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Chapter 10 Threaded Fasteners and Power Screws
10.1 Introduction
A layman might consider threaded fasteners (screws, nuts, and bolts) to be
the most mundane and uninteresting of all machine elements. In fact
fasteners are used everywhere. There are more than 2 million fasteners on an
airplane, cost over a million dollars.
Engineers are concerned with the selection and use of fasteners, and
they need to know choices available, governing factors and standards.
Power screws are also commonly used machine components. Their
engineering and design has much in common with that of threaded fasteners.
10.2 Thread Forms, Terminology, and Standards
Figure 10.1 (p. 386)
Helical threads of pitch p, lead L and lead angle λ.
Fig. 10.1 illustrates a helical thread wound around a cylinder, as used on
screw-type fasteners, power screws and worms.
L = lead, p = pitch, λ = lead angle, and hand of thread.
Virtually all bolts and screws have a single thread, power screws and worms
have double, triple and even quadruple threads. Unless otherwise noted, all
threads are assumed to be right-hand.
Fig. 10.2 shows the standard geometry of screw threads used on fasteners.
Standard sizes for the two, Unified and ISO systems are given in Tables 10.1
and 10.2.
Figure 10.2 (p. 386)
Unified and ISO thread geometry. The basic profile of the external thread is
Table 10.1a (p. 387)
Basic Dimensions of Unified Screw Threads (Continued on next two slides.)
Table 10.1b (cont.)
Unified threads are specified as size-threads per inch and series. For
example: ½ in.-20UNF, 1 in.-8UNC.
Metric threads are identified by diameter and pitch as M8 x 1.25.
Fig. 10.4 illustrates most of the standard thread forms used for power
Standard sizes are given in Table 10.3.
Table 10.3a (p. 390)
Standard Sizes of Power Screw Threads (Continued on next slide.)
Figure 10.4 (p. 389)
Power screw thread forms. [Note: All threads shown are external (i.e., on
the screw, not on the nut); dm is the mean diameter of the thread contact and
is approximately equal to (d + dr)/2.]
10.3 Power Screws
Power screws, sometimes called linear actuators or translation screws, are
used to convert rotary motion of either the nut or the screw to relatively slow
linear motion of the mating member along the screw axis. The purpose of
many power screws is to obtain a great mechanical advantage in lifting
weights, as in screw type jacks, or to exert large forces, as in presses and
tensile testing machines, home garbage compactors, and C-clamps. The
purpose of others, such as micrometer screws or the lead screw of a lathe, is
to obtain precise positioning of the axial movement.
Figure 10.5 (p. 391)
Weight supported by three screw jacks. In each screw jack, only the shaded
member rotates.
Fig. 10.5 shows a simplified drawing of three different screw jacks
supporting a weight. They are basically same, the torque is Fa.
Let’s take one turn of the screw thread, a triangle would be formed,
see Fig. 10.6:
Figure 10.6 (p. 391)
Screw thread forces.
tan λ = πdLm
λ = lead angle
L = lead
dm = mean diameter of thread contact
Take an infinitesimally small segment of the nut in Fig. 10.6, there are
load w, normal force n, friction force fn, and tangential force q. Note, torque
= q x dm /2. dm is the mean diameter of the thread contact and is approximately
equal to (d + dr)/2.
Summing tangential/horizontal forces,
∑Ft = 0: q – n(fcosλ + cosαnsinλ) = 0
Summing axial/vertical forces, αn is thread shape angle relative to
normal direction.
∑Fa = 0: w + n(fsinλ - cosαncosλ) = 0
cos α n cos λ − f sin λ
Combining Eqs. a and b, we have
f cos λ + cos α n sin λ
q = w cos
α n cos λ − f sin λ
Integration over the entire thread surface in contact results in the same
equation except that w and n become W and N
The equation for torque required to lift load W is
Wd m f cos λ + cos α n sin λ
2 cos α n cos λ − f sin λ
Note, T also = Fa in Fig. 10.5c.
Since L is usually known, according to tanλ = L/πdm, (10.2) changes
Wd m fπd m + L cos α n
2 πd m cos α n − fL
Most applications of power screws require a bearing surface or thrust collar
between stationary and rotating members. In Fig. 10.5 this function is served
T =Q
by the ball thrust bearing of diameter dc. If the coefficient between them is
fc, then the total torque to lift W is
T = Wd2 m
fπd m + L cos α n
πd m cos α n − fL
+ Wf2c d c
For the special case of the square thread, cos αn = 1, and Eq. 10.4,
Wd m fπd m + L
Wf c d c
2 πd m − fL
For the Acme thread, cos αn is so nearly equal to unity (αn =14.50, cos αn =
cos(14.5/180)=0.9968) that Eq. 10.4a can usually be used.
