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.
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.
Sudinbanerjee said:
8 years ago
Your explanation is absolutely correct, Thanks @Ganesh.
(1)
Satya said:
7 years ago
Thank you @Ganesh.
Reddy said:
7 years ago
Thank you @Ganesh.
Bittu said:
7 years ago
Thanks @Ganish.
(1)
Munni said:
7 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.
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