# Civil Engineering - Strength of Materials - Discussion

### Discussion :: 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

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

Explanation:

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.