Civil Engineering - Theory of Structures - Discussion
Discussion Forum : Theory of Structures - Section 1 (Q.No. 16)
16.
A load of 1960 N is raised at the end of a steel wire. The minimum diameter of the wire so that stress in the wire does not exceed 100 N/mm2 is :
Discussion:
13 comments Page 1 of 2.
Sanjeev said:
1 decade ago
Stress = load/area.
Jack said:
1 decade ago
Given:
Load = 1960N.
Area of Wire = ?
Allowable Stress = 100 N/mm2.
So Stress = P/A.
A= P/Stress.
A= 1960N/100N/mm2.
A= 19.6mm2.
Dia = Pid2/4.
19.6mm2 = 3.14d2/4.
(19.6*4/3.14)^.5 = d = 5.00mm.
Load = 1960N.
Area of Wire = ?
Allowable Stress = 100 N/mm2.
So Stress = P/A.
A= P/Stress.
A= 1960N/100N/mm2.
A= 19.6mm2.
Dia = Pid2/4.
19.6mm2 = 3.14d2/4.
(19.6*4/3.14)^.5 = d = 5.00mm.
(1)
Arghya said:
1 decade ago
The answer is 4.9 we have to take the approx value which is 5.00.
Vikram singh said:
9 years ago
Very good explanation @Jack.
KRISHNA said:
9 years ago
Nice explanation. Thanks.
Maitri said:
8 years ago
Thanks @Jack.
Harsha said:
8 years ago
Stress = force/area.
Area = force/stress,
Area = 1960/100=19.6.
πdd/4=19.6,
dd/4=19.6/3.14,
dd=6.24*4,
d= √24.96,
d=5mm.
Area = force/stress,
Area = 1960/100=19.6.
πdd/4=19.6,
dd/4=19.6/3.14,
dd=6.24*4,
d= √24.96,
d=5mm.
(2)
Tirusew S. said:
8 years ago
Thank you all for the given explanation.
Seenu said:
8 years ago
2*PI*R = 2*π*d/2*d/2.
Monika said:
7 years ago
Thanks for the explanation.
(1)
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