Civil Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 2 (Q.No. 25)
25.
While testing a cast iron beam (2.5 cm x 2.5 cm) in section and a metre long simply supported at the ends failed when a 100 kg weight is applied at the centre. The maximum stress induced is :
Discussion:
15 comments Page 2 of 2.
Ajeet said:
7 years ago
It is also solved by the following method.
qmax=3/2q.avg.
So, qmax=1.5*100/(2.5*2.5).
qmax=3/2q.avg.
So, qmax=1.5*100/(2.5*2.5).
(1)
Sarang said:
6 years ago
Asked about maximum but not clearly defined which shear or bending stress. Kinda tough situation:
Shear stress is 1200.
Bending stress is 960.
Shear stress is 1200.
Bending stress is 960.
Khait kanwal said:
6 years ago
Good explanation.
(1)
Tanmoy Ghosh said:
5 years ago
M = WL/4 (100*100)/4 = 2500.
Z = I/Y (bd^2/6) = 2.604.
M/I = F/Y.
F = M/Z (2500/2.604).
F = 960Kg-cm^2.
Z = I/Y (bd^2/6) = 2.604.
M/I = F/Y.
F = M/Z (2500/2.604).
F = 960Kg-cm^2.
(8)
Chem said:
4 years ago
Bd 2square/6 is the moment of inertia for square?
Is it not bd^2/12? Please correct me, if I'm wrong.
Is it not bd^2/12? Please correct me, if I'm wrong.
(1)
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