Mechanical Engineering - Machine Design - Discussion
Discussion Forum : Machine Design - Section 11 (Q.No. 20)
20.
Two helical springs of the same material and of equal circular cross-section, length and number of turns, but having radii 80 mm and 40 mm, kept concentrically (smaller radius spring within the larger radius spring), are compressed between two parallel planes with a load W. The inner spring will carry a load equal to
Discussion:
10 comments Page 1 of 1.
Zakir said:
4 years ago
Here it's using two different springs then how can you calculate it on basis of an inverse relationship? It could be used if he reduces diameater.
Ganeshr said:
5 years ago
Option D is the correct answer.
Vinay BEL said:
5 years ago
8 W/9 is right I think.
Panja said:
6 years ago
Not R but inversely proportional to r^3.
Sch said:
7 years ago
W = W1 + W2.
but, deflection, x1 = x2.
therefore, w2 = 8W/9 (Ans. D).
but, deflection, x1 = x2.
therefore, w2 = 8W/9 (Ans. D).
Biren desai said:
9 years ago
No, the given answer is right.
Ashutosh said:
9 years ago
The answer should be option (D)
w is inversely proportional to R^3.
R1 = 80 & R2 = 40.
W1/W2 = 1/8 and W1 + W2 = W.
W1 = W/9 & W2 = 8W/9.
w is inversely proportional to R^3.
R1 = 80 & R2 = 40.
W1/W2 = 1/8 and W1 + W2 = W.
W1 = W/9 & W2 = 8W/9.
ASHUTOSH said:
9 years ago
The answer is D.
R1 = 40 and R2 = 20.
W inversely proportional to R^3 therefore, w1/w2 = 1/8 and w1 + w2 = w,
So, w2 = w*8/9 and w1 = w*1/9.
Therefore, the answer must be option D.
R1 = 40 and R2 = 20.
W inversely proportional to R^3 therefore, w1/w2 = 1/8 and w1 + w2 = w,
So, w2 = w*8/9 and w1 = w*1/9.
Therefore, the answer must be option D.
Ashok prajapati said:
1 decade ago
That option is wrong b/c inner spring stiffness is k than outer spring stiffness is 125k. So that inner spring take 8w/9 force.
Bharat said:
1 decade ago
How is it option B? I got option D.
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