Civil Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 1 (Q.No. 19)
19.
The ratio of the maximum deflections of a beam simply supported at its ends with an isolated central load and that of with a uniformly distributed load over its entire length, is
1
Discussion:
24 comments Page 2 of 3.
Shubham said:
8 years ago
The answer is 8L/5.
Kallan said:
8 years ago
8/5 = 1.6,
24/15 = 1.6,
Both are same.
24/15 = 1.6,
Both are same.
Ezaaz Ahmed said:
8 years ago
Maximum Deflection at the mid-span of a SSB with an isolated central load is WL3/48EI--> i
Maximum deflection at the mid-span of a SSB with UDL through the entire span is 5WL3/384EI--> ii.
i/ii=8/5.
So the correct answer is 8/5.
Maximum deflection at the mid-span of a SSB with UDL through the entire span is 5WL3/384EI--> ii.
i/ii=8/5.
So the correct answer is 8/5.
(1)
Prasanth said:
8 years ago
But some books said 5/8, they took udl deflection divided by point load deflection.
Deepak Singh Bisht said:
7 years ago
8/5l is the right answer.
Mekonnen S. said:
7 years ago
8/5 is the right answer.
Therefore, 8/5 * 3/3 = 24/15 is correct.
Therefore, 8/5 * 3/3 = 24/15 is correct.
Shankar gubbala said:
7 years ago
The isolated load is nothing but point load. So,
For simply supported beam deflection is wl^3/48EI.
For UDL 5WL^4/384EI,
So the ratio of these two is 24/15,
Again 24 and 15 cancel in 3 table,
Finally we get 8/5.
For simply supported beam deflection is wl^3/48EI.
For UDL 5WL^4/384EI,
So the ratio of these two is 24/15,
Again 24 and 15 cancel in 3 table,
Finally we get 8/5.
(2)
RAJEEV KUMAR said:
6 years ago
Yes, it is 8÷5. I too agree.
(1)
Rahul said:
6 years ago
8/5 * 3 = 24/15.
(1)
THIRU said:
6 years ago
8/5 original answer but this answer multiply with 3/3 then get options answer.
(8/5)*(3/3) = 8*3/5*3 = 24/15.
(8/5)*(3/3) = 8*3/5*3 = 24/15.
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