on 2(a) : surveyor's tape is to be used to measure the length of a line. The of 2 cm by 0.5 cm and a length of 100 cm when T, and the te tape is AA N. Determine the true length of the line if the tape sho when used v Formula BarN at T,The ground on which it 63 °C P • 93 °C 20 2 cm 463.28 ст C = 35 0.5 cm N 12(10“)/°C = 200 GPa on : α ΔΤ L = cm PL 8 : AE Page 1 cm L = 463.28 + on 2(b) :

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Chapter9: Moments And Products Of Inertia Of Areas
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Question 2(a) :
A steel surveyor's tape is to be used to measure the length of a line. The tape has a rectangular
section of 2 cm by 0.5 cm and a length of 100 cm when T, and the tension or pull
on the tape is AA N. Determine the true length of the line if the tape shows the reading to be
BB cm when used v Formula Bar N at
T,The ground on which it is placed is flat.
Take:
T =
63
°C
P
P
93
°C
T,
АА -
20
N
2 ст
BB =
463.28
ст
СС -
35
N
0.5 ст
12(10*)/°C
-6
Et
200 GPa
Solution :
8, =
Ξα ΔΤL=
cm
PL
8 =
AE
Page 1
ст
L =
463.28
+
cm
Ans
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
Сорper
Est = 200 GPa
ast = 12(10¬)/°C _@br = 21(10¬6)/°C acu =
Epr = 100 GPa
Take :
Ecu = 120 GPa
17(10-6)/°C
T =
12
°C
Acu = 515 mm²
21
°C
|Ast = 200 mm² Abr = 450 mm²
T,
303
A =
тт
A mm
200 mm
Solution :
(+-)0=8, -8,
0 =
100 mm
9 (12)(10“)A+
9 (21)(10 )(0.2)+ 9 (17)(10“)(0.1)...
AF
0.2F
0.1F
200(10- 200(10°) 450(10^ho0(10°) 515(10^)120(10")
F =
Transcribed Image Text:Question 2(a) : A steel surveyor's tape is to be used to measure the length of a line. The tape has a rectangular section of 2 cm by 0.5 cm and a length of 100 cm when T, and the tension or pull on the tape is AA N. Determine the true length of the line if the tape shows the reading to be BB cm when used v Formula Bar N at T,The ground on which it is placed is flat. Take: T = 63 °C P P 93 °C T, АА - 20 N 2 ст BB = 463.28 ст СС - 35 N 0.5 ст 12(10*)/°C -6 Et 200 GPa Solution : 8, = Ξα ΔΤL= cm PL 8 = AE Page 1 ст L = 463.28 + cm Ans 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 Сорper Est = 200 GPa ast = 12(10¬)/°C _@br = 21(10¬6)/°C acu = Epr = 100 GPa Take : Ecu = 120 GPa 17(10-6)/°C T = 12 °C Acu = 515 mm² 21 °C |Ast = 200 mm² Abr = 450 mm² T, 303 A = тт A mm 200 mm Solution : (+-)0=8, -8, 0 = 100 mm 9 (12)(10“)A+ 9 (21)(10 )(0.2)+ 9 (17)(10“)(0.1)... AF 0.2F 0.1F 200(10- 200(10°) 450(10^ho0(10°) 515(10^)120(10") F =
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