The preceding analysis pertained to raising a load or to turning the rotating
member “against the load.” The analysis for lowering a load, or turning a
rotating member “with the load” is exactly the same except that the
directions of q and fn (Fig. 10.6) are reversed. The total torque required to
lower the load W is
T = Wd2 m
fπd m − L cos α n
πd m cos α n + fL
+ Wf2c d c
For square thread,
T = Wd2 m
fπd m − L
πd m + fL
+ Wf2c d c
10.3.1 Values of Friction Coefficients
When a ball or roller thrust bearing is used, fc is usually low enough that
collar friction can be neglected, thus elimination the second term from
preceding equations.
When a plain thrust collar is used, values of fc vary generally between about
0.08 and 0.20.
10.3.2 Values of Threaded Angle in the Normal Plane
Fig. 10.7 shows the thread angle measured in the normal plane αn and in the
axial plane α. From Fig. 10.7 it can be seen,
tanαn = tanα cosλ
For small helix angles, cosλ is often taken as unity.
Figure 10.7 (p. 394)
Comparison of thread angles measured in axial and normal planes (α and αn).
10.3.3 Overhauling and Self-Locking
A self-locking screw is one that requires a positive torque to lower the load;
an overhauling screw is one that has low enough friction to enable the load
to lower itself. If collar friction can be neglected, Eq. 10.5 shows that a
screw is self-locking if
f ≥ L cos
πd m
For a square thread, this simplifies to
f ≥ πd m
, or f ≥ tanλ
10.3.4 Efficiency
The work output from a power screw for one revolution of the rotating
members is the product of force times distance, or WL. Corresponding work
input is 2πT. The ratio WL/2πT is efficiency. Substituting T in Eq. 10.4,
with collar friction neglected,
L πd m cos α n − fL
πd m πfd m + L cos α n
L πd m − fL
πd m πfd m + L
For the case of square thread
Considering tan λ = L/πdm (10.1), the Eq. 10. 8 gives
cos α n − f tan λ
cos α n + f cot λ
For the square thread,
tan λ
e = 11−+ ff cot
Figure 10.10 (p. 398)
Screw jack lifting a nonrotating load.
Example 1. A screw jack, Fig. 10.10, with a 1-in., double-thread Acme
screw is used to raise a load of 1000 lb. A plain thrust collar of 1 ½-in. mean
diameter is used. Coefficients of running friction are estimated as 0.12 and
0.09 for f and fc, respectively. 1) Determine the screw pitch, lead, thread
depth, mean pitch diameter and helix angle. 2) Estimate the starting torque
for raising and for lowering the load. 3) Estimate the efficiency of the jack
when raising the load.
1) From Table 10.3, there are five threads per inch, hence p = 0.2 in.
Because of the double thread, L = 2p = 0.4 in.
From Fig. 10.4a, thread depth = p/2 = 0.1 in.
From Fig. 10.4a, dm = d – p/2 = 1 – 0.1 = 0.9 in.
From Eq. 10.1, λ = tan-1 (L/π dm) = tan-1 (0.4/π0.9) = 8.050
2) For starting, increase the given coefficients of friction by 1/3, giving f =
0.16, and fc = 0.12. From Eq. 10.6 to find αn
αn = tan-1(tanα cosλ) = tan-1(tan14.50 cos8.50) = 14.360
Substituting in Eq. 10.4, we have
Wd m fπd m + L cos α n
Wf c d c
2 πd m cos α n − fL
= 10002×0.9
0.16π ×0.9 + 0.4 cos 14.360
0.9π cos 14.360 − 0.16×0.4
+ 1000×02.12×1.5
= 141.3 + 90 = 231.3lb • in.
For lowering the load using Eq. 10.5:
Wd m fπd m − L cos α n
Wf c d c
2 πd m cos α n + fL
= 10002×0.9
0.16π ×0.9 − 0.4 cos 14.360
0.9π cos 14.360 + 0.16×0.4
+ 1000×02.12×1.5
= 10.4 + 90 = 100.4lb • in.
3) Efficiency is the ratio of friction-free torque to actual torque. In Eq. 10.4
if take f and fc as zero then friction free T = 63.7 + 0 = 63.7 lb·in; if take f =
0.12 and fc = 0.09, then actual torque T = 121.5 + 67.5 = 189 lb·in.
The efficiency to raise the load is
e = 63.7/189 = 33.7%
Home work: Fourth edition: 10,4, 10.7, 109, 10.10, 10.11.