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Unformatted text preview: hrough 10 are valid, then T, G, and J are constant between x1 and x2, and from Equation (5.11) we obtain T ( x2 – x1 )
φ 2 – φ 1 = ----------------------GJ (5.12) In Equation (5.12) points x1 and x2 must be chosen such that neither T, G, nor J change between these points. 5 .2.4 Sign Convention for Internal Torque The shear stress was replaced by a statically equivalent internal torque using Equation (5.1). The shear stress τxθ is positive
on two surfaces. Hence the equivalent internal torque is positive on two surfaces, as shown in Figure 5.21. When we make the
imaginary cut to draw the free-body diagram, then the internal torque must be drawn in the positive direction if we want the formulas to give the correct signs.
Positive x Positive x
Positive T x Outward
normal Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm Positive T Figure 5.21 Sign convention for positive internal torque. Outward
normal Sign Convention: Internal torque is considered positive counterclockwise with respect to the outward normal to the
imaginary cut surface.
T may be found in either of two ways, as described next and elaborated further in Example 5.6. 1. T is always drawn counterclockwise with respect to the outward normal of the imaginary cut, as per our sign convention.
The equilibrium equation is then used to get a positive or negative value for T. The sign for relative rotation obtained from
Equation (5.12) is positive counterclockwise with respect to the x axis. The direction of shear stress can be determined
using the subscripts, as in Section 1.3.
January, 2010 M. Vable Mechanics of Materials: Torsion of Shafts 5 219 2. T is drawn at the imaginary cut to equilibrate the external torques. Since inspection is used to determine the direction
of T, the direction of relative rotation in Equation (5.12) and the direction of shear stress τxθ in Equation (5.10) must
also be determined by inspection. 5 .2.5 Direction of Torsional Stresses by Inspection. The significant shear stress in the torsion of circular shafts is τxθ. All other stress components can be neglected provided the ratio
of the length of the shaft to its diameter is on the order of 10 or more.
Figure 5.22a shows a segment of a shaft under torsion containing point A. We visualize point A on the left segment and consider the stress element on the left segment. The left segment rotates clockwise in relation to the right segment. This implies that
point A, which is part of the left segment, is moving upward on the shaded surface. Hence the shear stress, like friction, on the
shaded surface will be downward. We know that a pair of symmetric shear stress components points toward or away from the
corner. From the symmetry, the shear stresses on the rest of the surfaces can be drawn as shown.
T T T T Figure 5.22 Direction of shear stress by inspection. (a) (b) Suppose we had considered point A on the right segment of the shaft. In such a case we consider the stress element as part of the
right segment, as shown in Figure 5.22b. The right segment rotates counterclockwise in relation to the left segment. This implies that
point A, which is part of the right segment, is moving down on the shaded surface. Hence the shear stress, like friction, will be upward.
Once more using the symmetry of shear stress components, the shear stress on the remaining surfaces can be drawn as shown.
In visualizing the stress surface, we need not draw the shaft segments in Figure 5.22. But care must be taken to identify the
surface on which the shear stress is being considered. The shear stress on the adjoining imaginary surfaces have opposite direction. However, irrespective of the shaft segment on which we visualize the stress element, we obtain the same stress element, as
shown in Figure 5.22. This is because the two stress elements shown represent the same point A.
An alternative way of visualizing torsional shear stress is to think of a coupling at an imaginary section and to visualize the
shear stress directions on the bolt surfaces, as shown in Figure 5.23. Once the direction of the shear stress on the bolt surface is
visualized, the remaining stress elements can be completed using the symmetry of shear stresses
T x x
z xy x
xz Figure 5.23 (a)
Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm x Torsional shear stresses.
(b ) After having obtained the torsional shear stress, either by using subscripts or by inspection, we can examine the shear
stresses in Cartesian coordinates and obtain the stress components with correct signs, as shown in Figure 5.23b. This process of
obtaining stress components in Cartesian coordinates will be important when we consider stress and strain transformation equations in Chapters 8 and 9, where we will relate stresses and strains in different coordinate systems.
The shear strain can be obtained by dividing the shear stress by G, the shear modulus of elasticity. Consolidate your knowledge
1. Identify five examples of circular shafts from your daily life.
2. With the book closed, derive Equations 5.10 and 5.12, listing all the assumptions as you go along. January, 2010 M. Vable Mechanics of Materials: Torsion of Shafts 5 220 EXAMPLE 5.5
The two shafts shown in Figure 5.24 are of the same material and have the same amount of material cross-sectional areas A. Show that
the hollow shaft has a larger polar moment of inertia than the solid shaft. RS RH
2RH Figure 5.24 Hollow and solid...
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This note was uploaded on 04/08/2010 for the course ENGR 232 taught by Professor Smith during the Spring '10 term at Aarhus Universitet.
- Spring '10