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.
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.
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.
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.
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.
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?
Deepak Singh Bisht said:
7 years ago
8/5l is the right answer.
Prasanth said:
8 years ago
But some books said 5/8, they took udl deflection divided by point load deflection.
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.
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.
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