Civil Engineering - Strength of Materials - Discussion


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

[A]. 2 cm
[B]. 4 cm
[C]. 6 cm
[D]. 8 cm
[E]. 10 cm.

Answer: Option D


No answer description available for this question.

Charles said: (Sep 20, 2015)  
Please show solution.

Ganesh said: (Nov 23, 2015)  
1600/x = 400/10-x.

Remember, stress diagram in RCC.

Akscivilian said: (Sep 23, 2016)  
It is mc/t = n/(d - n). But we don't have m here.

Baloch said: (May 9, 2017)  
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.

Spsg said: (Aug 8, 2017)  
E/R is same so equate bending stress/distance from nutral axis for both tensile and compressive.

Sudinbanerjee said: (Aug 13, 2017)  
Your explanation is absolutely correct, Thanks @Ganesh.

Satya said: (May 25, 2018)  
Thank you @Ganesh.

Reddy said: (Jul 29, 2018)  
Thank you @Ganesh.

Bittu said: (Aug 5, 2018)  
Thanks @Ganish.

Munni said: (Dec 25, 2018)  
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.

Pawan said: (Mar 13, 2019)  
Thank you very much @Ganesh.

Protul said: (Jun 26, 2019)  
Thanks all for explaining it.

Rabin Das said: (Feb 5, 2020)  
Thanks @Ganesh.

Raj Keshri said: (Sep 12, 2020)  
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).

Srikanta said: (Nov 18, 2020)  
1600/x = 400/(10-x).
5x = 40.
X = 8 cm.
Where x = Depth of neutral axis from top.

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