STRUCTURES: PURPOSE AND FUNCTION
2.11 MOMENTS
is +500 ft-lb.
In Figure 2.23, the 100-lb force has a moment arm of 3 ft with respect to point B. With respect to point B, the force has a counterclockwise mo- ment, determined to be 100 ×3 = –300 ft-lb.
Increasing Moments
A moment may be increased by increasing the magnitude of the force or by increasing the distance of the moment arm. For the wrench in Figure 2.24, the limit for rotational effort in terms of moment on the bolt head is limited by the effective wrench length and the force exerted on the han- dle. Additional twisting moment on the bolt can be developed by in- creasing the force. However, for a limited force, the wrench length might be extended by slipping a pipe over the wrench handle, thus producing a larger moment with the same force.
If a given moment is required, various combinations of force and moment arm may be used to produce the moment. For example, if the combination of the given force of 50 lb was found to be just sufficient to twist the nut in Figure 2.24 with the pipe over the wrench handle, what force would have been required if the pipe was not used? With the pipe, the moment is 50 ×25 = 1250 in.-lb. If the pipe is not used, the required force is thus found as 1250 / 10 = 125 lb.
Figure 2.24 Effect of change in the moment arm.
Moment of a Mechanical Couple
A mechanical couple is a means for visualization of a pure rotational ef- fect. As produced by a couple, it takes a form as shown in Figure 2.25, with two parallel forces (the couple) acting in opposite directions at some distance apart. If the two forces are equal in magnitude, the resultant of the forces is zero as a force magnitude. However, the resultant effect of the forces produces a moment, which is the true resultant of the force system: a mechanical couple. The magnitude of the moment is simply the product of one of the forces times the distance between the separated lines of action of the parallel forces. In the illustration, the sense of the moment is counterclockwise.
An example of a mechanical couple is that produced when a person uses two hands to turn a steering wheel. The result of this push-pull ef- fort is neither a net push or a net pull on the wheel, but rather a pure ro- tation of the steering column. This is directly analogous to the development of internal bending resistance in structural members, where opposed tension and compressive stresses produce pure rotational effort.
This phenomenon is discussed for beams in Chapter 11.
Force Required to Produce Motion
Figure 2.26ashows a wheel under the action of a horizontal force that is attempting to roll the wheel over a fixed block. In order to produce mo- tion, the force must be slightly greater than that required for equilibrium.
Pushing on the wheel produces a set of forces consisting of the weight of the wheel, the pushing force, and the force of the corner of the fixed block that pushes back on the wheel. The combination of these three
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Figure 2.25 A mechanical couple.
forces is shown in the free-body diagram of the wheel in Figure 2.26b.
They constitute a concentric force system for which a force polygon is shown in Figure 2.26c.
If the wheel weighs 400 lb and the vector for this force is drawn to a scale in proportion to the 400-lb magnitude (caon the force polygon), the force required for equilibrium may be found by measuring the vector bc on the polygon. A graphic solution that begins with the scaled layout of the wheel, the block, and the pushing force (Figure 2.26a) to determine the angle of force CA, will determine that the pushing force at the point of motion must exceed a value of approximately 330 lb. An algebraic so- lution can also be performed, for example, a summation of moments about the contact point between the wheel and the fixed block.
Example 8. Figure 2.27ashows a masonry pier that weighs 10,000 lb.
Determine the magnitude of the horizontal force applied at the upper left corner that will be required to overturn the pier.
Solution: Tipping of the pier will occur with rotation about the lower right corner of the pier. The forces on the pier at the point of tipping will consist of the pier weight, the horizontal push at the top, and the force ex- erted by the ground at the bottom right corner. A free-body diagram of the pier under the action of these three forces is shown in Figure 2.27b.
Figure 2.27cshows a force polygon for these forces that includes a mag- nitude for the pushing force at the moment of the beginning of tipping. A slight increase in the tipping force above this value will produce tipping (more often described as overturningin engineering).
As with the wheel in the preceding illustration, a scaled layout may be used to determine the magnitude of the pushing force. However, a sim-
Figure 2.26 Force required to produce motion; graphical solution.
ple algebraic solution may be performed using a summation of moments about the lower right corner (point Oin Figure 2.27b). As the line of action of the force at this point has no moment in this summation, the equation for moments is reduced to that involving only the pushing force and the weight of the pier. Thus,
ΣMo= +(BC×8) –(AB×2)
Entering the known value of 10,000 lb for ABin this equation will pro- duce an answer of 2500 lb for the pushing force. Any force exceeding 2500 lb will tend to tip the pier.
Problem 2.11.A
Using a graphical solution, find the horizontal force Prequired to roll the cylinder in Figure 2.28aover the fixed block. The cylinder is 20 in. in di- ameter and weighs 500 lb.
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Figure 2.27 Example 8.
Figure 2.28 Problems 2.11.A–C.
Problem 2.11.C
If the pier in Figure 2.11bweighs 5000 lb, find the magnitude required for force Pto cause overturning.