Mechanical Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 1 (Q.No. 11)
11.
For a beam, as shown in the below figure, the maximum deflection is 
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. Discussion:
67 comments Page 3 of 7.
Venu chowdary said:
6 years ago
It's simply supported beam maximum deflection occurs at the midpoint of load applied.
Nandhakumar said:
1 decade ago
In a simply supported beam maximum deflection always occur at the center of the beam.
Skd said:
1 decade ago
Its a simple case of concentrated load and can be solved by Macaulay's method easily.
Rakesh Kumar said:
3 years ago
Wa^2b^2/3EIL is deflection under the applied load, it is not the maximum deflection.
(1)
Prakash said:
1 decade ago
When simply supported beam always maximum deflection occurs at centre of the beam.
Jayanaidu said:
1 decade ago
Above question is satisfying the equation when a=l/2, b=l/2 in the w*a^2*b^2/48EI.
Hanu said:
1 decade ago
In simply supported beam max deflection occurs at center of the load applied.
Rajesh said:
1 decade ago
Ans: A.
Maximum deflection at the center = (wl^3)/(192EI) if a = b = l/2.
Maximum deflection at the center = (wl^3)/(192EI) if a = b = l/2.
Pranav Gijare said:
8 years ago
Here, it is a simply supported beam so we can directly apply BM= Wl^2/8.
Ash said:
1 decade ago
Max Deflection = Wa/9underroot3 EIL * (Lsquare - a square)rest to 3/2.
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