Question

Problem. (20 points) A helical compression spring will be used in a pressure relief valve. When the valve is closed the spring leneth is 50 mm, and the spring force is to be 7 N. The spring uses a wire diameter of d 1.5 mm and a mean spring diameter of D = 10 mm, and has a solid length ofl.- 30 mm. Use hard-drawn ASTM A227 wire, squared and ground ends, and ignore buckling since the spring is in a cage. Use G-79.3 GPa. Perform the following: a) Find the force needed to compress the spring to its solid length, b) Find the number of active coils and spring index, c) Find the spring rate, d) Find the spring's free length, and e) Find the safety factor against torsional yielding Solution:

Answer 1

Probiem = 03 Griven :- Helical Compression spring fon poces ule realicf valve When valve is closed spring length l= 50 mm ts= IN wine diametu d=1.5 mm quing sping Sulid lemgth is= 30mm diametay D= jo mm meom G4 = 79.3 GPa Spaing is mode of Consiluing hand-dnawn - ASTM -A227 wire 'min.tenstle strees = 1390mPa atd=1.5mm Talke 0.30 to 0,5o Sut T= 0.45 Set Fun cold drawneol mor, allowable s7= 625,5 mPq Tonsional sfrees Salid = total no. of cold x wine diq length Is n+= Nactive t ninocttve Is = (Moctive t2) ×1.5 30 = (nacttve +2)x1.5 ninoctive =2 Fon squaed amd ground onds 18" Nactire

Find deflection when valve is cosed fs=7N 2=5omm 8* Fs * D°na G1x d4 I 8Y7X( 10)°x18 79.3 x10 x(1:5)4 = 2.510 85 mm this defleetion will orcurs when fonce is IN NOW when noi no fonce is octing Spimg the longth will be free longth of sping sping = 50 + 2.5I085 = 52.5108Smm Mari'mum deflection Free lorgth y = ds + Smorx + To tolgap Hele Totalgar (0.5 to 2 mm) (n1-1) * gcip, betucen two cadjacentcoils Grenenally takem 15% af Smax) %3D in some cares Totalgap = 19x (0-5ite 2mm) of taking Is%. "In qun cale we are raup. 1. lf = ds + Smdx +o.15 Smox 52.51085 = 30 + Snocix (1. 15)

dmoz = 19.57 mm fmor Now Smar * K = Smor x Gid9 8D3n fmoy= 19.57X 2.7879 fmex = S4.559 N fon forton of safety = 0.615 :-2=6.666 where c=: yeaü + K= 1.224 1.224X 8X S4.55996-666 mot T= 504.06 M Pa Tallewable = 625.smPq FOS = Focton of Safety = Lalowable (Tmox) design %3D 625. S %3D S04.06 Fos= 1.24

a) fonce needed to compres the spelng to its sald length Fmery = 54. SS9N b) Numbet af active tunn = 18 Sping index C=6.66 ) spring nate R= Gidy 803n 2.7879 d) Speings free bongth= 5a,510g man e) Satety factor = 1.24