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A vertical, solid steel post 25 cm in diameter and 2.50 m long is required to support a load of 8000kg. You can ignore the weight of the post. What are (a) the stress in the post; (b) the strain in the post; and (c) the change in the post’s length when the load is applied?

Short Answer

Expert verified

(a)-1.60×106Pa

(b)-8×10-6

(c)-2.0x10-5m

Step by step solution

01

Given information

Length lo=2.50m, diameter =2.5×10-2m, Load m=8000kg

02

Concept/Formula used

Y=l0FA∆l

Where, Y is Young’s modulus, l0 is length of muscle, F is muscle force, A is cross-sectional area and∆l is elongation.

03

Cross –sectional area of the post

A=ττd24=ττ2.50×10-2m24=0.05m2

04

Force application on the post

F=mg=(8000kg)(9.80mIs2)=7.84×104N

Young’s Modulus for steel isY=2.0×1011Pa

05

Stress in the post

(a)

stress=FA=-7.84×104N0.05m2=-1.60×106Pa

Note: negative sign shows that the stress is compressive.

06

Strain in the post   

(b)

stress=stressY=-1.60×106Pa2×1011Pa=-8×106

Note: negative sign shows that the length decreases.

07

Change in the post length

(C)

∆l=l0strain=2.5m-8×10-6=-2.0×10-5m

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