Principles of Dairy Machine Design

Module 7. Springs

Lesson 32
SOLVING NUMERICAL

32.1 Numerical

A helical spring of wire diameter 6mm and spring index 6 is acted by an initial load of 800N. After compressing it further by 10mm the stress in the wire is 500MPa. Find the number of active coils. G = 84000MPa.

Solution:

D = Spring index (C) x d = 36mm

τ_max = (K_w) x ( 8FD / Πd³ )

K_w = (4C-1 / 4C-4) + ( 0.615/ C) = 1.2525

Or

500 = 1.2525 x (8F x 36/ π x F)

Therefore F = 940.86 N

K = F / δ
= (940.6 - 800) / 10
= 14 N/mm
K = (Gd⁴) / (8D⁴N)
Or
N = (Gd⁴) / (K8D³N)
= (84000 x 6⁴) / 14 x 8 x 3 x 6^{3
= ~ 21tuns}

32.2 Problem

Design a leaf spring for the following specifications: Total load = 140 kN ; Number of springs supporting the load = 4 ; Maximum number of leaves= 10; Span of the spring = 1000 mm ; Permissible deflection = 80 mm.Take Young’s modulus, E = 200 kN/mm² and allowable stress in spring material as 600 MPa.

Solution:

Given: Total load = 140 kN ; No. of springs = 4; n = 10 ; 2L = 1000 mm or

L = 500 mm; δ = 80 mm; E = 200 kN/mm² = 200 × 103 N/mm²; σ = 600 MPa = 600 N/mm²

We know that load on each spring,

2W =Total load/No. of springs

=140/4

=35 kN

There fore,

W = 35 / 2 = 17.5 kN = 17500 N

Let t = Thickness of the leaves, and

b = Width of the leaves.