Mechanical Engineering - Machine Design - Discussion
Discussion Forum : Machine Design - Section 9 (Q.No. 40)
40.
A long helical spring having a spring stiffness of 12 kN/m and number of turns 20, breaks into two parts with number of turns 10 each in both the parts. If the two parts are connected in series, then the softness of the resultant spring will be
Discussion:
3 comments Page 1 of 1.
Jehan aamir said:
8 years ago
Each spring will have a spring stiffness of Nk ...now according to question its is cut into 2 parts then each vill have;
**** 2 x 12 = 24 kN/m,
Now if they are connected in series,
Resultant stiffnes is;
1/R = 1/R1 + 1/ R2.
r1 and r2 are same.
Solving we get R = 12 R= k.
**** 2 x 12 = 24 kN/m,
Now if they are connected in series,
Resultant stiffnes is;
1/R = 1/R1 + 1/ R2.
r1 and r2 are same.
Solving we get R = 12 R= k.
AMAL said:
10 years ago
Stiffness is inversely proportional to no. of turns.
As it becomes half, stiffness is doubled.
Now they are connected in series so:
[(1/12)+(1/12)]^-1 = 12 kN/m^2.
As it becomes half, stiffness is doubled.
Now they are connected in series so:
[(1/12)+(1/12)]^-1 = 12 kN/m^2.
Sam said:
9 years ago
@Amal.
You are right, only the difference is (K=24 for each piece) then the answer will 12.
You are right, only the difference is (K=24 for each piece) then the answer will 12.
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