A hollow copper pipe undergoes a 1% change in length under a tensile load. If the Young’s modulus of copper is 110 GPa, the pipe’s outer diameter is 2 cm and its wall thickness is 5 mm, what is the magnitude of the tensile load? Consider the whole change happening in the elastic region.

Question
Asked Sep 28, 2019
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A hollow copper pipe undergoes a 1% change in length under a tensile load. If the Young’s modulus of copper is 110 GPa, the pipe’s outer diameter is 2 cm and its wall thickness is 5 mm, what is the magnitude of the tensile load? Consider the whole change happening in the elastic region.

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Expert Answer

Step 1

We have been given the following data:

The percentage change in length of the hollow copper pipe, ΔL/L = 1% = 0.01.

The Young’s modulus of copper, E = 110 Gpa.

The outer diameter of the pipe, Do = 2 cm.

The wall thickness, t = 5 mm. 

Step 2

Calculate the inner diameter of the pipe and the cross-sectional area of the pipe.

AD-D)
(2-1)
D D, -2t
4
D, 2-2(0.5)
A=
=2 1
A 2.356 cm2
A 2.356x10 m2
=1 cm
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AD-D) (2-1) D D, -2t 4 D, 2-2(0.5) A= =2 1 A 2.356 cm2 A 2.356x10 m2 =1 cm

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Step 3

Calculate the tensile load by applying the longitudinal stres...

AL
= E
L
A
AL
P AE
L
Plugging in allthe respective values.
P (2.356x10 m2)(110 GPa)(0.01)
=(2.356x 10 m2110x10 Pa) (0.01)
-259.18 x103 N
=259.18 kN
help_outline

Image Transcriptionclose

AL = E L A AL P AE L Plugging in allthe respective values. P (2.356x10 m2)(110 GPa)(0.01) =(2.356x 10 m2110x10 Pa) (0.01) -259.18 x103 N =259.18 kN

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