Question

THECHNICAL SPECIFICATION Maximum Lift capacity 2000 lbs. Min. Lift Height 3-5/16 in. Max. Lift 15 in. Safety factor 2-3 Durable screw-drive system raises and lower jack with ease Operation : Manual

design a compact scissor jack with the specifcations. just need calculations please

it's for a machine design course

Answer 1

Coefficient of friction between threads= 0.25.

The whole calculation is made by assuming material of cast steel.

Let L1, L2, L3, L4 be the length of each link. Such that L1 =L2 =L3 =L4 =160mm and

W1+W2+W3 be the Length of power Screw So, W1 = W3 = 150 mm W2 = 50 mm Max. Lift = (h1+h2) = 375 mm

θ is the angle between Link with the horizontal when jack is at its lowest position.

W = (Mass * g) = 1 lb= 4.448 N

Therefore 2000 lbs = 9KN ( approx)

= 9 KN The tension T acting on the power screw.

Cos θ = (175-25)/160 = 20.36˚

Tension, T = W/2*tan θ

Total tension = 2*T = W/tan θ

Since we have assumed that the material for power Screw is a Cast Steel, so σt = 130N/mm2

Let dc be the core diameter of the screw.

Load = (π/4)* dc² * σt 2*T = W/Tan θ = (π/4)* dc² * σt

2*T = 9000/Tan (20.36˚) = 24252.08 N dc² = [W/ Tan (θ)*4]/ [(π* σt)] Hence, dc = 12.58 mm Since the screw is subjected to torsion shear stress we take, dc = 12 mm, pitch P = 1.5mm Mean diameter, d = do - P/2 = 13.5-1.5/2 = 12.75 mm Outer diameter, do = dc + P = (12+1.5) = 13.5 mm

Check for self-locking Tan (α) = Lead/π*d; α= helix angle

Lead L = 2*P; since the screw has a double start square thread.

Tan (α) = 2*p/π*d = 2*1.5/ π*12.75 = 0.0606

Helix angle; α= 3.47˚

Coefficient of friction; μ = Tan ϕ = 0.25; friction angle: ϕ= 14.03˚ Since, ϕ >α hence the screw is self-locking Effort required to support the load = 2*T Tan (θ+α).

Since, ϕ >α hence the screw is self-locking

Effort required to support the load = 2*T Tan (θ+α) = 24252.08 (Tan (α) + Tan (θ))/ (1- (Tan (α) * Tan (θ))) = 7648.5N Torque required to rotate the screw = effort*d/2 = 48759.18 N-mm

Tensile stress σ t = 2*T/ (π/4) * dc² = 24252.08/ (π/4) * 12 2 = 214.43 N

Shear stress in the screw due to torque ζ= 16*T/ (π* dc³) = 23.02  N/mm2

Maximum shear stress ζ max = √ (σt² +ζ²)/2 = 107.83 N/mm2

Maximum principal stress σ t max = σ t /2 + √ (σt² + ζ²)/2 = 215.045 N/mm2

Since the maximum stresses σ t max and ζ max within the safe limits.