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ADVANCED STRENGTH OF MATERIALS
By J. P. Den Hartog Dover Publications, Inc.
Copyright © 1952 McGraw-Hill Book Company, Inc.
All rights reserved.
ISBN: 978-0-486-13872-5
CHAPTER 1
TORSION
1. Non-circular Prisms. The most useful and common element of construction subjected to torsion is the shaft of circular cross section, either solid or hollow. For this element the theory is quite simple, and we remember that the shear stress in a normal cross section is directed tangentially and its magnitude is given by the expression
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (1a)
where R is the outside radius of the shaft, r (smaller than or equal to R) is the radius at which the stress is measured, and M, is the twisting moment on the shaft. We also remember that the angle of twist θ is determined by
θ = Mil/GIv (1b)
In the derivation of these two equations it was assumed (or rather it was shown by an argument of symmetry) that plane cross sections in the untwisted state remain plane when the twisting torque is applied and also that these cross sections remain undistorted in their own plane.
We now propose to find formulae for the twisting of shafts that are not circular in cross section but that are still prisms, i.e., their cross sections are all the same along the length. For such shafts it is no longer possible to prove that plane cross sections remain plane or that they remain undistorted in their own plane. In the proof of these properties for a circular cross section the rotational symmetry of that section about its central point is essential.
If plane cross sections remain plane and undistorted, it follows logically that the shear stress must be along a set of concentric circles, as in the round shaft. Now it can be easily seen that this cannot be true for a non-circular shaft, because then the stress (Fig. 1) would not be tangent to the boundary of the cross section, and would have a component perpendicular to that boundary. Such a component would be associated with another shear stress on the free outside surface of the shaft, which does not exist. The shear stresses on a cross section thus must be tangent to the periphery of the cross section. As a special case of this we see that the shear stress in the corners of a rectangular cross section must be zero, because neither one of its two perpendicular components can exist.
Figure 2 shows a circular shaft and a square shaft being twisted. In both cases longitudinal lines on the periphery which were originally straight and parallel to the shaft's center line become spirals at a small angle γ as a result of the twisting. In the round shaft (Fig. 2a), an element dr rdθ dl, in which all angles are 90 deg in the untwisted state, then acquires angles of 90 - γ, as shown in Fig. 2c, because the plane cross section remains plane. This angle of twist γ of the particle is associated with the shear stress of twist. Now let us look at the corner particle dx dy dl of the square shaft. Again the spiral effect of the twisting couple causes the small angle γ, and if plane cross sections would remain plane, then the angle γ would necessarily be associated with the shear stress of Fig. 2c. But, as we have seen, this is impossible because the shear stress Ssn on the free outside surface does not exist. Hence the only possibility is shown in Fig. 2d; the upper face of the cross section also must turn through an angle γ, to keep the angle at 90 deg, so that no shear stress occurs. This means that the corner element of area of the cross section is perpendicular to the spiraled longitudinal edge, and since this must be the case at all four edges, the plane cross section is no longer plane but becomes warped vertically.
For the circular cross section it was shown that plane cross sections remain undistorted in their own plane. This means that if we draw on that normal section a network of lines at right angles (such as a set of concentric circles and radii or also a square network of parallel x and y lines), then these right angles remain 90 deg when the torsional couple is applied. We cannot prove that this property remains true for non-circular cross sections. However, in Fig. 3 we see what a distortion of the normal cross section implies. With such a distortion shear stresses appear in sections parallel to the longitudinals, while no shear stress in a normal section is necessary. Only these latter stresses can possibly add up to a twisting torque, and the stresses of Fig. 3 are useless for resisting a twisting torque. Later, in Chap. VII we shall see that such useless stresses never appear. Nature opposes a given action (here a twisting torque) always with the simplest possible stresses: to be precise, the resisting stresses are so that they contain a "minimum of elastic energy." The stresses of Fig. 3 add to the stored elastic energy in the bar, while they do not oppose the imposed twisting couple. Although this argument does not constitute a proof, it makes it plausible that normal cross sections do not distort in their own plane.
