Page 428 - The Mechatronics Handbook

0066_Frame_C19 Page 50 Wednesday, January 9, 2002 5:27 PM

A nonzero resultant torque will cause the body to undergo a proportional angular acceleration, found,
by application of Newton’s second law, from

T r = Ia (19.59)

2
where I, having units of kilogram square meter (kg m ), is the moment of inertia of the body around
the axis (i.e., its polar moment of inertia). Equation (19.59) is applicable to any body regardless of
its state of motion. When α = 0, Eq. (19.59) shows that T r is also zero; the body is said to be in
equilibrium. For a body to be in equilibrium, there must be either more than one applied torque, or
none at all.
Stress, Rigidity, and Strain
Any portion of a rigid body in equilibrium is also in equilibrium; hence, as a condition for equilibrium
of the portion, any torques applied thereto from external sources must be balanced by equal and direc-
tionally opposite internal torques from adjoining portions of the body. Internal torques are transmitted
between adjoining portions by the collective action of stresses over their common cross-sections. In a
solid body having a round cross-section (e.g., a typical shaft), the shear stress τ varies linearly from zero
at the axis to a maximum value at the surface. The shear stress, τ m , at the surface of a shaft of diameter,
d, transmitting a torque, T, is found from

t m = 16T (19.60)
---------
pd 3
Real materials are not perfectly rigid but have instead a modulus of rigidity, G, which expresses the
finite ratio between τ and shear strain, γ. The maximum strain in a solid round shaft therefore also exists
at its surface and can be found from

16T
g m = ----- = ------------- (19.61)
t m
G pd G
3
Figure 19.45(b) shows the manifestation of shear strain as an angular displacement between axially
separated cross sections. Over the length L, the solid round shaft shown will be twisted by the torque
through an angle φ found from

32 LT
f = -------------- (19.62)
4
pd G
Work, Energy, and Power
If during the time of application of a torque, T, the body rotates through some angle θ, mechanical work

W = Tq (19.63)

is performed. If the torque acts in the same CW or CCW sense as the displacement, the work is said to
be done on the body, or else it is done by the body. Work done on the body causes it to accelerate, thereby
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appearing as an increase in kinetic energy (KE = Iω /2). Work done by the body causes deceleration with
a corresponding decrease in kinetic energy. If the body is not accelerating, any work done on it at one
location must be done by it at another location. Work and energy are each measured in units called a
joule (J). Equation (19.63) shows that 1 J is equivalent to 1 Nm rad, which, since a radian is a dimen-
sionless ratio, ≡ 1 Nm. To avoid confusion with torque, it is preferable to quantify mechanical work in
units of mN, or better yet, in J.

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