Civil Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 1 (Q.No. 25)
25.
A rectangular beam 20 cm wide is subjected to a maximum shearing force of 10, 000 kg, the corresponding maximum shearing stress being 30 kg/cm2. The depth of the beam is
Discussion:
26 comments Page 2 of 3.
Hemanth Ganesh Erla said:
5 years ago
Thanks @Arpita.
(1)
Mounika Reddy said:
1 decade ago
T = SQ/Ib.
S=10000, T=30, Q=bh^2/8, I=bh^3/12.
Then h = 25cm,
or
Tmax = (3*Vmax)/(2*A).
h = 25cm, how will be 15cm.
S=10000, T=30, Q=bh^2/8, I=bh^3/12.
Then h = 25cm,
or
Tmax = (3*Vmax)/(2*A).
h = 25cm, how will be 15cm.
Chintu said:
8 years ago
Excellent explanation @Arpita.
Abhay said:
10 years ago
Maximum shearing stress σ max = 3P/2(bd).
where P = Shearing force.
30 = 3*10000/(2*20*d) => d = 25 cm.
where P = Shearing force.
30 = 3*10000/(2*20*d) => d = 25 cm.
Krishna said:
9 years ago
Convert the given max to average shear stress and then proceed using,
T(max) = 2/3 ( T(avg)).
T(max) = 2/3 ( T(avg)).
Wakill Ahmad said:
4 years ago
Thanks all for explaining.
Prasanta Kumar sahu said:
5 years ago
Thanks all.
Vipin said:
5 years ago
Well said Thanks @Arpita Saxena.
Dipak M said:
5 years ago
Thanks all for explaining the answer.
Protap said:
7 years ago
Thanks all for explaining the answer.
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