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 :
960 kg/cm2
980 kg/cm2
1000 kg/cm2
1200 kg/cm2.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
15 comments Page 1 of 2.

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.
(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.
(8)

Khait kanwal said:   6 years ago
Good explanation.
(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.

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).
(1)

Satya said:   7 years ago
Excellent explanation @Prabhu.

Priya said:   8 years ago
Yes, its not related bending stress.
(1)

Anil said:   8 years ago
It is not stated that bending or shear stress.

Tailam said:   8 years ago
Superb @Prabhu.
(1)

Prabhu said:   8 years ago
The maximum stress (f) = M/Z,
M = WL/4= (100*100)/4=2500 in question a meter is given, it means 1-meter length converted to cm ans is in cm.
Z=bd^2/6=(2.5*2.5^2)/6=2.6042.
f = M/Z=2500/2.6042 = 959.98 nearly equal to 960 kg/cm^2.
(4)


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