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.
Dipak M said:
5 years ago
Thanks all for explaining the answer.
Protap said:
7 years ago
Thanks all for explaining the answer.
Manny said:
7 years ago
Thanks for the explanation.
Kumar said:
7 years ago
Thank you all for explaining it.
Sheela said:
7 years ago
Thanks @Arpita.
Senthilkumar said:
7 years ago
Good explanation. Thank you all.
Jayram said:
8 years ago
Thanks.
Protap said:
8 years ago
Thanks all.
Katakam saketh said:
8 years ago
Here, qmax=(1.5)*(F/bd).
F = 10000,b=20,qmax=30,d=?.
30 = 1.5*10000/(20*d),
d = 25cm.
F = 10000,b=20,qmax=30,d=?.
30 = 1.5*10000/(20*d),
d = 25cm.
(1)
Tailam mahesh said:
8 years ago
Excellent work @Arpita.
(1)
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