Question

Problem 3. A plate is resting on four identical helical compression springs at its four corners. The plate is loaded centrally so that each of these springs undergoes same amount of deflection. For safety, a much harder spring is placed at the center of the plate. The free lengths of the softer springs are 20 mm more than the harder one. Corner Springs Center Spring Mean Diameter, D 40 mm 25 mm Wire Diameter, d 5 mm 5 mm Active Coils, Na 15 12 End Conditions Square and Ground Square and Ground Material for all the springs: Carbon Steel with G=83 GPa and p= 7800 kg/m Find the following: (A) Spring Rate at the initial loading condition (when only the corner springs are active). (B) Total load required to compress the corner springs by x mm (and hence compressing the central spring by x-20 mm). (C) Maximum shear stress at the center spring when it is compressed by 20 mm as in part B. 3 x=38 Four Corner Springs TOP VIEW Center Spring P Plate 20 mm FRONT VIEW

Answer 1

Helical spring Heet Load P = P1+Pz +P34P4 - identical Spring PI= P2=13= P4 = P14 Governing Equation for Helical spring maximum shear stress temax) Р Tmax= 8P DAW -a) Td3 • maximum Deflection comax ). Smax = 8PD 3. Na -(2) Gdf bo Di De mean Dia Do= oleter Dia DI= Inner Dia d= wire dia Na Number of active coil C= b sbring jodex ka Spring constant Spring rate KW = Wahl's factor Kw = 464 +615 41-4 С P=k. Smax Note! for EqOG. and Kw Orefer Helical sbring Chapter of your textbook. k= Gd4 8 D? Na 3 (A) K = 838103x 54 8 (40)3.15 k=6.75 N/mm TK = 6.75 kN/m] a spring rate.

Defaction Ean for centre spring sboing Required Loga, 2 (B) Y = 20 mm for coonor from E90 Pie Smax. Gd 4 8 D3 Ha Pe 20x 83X103x54 8x4013. 15 P = 135.09 N (Total Load) P = P, APL +P3 +P4 po 540.36 N (C) centre spring (I=20mm); it means cornerspring are 8Pp3 Na uc= Gd4 XG dt Pa 20x83x/03 x 54 8 D? Na 8(25)3.12 IP=691066 N/ & Load acting on C entre spring, How from EaD maximum shear stress at a centre spring c= b = as = 5 + 5 Kw=4851 ФҮГ-ч (T max centre = 8P Dok w Td3 (Tmax ) = 8 PG .kw Td2 Emax=462.34 MPa Kw z 1.3125