Civil Engineering - Strength of Materials - Discussion

Discussion Forum : Strength of Materials - Section 4 (Q.No. 41)
41.
The section modulus of a rectangular section is proportional to
area of the section
square of the area of the section
product of the area and depth
product of the area and width
half moment of inertia of the section.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
21 comments Page 2 of 3.

Jaz said:   6 years ago
Z=I/Y and I=AK^2.

So, that z proportional to area and radius of Gyration.

Roshan said:   7 years ago
Z=I/y, I = the second moment of area = Ai*yi.

Z=(Ai*yi)/y.

Z proportional to the summation of the area of the element.

Jimmy said:   1 decade ago
How? As per calculation it is A*d/6. Answer should be C.

Sehra said:   8 years ago
Z=I/y max=bd^2 /6
Z=Ad/6 A=bit
Z-A or Z- d.

Thebaloch said:   8 years ago
S = m/z
Z = M/S
S = F/A

So Z=AM/F.
So, A is correct.

Baloch said:   9 years ago
E seems right,

The second moment of area of rectangular sec = bh^3 / 12 , and its Z = bh^2/6 (half).

Yogesh said:   9 years ago
Z = bd^2/6.

Z ^ bd^2.
Z^(bd) x (d).
Z ^ (Area) x (Depth).

Answer should be 'C'.

Suwami said:   9 years ago
Z = I/y = Ak2/y.

Rakesh kumar sah said:   9 years ago
Section modulus = Z/A.

Hence option A is correct.

S s said:   9 years ago
You are right @Jimmy, its bd2/6.


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