Chapter-12
Spring
S K Mondal’s
The number of coils in the spring 1,2 and 3 is 10, 12 and 15 mean diameter of spring 1,2 and 3 in the ratio of 1 : 1.2 : 1.4 Find out distance x so that rod remains horizontal after loading. Since the rod is rigid and remains horizontal after the load p is applied therefore the deflection of each spring will be same
δ1 = δ2 = δ3 = δ
(say)
Spring are made of same material and out of the rods of equal diameter
G1 = G2 = G3 = G and d1 = d 2 = d3 = d
Load in spring 1
P1 =
Gd 4 δ Gd 4 δ Gd 4 δ = = 64R13n1 64R13 × 10 640R13
.....(1)
Load in spring 2
P2 =
Gd 4 δ Gd 4 δ Gd 4 δ = = 64 × R32n2 64 × (1.2)3 × 12R13 1327.10R13
.....(2)
Load in spring 3
P3 =
Gd 4 δ Gd4 δ Gd 4 δ = = 64R33n3 64 × (1.4)3 × 15R13 2634.2R13
.....(3)
From eqn (1) & (2)
640 P1 1327.1 P2 = 0.482 P1 P2 =
from eq n (1) & (3) 640 P3 = P1 = 0.2430 P1 2634.2 Taking moment about the line of action P1 P2 × L + P3 × 2L = P.x 0.4823 P1L + 0.2430 P1 × 2L = P.x. x=
( 0.4823 + 0.486 ) P1L
P total load in the rod is P=P1 +P2 +P3
........(4)
P = P1 + .4823P1 + 0.2430P1 P = 1.725 P1
......(5)
Equation (4) & (5) x=
0.9683L 0.9683L = = 0.5613L 1.725 P1 / P1 1.725 x = 0.5613 L
Conventional Question ESE-2008 Question:
Answer:
A close-coiled helical spring has coil diameter to wire diameter ratio of 6. The spring deflects 3 cm under an axial load of 500N and the maximum shear stress is not to exceed 300 MPa. Find the diameter and the length of the spring wire required. Shearing modulus of wire material = 80 GPa.
Gd 4 P Stiffness, K = = δ 8D 3 n
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