A square power screw is to be developed to lift a car The sc

A square power screw is to be developed to lift a car. The screw is to be made of AISI 1050 CD steel. The load on the jack is 4kN. The lift must operate from There is not a collar used for the design as pictured below. Although the screw has a right handed and left handed portion and the two nuts are either right handed or left handed on either side, there is not a need to consider this in your analysis. The coefficient of friction is 0.09. The screw should be designed for a minimum safety factor of 1.5 for all stresses.

a) Determine the root diameter required based on yielding

b) Determine the pitch based on thread stresses

c) Determine the nominal diameter and mean diameter

d) Determine the maximum torque required to lift the weight.

(Note the force in the screw is not the weight that’s being lifted.)

e) Is this a self locking design? Explain your answer.

you can neglect the torsion at first when calculating the root area. Also use preferred sizes in metric for the root diameter, pitch and determine the required sizes. When you calculate the thread stresses substitute use nt=2.6 for a conservative estimate.

Solution

GIVEN:- Material = AISI 1050 CD Steel, W = 4000N, uf = 0.09, Factor of Safety(fs) = 1.5, Yield strength of AISI 1050 CD Steel (Ys = 580 N/mm2), Tensile strength (Ts = 690 N/mm2), Angle is not given (Assume = 20.36) Standard angle for scissor jack.

TO FIND:- (a) Root diameter dr = ?, (b) Pitch p = ?, (c) Nominal diameter dn = ?, & Mean diameter dm = ?, (d) Maximum torque Tmax = ?, (e) To determine if design is self-locking or not = ?

SOLUTION:-

(a) By formula, 2xT = W / (tan) = 3.142 / 4 x dr^2 x Ys / fs

So, 4000 / tan(20.36) = 3.142 / 4 x dr^2 x 580 / 1.5 that is dr^2 = 35.59

So, dr = 5.96 = 6 mm, So Root diameter dr = 6mm (ANSWER)

(b) Area At for tensile stress considering fs =1.5 is given by At = [W/(tan) x Fs] / Ts = 23.5 mm2

Now, by standard formula At = 3.142/4 [dr - 0.94 x p]^2, So 23.5 = 3.142 / 4 [ 6 - 0.94 x p]^2

So p = 0.6 = 1 mm (approx), So Pitch p = 1mm (ANSWER)

(c) dn = dr + p = 6+1 = 7 mm and dm = dn x p / 2 = 7 x 1/2 = 3.5 mm

So Nominal diameter dn = 7 mm and Mean diameter dm = 3.5 mm (ANSWER)

(d) tan() = 2 x p / 3.142 x dm = 2 x 1 / 3.142 x 3.5 = 0.2, uf = tan() = 0.09

So Effort required to support load = 2 x T x tan(+) = 10810.81 x tan(5.14+11.31) = 3192.05 N

Max torque Tmax = Effort x dr/2 = 3192.05 x 6/2 = 9576.15 N.mm

Maximum torque considering root diameter = 9576.15 N.mm (ANSWER)

(e) = 5.14 and = 11.31

Since < , the power screw is not self-locking (ANSWER)