Civil Engineering - Strength of Materials - Discussion

Discussion Forum : Strength of Materials - Section 5 (Q.No. 26)
26.
The cross sections of the beams of equal length are a circle and a square whose permissible bending stress are same under same maximum bending. The ratio of their flexural weights is,
1.118
1.338
1.228
1.108
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
13 comments Page 1 of 2.

Ankit said:   9 years ago
How its possible?

Asha said:   9 years ago
Anyone knows exact solution?

Rahul meena said:   9 years ago
Square/circle = 1/1.12.

Naren said:   8 years ago
What is meant by flexural weight?

Mohit said:   7 years ago
0.894 is the approximate value.

Usman Khan said:   7 years ago
Please anyone explain the solution.

Kiran said:   6 years ago
Can anyone explain the solution?

Devendra singh said:   6 years ago
Stress the same relationship sectional modulus ratio.

Circle sectional modulus= &pi& * d^3/16.sovle .19625.
Square z = d^3/6. Solve 1/6=.1667.
Ratio = .19625/.1667 = 1.118.

Sunn said:   6 years ago
Could not understand it. Please explain in detail.

Roshni said:   6 years ago
Flexural weight = density *volume.

Density and length are the same, so the ratio of flexural weight comes to be the ratio of their area.

Wcircle/Wsquare = ( (pi*d^2) /4) / a^2.

Now, to find the ratio of d^2/a^2, we use the condition same bending stress.

Bending stress=My/I, where M is the same for both, so this becomes section modulus of the circle equals section modulus of the square. And find the relation among d and a, substitute in the weight equation above the answer comes to be the option (A).
(1)


Post your comments here:

Your comments will be displayed after verification.