Page 444 - Shigley's Mechanical Engineering Design

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Screws, Fasteners, and the Design of Nonpermanent Joints 419
the coordinate system of Fig. 8–8, we note

6F
σ x = τ xy = 0
πd r n t p
4F 16T
σ y =− τ yz =
πd 2 r πd r 3
σ z = 0 τ zx = 0

then use Eq. (5–14) of Sec. 5–5.
The screw-thread form is complicated from an analysis viewpoint. Remember the
origin of the tensile-stress area A t , which comes from experiment. A power screw lift-
ing a load is in compression and its thread pitch is shortened by elastic deformation.
Its engaging nut is in tension and its thread pitch is lengthened. The engaged threads
cannot share the load equally. Some experiments show that the first engaged thread
carries 0.38 of the load, the second 0.25, the third 0.18, and the seventh is free of load.
In estimating thread stresses by the equations above, substituting 0.38F for F and set-
ting n t to 1 will give the largest level of stresses in the thread-nut combination.

EXAMPLE 8–1 A square-thread power screw has a major diameter of 32 mm and a pitch of 4 mm
with double threads, and it is to be used in an application similar to that in Fig. 8–4.
The given data include f = f c = 0.08, d c = 40 mm, and F = 6.4 kN per screw.
(a) Find the thread depth, thread width, pitch diameter, minor diameter, and lead.
(b) Find the torque required to raise and lower the load.
(c) Find the efficiency during lifting the load.
(d) Find the body stresses, torsional and compressive.
(e) Find the bearing stress.
( f ) Find the thread bending stress at the root of the thread.
(g) Determine the von Mises stress at the root of the thread.
(h) Determine the maximum shear stress at the root of the thread.

Solution (a) From Fig. 8–3a the thread depth and width are the same and equal to half the
pitch, or 2 mm. Also
d m = d − p/2 = 32 − 4/2 = 30 mm

Answer d r = d − p = 32 − 4 = 28 mm
l = np = 2(4) = 8mm

(b) Using Eqs. (8–1) and (8–6), the torque required to turn the screw against the load is

Fd m l + π fd m Ff c d c
T R = +
2 πd m − fl 2
6.4(30) 8 + π(0.08)(30) 6.4(0.08)40

= +
2 π(30) − 0.08(8) 2
Answer = 15.94 + 10.24 = 26.18 N · m