Civil Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 4 (Q.No. 26)
26.
If a steel rod of 20 mm diameter and 5 metres long elongates by 2.275 mm when subjected to an axial pull of 3000 kg, the stress developed, is
Discussion:
6 comments Page 1 of 1.
Attaurrehman said:
7 years ago
Stress = F/A.
F = 3000kg.
Area =π/4*d^2.
π=3.14.
D=20mm=2cm, so.
Area=3.14/4 *(2)^2=3.14.
Now, the stress = 3000/3.14.
Stress = 955.41 kg/cm2.
F = 3000kg.
Area =π/4*d^2.
π=3.14.
D=20mm=2cm, so.
Area=3.14/4 *(2)^2=3.14.
Now, the stress = 3000/3.14.
Stress = 955.41 kg/cm2.
(2)
Bikash kabiraj said:
8 years ago
Stresd = f/a, f = 3000,a=22*2*2/7*4=22/7=3.14.
3000/3.14 = 955.41.
3000/3.14 = 955.41.
(1)
Newton said:
10 years ago
Stress = f/a = 3000/3.14*1*1 = 955.41 kg/cm^2.
Yogesh said:
9 years ago
3000/(φ/4 * 2 * 2).
Note: dia is in terms of cm.
Note: dia is in terms of cm.
Priya said:
8 years ago
We know that stress = force /area.
3000/pai/4*(d) 2.
3000/pai/4*20*20.
9.554kg/m2.
3000/pai/4*(d) 2.
3000/pai/4*20*20.
9.554kg/m2.
Noor ahmad said:
3 years ago
The right answer 9.55kg/mm2.
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