Question 2(b) : Three bars each made of different materials are connected together and placed between two walls when the temperature is T . Determine the force exerted on the rigid supports when the temperature becomes bar are given in the figure. T The material properties and cross-sectional area of each Steel Brass Copper Est 200 GPa Epr = 100 GPa 12(10-")/°C abr = 21(10¬")/C acu = 17(10-)/°C Ecu = 120 GPa Take : ast = T = 12 °C Acu = 515 mm² °C |Ast = 200 mm² Abr = 450 mm? T, = 15 297 A = mm A mm 200 mm Solution : mm (+ +)0 = 8p – 8p 0 = 3 (12)(10-)A+ 3 (21)(10 )(0.2)+ 3 (17)(10)(0.1)... AF 0.2F 0.1F 200(10- 200(10°) 450(10 100(10°) 515(10-6 1 20(10°) F = N

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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Question 2(b) :
Three bars each made of different materials are connected together and placed between two
walls when the temperature is T . Determine the force exerted on the rigid supports when
the temperature becomes
bar are given in the figure.
T
The material properties and cross-sectional area of each
Steel
Brass
Copper
Take :
Est
200 GPa
Epr = 100 GPa
12(10¬6)/°C abr = 21(10¬")/C acu = 17(10-)/°C
Ecu = 120 GPa
ast =
T =
12
°C
Acu = 515 mm²
T, =
°C
|Ast = 200 mm² Abr = 450 mm?
15
297
A =
mm
A mm
200 mm
Solution :
mm
(+ +)0 = 8p – 87
0 =
3 (12)(10-)A+
3 (21)(10 )(0.2)+ 3 (17)(10)(0.1)...
AF
0.2F
0.1F
200(10- 200(10°)
450(10- 100(10°)
515(10-6120(10°)
F =
N
Transcribed Image Text:Question 2(b) : Three bars each made of different materials are connected together and placed between two walls when the temperature is T . Determine the force exerted on the rigid supports when the temperature becomes bar are given in the figure. T The material properties and cross-sectional area of each Steel Brass Copper Take : Est 200 GPa Epr = 100 GPa 12(10¬6)/°C abr = 21(10¬")/C acu = 17(10-)/°C Ecu = 120 GPa ast = T = 12 °C Acu = 515 mm² T, = °C |Ast = 200 mm² Abr = 450 mm? 15 297 A = mm A mm 200 mm Solution : mm (+ +)0 = 8p – 87 0 = 3 (12)(10-)A+ 3 (21)(10 )(0.2)+ 3 (17)(10)(0.1)... AF 0.2F 0.1F 200(10- 200(10°) 450(10- 100(10°) 515(10-6120(10°) F = N
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