PROBLEM E. Determine the equivalent nodal forces in addition to the load applied as shown, member 2 undergoes a temperature increase of 60°F. All bar areas are 2 in², E = 29 x 105 psi and α = 6.5 x 106 in/in/°F. Equivalent Nodal Forces For Member 1 Fras- 10000 #/in 10 in 2 15000 # 15 in 10000 #/in 10 in Equivalent Nodal Forces 50000 Froz 50000# 22620# 15000# 42380# 22620#

Structural Analysis
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Author:KASSIMALI, Aslam.
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Chapter2: Loads On Structures
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PROBLEM E. Determine the equivalent nodal forces in addition to the load applied as
shown, member 2 undergoes a temperature increase of 60°F. All bar areas are 2 in²,
E = 29 x 106 psi and a = 6.5 x 106 in/in/°F.
Equivalent Nodal Forces
For Member 1
For Member 2
Fegz
10000 #/in
FEQ1
2
10 in
=
EQ2
=
2
P₂L
2
15000 #
15 in
10000 #/in
=
10 in
10000 (10)
2
15 in
FEQ2 = FEQ3 = AEa(AT)
►
Equivalent Nodal Forces
= 50000 #
50000#
= 2(29 x 106) (6.5 x 10-6)(60) = 22620#
FrQ3
50000#
10 in
22620#
15000#
42380#
15 in
22620#
Transcribed Image Text:PROBLEM E. Determine the equivalent nodal forces in addition to the load applied as shown, member 2 undergoes a temperature increase of 60°F. All bar areas are 2 in², E = 29 x 106 psi and a = 6.5 x 106 in/in/°F. Equivalent Nodal Forces For Member 1 For Member 2 Fegz 10000 #/in FEQ1 2 10 in = EQ2 = 2 P₂L 2 15000 # 15 in 10000 #/in = 10 in 10000 (10) 2 15 in FEQ2 = FEQ3 = AEa(AT) ► Equivalent Nodal Forces = 50000 # 50000# = 2(29 x 106) (6.5 x 10-6)(60) = 22620# FrQ3 50000# 10 in 22620# 15000# 42380# 15 in 22620#
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