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.

MAYANK JAIN 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)] = 24/15.

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.

Bhakti deshmukh said:   1 decade ago
@Bhakti deshmukh:

1) Max. Deflection of ssb subjected to central load = WL^3/48 EI.

2) Max. Deflection of beam subjected to udl = (5wL^4)/(384EI).

Hence ratio of these two is 8/5.

Mayuri said:   10 years ago
Ratio is 8/5, hence multiplying by 3 we will get 24/15 which is given in options.

Manoranjan said:   9 years ago
I am not getting this, Please, Properly discuss this question.

Sanjeev kumar said:   9 years ago
Answer is 8/5. Why we need to multiply the factor 3?

Vicky said:   8 years ago
Question is asking max deflection at ends and these formulas are for max deflection at centre.

Clear it please.

Jyothi said:   8 years ago
Max Deflection where occurs in the middle only. At supports, it is zero so 8/5 is the answer. But why 24/15?

Nikhil said:   8 years ago
24/15 is equal to 8/5 so 24/15 is correct answer.

Indu said:   8 years ago
How 24/15 is correct, could you explain me please clearly?


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