How much things stretch, twist or bend is important in designing machines and structures
Ut tensio, sic vis, asserted Robert Hooke (1635-1703): as the extension, so the force. To a good approximation, the deformation of a rigid body is proportional to the force, and when the force is removed, the body resumes its original shape. Some bodies break, others flow, but the observation is a useful one for most materials of construction under working loads. The amount of extension is measured by the change in a dimension divided by the dimension itself, called the strain e = Δl/l, and the force is measured by the force divided by the area on which it acts, called the stress p = F/A. When this is properly done, the ratio of stress to strain is a constant depending only on the material, not on its shape or size. This can only be managed in simple cases, but the result in more complicated cases, such as bending or torsion, can always be worked out from the simple ones. The constant is called the modulus, which has the same dimensions as the stress. It is the stress that would be required for unit strain, if unit strain could be produced elastically. However, practical strains are usually only a few percent at the most.
Hooke's Law implies that the strain produced by several forces is the sum of the strains that would be produced by each force separately, and that the relation between stress and strain is the same for a force in either direction. These things are true only when the strains are less than the elastic limit for the material. A material may, indeed, have no elastic region at all. The typical relation between stress and strain for a steel rod subjected to tension is shown in the Figure. I have shown the stress and strain calculated on the basis of the original measurements. For steel, there is little change in the original measurements except near the breaking point, where the tension sample necks down. In the Figure, the elastic limit, the yield point, which is rather arbitrary, the ultimate or maximum strength, and the breaking strength are labelled. For steel, the elastic region is long, and the distance from the elastic limit to the yield point is short. For cast iron, it is exactly the reverse. There may be no permanent set when the load is removed, even between the elastic and yield points, as occurs in cast iron, but there is usually permanent set beyond the yield point.
The relation between the results of a tension test and the allowable stresses in a structure and machine is not as close as might be supposed. There is still argument and confusion over just what stresses result in failure, and the criteria differ in different types of materials. Traditionally, the ratio of the breaking or ultimate stress to the maximum design stress in a member, which is called the factor of safety, was taken as some number that was larger the more poorly known were the actual stresses, perhaps between 2 and 10, for which failures or other signs of distress were not observed in practice. Advantage was taken of the plasticity of steel to even out excessive stresses. The tensile test then became a test of the quality of the steel. Deflections, buckling, crack propagation and fatigue must all be taken into consideration in design, not just working stresses.
If we consider a small piece of the material under consideration, say a small cube which we mark off mentally deep in the body, things become simpler. The force on any small surface, such as one of the faces of the cube, that is necessary to represent the action of the material around the cube if we imagine this material removed, can be resolved into three mutually perpendicular components, a normal compression or tension, and two shearing forces parallel to the surface. Dividing by the area of the surface, we get the corresponding stresses. Such stresses act on each of the six faces of the cube. Since the cube must not move or rotate bodily, there are necessary relations between the stresses. Only six can be independently specified, three elongations and three shears. The strain of the cube in the stressed condition can similarly be resolved into stretches perpendicular to the faces, and distortions of the faces into rhombuses, and six strains identified. In general, each of the six strains is linearly dependent on each of the six stresses, so there could be 36 different elastic constants, a daunting prospect. In the case of crystals, there are still 21 elastic constants for the least symmetric classes.
Fortunately, most materials possess some symmetry in their properties, and this symmetry can be used to reduce the number of independent elastic constants. We will only consider the most symmetric materials, those that are isotropic, or the same in all directions. This is typical of polycrystalline or amorphous substances of uniform composition. Other cases are exercises in tensor analysis that we will avoid here. The stress and the strain can each be written in the form of a matrix of three rows and three columns referred to the three coordinate directions in space. These matrices are symmetric (can be reflected in the major diagonal), which gives them the property of being reduced to a diagonal of three elements by a proper choice of the orientation of the axes. The same axes diagonalize the stress and the strain, so now we have only three normal strains depending on three normal stresses (the shears have disappeared in the diagonalization). These are called the principal stresses and strains. At most, we could have nine elastic constants, but it gets even better than that, because each of the axes must be equivalent.
