Question

A square-threaded power screw with single thread is used to raise a load of 25,000 lb. The screw has a mean diameter of 1-in. and four threads per inch. The collar mean diameter is 1.5 in. The coefficient of friction is estimated as 0.1 for both the thread and the collar. Assume the length of thread engagement is 1.5 times the outside diameter of the screw.

1. Calculate the safety factor for the power screw to fail in torsion. Assume the screw is made of steel.

2. Calculate the safety factor for the nut to fail in shear. Assume the screw and nut are both made of the same steel.

Answer 1

Solution:

Schematic and Given Data: Load 25,000 lb dm = 1 in. 4 threads/in. de = 1.5 in. f=f, = 0.1

solution: Assumption: The Screw is not exposed to viboution. (1) Calculate the safety factor for the power segew to fail in torsion. → Torque Required to lift the loud, To TheX E xde) Where to sheus stress due to tosque T de = coller meam diameter .: IP 16T TT (d) here, for que T is given by, anth + Z To Widmf. Tom +17, wfc.de - L Ñ .dm - fL 2 = (25000) (1) (0.1) TT (1) +0.25 7 | TICI) -(0.10 70-25) + (25000) (0-1)(15) 2 = 2263 + 1875 = 4138 lb.in 16T I TCC) 16 x 4138 TIX (1.509 :, I = 66208 10.5g = 6251 lb 12 → vne of Shens stsess should not be exceed 62 51e

(2) Calculate the sufety factor for the nut to fuil in Shells. in nut is generated given by → sheud stress w . Icyt) * Tin.(do). + here, do = dmt Pl2 = 1+0.25 i do = 1.125 inch screw = of P. : t= thickness . t = 0.125 inch i Inyt) Tim. (do). t (do).+ 25,000 3.14% 1.51250.125 = 25,000 0.6623 Tenut) = 37, 747.244 (46)/cin) ² Shear stress Produced in nut shond not be exceed 37,747.244 Leb)/cinje. -