Mechanical Engineering - Strength of Materials - Discussion

Discussion Forum : Strength of Materials - Section 1 (Q.No. 27)
27.
The simply supported beam 'A' of length l carries a central point load W. Another beam 'B' is loaded with a uniformly distributed load such that the total load on the beam is W. The ratio of maximum deflections between beams A and B is
5/8
8/5
5/4
4/5
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
13 comments Page 1 of 2.

Maunankbhavsar said:   1 decade ago
((wl^3)/(48EI))/((5wl^3)/(384EI)) = 384/(5*48) = 8/5.

J B CHAUHAN said:   1 decade ago
POINT LOAD = WL^3/48EI.

UDL = 5WL^3/384EI.

ABHISHEK SINHA said:   1 decade ago
Deflection for center load = WL^3/48EI.
Deflection for udl load = 5WL^3/384EI.

Ratio of both load = 8/5.

Naresh Goggela said:   1 decade ago
At point load = WL^3/48EI.

At uniformly distributed = 5WL^3/384EI, So.
Ratio is= 8/5.

LAVANYA said:   1 decade ago
Point load WL^3/3EI.

UDL 5WL^3/384EI.

RATIO IS 8/5.

Taolin said:   10 years ago
Maximum deflection of a uniform loaded simple beam is (5FL^4)/384EI not (5FL^3)/384EI.
(1)

Ajeet said:   10 years ago
Ya deflection on UDL beam is (5wl^4)/384EI.

Naveen said:   9 years ago
Yes, but according to the question, W = wl which means total load.

Now, W becomes 5wl^3/384EI.

D g maru said:   9 years ago
Deflection for center load = WL^3/48EI.

Deflection for udl load = 5WL^3/384EI.

Ratio of both load = 8/5.

Manoj meena said:   9 years ago
Ratio of both = 8/5.


Post your comments here:

Your comments will be displayed after verification.