Question

why i section beam has higher strength compared to circular beam

Answer 1

In structural mechanics, “more efficient” against bending usually means the particular cross-section will develop lower bending stress (σσ) under the action of the same bending moment (MM). The formula for determining the bending stress in a beam under simple bending is:
σ=My/Iσ=My/I ; (Ref.)
Where II = moment of inertia of the particular cross-section,
yy = distance of a horizontal plane on the cross-section from the neutral axis plane (usually the centroidal axis plane)

Therefore, for the same MM and same yy in the above formula, lowering the bending stress (σσ) will require the moment of inertia (II) to be higher. Now, let’s check whether the circular section or the square section has the higher value of II:

The area (AA) of a square section with side hh is A=h2A=h2 and its I=h4/12I=h4/12
The area (AA) of a circular section with radius rr is A=πr2A=πr2 and its I=πr4/4I=πr4/4
Given that the two areas are equal: h2=πr2h2=πr2 or, r=h/π−−√r=h/π
Plugging this value of rr into the circular I=πr4/4I=πr4/4 will give: I=h4/4π=h4/12.57I=h4/4π=h4/12.57 which is smaller than the II for square section: I=h4/12I=h4/12

Therefore, the higher value of II for the square section means the bending stress will be lower at any point on the square cross-section compared to the circular section and the square section will be more efficient.

hope it's help you.