1. Screws, Fasteners, and the Design of Nonpermanent Joints
Chapter 8
Screws, Fasteners, and the Design of Nonpermanent Joints

2. Chapter Outline
8-1 Thread Standards and Definitions 8-2 The Mechanics of Power Screws 8-3 Strength Constraints Joints-Fasteners Stiffness
8-5 Joints-Member Stiffness 8-6 Bolt Strength 8-7 Tension Joints-The External Load 8-8 Relating Bolt Torque to Bolt Tension
8-9 Statically Loaded Tension Joint with Preload
8-10 Gasketed Joints
8-11 Fatigue Loading of Tension Joints
8-12 Shear Joints
8-13 Setscrews
8-14 Keys and Pins
8-15 Stochastic Considerations

3. Announcements HW #5 Ch. 18, on WebCT
Due Date for HW #5 is Mon. DEC. 31, 2007
Quiz on Ch. 18, Mon. DEC. 31, 2007 ?????

4. LECTURE 34
8-2 The Mechanics of Power Screws

5. Example-1
A power screw is 23 mm in diameter and has a thread pitch of 7 mm. (a) Find the thread depth, the thread width, the mean and root diameters, and the lead, provided square threads are used. (b) Repeat part (a) for Acme threads.
Given: Diameter of the power screw, d = 23 mm Thread pitch, p = 7 mm

8. The Mechanics of Power Screws
A power screw is a device used in machinery to change the angular motion into linear motion, and usually, to transmit power.
Applications:
Lead screws of lathes
Screws for vises, presses and jacks
Figure 8-4
The Joyce worm-gear screw jack.

9. The Mechanics of Power Screws
In Figure 8-5 a square threaded power screw with single thread having a mean diameter dm, a pitch angle p, and a lead angle λ, and a helix angle ψ is loaded by the axial compressive force F.
We wish to find an expression for the torque required to raise this load, and another expression for the torque required to lower the load.
Figure 8-5 Portion of a power screw (Square)

10. Figure 8-6 Force Diagrams (a) Lifting the load; (b)lowering the load
Imagine that a single thread of the screw is enrolled or developed (Fig. 8-6) for exactly a single turn. Then on edge of the thread will form the hypotenuse of a right triangle whose base is the circumference of the mean-thread- circle and whose height is the lead. The angle λ is the lead angle of the thread . For raising the load a force PR acts to the right and to lower the load, PL acts to the left.

11. For raising the load
(a)
(b)
For lowering the load

12. Eliminating N from the previous equations and solving for P gives
For raising the load
(c)
(d)
For lowering the load

13. Next, divide the numerator and the denominator of these equations by cos λ and use the relation
For raising the load
(f)
For lowering the load

14. The torque is the product of the force P and the mean radius
Torque required for raising the load
to overcome thread friction and to raise the load
(8-1)
Torque required for lowering the load
to overcome part of the thread friction in lowering the load
(8-2)

15. Self Locking Condition
If the lead is large or the friction is low, the load will lower itself by causing the screw to spin without any external effort. In such cases the torque from Eq. (8-2) will be negative or zero.
When a positive torque is obtained from this equation, the screw is said to be self locking
Condition for Self Locking:
Dividing both sides of the above inequality by and recognizing that , we get
(8-3)

16. Self Locking Condition
The critical coefficient of friction for the lead concerned,
If f = fcr the nut is on the point of moving down the thread without any torque applied.
If f > fcr then the thread is self-locking in that the nut cannot undo by itself, it needs to be unscrewed by a definite negative torque; Clearly self-locking behavior is essential for threaded fasteners.
Car lifting jacks would not be of much use if the load fell as soon as the operating handle was released.

17. Power Screw-Overhauling
If f < fcr then the thread is  overhauling in that the nut will unscrew by itself under the action of the load unless prevented by a positive tightening torque.
Some applications of power screws require overhauling behavior.
The Archimedean drill
2. Pump action screwdrivers
(Yankee screw drivers)
These devices incorporate very
large lead angles
Increasing lead (angle)  overhauling

18. Power Screw-Overhauling
Sensitive linear actuators may incorporate recirculating ball screws such as that illustrated here to reduce thread friction to levels which go hand-in-hand with overhauling.
decreasing thread friction  overhauling

19. Power Screw-Overhauling
Sensitive linear actuators may incorporate recirculating ball screws such as that illustrated here to reduce thread friction to levels which go hand-in-hand with overhauling.
decreasing thread friction  overhauling

20. Efficiency If we let in Eq. (8-1), we obtain
(g)
which, is the torque required to raise the load.
The efficiency is therefore
(8-4)

21. Efficiency f

22. Power Screw- ACME Thread
F is parallel to screw axis i.e. makes angle α= 14.5° with thread surface ignoring the small effect of l, the resultant normal force N is F/cos α . The frictional force = f N is increased and thus friction terms in Eq. (8.1) are modified accordingly:
Torque required to raise load F
(8-5)
ACME thread is not as efficient as square thread because of additional friction due to wedging action but it is often preferred because it is easier to machine.

23. Power Screw with Collar
In most of power screw applications (load lifting) a collar is to be designed. The presence of collar increases the friction torque. A thrust collar bearing must be employed between the rotating and stationary members in order to carry the axial component

24. Power Screw with Collar

25. Power Screw with Collar
If is the coefficient of collar friction, the torque required is
fc= collar friction coefficient
dc = collar mean diameter
(8-6)

26. Power Screws-friction coefficients
Friction wears thread surface for safe applications Max thread bearing pressure is given in Table 8-4.

27. Power Screws-friction coefficients
Table 8-5 Coefficients of friction f for Threaded Pairs

28. Power Screws-friction coefficients
Table 8-6 Thrust Collar friction coefficient, fc
Coefficients of friction around 0.1 to 0.2 may be expected for common materials under conditions of ordinary service and lubrication.

29. Example-2 P Larm Problem # 8.8 (modified) Given:
5/8”-6ACME? i.e. d=5/8” and N=6
f=fc= 0.15
dc=7/16 in
P = 6 lb
Larm=2 3/4 in
Required:
F, efficiency, Self-Lock?

30. Example-2 (Cont.’d)
Lever torque d
p/2 =1/2N
l =1/N R
Clamping force

31. Example-2 (Cont.’d) Efficiency Self-lock
which is clear that it is self lock