# Civil Engineering - Steel Structure Design - Discussion

Discussion Forum : Steel Structure Design - Section 2 (Q.No. 39)

39.

If

*f*is the maximum allowable bending stress in a tension member whose radius of gyration is*r*and depth is 2*y*, the required cross sectional area*A*is given byDiscussion:

4 comments Page 1 of 1.
Sanket said:
2 years ago

Here, I = A * R^2.

(1)

Arifkhan said:
4 years ago

We have: F=My/I.

Also, I=√(I/A)=> I= r^2*A.

Now F =My/r^2*A => A= My/fr^2.

Also, I=√(I/A)=> I= r^2*A.

Now F =My/r^2*A => A= My/fr^2.

(1)

Chhaya said:
7 years ago

F/Y = M/I,

F = MY/I,

r = sqrt I/A,

I = Ar2.

By substituting the values, you will get the answer.

F = MY/I,

r = sqrt I/A,

I = Ar2.

By substituting the values, you will get the answer.

Kush said:
8 years ago

Pure bending equation:

(F/Ymax) = (M/I) = (E/R).

Put,

Ymax = 2y/2 = y.

I = Ar^2 in the above equation.

Then we get the answer A.

(F/Ymax) = (M/I) = (E/R).

Put,

Ymax = 2y/2 = y.

I = Ar^2 in the above equation.

Then we get the answer A.

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