Engineering Fundamentals: An Introduction to Engineering, 4th ed.c

(Steven Felgate) #1

280 Chapter 10 Force and Force-Related Parameters


Note that for this problem we could have started with the equation that relatesF 2 toF 1 ,
and then simplified the similar quantities such as pandgin the following manner:

This approach is preferred over the direct substitution of values into the equation right away
because it allows us to change the value of a variable, such asm 1 or the dimensions of the pis-
tons, and see what happens to the result. For example, using the second approach, we can see
clearly that ifm 1 is increased to a value of 200 kg, thenm 2 changes to 1800 kg.

Stress


Geometric properties of an object such as its length, area, and first and second moment of area
also play significant roles in the way the object responds to a force. As explained previously,
when an object is subjected to an external force, internal forces are created inside the material
to hold the material and the components together. Stress provides a measure of the intensity of
internal forces acting over an area. Consider the situation shown in Figure 10.26. The plate
shown is subjected to the compressive force. Because the force is applied at an angle, it has two
components: a horizontal and a vertical component. The tendency of the horizontal compo-
nent of the force is to shear the plate, and the tendency of the vertical component is to com-
press the plate.
The ratio of the normal (vertical) component of the force to the area is called the normal
stress, and the ratio of the horizontal component of the force (the component of the force that
is parallel to the plate surface) to the area is called shear stress. The normal stress component
is often calledpressure.

m 2 


1 R 22
2

1 R 12
2 m^1 

1 15 cm 2
2

1 5 cm 2
2 1 100 kg^2 900 kg

F 2 


A 2


A 1


F 1 m 2 g


p 1 R 22
2

p 1 R 12
2 1 m^1 g^2

m 1 = 100 kg Load


R 1 = 5 cm R 2 = 15 cm


■Figure 10.25 The hydraulic system of Example 10.12.


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