Mechanical Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 1 (Q.No. 18)
18.
Two closely coiled helical springs 'A' and 'B' are equal in all respects but the number of turns of spring 'A' is half that of spring 'B' The ratio of deflections in spring 'A' to spring 'B' is
Discussion:
9 comments Page 1 of 1.
Apoorva said:
5 years ago
δ= (8P(D^3)N)/(Gd^4).
δ of A/ δ of B= N of A/ N of B.
=(N of B/2)/ N of B = 1/2.
δ of A/ δ of B= N of A/ N of B.
=(N of B/2)/ N of B = 1/2.
Basar sk said:
6 years ago
The maximum diameter of the hole that can be punched from a plate of maximum shear stress 1/4th of its maximum crushing stress of punch, is equal to (where t = thickness of the plate).
Sourav majumder said:
7 years ago
Deflection in spring, X= (8w (D^3) n)/G (d^4).
Where w=axial load, D=mean coil diameter, n=no of active coils, G=modulus of rigidity, d=dia of spring wire,
Put the value of n for A and B and then divide deflection due to A by deflection due to B, and get 1/2.
Where w=axial load, D=mean coil diameter, n=no of active coils, G=modulus of rigidity, d=dia of spring wire,
Put the value of n for A and B and then divide deflection due to A by deflection due to B, and get 1/2.
(1)
Siram said:
7 years ago
Deflection= (8F(D*D*D)*N)/((d*d*d*d)*G).
Since deflection is directly proportional to N;
N(A)/N(B) = 1/2.
So, def(1)/def(2)=1/2.
Since deflection is directly proportional to N;
N(A)/N(B) = 1/2.
So, def(1)/def(2)=1/2.
B.chandrasekhar said:
9 years ago
Here those two springs having no of turns are,
A = 1/2.
B = 1.
A/B = 1/2.
A = 1/2.
B = 1.
A/B = 1/2.
Ilu said:
1 decade ago
A = 1/2*B.
Therefore A/B = 1/2.
Therefore A/B = 1/2.
Anil said:
1 decade ago
Deflection of spring is directly proportional to the number of turns.
Δ of A/δ of B = N/2N = 1/2.
Δ of A/δ of B = N/2N = 1/2.
(1)
SHASHANK said:
1 decade ago
For spring A = (64WR)^3/2 /(CD)^4 ..........(1).
For spring B = (64WR)^3 / (CD)^4 ..........(2).
from eqn 1 divided by eqn 2 will get,
ANS : A/B = 1/2.
For spring B = (64WR)^3 / (CD)^4 ..........(2).
from eqn 1 divided by eqn 2 will get,
ANS : A/B = 1/2.
Vinod arya said:
1 decade ago
Deflection in close coiled helical spring is given by 64WR^3n/cd^4.
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