Page 445 - Shigley's Mechanical Engineering Design

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420 Mechanical Engineering Design
Using Eqs. (8–2) and (8–6), we find the load-lowering torque is

Fd m π fd m − l Ff c d c
T L = +
2 πd m + fl 2
6.4(30) π(0.08)30 − 8 6.4(0.08)(40)
= +
2 π(30) + 0.08(8) 2
Answer =−0.466 + 10.24 = 9.77 N · m
The minus sign in the first term indicates that the screw alone is not self-locking and
would rotate under the action of the load except for the fact that the collar friction is
present and must be overcome, too. Thus the torque required to rotate the screw “with”
the load is less than is necessary to overcome collar friction alone.
(c) The overall efficiency in raising the load is

Fl 6.4(8)
Answer e = = = 0.311
2πT R 2π(26.18)
(d) The body shear stress τ due to torsional moment T R at the outside of the screw
body is
3
Answer τ = 16T R = 16(26.18)(10 ) = 6.07 MPa
3
πd 3 π(28 )
r
The axial nominal normal stress σ is
4F 4(6.4)10 3
Answer σ =− =− =−10.39 MPa
2
πd r 2 π(28 )
(e) The bearing stress σ B is, with one thread carrying 0.38F,

2(0.38F) 2(0.38)(6.4)10 3
Answer σ B =− =− =−12.9 MPa
πd m (1)p π(30)(1)(4)
( f ) The thread-root bending stress σ b with one thread carrying 0.38F is

6(0.38F) 6(0.38)(6.4)10 3
Answer σ b = = = 41.5MPa
πd r (1)p π(28)(1)4
(g) The transverse shear at the extreme of the root cross section due to bending is
zero. However, there is a circumferential shear stress at the extreme of the root cross
section of the thread as shown in part (d) of 6.07 MPa. The three-dimensional stresses,
after Fig. 8–8, noting the y coordinate is into the page, are

σ x = 41.5MPa τ xy = 0
σ y =−10.39 MPa τ yz = 6.07 MPa
σ z = 0 τ zx = 0
For the von Mises stress, Eq. (5–14) of Sec. 5–5 can be written as

1
2
2 1/2
2
2
Answer σ = √ {(41.5 − 0) + [0 − (−10.39)] + (−10.39 − 41.5) + 6(6.07) }

2
= 48.7MPa