Torque and Efficiency of a Power Screw Problem Determine the lifting and lowering torques and the efficiency of the power screw shown in Figure 15-4 (p. 866) using an Acme screw and nut. Is it self-locking? What is the contribution of collar friction versus screw friction if the collar has (a)

Question

Torque and Efficiency of a Power Screw

Problem Determine the lifting and lowering torques and the efficiency of the power screw shown in Figure 15-4 (p. 866) using an Acme screw and nut. Is it self-locking? What is the contribution of collar friction versus screw friction if the collar has (a) sliding friction, (b) rolling friction?

Given The screw is a single-start Acme 1.25–5. The axial load is 1 000 lb. The mean collar diameter is 1.75 in.

Assumptions The screw and nut are lubricated with oil. Sliding friction μ = 0.15, rolling friction μ = 0.02.

Accepted Answer

1 There are several aspects to this problem. We need to compute the lifting and lowering torques for two cases, one with a sliding-friction collar and one with a ballbearing collar. In both cases we will calculate the screw and collar contributions to torque and efficiency separately for comparison and also combine them. First do the case of the sliding collar.

2 Since it is a single-start thread, the lead [latex]L[/latex] equals the pitch [latex]p[/latex], which is [latex]1 / N[/latex] = 0.2. The pitch diameter of the thread [latex]d_{p}[/latex] is found in Table 15-3 (p. 867). The torque to lift the load is found from equation 15.5a (p. 869):

[latex] T_u=T_{s_u}+T_c=\frac{P d_p}{2} \frac{\left(\mu \pi d_p+L \cos \alpha ight)}{\left(\pi d_p \cos \alpha-\mu L ight)}+\mu_c P \frac{d_c}{2} [/latex] (15.5a)

[latex] \begin{aligned} T_u &=T_{s_u}+T_c=\frac{P d_p}{2} \frac{\left(\mu \pi d_p+L \cos \alpha ight)}{\left(\pi d_p \cos \alpha-\mu L ight)}+\mu_c P \frac{d_c}{2} \ &=\frac{1000(1.15)}{2} \frac{(0.15 \pi(1.15)+0.2 \cos 14.5)}{(\pi(1.15) \cos 14.5-0.15(0.2))}+0.15(1000) \frac{1.75}{2} \ T_u &=122.0+131.2=253.2 lb - in \end{aligned} [/latex] (a)

Note that the collar friction exceeds the screw friction.

3 The torque to lower the load is found from equation 15.5b:

[latex] T_d=T_{s_d}+T_c=\frac{P d_p}{2} \frac{\left(\mu \pi d_p-L \cos \alpha ight)}{\left(\pi d_p \cos \alpha+\mu L ight)}+\mu_c P \frac{d_c}{2} [/latex] (15.5b)

[latex] \begin{aligned} T_d &=T_{s_d}+T_c=\frac{P d_p}{2} \frac{\left(\mu \pi d_p-L \cos \alpha ight)}{\left(\pi d_p \cos \alpha+\mu L ight)}+\mu_c P \frac{d_c}{2} \ &=\frac{1000(1.15)}{2} \frac{(0.15 \pi(1.15)-0.2 \cos 14.5)}{(\pi(1.15) \cos 14.5-0.15(0.2))}+0.15(1000) \frac{1.75}{2} \ T_d &=56.8+131.2=188.0 lb - in \end{aligned} [/latex] (b)

4 The efficiency in the lifting mode will be less than for lowering and is found from equations 15.7 (p. 871). We choose the version shown as equation 15.7c in order to account for both screw and collar components.