Professor G. Lamé of the École Polytechnique (1795-1870) reasoned that each principal stress pi could depend only on its own principal strain ei, or on the sum of the principal strains, with the same moduli in each case. The sum of the principal strains is the dilatation, Δ = e1 + e2 + e3, or relative change in volume of our infinitesimal cube, ΔV/V, and is the only symmetrical linear combination of the three principal strains. There are, therefore, only two independent moduli, which are written λ and μ. Lamé defined them as follows:
pi = λΔ + 2μei,
where i = 1,2,3 for the three mutually perpendicular directions. Now we can work out the moduli for some interesting practical cases. First, consider the case of a hydrostatic pressure, or pi = p. Then, ei = Δ/3, so we have p = (λ + 2μ/3)Δ = κΔ, and κ is the bulk modulus. Prescott uses the constants m,n where λ = m - n and μ = n.
A very useful case is that of a bar or wire under tension or compression in one direction. Then, p2 = p3 = 0, and p1 = F/A, the tensile stress. Directions 2 and 3 give us e2 = e3 = e', the transverse strain, and e' = -λe1/(2λ + 2μ) = -σe1. σ, the ratio of the transverse to the longitudinal strain, is called Poisson's Ratio. It is normally around 1/3 or a little less. For an incompressible material (Δ = 0), it is 1/2. Then we have F/A = λ(1 - 2σ)e + 2μe = Ye, where Y = (3λ + 2μ)μ/(λ + μ) is called Young's Modulus. A more common symbol for it is E.
A third important case is pure shear, shown in the Figure. If the side of the cube is of unit length, then F is the shear stress, and x the shear strain. The cube distorts as shown in red (much exaggerated), and the diagonals become the principal axes. One is lengthened and the other shortened. There is no strain in the direction normal to the section shown. The displacement resolved along a principal axis is x cos45°, while the length of the axis is 1/cos45°, making the strain x/2. This is a compressive strain along one principal axis, and an equal tensile strain along the other. Therefore, x = 2e is the shear strain. Similarly, the forces can be resolved parallel and perpendicular to the principal axis. The component in the direction of an axis is F cos45°, while the area is cos 45°, so the principal stresses are of magnitude p = F. The dilatation Δ = 0, so F = p = 2μe = μ x. The 2 was introduced in the definition of the moduli so that μ would work out to be the shear modulus.
Young's modulus (Y or E), the shear modulus (μ or G), the bulk modulus and Poisson's ratio are the most commonly found elastic constants. If any two are known, the rest can be found. Young's modulus and the shear modulus are the ones most conveniently determined by experiment. The shear modulus is Y divided by 2(1 + σ), or about 0.38Y if σ = 0.3. In using the moduli, it must be remembered that the properties we have found assume an isotropic material, and things become more complicated in other cases. For steel, the density ρ = 7.849 g/cm3, Poisson's ratio is 0.310, and Y = 2.139 x 1012, μ = 8.19 x 1011, κ = 1.841 x 1012, all dy/cm2. For cast iron, the figures are 7.235, 0.267, 1.349 x 1012, 5.32 x 1011, 0.964 x 1012. For copper, the figures are 8.843, 0.378, 1.234 x 1012, 4.47 x 1011, 1.684 x 1012. For aluminium, ρ = 2.7 and Y = 0.734 x 1012. Rubber has Y = 107, roughly, so it is about 100,000 times as extensible as metals. Its Poisson's Ratio should be about 1/2, if its volume does not change much on stretching. This makes μ = Y/3. A piece of rubber can stretch up to 5 or 10 times its unstressed length, and the curve of stress vs. strain is not linear.
We are now equipped to answer three frequently-asked questions: how much does a wire stretch, how much does a shaft twist, and how much does a beam deflect? The first is dead easy, a simple application of Young's modulus. ΔL = FL/YA, proportional to the tension F and the length L, and inversely proportional to the area of the wire. We can measure Young's modulus by stretching a wire.