With this preliminary discussion we are now ready to start with the theory of twist of non-circular cylinders or prisms. This theory is due to Saint-Venant and was first published in 1855.
2. Saint-Venant's Theory. Let x and y be perpendicular coordinates in the plane of a normal cross section with their origin in the "center of twist," i.e., in the point about which the cross section turns when twisting. Let z be the coordinate along the longitudinal center line, and at z = 0 we place our section of reference which is not supposed to turn. Then Saint-Venant's assumption for the deformation (Fig. 4) can be written as
u = θ1z · y
v = -θ1z · x
w = f(x, y) (2)
Here u, v, and ω are the displacements of a point x, y, z from the untwisted state (A in Fig. 4) to the twisted state B, in the x, y, and z directions, respectively, and θ1 is the angle of twist of the shaft per unit length. The expressions for u and v state that a cross section at distance z from the base turns about the origin through an angle θ1z in a clockwise direction. The sign of v is negative because for positive x, a point moves in the negative y direction when it turns clockwise. The third expression w = f(x, y) states that a cross section warps by an amount ω in the longitudinal z direction; that this warping is different for different points x, y, following an as yet unknown pattern f(x,y). It further states that all cross sections warp in the same manner since ω is independent of z.
The next step is to express the assumed displacements, Eq. (2), into "strains," which are displacements of one point relative to a neighboring point. There are two kinds of strains: direct strains ε, which are extensions or contractions, and shear strains γ, which are angular changes.
The first two of Eqs. (2) express the assumption of no distortion in the normal cross section. Thus [member of]x = [member of]y = 0 and also γzy = 0. The third of Eqs. (2) states that all planes warp in the same manner; hence the z distances between two cross sections are the same for all points of that section, or [member of]z = const. If we rule out longitudinal tension in the shaft, we have, in particular, [member of]z = 0. Thus there are only two strains left: γxz and γyz. In order to express γxz we study a section in an x-z plane (y = constant) which is parallel to the longitudinals of the bar, shown in Fig. 5. An element ABCD goes to A'B'C'D' because of twisting, and the two originally plane sections z and z + dz become the warped ones shown in dashes. The horizontal distance between A and A' we have called u. The horizontal distance between C and C' can be written as u + ([partial derivative]u/ [partial derivative]z) dz, because point C differs from point A by the distance dz only, while A and C have the same x value. The horizontal distance between A' and C' then is ([partial derivative]u/[partial derivative]z) dz, and the (small) angle between A'C' and the vertical is [partial derivative]u/ [partial derivative]z. We now repeat this whole story for the vertical distances (instead of the horizontal ones) of the points A, B and A', B'. The reader should do this and draw the conclusion that the (small) angle between A'B' and the horizontal is [partial derivative]ω/ [partial derivative]x. Hence the difference between [angle] C'A'B' and [angle] CAB is [partial derivative]u/[partial derivative]z) + [partial derivative]ω/[partial derivative]x), and by definition this is the angle of shear γxz in the xz plane. For the yz plane the analysis is exactly the same, only the letter x is replaced by y (and consequently u is replaced by v) in all of the algebra as well as in Fig. 5. Thus we arrive at
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
Now Eqs. (2) state what the displacements are. Substituting Eqs. (2) into the above leads to
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (3)
With Hooke's law these strains are expressible in the stresses
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (4)
where the signs of the stresses are shown in Fig. 6.