[latex] W_{i n}=2 \pi T [/latex] (15.7a)

[latex] W_{ ext {out }}=P L [/latex] (15.7b)

[latex] e=\frac{W_{ ext {out }}}{W_{i n}}=\frac{P L}{2 \pi T} [/latex] (15.7c)

[latex] e=\frac{L}{\pi d_p} \frac{\pi d_p \cos \alpha-\mu L}{\pi \mu d_p+L \cos \alpha} [/latex] (15.7d)

[latex] e=\frac{\cos \alpha-\mu an \lambda}{\cos \alpha+\mu \cot \lambda} [/latex] (15.7e)

[latex] e=\frac{1-\mu an \lambda}{1+\mu \cot \lambda} [/latex] (15.7f)

[latex] \begin{aligned} &&e &=\frac{P L}{2 \pi T} \ ext {For the screw}&&e_{ ext {screw }} &=\frac{1000(0.2)}{2 \pi(122.0)}=0.26 \ ext{For both combined}&&e &=\frac{1000(0.2)}{2 \pi(253.2)}=0.13 \end{aligned} [/latex] (c)

5 Now recalculate the collar torque and total torque to lift the load with a ball-bearing thrust washer, using equation 15.4f.

[latex] T_c=\mu_c P \frac{d_c}{2} [/latex] (15.4f)

[latex] T_c=\mu_c P \frac{d_c}{2}=0.02(1000) \frac{1.75}{2}=17.5 lb - in [/latex] (d)

[latex] T_u=T_{s_u}+T_c=122.0+17.5=139.5 lb - ext { in } [/latex] (e)

6 The efficiency with the ball-bearing thrust washer is now:

[latex] e=\frac{P L}{2 \pi T}=\frac{1000(0.2)}{2 \pi(139.5)}=0.23 [/latex] (f)

The improvement in efficiency is significant and shows why it is poor practice to use anything but a rolling-element bearing as a thrust washer on a power screw.

7 The self-locking aspects of the screw are independent of the thrust collar friction and can be found from equation 15.6a (p. 870).

[latex] \mu \geq \frac{L}{\pi d_p} \cos \alpha \quad ext { or } \quad \mu \geq an \lambda \cos \alpha [/latex] (15.6a)

[latex] \begin{aligned} \mu & \geq \frac{L}{\pi d_p} \cos \alpha \ 0.15 & \geq \frac{0.2}{\pi(1.15)} \cos \left(14.5^{\circ} ight) \ 0.15 & \geq 0.06 \end{aligned} \quad [/latex] so it self - locks (g)

Note that the positive lowering torque in step 3 also indicates that the screw is selflocking. A negative lowering torque means that a braking torque of opposite sense to the lift torque must be applied to hold the load. The files EX15-01A and EX15-01B are on the CD-ROM.

Table 15-3 Principal Dimensions of American Standard Acme Threads See Reference 2 for More Complete Dimensional and Tolerance Information
Major Diameter (in)	Threads per inch	Thread Pitch (in)	Pitch Diameter (in)	Minor Diameter (in)	Tensile Stress Area (in²)
0.250	16	0.063	0.219	0.188	0.032
0.313	14	0.071	0.277	0.241	0.053
0.375	12	0.083	0.333	0.292	0.077
0.438	12	0.083	0.396	0.354	0.110
0.500	10	0.100	0.450	0.400	0.142
0.625	8	0.125	0.563	0.500	0.222
0.750	6	0.167	0.667	0.583	0.307
0.875	6	0.167	0.792	0.708	0.442
1.000	5	0.200	0.900	0.800	0.568
1.125	5	0.200	1.025	0.925	0.747
1.250	5	0.200	1.150	1.050	0.950
1.375	4	0.250	1.250	1.125	1.108
1.500	4	0.250	1.375	1.250	1.353
1.750	4	0.250	1.625	1.500	1.918
2.000	4	0.250	1.875	1.750	2.580
2.250	3	0.333	2.083	1.917	3.142
2.500	3	0.333	2.333	2.167	3.976
2.750	3	0.333	2.583	2.417	4.909
3.000	2	0.500	2.750	2.500	5.412
3.500	2	0.500	3.250	3.000	7.670
4.000	2	0.500	3.750	3.500	10.321
4.500	2	0.500	4.250	4.000	13.364
5.000	2	0.500	4.750	4.500	16.800

Question 15.1: Torque and Efficiency of a Power Screw Problem Determine the...

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