The shaft is a little more difficult. We consider a thin cylinder first, of radius r and thickness dr. The applied torque is dN, and the twist is an angle θ in a length L. Then, the shear stress is (dN/r)/2πrdr) and the shear strain is rθ/L. Thus, dN = (2πr3dr)θ/L. We imagine a solid circular shaft as composed of such cylinders, with the stress and strain proportional to the distance from the axis. In this case, to get the total torque we simply integrate the right-hand side from 0 to a, the radius of the shaft, since θ is the same for each cylinder. The result is N = μπa4θ/2L. J = πa4/2 is the polar moment of inertia of area of the wire. The twist is θ = NL/μJ, proportional to the torque and the length, and inversely proportional to the polar moment of inertia. For other shapes than cylindrical, one can try substituting the corresponding polar moment of inertia of area. By twisting a wire, we can measure the shear modulus.
Circular shafts have the property that a plane cross-section remains plane when the shaft is twisted. This is not true for noncircular shafts, and the cross-section warps when the shaft is twisted, so that the shear strain is no longer proportional to the distance from the axis. In fact, when an elliptical shaft is twisted, the maximum strain is at the ends of the minor diameter, not the major as might be expected. In addition, the stress is very high at re-entrant corners, such as at the base of keyways. These stress concentrations have provided unpleasant surprises for some designers. Stresses in noncircular shafts were at one time studied experimentally by analogy with a soap film with a boundary like that of the shaft, with an excess pressure on one side. The mathematical theory, by St Venant, is difficult to use analytically except in a few simple cases.
The beam presents the most difficult of the three questions, and there are many possible cases of beam shape, support and loading. Further details are given in the Appendixes. We will consider only the simplest, a rectangular beam of width b, height 2c and length between supports L, with a single concentrated load W at the centre. The upward reactions at the ends of the beam are each W/2, and the bending moment in the beam is Wx/2, where x is the distance from the left-hand support. The first step is to find the connection between the bending moment and the radius of curvature R of the beam. We assume that the stresses are longitudinal, compressive above the middle of the beam, and tensile below, and are proportional to the distance y from the middle, or neutral axis. Equating the moment of these forces to the applied bending moment, we find that the stress p = My/I, where I is the moment of inertia of area about the neutral axis, the integral of by2dy from -c to +c, or 2bc3/3. If e is the strain at y = c, then the radius of curvature R = c/e = IY/M. If z is the upward deflection of the beam, d2z/dx2 = 1/R = M/IY = Wx/2IY. The integral of this is dz/dx = Wx2/4IY + C. Since dz/dx = 0 under the load at x = L/2, C = -WL2/16IY, the slope at the end of the beam. Integrating once more, and setting z = 0 at x = 0, z = Wx(x2/3 - L2/4)/4IY. The deflection at the centre of the beam is δ = WL3/48IY. We have made a lot of assumptions, but the result agrees well with practice. For beams of other shapes, the results are the same if the proper moment of inertia of area is used.
The formula for the deflection of a beam can be put into a convenient form for comparison by introducing the maximum fibre stress σmax = Mc/I = WLc/2I. Then, δ = (σmax/Y)L2/c. For a given ratio of maximum stress to Young's modulus (about 1/1000 or so for steel), the deflection is proportional to the square of the span, and inversely proportional to the depth of the beam (2c in this case). This implies that for equal rigidity, a long span must be relatively deeper than a short span. If you pile four planks one on another to make a beam, the result is only 1/16 as strong, and 16 times as flexible, as if the planks were glued together so that they could not move relatively. The moment of inertia of area of N planks glued together is N3 times as large as the moment of one plank.
The strength of beams is a special study, and many factors must be taken into consideration. A beam may fail by other modes than tensile or compressive failure at the top or bottom. A short, deep beam may fail in shear (like a pile of planks), and beams built up from plates may buckle, either in the web or the flanges. The whole span may buckle or rotate, or stress concentrations may cause local failures that then spread. Bad connections are the chief cause of failure in wooden structures. There are many routes to disaster!
