Mechanical Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 9 (Q.No. 45)
45.
The extension of a circular bar tapering uniformly from diameter d1 at one end to diameter d2 at the other end, and subjected to an axial pull of P is __________ the extension of a circular bar of diameter d1 d2 subjected to the same load P.
Discussion:
9 comments Page 1 of 1.
Balram said:
3 years ago
In the above questions, the load and E of both bars are the same, so elongation is the same.
So equate both the elongation formula.
So equate both the elongation formula.
(1)
Ahsan said:
4 years ago
The answer is B because the product of dia's of both the bars is different and dia's are inversely proportional to elongation.
(2)
Bhavik ahir said:
5 years ago
Thanks @Srilaksmi.
(1)
Ravikiran Gokul Darade said:
5 years ago
Well said, thanks @Srilakshmi.
Krishna said:
5 years ago
Very nice, Thanks @Srilaksmhi.
Aqib javed said:
7 years ago
You are absolutely right @Sri Lakshmi.
Srilakshmi said:
8 years ago
Extension of tapering bar is 4pl/pie*e*E*d1*d2.
Extension of circular bar pl/aE.
As given in question circular bar dia sqe.root d1*d2 so;
Area of circle pie/4*D^2.
pie/4*(sqe.rootd1*d2).
So ans is 4pl/pie*Ed1d2.
Extension of circular bar pl/aE.
As given in question circular bar dia sqe.root d1*d2 so;
Area of circle pie/4*D^2.
pie/4*(sqe.rootd1*d2).
So ans is 4pl/pie*Ed1d2.
(2)
Srilakshmi said:
8 years ago
Extension of circular tapering bar is;
= 4pl/π*E*d1*d2.
= 4pl/π*E*d1*d2.
Sozharajan said:
8 years ago
Extensions only depand on Young's modulus but not depend on area.
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