Figure 13.13 shows a simple screw jack that consists of a nut and a screw fitted into it. The load W to be lifted is placed on the head of the screw. The load W can be raised or lowered by rotating the screw by a handle attached to its head.
Figure 13.13 (a) Screw Jack, (b) Screw, and (c) Development of Screw Surface
Screw is formed by cutting a helical groove on a solid cylinder of metal as shown in Figure 13.13 (b). The solid material separating the successive turns of the groove forms the thread of the screw. In Figure 13.13, p denotes the pitch, d denotes the diameter of screw, α denotes the helix angle. A nut is formed by cutting a helical groove in a cylindrical hole in a solid material. The cross-section of the groove must be of the same form as the cross-section of the thread of the screw with which it is intended to work.
When the screw in mesh with its nut is rotated once, it moves in the direction of its axis, since the position of the nut is fixed. The distance moved in one rotation of the screw is equal to its pitch length. If the screw is double threaded, the distance moved by the screw is equal to double of its pitch which is known as lead of the screw.
Let the effort applied at the circumference of the screw to lift the load W be P1, α be the helix angle, ɸ be the friction angle.
P1 = W tan (α + ɸ)
Now the effort is actually applied at the end of the handle. Let l be the length of the handle, d be the mean diameter of the screw, and P be the effort required. According to moment equation, moment of P1 about the axis must be equal to the moment of P.
where 
and tan ɸ = μ, μ is coefficient of friction.
In Case of Lowering the Load
where and tan ɸ = μ, μ is coefficient of friction.
Maximum Efficiency of Screw Jack for Raising the Load W: Efficiency of the screw jack can be given by the ratio of ideal effort to the actual effort to lift the load W.
α + ɸ = α gives ɸ = 0 which is not acceptable.
Hence,
Putting the value of α in the equation, 
, we get
Efficiency in terms of MA and VR
Example 13.14: A screw jack carries a load of 5,000 N. The mean diameter of the screw rod is 60 mm and the pitch of the square thread is 15 mm. If the coefficient of friction is 0.25, find the torque required to lift the load and the efficiency of the machine. Also, find the torque required to lower the load.
Solution:
Given: W = 5,000 N, d = 60 mm, p = 15 mm, μ = 0.25
(From Figure 13.14)
tan ɸ = μ = 0.25 ⇒ ɸ = 14.03°
Figure 13.14 Pitch of a Screw
Example 13.15: A screw jack has a mean diameter of 100 mm and a pitch of 20 mm. The coefficient of friction between the screw and nut is 0.1. Find the effort required to be normally applied at the end of the operating lever of length 500 mm to (i) lift the load of 25 kN, and (ii) to lower the same load.
Solution:
Given: W = 25 × 103 N, L = 500 mm, p = 20 mm, d = 100 mm, and μ = 0.1
- When load to be lifted,
- When load to be lowered,
Example 13.16: The efficiency of a screw jack is 45%, when a load of 1,500 N is lifted by an effort applied at the end of a handle of length 50 cm. Determine the effort applied if pitch of screw thread is 1 cm.
Solution:
Given: η = 0.45, W = 1,500 N, L = 50 cm, p = 1 cm
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