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:
5 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:
5 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)
Sudinbanerjee said:
8 years ago
Your explanation is absolutely correct, Thanks @Ganesh.
(1)
Bittu said:
7 years ago
Thanks @Ganish.
(1)
Charles said:
10 years ago
Please show solution.
Ganesh said:
10 years ago
1600/x = 400/10-x.
Remember, stress diagram in RCC.
Remember, stress diagram in RCC.
Akscivilian said:
9 years ago
It is mc/t = n/(d - n). But we don't have m here.
Baloch said:
8 years ago
Here, n/d = m x Compressive / m x Comp. + tensile.
M = modular ratio, not given you can take it 18 approx.
Solving will give 9.8 cm almost 8.
M = modular ratio, not given you can take it 18 approx.
Solving will give 9.8 cm almost 8.
Spsg said:
8 years ago
E/R is same so equate bending stress/distance from nutral axis for both tensile and compressive.
Satya said:
7 years ago
Thank you @Ganesh.
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