Civil Engineering - Strength of Materials - Discussion

Discussion Forum : Strength of Materials - Section 2 (Q.No. 11)
11.
A simply supported beam of span L carries a uniformly distributed load W. The maximum bending moment M is
Answer: Option
Explanation:
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Discussion:
18 comments Page 1 of 2.

Akash said:   5 years ago
The correct answer is WL/8.

See the W is capital that means total load.
In these types of cases, W represents total load.
But we know the maximum bending moment is wl^2/8.

These cases w=W where w is udl intensity and W is a total load.
Thanks.
(1)

VINAYAK said:   1 decade ago
Due to symmetric loading, reactions are equal having magnitude = WL/2.

The maximum moment will occurs at center of span because of symmetric loading.

Take moment at this point,
Maximum bending moment = (WL/2) (L/2) - (WL/2) (L/4) = WL^2/8.

Mounika reddy said:   1 decade ago
Support reactions are Vay = wl/2 = Vby.

Cut s/n and take the moment at pont zero.

M = - (wx^2/2) +wlx/2.

Max moment at mid pnt l/2.

M = Wl^2/8.

For uniformly distributed load moment varies parabolically not linearly.

P.kishore said:   9 years ago
S L be the length of span kn/m total udl then we calculate maximum bending moment = force * distance.
And also half of the total udl distance then wl^2/8.

Faruque said:   6 years ago
It should said that total uniformly distributed load is W.

Sachin said:   9 years ago
According to me, the correct answer is WL^2/8 not wl/8.

Baloch said:   9 years ago
I agree that the correct answer is wl^2 / 8.

Avinash said:   7 years ago
Thanks for explaining @Mounika Reddy.

Rakesh yadav said:   2 years ago
Agree, the right answer is WL/8.

Kaku said:   4 years ago
The Correct answer is wl^2/8.


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