Civil Engineering - Strength of Materials - Discussion

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.

@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.