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 3 of 3.
Vijay kumar s said:
5 years ago
According to me, max shear stress = 3P/2BD.
(1)
Wakill Ahmad said:
4 years ago
Thanks all for explaining.
Prasanna Gosavi said:
4 years ago
Thanks all for explaining.
(2)
Madhu said:
3 years ago
Thanks all for explaining the answer.
(3)
Aravind said:
3 years ago
But how will max shear force create average shear stress instead of max shear stress? Please explain me.
(4)
Ajay Reddy said:
1 year ago
Let's
Maximum shear stress in rectangular beam = 3 ÷ 2 avg shear stress
30 = 3 ÷ 2(10000÷20×d)
30 = 750 ÷ d.
d = 750 ÷ 30.
D = 25cm.
Maximum shear stress in rectangular beam = 3 ÷ 2 avg shear stress
30 = 3 ÷ 2(10000÷20×d)
30 = 750 ÷ d.
d = 750 ÷ 30.
D = 25cm.
(5)
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