Civil Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 4 (Q.No. 50)
50.
The maximum compressive stress at the top of a beam is 1600 kg/cm2 and the corresponding tensile stress at its bottom is 400 kg/cm2 . If the depth of the beam is 10 cm, the neutral axis from the top, is
Discussion:
15 comments Page 1 of 2.
Srikanta said:
4 years ago
1600/x = 400/(10-x).
5x = 40.
X = 8 cm.
Where x = Depth of neutral axis from top.
5x = 40.
X = 8 cm.
Where x = Depth of neutral axis from top.
(3)
Raj Keshri said:
4 years ago
Since we know that, sigma/y in bending equation is equal for both compressive and tensile stresses, So,
Let, a distance of neutral axis from the top be x.
Therefore the distance of the bottom layer from the neutral axis is 10-x.
Therefore; (Compressive stress/x) = (tensile stress/10-x).
Let, a distance of neutral axis from the top be x.
Therefore the distance of the bottom layer from the neutral axis is 10-x.
Therefore; (Compressive stress/x) = (tensile stress/10-x).
(2)
Rabin Das said:
5 years ago
Thanks @Ganesh.
Protul said:
6 years ago
Thanks all for explaining it.
Pawan said:
6 years ago
Thank you very much @Ganesh.
Munni said:
6 years ago
Max tensile stress/max. Compressive stress=m (d-h/h) , considering m which is the modular ratio=1 we get 400/1600=1 (10-h/h), by solving we get h=8cm.
Bittu said:
7 years ago
Thanks @Ganish.
(1)
Reddy said:
7 years ago
Thank you @Ganesh.
Satya said:
7 years ago
Thank you @Ganesh.
Sudinbanerjee said:
8 years ago
Your explanation is absolutely correct, Thanks @Ganesh.
(1)
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