Civil Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 3 (Q.No. 1)
1.
The ratio of the moments of resistance of a solid circular shaft of diameter D and a hollow shaft (external diameter D and internal diameter d), is
none of these.
Discussion:
10 comments Page 1 of 1.
Deepak raj joshi said:
1 month ago
Moment of resistance is proportional to the section modulus:
For solid shaft: Z = MOI/y = (π/64) *D^4 / (D/2).
For hollow shaft: Z = MOI/y = (π/64) * (D^4-d^4)/(D/2).
(Z)solid/(Z) hollow = D^4/(D^4 - d^4) ans.
For solid shaft: Z = MOI/y = (π/64) *D^4 / (D/2).
For hollow shaft: Z = MOI/y = (π/64) * (D^4-d^4)/(D/2).
(Z)solid/(Z) hollow = D^4/(D^4 - d^4) ans.
R.k said:
5 years ago
M= σ *Z = π *d^3* σ /32 (for circular).
Nabajit Sarkar said:
5 years ago
Deflection for UDL = (w/L)*(L^4/8EI).
=wL^3/8EI.
Deflection of Pt. Load= wL^3/3EI.
Max. Deflection = -(5/24)*(wL^3/EI).
=wL^3/8EI.
Deflection of Pt. Load= wL^3/3EI.
Max. Deflection = -(5/24)*(wL^3/EI).
(1)
Amu bokato said:
6 years ago
Thanks all for explaining the answer.
Utsav said:
7 years ago
It's solid to hollow.
A is correct.
A is correct.
Deepak said:
7 years ago
For solid shaft:
Tmax = (π/ 16) τmax D3.
For a hollow shaft:
Tmax = (π/ 16) τmax (D4 - d4)/D.
Tmax = (π/ 16) τmax D3.
For a hollow shaft:
Tmax = (π/ 16) τmax (D4 - d4)/D.
(3)
Naveen kashyap said:
8 years ago
The answer should be C.
Priya said:
8 years ago
Yes, A right. I agree.
Ranjitha said:
8 years ago
I(solid shaft)= π D^4/64.
I(hollow shaft)= π (D^4-d^4)/64.
So, D^4/(D^4-d^4) is the answer.
Option A is correct.
I(hollow shaft)= π (D^4-d^4)/64.
So, D^4/(D^4-d^4) is the answer.
Option A is correct.
(1)
Taba Amith said:
9 years ago
The correct answer should be C.
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