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Lesson 32. SOLVING NUMERICAL
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 / Π 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 = (Gd^{4}) / (8D^{4}N) Or N = (Gd^{4}) / (K8D^{3}N) = (84000 x 6^{4}) / 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/mm^{2} 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/mm^{2} = 200 × 103 N/mm^{2}; σ = 600 MPa = 600 N/mm^{2} 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. |