Civil Engineering - Strength of Materials - Discussion

Discussion Forum : Strength of Materials - Section 1 (Q.No. 1)
1.
A rectangular bar of width b and height h is being used as a cantilever. The loading is in a plane parallel to the side b. The section modulus is
none of these.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
45 comments Page 1 of 5.

Mukesh yadav said:   1 decade ago
Section modulus = I/Y(max).

Then I = (b^3*h)/12.

And y = b/2.

Then z = (b^2*h)/6.

Mahesh said:   1 decade ago
Section modulus = I/y(max).

Here the load is parallel to the width b so,
I = (b^3*h)/12.

And y = b/2.

Then z = (b^2*h)/6.

C.mallireddy said:   1 decade ago
Section modulus = I/Y (this is max).

Moment of inertia (I) = (d*b^3)/12.

= (h*b^3)/12.

In this problem width = b.

Depth d = h.

And y = b/2.

:- Z= ( (h*b^3)/12)/(b/2).

Z = (h*b^2)/6.

Prem kumar meena said:   1 decade ago
Z = I/Y(max.) I = bd^3/12 ;

y(max.) = d/2 load is parallel to side b then b=h and d=b ;

Z = hb^3/12/b/2 ;

Z = hb^2/6.

Deepak Singh said:   1 decade ago
We know,

Z = I/Y [where, y=y(max)].
But, I = db^3/12.
y = b/2.

Now, Z = 2db^3/12d,
= db^2/6 [A/C to question d=h],
Z = hb^2/6.

Raj said:   1 decade ago
Can somebody explain me "plane parallel to side b"?

Josh said:   1 decade ago
Its like load is normal to height h which means parallel to b.

Ithi said:   1 decade ago
(hb^3)% (0.5b) = (hb^2%6).

Rima said:   1 decade ago
Section modulus = I/Ymax.

Here I = h*b^3/12.

And Ymax = b/2, then section modulus= (h*b^3/12)/(b/2) = h*b^2/6.

GYAN PRAKASH TIWARI said:   1 decade ago
If b is width and h is height then section of modulus.

z = I/Y ; I = hb^3/12 and y = b/2 then.

z = (hb^3/12)/(b/2).

Then z = (b^2/6) is answer. Because the load is in a plane to the side b.


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