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.
Mahesh said:
2 years ago
Stress = p/A.
Stress = 100N/mm2.
P = 1960N.
100 = 1960/A.
A = 196mm2.
3.14 * d2/4 = 196.
d = 5mm.
Stress = 100N/mm2.
P = 1960N.
100 = 1960/A.
A = 196mm2.
3.14 * d2/4 = 196.
d = 5mm.
(4)
Rajiv said:
4 years ago
P= 1960 N.
Stress= 100N/mm^2.
Formula(Sress= P/A).
100 = 1960/A.
A = 1960/100.
Where A=π/4d^2.
Then :- 22/7*4*d^2=196/10.
d^2= 28/22*196/10.
d= √25.
d= 5 =>Answer.
Stress= 100N/mm^2.
Formula(Sress= P/A).
100 = 1960/A.
A = 1960/100.
Where A=π/4d^2.
Then :- 22/7*4*d^2=196/10.
d^2= 28/22*196/10.
d= √25.
d= 5 =>Answer.
(1)
K.vignesh. said:
6 years ago
Good explanation, Thanks @Jack.
Monika said:
6 years ago
Thanks for the explanation.
Seenu said:
7 years ago
2*PI*R = 2*π*d/2*d/2.
Tirusew S. said:
7 years ago
Thank you all for the given explanation.
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)
Maitri said:
8 years ago
Thanks @Jack.
KRISHNA said:
8 years ago
Nice explanation. Thanks.
Vikram singh said:
9 years ago
Very good explanation @Jack.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers