Civil Engineering - Strength of Materials - Discussion

Discussion Forum : Strength of Materials - Section 2 (Q.No. 36)
36.
A simply supported wooden beam 150 cm long and having a cross section 16 cm x 24 cm carries a concentrated load, at the centre. If the permissible stress ft = 75 kg/cm2 and fs = 10 kg/cm2 the safe load is
3025 kg
3050 kg
3075 kg
3100 kg.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
19 comments Page 1 of 2.

Asha said:   4 years ago
Permissible stress should be lesser of two

Why did we take 75, it should be 10 right?

Civi said:   5 years ago
@Pronay.

M- MOMENT OF INERTIA;
C- a distance of extreme fibre from the neutral axis (H/2)
but it for I section beam why they used it in rectangular? please explain me.

BHASKAR said:   5 years ago
f/y = M/I.
M = fI/y.
M = 75*((16 * 24^3)/12) * (2/24).
M = 115200.

WL/4 = 115200.
W = (115200*4)/150.
W = 3072.
(4)

Pronay said:   5 years ago
@Anomi.

What is M.C and where you get the value of 12?

Anomi said:   5 years ago
Ft = tensile stress
Ft = M.C/I where i=bh3/12 = 18432.
75 = M. 12/18432.
M = 115200.
M = PL/4.
115200 = p 150/4.
P= 3072.

Answer is 3075,i.e Option C.
(1)

Abhik said:   6 years ago
@Kailash.

3071 is the correct answer. Then how it will be 3050?

Deepak said:   7 years ago
Permissible stress is 3075 but after applied fos 10, the safe load is 307.5 kg.

Kailash said:   7 years ago
The answer should be 3050 because max permissible value is 372 then how this structure can withstand 375 load?

Praveen Balaji said:   8 years ago
Calculate Z and M = f x z..
After calculating M = 115200kgcm.
Then M =wl/4.
From this w= 3072kg.

Sandeep said:   8 years ago
@Dalisha.

How to get the answer, can you explain, What is the value of w?


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