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,
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.
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).
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)
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