Question

Question 1 (10 pts) A square-threaded, single thread power screw with a mean diameter of 1 in and four threads per inch is used to raise a load of 25,000 lb. The collar mean diameter is 1.5 in. The coefficient of friction is 0.1 for both the thread and the collar. Determine the following: Load 25,000 lb Q1.1. Determine the major diameter of the screw (2pts) Q1.2. Determine the screw torque required to raise the load (2pts) Q1.3. Determine the minimum thread coefficient of friction allowed to prevent the screw from overhauling if collar friction is eliminated. (2pts) Q1.4. If the screw is lifting the load at the rate of 50 in/min, determine the rotating speed of the power screw (1.5pts) Q1.5. Also determine the efficiency of the power screw (1pts) Q1.6. Does the efficiency increase or decrease if the collar is eliminated? Justify your answer (1.5pts)

Answer 1

Given: Square thread, single start (n=1) dm = lin , 4 throad /inch (lead = ) W = 25,00016 ,dc= 1.5 in f = fe=0.1 Solution: 1. 1.) Major diameter (d) pitch (p) =(4) lead & no. of starts(n) p= 4xl 7 p=0.25 in d = dm + b 2 = 1 +0.25 d= 0.125 +1 = d=1.425 in 1.2) Torque required to raised the load (7) T= Wdm fromth Tam -fL JT T= 2500001 (0.1) T (1) +0.25 725000 (0.1) (15) 2 LT (1) -(0-1)(0.25) T-2263 + 1875

| T=4138 lb.in 1.3) fmin = ? to prevent overhauling, fe neglect Overhauling is the condition when load itself begin to turn the scrow and descond down So to prevent this we make self-locking scrow. And the scrow is sell locking fZ L r dm f2 0.25 I (1) If 20.08 1.4) Rotating Speed (N) if V = so unl min Rotating speed of power screw is a function of linear speed and lead of scrow. Rotating speed (N) = V Lead (L) N = 50 = 200 0.25 N = 200 rpm

1.5) Efficiency of power screw (h) n = WL 2T T 25,000 x 0.25 21 4 4 138 [2= 24.038 % 1.6) Efficiency when collar friction is neglected le. fc=0 T'= Wam [ predmth 7+Wfche 2 1 tdm-fL J K T'= 2263 lb-in bi - WL 25000 X 0.25 211 x 2263 ant' n = 43.95% So, the efficiency increases when callar friction is neglected.