Microsoft Word - Cengel and Boles TOC _2-03-05_.doc

Chapter 2 | 67

T = 200 N • m

n· = 4000 rpm

FIGURE 2–29

Schematic for Example 2–7.

Rest

position dx

x F

FIGURE 2–30

Elongation of a spring under the

influence of a force.

EXAMPLE 2–7 Power Transmission by the Shaft of a Car

Determine the power transmitted through the shaft of a car when the torque

applied is 200 N · m and the shaft rotates at a rate of 4000 revolutions per

minute (rpm).

Solution The torque and the rpm for a car engine are given. The power

transmitted is to be determined.

Analysis A sketch of the car is given in Fig. 2–29. The shaft power is deter-

mined directly from

Discussion Note that power transmitted by a shaft is proportional to torque

and the rotational speed.

Spring Work

It is common knowledge that when a force is applied on a spring, the length
of the spring changes (Fig. 2–30). When the length of the spring changes by
a differential amount dxunder the influence of a force F, the work done is

(2–27)

To determine the total spring work, we need to know a functional relation-
ship between Fand x. For linear elastic springs, the displacement xis pro-
portional to the force applied (Fig. 2–31). That is,

(2–28)

where kis the spring constant and has the unit kN/m. The displacement xis
measured from the undisturbed position of the spring (that is,x
0 when
F
0). Substituting Eq. 2–28 into Eq. 2–27 and integrating yield

(2–29)

where x 1 and x 2 are the initial and the final displacements of the spring,
respectively, measured from the undisturbed position of the spring.
There are many other forms of mechanical work. Next we introduce some
of them briefly.

Work Done on Elastic Solid Bars

Solids are often modeled as linear springs because under the action of a
force they contract or elongate, as shown in Fig. 2–32, and when the force
is lifted, they return to their original lengths, like a spring. This is true as
long as the force is in the elastic range, that is, not large enough to cause
permanent (plastic) deformations. Therefore, the equations given for a linear
spring can also be used for elastic solid bars. Alternately, we can determine

Wspring
^12 k 1 x^22 x^212 ¬¬ 1 kJ 2

F
kx¬¬ 1 kN 2

dWspring
F dx

83.8 kW¬¬ 1 or 112 hp 2

W

#

sh
^2 pn

#

T
 12 p2a 4000

1

min

b1200 N#m2a

1 min

60 s

ba

1 kJ

1000 N#m

b

Wsh = 2πnT

r

F

n

Torque = Fr

· ·

·

FIGURE 2–28

Shaft work is proportional to the

torque applied and the number of

revolutions of the shaft.