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 1 of 3.
Dheeraj said:
5 years ago
I do not know why they are providing like (24/15) in option. Whereas it is clearly visible that the answer should come (8/5).
It is clearly time-wasting of students.
It is clearly time-wasting of students.
(4)
Fakhar Naveed said:
1 year ago
It's correct.
384/48×5 = 384/ 240 = by dividing you will get the answer 24/15.
384/48×5 = 384/ 240 = by dividing you will get the answer 24/15.
(2)
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)
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)
Suresh said:
4 years ago
It must be 8/5 because ratios are that one which cannot be further divisible with common number how you can make it 24/15 because it is divisible of 3.
(1)
Rahul said:
6 years ago
8/5 * 3 = 24/15.
(1)
RAJEEV KUMAR said:
6 years ago
Yes, it is 8÷5. I too agree.
(1)
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.
Souryadeep Chaki said:
1 decade ago
Maximum deflections of a beam simply supported at its ends with an isolated central load = (WL^3)/(48EI).
Maximum deflections of a beam simply supported at its ends with uniformly distributed load= (5wL^4)/(384EI) {W=wL}.
So [(WL^3)/(48EI)] / [(5wL^4)/(384EI)] = 8/5.
Maximum deflections of a beam simply supported at its ends with uniformly distributed load= (5wL^4)/(384EI) {W=wL}.
So [(WL^3)/(48EI)] / [(5wL^4)/(384EI)] = 8/5.
Fakhar Navee said:
2 years ago
Here, it should be wl³/48EI × 384EI /5wl⁴ = 8/5.
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