The next step is the derivation of the equation of equilibrium. It is clear that the shear stresses are not constant across a normal cross section xy but differ from point to point. (In the circular section the stresses are zero at the center and grow linearly with the distance from it.) Thus the stresses on opposite faces of the dx dy dz element of Fig. 7 are not exactly alike but differ from each other by small amounts. If, for example, the stress on the dy dz face for the smaller of the two x values is denoted by (ss)xz, shown dotted in the figure, then the stress on the opposite dy dz face (which is distance dx farther to the right) is somewhat different and can be written as
(ss)xz + d(ss)xz
or more precisely as
The extra, unbalanced, upward stress on the pair of dy dz faces thus is [partial derivative]/[partial derivative]x (ss)xz dx, and since this stress acts on an area dy dz, the unbalanced force is [partial derivative]/[partial derivative]x (ss)xz dx dy dz. The reader should now repeat this argument for the two dx dz faces (fore and aft) and conclude that the upward unbalanced forces there is [partial derivative]/[partial derivative]y (ss)yz dy dx dz. There are no other vertical forces or stresses on the element, and an element in a twisted bar is obviously in equilibrium. Setting the net upward force equal to zero and dividing by the volume element dx dy dz gives the equilibrium equation
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (5)
This is a partial differential equation in terms of two unknown functions (ss)xz and (ss)yz, both depending on two variables x and y. The problem of finding a solution would be very much simpler if we had to deal with one single function of (x, y) instead of with two. Here we come to the vital step in Saint- Venant's analysis. He assumes that there is a function Φ(x, y), such that the stresses can be found from it by differentiation, thus:
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (6)
The definition (6) of the new function Φ has been chosen cleverly: by substituting (6) into (5) we see that Eq. (5) is automatically satisfied for any arbitrary function Φ, provided that
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
which is always true if Φ is a continuous function. The function Φ is called the "stress function" of the problem; here it is "Saint-Venant's torsion stress function." Other examples of stress functions will be seen later, on pages 42 and 174.
Now we can begin to visualize the situation geometrically. The value Φ can be plotted vertically on an xy base and thus forms a curved surface. Then Eqs. (6) state that the (ss)yz stress, i.e., the stress in the y direction, is the slope of the Φ) surface in the x direction, and vice versa, that the shear stress in the x direction is the (negative) slope of the Φ surface in the y direction. We shall now prove that this statement can be generalized and that the shear stress component in any direction equals the slope of the Φ surface in the perpendicular direction. Before we proceed to the proof, we notice that by Fig. 1 (page 2) the shear stress normal to the periphery of the shaft is zero; hence the Φ slope along the periphery must be zero, which means that the Φ height all along the periphery must be constant. The Φ surface then can be visualized as a hill, and if we cut this hill by a series of horizontal planes to produce "contour lines," then the shear stress follows those contour lines. For a circular cross section the contour lines are concentric circles, and the Φ hill is a paraboloid of revolution.
Now to the proof. Consider in Fig. 8 an element at point A with the stresses (ss)xz and (ss)yz. Draw through A the line AB in an arbitrary direction α, and let AB be dn, the element of the "normal" direction. Perpendicular to AB is the line CD. When AB = dn, then AE = dx and EB = dy, and we see from the figure that dx/dn = cos α and dy/dn = sin α. Over all these points is the hilly surface of the stress function Φ, and we have in general
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
or
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
Transcribing this by means of Eqs. (6) leads to
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
Now, looking at Fig. 8 we see that
dΦ/dn = AD - AC
or equal to the component of stress in the direction DAC, which is perpendicular to AB. But dΦ/dn is the slope of the Φ surface in the AB direction, which completes our proof.
Now we possess the single stress function Φ, with which we shall operate instead of with the pair of stresses (ss)xz and (ss)yz. The first thing to do is to rewrite all our previous results in terms of Φ. Turning back, we first find Eq. (5), which we have seen is automatically satisfied by the judicious definition of Φ. Next we find Eqs. (4), which now become
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (7)
In these the quantity ω is the warping of the cross section. We do not know what w looks like, but it is certain that ω and its derivatives are continuous functions of x and y: the warping will have no sudden jumps or cracks in it. One mathematical consequence of this continuity is
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
(Continues...)
Excerpted from ADVANCED STRENGTH OF MATERIALS by J. P. Den Hartog. Copyright © 1952 McGraw-Hill Book Company, Inc.. Excerpted by permission of Dover Publications, Inc..
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