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 4 of 5.
GHANSHYAM said:
9 years ago
Please explain that what is y and how y = b/2?
And how calculation is z = {(h*b^3/12)/(b/2).
And how calculation is z = {(h*b^3/12)/(b/2).
Aftershock said:
9 years ago
It says parallel to the plan b, that means.
I = bh^3/12.
y = b/2,
Hence Z= b^2h/6.
I = bh^3/12.
y = b/2,
Hence Z= b^2h/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.
Rehan rufead said:
9 years ago
Section modulus for the rectangular section is 1/6 bd^2. ie I/y.
I = bd^3/12.
Y = d/2 bd^2/6.
Then, how the answer is option C.
I = bd^3/12.
Y = d/2 bd^2/6.
Then, how the answer is option C.
Ooha said:
1 decade ago
We are using different methods but same concept. So your process is correct no comment.
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.
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.
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.
Here I = h*b^3/12.
And Ymax = b/2, then section modulus= (h*b^3/12)/(b/2) = h*b^2/6.
Ithi said:
1 decade ago
(hb^3)% (0.5b) = (hb^2%6).
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"?
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