Civil Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 3 (Q.No. 14)
14.
For the beam shown in below figure, the maximum positive bending moment is nearly equal to negative bending moment when L1 is equal to
Discussion:
19 comments Page 1 of 2.
PREM ROSHAN said:
3 years ago
For the maximum positive bending moment to be nearly equal to negative bending moment L should be greater than 2L1. So, L1 is less than 0.5L. Therefore the answer is .35L.
ARUN kumar said:
4 years ago
For equal moment L1 = 0.35 L,
For max bending moment is minimum L1 = 0.207 L,
If point load at both ends & for equal deflection condition L1= 0.207L.
For max bending moment is minimum L1 = 0.207 L,
If point load at both ends & for equal deflection condition L1= 0.207L.
ARUN kumar said:
4 years ago
For equal moment L1 = 0.35 L.
For equal deflection L1 = 0.207 L.
For equal deflection L1 = 0.207 L.
Shubham said:
4 years ago
BM at centre = WL1^2/8 + wl1L1^2/2.
Bm at support = WL1^2/2.
WL1^2/8 + wl1L1^2/2 = WL1^2/2.
= 0.207.
Bm at support = WL1^2/2.
WL1^2/8 + wl1L1^2/2 = WL1^2/2.
= 0.207.
Anitha said:
4 years ago
@Kajal.
How come 4a^2?
How come 4a^2?
(1)
Akshay said:
7 years ago
Thanks @Kajal.
Kajal Tomar said:
7 years ago
Correct ans is D; see how.
Max. Positive B.M = W/8 ( L^2 - 4a^2).
Max. Negative B.M = Wa^2 /2.
W/8( L^2 - 4a^2 ) = Wa^2 /2 ( acc. to quest);
Sove nd get a = 0.35 L.
Here (L1 = a = overhang).
Max. Positive B.M = W/8 ( L^2 - 4a^2).
Max. Negative B.M = Wa^2 /2.
W/8( L^2 - 4a^2 ) = Wa^2 /2 ( acc. to quest);
Sove nd get a = 0.35 L.
Here (L1 = a = overhang).
(1)
Garry said:
7 years ago
Option D is correct the answer will be point 353 times L.
Priya said:
7 years ago
Please describe it briefly.
A. AFARID said:
8 years ago
The corret option D.
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