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.
Discussion:
45 comments Page 3 of 5.
Poluri Naresh said:
5 years ago
Options B is the correct answer.
Avi said:
7 years ago
I think, option B is correct.
Josh said:
1 decade ago
Its like load is normal to height h which means parallel to b.
Raj said:
1 decade ago
Can somebody explain me "plane parallel to side b"?
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.
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.
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.
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.
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.
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.
Kumari Podala said:
11 months ago
Yes, According to me, the correct answer is option B.
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.
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.
Devendranaik said:
9 years ago
Section modules (z) =I/Y.
MOMENT OF INERTIA (I) =B^3h/12.
Y = b/2.
Therefore = Z = I/Y = B^2h/6.
MOMENT OF INERTIA (I) =B^3h/12.
Y = b/2.
Therefore = Z = I/Y = B^2h/6.
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