Civil Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 1 (Q.No. 2)
2.
As compared to uniaxial tension or compression, the strain energy stored in bending is only
Discussion:
26 comments Page 2 of 3.
Surya said:
8 years ago
Please give the explanation.
Hiralal Singha said:
8 years ago
Not getting this, Explain it please.
Deepak said:
8 years ago
How to solve this? please explain.
Maneesha.puvvala said:
8 years ago
Can you give me a deep explanation?
R N yadav said:
8 years ago
Please explain it clearly.
Ajay munde said:
8 years ago
Not getting this. Please explain.
Shakil said:
8 years ago
Due to avail stress =σ2/2E.vol.
Due bending = Σ2/6E.vol.
Due bending = Σ2/6E.vol.
(2)
Jeldi said:
8 years ago
U= (σ^2*v)/2E -----> (axial loading)
U=(σ^2*v)/6E -----> (bending)
So ans is (1/3)
For bending
U=Ʃ[ (M^2)/2E ] dx{hear integration frm 0 to L}
U=[ (M^2)/2E ]*Ʃ1dx
U=[ (M^2)/2E ]*L
Now M/I = E/R=σ/y
So M=σ*I/y
On simplifying
U=(σ^2*v)/6E
U=(σ^2*v)/6E -----> (bending)
So ans is (1/3)
For bending
U=Ʃ[ (M^2)/2E ] dx{hear integration frm 0 to L}
U=[ (M^2)/2E ]*Ʃ1dx
U=[ (M^2)/2E ]*L
Now M/I = E/R=σ/y
So M=σ*I/y
On simplifying
U=(σ^2*v)/6E
(1)
Shalu said:
7 years ago
Can't understand can you explain?
(1)
Shiney said:
7 years ago
Will you please explain it clearly?
(1)
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