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
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
24 comments Page 1 of 3.

Fakhar Naveed said:   1 year ago
It's correct.

384/48×5 = 384/ 240 = by dividing you will get the answer 24/15.
(2)

Fakhar Navee said:   2 years ago
Here, it should be wl³/48EI × 384EI /5wl⁴ = 8/5.

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)

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.
(4)

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.

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)

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.
(2)

Mekonnen S. said:   7 years ago
8/5 is the right answer.

Therefore, 8/5 * 3/3 = 24/15 is correct.

Deepak Singh Bisht said:   7 years ago
8/5l is the right answer.


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