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The helical Spring 'B' is placed inside the coils of second helical spring 'A' having the same number of coils and free axial length and material. The Mean coil diameter of A and B are in the ratio 3:2 and wire diameter are 12 mm and 7 mm resp. Calculate the mean coil diameter of both the springs and load capacity of the assembly if the maximum shear stress in the spring A is limited to 20MPa.

User Matthew M
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1 Answer

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The ratio of deflection in spring A to B is 1/8. The correct answer is option 1. 1/8.

Here's why:

Spring Deflection Formula: The deflection of a helical spring is directly proportional to the applied load (W) and inversely proportional to the spring constant (k). It can be expressed by the formula:

δ = W / k

Spring Constant and Mean Diameter: The spring constant of a helical spring is inversely proportional to the cube of the mean diameter (D) and can be expressed by the formula:

k = Gd⁴ / 8D³n

where:

G is the modulus of rigidity of the spring material

d is the diameter of the wire

n is the number of active coils

Given Information: We are given that both springs have the same number of active coils (n) and the same wire diameter (d). The mean diameter of spring A (DA) is half that of spring B (DB). This means DA = DB/2.

Comparing Spring Constants: Using the formula for spring constant, we can compare the spring constants of A and B:

KA = GBd⁴ / 8(DA)³n

KB = GBd⁴ / 8(DB)³n

Dividing these two equations, we get:

KA / KB = (DB)³ / (DA)³ = (2DA)³ / DA³ = 2³ = 8

Therefore, the spring constant of spring A is 8 times higher than the spring constant of spring B.

Ratio of Deflections: Using the formula for deflection and substituting the information we have:

δA / δB = (KB / KA) = 1 / 8

Option 1 is the right choice.

Question:-

If two closely coiled helical springs A&B with the mean diameter of spring A is half of that of spring B and having equal number of active coils and same wire diameter are subjected to same axial load of W., then the ratio of deflection in spring A to B.

1. 1/ 8

2. 1/ 4

3. 2

4. 8

User Dave Satch
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