displacement at B due to the applied loads.
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• Solve for the displacement at B due to the applied
loads with the redundant constraint released,
EEA
LP
LLLL
AAAA
PPPP
i ii
ii9
L
4321
2643
2621
34
3321
10125.1
m 150.0
m10250m10400
N10900N106000
d
• Solve for the displacement at B due to the redundant
constraint,
i
B
ii
iiR
B
E
R
EA
LPδ
LL
AA
RPP
3
21
262
261
21
1095.1
m 300.0
m10250m10400
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• Require that the displacements due to the loads and due to
the redundant reaction be compatible,
kN 577N10577
01095.110125.1
0
3
39
B
B
RL
R
E
R
Ed
ddd
• Find the reaction at A due to the loads and the reaction at B
kN323
kN577kN600kN 3000
A
Ay
R
RF
kN577
kN323
B
A
R
R
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Thermal Strains and Stresses • A temperature change results in a change in length or
thermal strain. There is no stress associated with the
thermal strain unless the elongation is restrained by
the supports.
coef.expansion thermal
ddAE
PLLT PT
• Treat the additional support as redundant and apply
the principle of superposition.
0
0
AE
PLLT
PT
ddd
• The thermal deformation and the deformation from
the redundant support must be compatible.
TEA
P
TAEPPT
ddd 0
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Example Problem 3
• The change in length resulting from the temperature
change is
Solution:
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Example Problem 3
• The stress needed to resist a change in length 0f 5.95 mm is
• The Internal force on the cross section of the rail will be
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Example Problem 4
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Example Problem 4
Pc PA
PB
PA
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Example Problem 4