Civil Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 1 (Q.No. 14)
14.
A closely coiled helical spring of radius R, contains n turns and is subjected to an axial load W. If the radius of the coil wire is r and modulus of rigidity of the coil material is C, the stress developed in the helical spring is
Discussion:
11 comments Page 1 of 2.
Pavani said:
5 years ago
T=16wR/πd^3 is the formula for the stress developed in the helical spring. put d=r you will get the answer.
(3)
Harika said:
6 years ago
Not getting this, Please explain me in detail.
(1)
Litu Malik said:
6 years ago
Not understanding, Please Explain in details.
Praveen Pawar said:
7 years ago
Thanks @Praveen Pawar.
Roy said:
8 years ago
Thanks for explaining it @Praveen.
Prakhar said:
8 years ago
Well said, @Praveen!
(1)
Yash said:
8 years ago
Well said, thanks @Praveen.
Praveen Pawar said:
9 years ago
Maximum stress developed by the spring :
Tmax = (16T)/(nD^3).
D = 2r, T = WR.
Tmax = (16WR)/(n8r^3).
= 2WR/nr^3.
Tmax = (16T)/(nD^3).
D = 2r, T = WR.
Tmax = (16WR)/(n8r^3).
= 2WR/nr^3.
(6)
KJICK said:
9 years ago
Not understand, Please explain in detail.
(1)
Mahesh said:
9 years ago
Twisting moment T = W * R and torsional stress applicable then T/J= ts/r.
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