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
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
26 comments Page 1 of 3.

Samrat said:   1 decade ago
How it is calculated?

Dipankar said:   1 decade ago
Due to middle-third rule.

SUBINSEKHAR said:   1 decade ago
Can you explain?

Syed aasif said:   1 decade ago
We know strain energy stored is direct stress square/2.

Akash said:   1 decade ago
Step 1: Calculate strain energy due to uniaxial loading.

Viz. U1 = {(f^2)xvol}/(2xE).

Step 2: Calculate strain energy due to pure bending.

Viz. U2 = 0.5xMxtheta.

= 0.5x[(fxI)/y]x[(Mxl)/(ExI)].

(from simple bending equation & moment area method resp.)

Solve it U2 = {(f^2)xvol}/(6xE).

i.e U2 = 0.3XU1.

:) :) :) :).

Subhra said:   10 years ago
@Akash,

Please explain me what is l?

Binit said:   10 years ago
Need more explanation.

Dheeraj hindu said:   9 years ago
We know that,
FOR COMPRESSIOM OR TENSIOIN.

Strain energy(SE) = (1/2) * force * deformation = 1/2) * P * PL/AE = P^2L/2AE.

FOR BENDING.

SE = (P^2L^3)/6AE.

So bending (SE)= (L^2)/3 tension(SE).

Naren said:   9 years ago
Here, [SE] FOR COMPRESSION OR TENSION IS.

= 1/2 * force * deformation.

And what about [SE] for BENDING.

@Dheeraj Hindu.
Could you please explain this one.

Srinivas said:   9 years ago
How it is calculated?

Please explain me.


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