To understand a beam--or, for that matter, bridges and structures in general, a clear picture of the forces acting is essential, and the best way to grasp the picture is graphically. The first step is to sketch the forces acting, which consist of the loads and the reactions. For a statically determinate system, all the forces are known at the beginning. The influence of these forces can then be shown in shear and moment diagrams, which have to be calculated. The shear diagram shows the shear force V, which is the integral of the forces to the left, and positive upwards. Start at the left, and simply add the forces up as they come. A concentrated force means a sudden jump, a uniform load a linear change. Similarly, the bending moment M is the integral of the shear force V, positive anticlockwise. It is much easier to find M from the shear diagram than by considering each force by itself. The bending moment is a maximum or minimum where the shear is zero, by the usual rule for finding extremal values (care is necessary at the ends, and with concentrated forces). In the Figure, note this connection between shear and moment.
Bridge stresses were once determined by graphical constructions. In fact, a graphical analysis was often demanded on the grounds of safety. Graphical procedures clearly show when results are reasonable or unreasonable, unlike numerical calculations where a number is liable to be accepted as valid if careful checks have been made to avoid human error in computing. With a digital computer, numerical results are very likely to be accepted without question. In stress analysis, exact numerical results are not required, so that graphical and slide-rule methods were perfectly adequate. In stress calculations, precision is of little value; accuracy, however, is paramount.
The Figure shows some shear and moment diagrams for several cases of loading. The one on the right is for the elementary case of a beam loaded at the centre and supported at the ends, and clearly shows that the maximum bending moment occurs under the load. The two in the middle are cantilevers, beams not supported at one end. In this case, the support must resist bending moment, and two ways of representing this are shown. On the right, a moment is shown applied at the support in addition to a force, but on the left the moment is supplied by two forces. The case furthest on the left is a beam with equal moments applied at the ends, here by couples. The moment is uniform and positive along the beam, the case known as simple bending.
Prescott gives a very instructive example relating to simple bending in Sec. 38, pp. 34-41. It begins by assuming the strain displacements u,v,w in the x,y,z directions: u = axz, v = -σayz, w = -(a/2)(x2 + σz2 - σy2). Of course, σ is Poisson's Ratio, and the displacements depend on the single parameter a. There is nothing shady about assuming a solution, so long as its consequences are closely examined to make sure it is reasonable. These displacements contain the second powers of the coordinates, the lowest order of approximation, and should be interpreted as small compared to the dimensions of the body, so that its shape is not significantly changed. Nothing is said of the shape of the body in making the assumptions.
The dilatation Δ = du/dx + dv/dy + dw/dz = (1 - 2σ)az. (Derivatives are to be considered partials). We first check that the body stresses are zero. In terms of Prescott's elastic constants m,n, we evaluate the equation of equilibrium m(dΔ/dx) + n div grad u + ρX = 0 and find X = 0. Similarly, Y = Z = 0. The assumption requires no body forces to be acting. Now the normal and shear stresses are evaluated. The normal stress in the x direction is P1 = (m - n)Δ + 2n(du/dx) = Eaz. All other normal and shear stresses are found to be zero. Therefore, the assumptions do not require any surface forces to be acting, except on the ends of the body in the x-direction. We really conceive of the body as a beam of uniform cross-section, initally straight, along the x-axis.
The net force acting on the end of the beam is T = ΣP1dA (Σ stands for the integral) = Eaz*, where z* is the centroidal distance of the cross-section from z = 0, or z*A = ΣzdA. For no net longitudinal force, we must choose z* = 0. That is, the neutral axis passes through the centroid of the cross-section. The moment of the longitudinal force about the neutral axis is M = ΣP1zdA = EaI, where I is the second moment of the area about an axis through the centroid, or the moment of inertia of the area. It is significant that M is a constant, and that a is proportional to M. The moment about the y-axis is M' = -ΣP1ydA = -EaIyz, in terms of the product of inertia Iyz. If the y and z axes are principal axes of the area, the products of inertia vanish (this determines the principal axes), and so does the sideways moment M'. If the y and z axes are axes of symmetry of the cross-section, they are then principal axes. It is always most convenient to resolve the moments with respect to principal axes. In the usual case, the z-axis is an axis of symmetry, and the body does not bow out sideways. However, this behaviour should be remembered.
