5
205
Mechanics of Materials: Torsion of Shafts
M. Vable
Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm
January, 2010
CHAPTER FIVE
TORSION OF SHAFTS
Learning objectives
1.
Understand the theory, its limitations, and its applications in design and analysis of torsion of circular shafts.
2.
Visualize the direction of torsional shear stress and the surface on which it acts.
_______________________________________________
When you ride a bicycle, you transfer power from your legs to the pedals, and through shaft and chain to the rear wheel. In a car,
power is transferred from the engine to the wheel requiring many shafts that form the drive train such as shown in Figure 5.1
a
. A
shaft also transfers torque to the rotor blades of a helicopter, as shown in Figure 5.1
b
. Lawn mowers, blenders, circular saws, drills—
in fact, just about any equipment in which there is circular motion has shafts.
Any structural member that transmits torque from one plane to another is called a
shaft
. This chapter develops the simplest
theory for torsion in circular shafts, following the logic shown in Figure 3.15, but subject to the limitations described in Section
3.13. We then apply the formulas to the design and analysis of statically determinate and indeterminate shafts.
5.1
PRELUDE TO THEORY
As a prelude to theory, we consider several numerical examples solved using the logic discussed in Section 3.2. Their solution
will highlight conclusions and observations that will be formalized in the development of the theory in Section 5.2.
•
Example 5.1 shows the kinematics of shear strain in torsion. We apply the logic described in
Figure 3.15
, for the case of
discrete bars attached to a rigid plate.
•
Examples 5.2 and 5.3 extend the of calculation of shear strain to continuous circular shafts.
•
Example 5.4 shows how the choice of a material model affects the calculation of internal torque. As we shall see the
choice affects only the stress distribution, leaving all other equations unchanged. Thus the strain distribution, which is
a kinematic relationship, is unaffected. So is static equivalency between shear stress and internal torque, and so are the
equilibrium equations relating internal torques to external torques. Though we shall develop the simplest theory using
Hooke’s law, most of the equations here apply to more complex models as well.
(a)
(b)
Figure 5.1
Transfer of torques between planes.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
5
206
Mechanics of Materials: Torsion of Shafts
M. Vable
Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm
January, 2010
EXAMPLE 5.1
The two thin bars of hard rubber shown in Figure 5.2 have shear modulus
G
=
280 MPa and crosssectional area of 20 mm
2
. The bars are
attached to a rigid disc of 20mm radius. The rigid disc is observed to rotate about its axis by an angle of 0.04 rad due to the applied
torque
T
ext
. Determine the applied torque
T
ext
.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '10
 SMITH
 Shear Stress, Torsion, M. Vable

Click to edit the document details