Civil Engineering - Strength of Materials - Discussion

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).

Since the density and length is same for two beams, the flexural weight now varies with their X-sectional area. So, we now have to find the ration of Ac/As.

Also we have,
M=Z*sigma.

M and sigma are the same for two beams, so the section modulus Z also should be the same for two beams.

i.e, Zcircle = Zsquare.
or, Pi*r^3 /4 = b^3 /6.
or, pi* {sq.root(Ac/pi)}^3 /4 = {sq.root (As)}^3 /6.
after simplifying, we get;
Ac/As = (2/3*pi^1/3)^2/3 = 1.11769.

Which makes (A) the right answer.

Section Modulus of Circle/Square gives the ratio of flexural weight.

Note: The C/S areas are same, So take Circle Diameter = 1.27m and Side of Sq = 1m, shall have the same area of cross-section
Zcircle = &P;i d^3/32.

Zsquare = bd^2 / 6.
Finally, Zcircle/Zsquare = 1.1189.

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.

@Devendra Singh

The SECTION MODULUS FOR CIRCLE IS (π*d^3/32) polar sec mod is (π*d^3/16) so the correct answer is 0.589.

Could not understand it. Please explain in detail.

Please anyone explain the solution.

What is meant by flexural weight?

Can anyone explain the solution?

0.894 is the approximate value.