The axis of the beam is deflected into a parabola in the yz-plane, convex upwards, z = -ax2/2. The radius of curvature R = -(1 + z'2)3/2/z" = (1 + a2x2)3/2/a, or approximately R = 1/a. A little reflection shows that with the moment M constant along the beam, it is actually bent into a circle, not a parabola, which is only the way we have represented a circle in the assumptions, so that R = 1/a indeed. This gives M = EI/R, a very useful relation between curvature and moment.
If the cross-section of the beam is a rectangle of width 2b and height 2c, it remains plane in the bent beam, but its upper and lower sides are bent into arcs of radius R/σ. The stretched side of the beam becomes narrower and is dished in, while the compressed side becomes wider and bulges out. Since the displacement of any point should remain small compared to the original dimensions, b2 << Rc. Recall the great difference in bending behaviour between a flat strip and a curved steel tape measure.
A beam that is continuous across a support can have a bending moment there, unlike a beam that is merely supported. As an example, consider a prop placed at the centre of a beam that is simply supported at both ends. The amount of weight taken by the prop depends on how long it is, and the elasticity of the beam, and cannot be found by the application of the equations of statics. Such a system is called statically indeterminate. A remarkable relation between the bending moments at three supports in a row, that depends only on the loads between the three supports, was discovered by Clapeyron, and is called the Theorem of Three Moments. It is derived by integrating the equation for the elastic curve, EI(d4η/dx4) = w(x), where η is the deflection, and w(x) is the load per unit length. There are enough conditions to eliminate the constants of integration, and the result is the desired theorem, as shown in the Figure. Prescott gives a derivation. The functions f(x) and F(x) are found by integrating the load density with respect to x from x = 0 four times, setting the constants of integration to zero. For a uniform load w, f(x) = wx4/24. The quantities y are the upward displacements of the supports 1 and 2.
As beams are normally used, there is a vertical shear V in addition to the bending moment M, and both produce stresses in the beam. However, in most cases, the shear stress is much less than that caused by bending, and it is easy to see why. The shear V is of the magnitude of the applied forces, while M is the magnitude of the forces times the length of the beam. In a rectangular beam, the maximum shear stress is S' = 3V/2A, while the maximum tensile or compressive stress is S = 3M/cA, where c is half the depth of the beam, and A its area. Therefore if S' is of the order of the force F, and M is about Fl, where l is the span, then S'/S = c/2l, a ratio that is small for most beams.
The shear stress in a rectangular beam of constant width 2b is found by considering the equilibrium of a small slice of beam of width 2b, height dz, and length dx, as shown in the Figure. The reason for the shear stress is seen to be the rate of change of bending moment, and this is proportional to the shear V. The shear is parabolically distributed, zero on the top and bottom faces, and a maximum at the neutral axis, where it is 3/2 of its average value. The derivation is easily done if 2b varies with z, and is not constant, as in an I-beam. Most of the shear is taken by the web of the beam, very little by the top and bottom flanges.
Any good text on Strength of Materials or Elasticity will cover these matters in detail, usually correctly. The methods of Cartesian tensors are very useful in elasticity, and are indispensable for advanced study of nonisotropic materials. The basic matters discussed above, however, should be in the toolkit of every engineer. Much is even included in good school physics.
H. Lamb, The Dynamical Theory of Sound, 2nd ed. (London: Edward Arnold, 1925), Chapter IV, gives an excellent short account of elasticity, as a background for studying mechanical vibrations.
J. Prescott, Applied Elasticity (London: Longmans, Green & Co., 1924. Reprint by Dover 1961). Lamb's reference for further details. Contains many solutions of the equations of elasticity.
The curious use of soap films in stress analysis, originated by Prandtl, is described by G. A. Taylor in Drysdale, et. al., The Mechanical Properties of Fluids (London: Blackie & Son, 1925), Chapter VII.
Composed by J. B. Calvert
Created 24 June 2000
Last revised 19 November 2007