Civil Engineering - Strength of Materials - Discussion

Discussion Forum : Strength of Materials - Section 2 (Q.No. 31)
31.
The tensile force required to cause an elongation of 0.045 mm in a steel rod of 1000 mm length and 12 mm diameter, is (where E = 2 x 106 kg/cm2)
166 kg
102 kg
204 kg
74 kg
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
9 comments Page 1 of 1.

Sagun said:   2 years ago
Change in length dl = 0.045 mm
Length = 100 mm.
D = 12 mm.
Radius (r) = 12/2 = 6mm.
E = 2*10^6.
Formula is dl = (p*L)/(A*E).

Area of cross section=steel cross section is circle area;
=πr^2=π×6^2=22/7) ×6×6.
= 113.142 mm^2.

dl = (p*L)/(A*E).
=> dl*E*A=P*L.
=> (dl*E*A) /L=P.
=> P= (0.045×2×10^4×113.14) /1000.
=>P=101.82.
=>P=Nearly 102 Kg.
(1)

Manish Ravindra nerkar said:   4 years ago
All dimensions convert in cm.

Elongation 0.0045 cm.
Area= π /4 *1.2*1.2.
= 1.13 cm^2.
l=1000mm=100cm

dl = Pl/AE.
0.045= P* 100/1.13*2*10^6.
P = 101.7 = 1100g.
(3)

Roja rani said:   5 years ago
How we got A as 10.28? Please explain me.

Sathya h d said:   6 years ago
A=x1. 2^2 /4.

A=1.131 cm^2,
Strain = 0.0045/100cm =4.5x10^-5,
E = p/e.
p = stress = E x e = 2x10^6x4.5x10^-5=90kg /cm^2,
p = P/A, P=pxA.
P = 90x1.131 = 101.79= 102 kg.

Raj said:   8 years ago
Pl/AE = .045*1000/10.28*2*10^9.
= 102.

Tailam mahesh said:   8 years ago
p/A= E * ε.

From Hool Law, Stress directly proportional to stran.
Stress =P/A.
Strain dl/L.

Equate then solve.

Uday said:   8 years ago
Can Anyone explain clearly? please.

Baloch said:   9 years ago
Thanks @Talari Hari.

Talari hari said:   1 decade ago
Given data is:

Change in length dl = 0.045 mm or 0.0045 cm.

Length = 100 mm.

D = 12 mm.

E = 2*10^6.

Formula is dl = (p*l)/(A*E).

Substitute given values and you will get answer.

Post your comments here:

Your comments will be displayed after verification.