Civil Engineering - UPSC Civil Service Exam Questions - Discussion
Discussion Forum : UPSC Civil Service Exam Questions - Section 24 (Q.No. 16)
16.
A close-coiled helical spring has 100 mm mean diameter and is mada of 20 turns to 10 mm diameter steel wire. The spring carries an axial load of 100 N Modulus of rigidity is 84 GPa. The shearing stress developed in the spring in N/mm2 is
Discussion:
3 comments Page 1 of 1.
Tiru said:
8 years ago
Shear stress = 16t/π*d cube.
T=w r.
d is dia of spring.
T=w r.
d is dia of spring.
Priyanka said:
5 years ago
Stress =16PR/pi D*3
Stress =160/pi
Stress =160/pi
Rahul verma said:
2 years ago
=> 16T/pi D*3.
T = PD/2 or PR
= 16PR/π D*3.
= 16*100*50/π10^3.
= 80/π.
T = PD/2 or PR
= 16PR/π D*3.
= 16*100*50/π10^3.
= 80/π.
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