Stresses in Beams

STRESSES IN BEAMS Introduction Understanding of the stresses induced in beams by bending loads took many years to develop. Galileo worked on this problem, but the theory as we use it today is usually credited principally to the great mathematician Leonard Euler (1707–1783). As will be developed below, beams develop normal stresses in the lengthwise direction that vary from a maximum in tension at one surface, to zero at the beam’s midplane, to a maximum in compression at the opposite surface. Shear stresses are also induced, although these are often negligible in comparision with the normal stresses when the length-to-height ratio of the beam is large. The procedures for calculating these stresses for various loading conditions and beam cross-section shapes are perhaps the most important methods to be found in introductory Mechanics of Materials, and will be developed in the sections to follow. This theory requires that the user be able to construct shear and bending moment diagrams for the beam, as developed for instance in Module 12. Normal Stresses A beam subjected to a positive bending moment will tend to develop a concave-upward curva­ ture. Intuitively, this means the material near the top of the beam is placed in compression along the x direction, with the lower region in tension. At the transition between the compressive and tensile regions, the stress becomes zero; this is the neutral axis of the beam. If the material tends to fail in tension, like chalk or glass, it will do so by crack initiation and growth from the lower tensile surface. If the material is strong in tension but weak in compression, it will fail at the top compressive surface; this might be observed in a piece of wood by a compressive buckling of the outer fibers. We seek an expression relating the magnitudes of these axial normal stresses to the shear and bending moment within the beam, analogously to the shear stresses induced in a circular shaft by torsion. In fact, the development of the needed relations follows exactly the same direct approach as that used for torsion: 1. Geometrical statement: We begin by stating that originally transverse planes within the beam remain planar under bending, but rotate through an angle θ about points on the neutral axis as shown in Fig. 1. For small rotations, this angle is given approximately by 1 the x-derivative of the beam’s vertical deflection function v(x)1 : u = −yv,x (1) where the comma indicates differentiation with respect to the indicated variable (v,x ≡ dv/dx). Here y is measured positive upward from the neutral axis, whose location within the beam has not yet been determined. Figure 1: Geometry of beam bending. 2. Kinematic equation: The x-direction normal strain �x is then the gradient of the displace­ ment: du = −yv,xx (2) dx Note that the strains are zero at the neutral axis where y = 0, negative (compressive) above the axis, and positive (tensile) below. They increase in magnitude linearly with y, much as the shear strains increased linearly with r in a torsionally loaded circular shaft. The quantity v,xx ≡ d2 v/dx2 is the spatial rate of change of the slope of the beam deflection curve, the “slope of the slope.” This is called the curvature of the beam. �x = 3. Constitutive equation: The stresses are obtained directly from Hooke’s law as σx = E�x = −yEv,xx (3) This restricts the applicability of this derivation to linear elastic materials. Hence the axial normal stress, like the strain, increases linearly from zero at the neutral axis to a maximum at the outer surfaces of the beam. 4. Equilibrium relations: Since there are no axial (x-direction) loads applied externally to the beam, the total axial force generated by the normal σx stresses (shown in Fig. 2) must be zero. This can be expressed as 1 The exact expression for curvature is d2 v/dx2 dθ = ds [1 + (dv/dx)2 ]3/2 This gives θ ≈ dv/dx when the squared derivative in the denominator is small compared to 1. 2 � � Fx = 0 = � A σx dA = A −yEv,xx dA which requires that � A y dA = 0 The distance ȳ from the neutral axis to the centroid of the cross-sectional area is � y dA A dA ȳ = �A Hence ȳ = 0, i.e. the neutral axis is coincident with the centroid of the beam cross-sectional area. This result is obvious on reflection, since the stresses increase at the same linear rate, above the axis in compression and below the axis in tension. Only if the axis is exactly at the centroidal position will these stresses balance to give zero net horizontal force and keep the beam in horizontal equilibrium. Figure 2: Moment and force equilibrium in the beam. The normal stresses in compression and tension are balanced to give a zero net horizontal force, but they also produce a net clockwise moment. This moment must equal the value of M (x) at that value of x, as seen by taking a moment balance around point O: � � MO = 0 = M + � M= A A σx · y dA (yEv,xx ) · y dA = Ev,xx � A y 2 dA Figure 3: Moment of inertia for a rectangular section. 3 (4) � The quantity y 2 dA is the rectangular moment of inertia with respect to the centroidal axis, denoted I. For a rectangular cross section of height h and width b as shown in Fig. 3 this is: � I= h/2 −h/2 y 2 b dy = bh3 12 (5) Solving Eqn. 4 for v,xx , the beam curvature is v,xx = M EI (6) 5. An explicit formula for the stress can be obtained by using this in Eqn. 3: σx = −yE M −M y = EI I (7) The final expression for stress, Eqn. 7, is similar to τθz = T r/J for twisted circular shafts: the stress varies linearly from zero at the neutral axis to a maximum at the outer surface, it varies inversely with the moment of inertia of the cross section, and it is independent of the material’s properties. Just as a designer will favor annular drive shafts to maximize the polar moment of inertia J, beams are often made with wide flanges at the upper and lower surfaces to increase I. Example 1 Figure 4: A cantilevered T-beam. Consider a cantilevered T-beam with dimensions as shown in Fig. 4, carrying a uniform loading of w N/m. The maximum bending moment occurs at the wall, and is easily found to be Mmax = (wL)(L/2). The stress is then given by Eqn. 7, which requires that we know the location of the neutral axis (since y and I are measured from there). The distance y from the bottom of the beam to the centroidal neutral axis can be found using the “composite area theorem” (see Prob. 1). This theorem states that the distance from an arbitrary axis to the centroid of an area made up of several subareas is the sum of the subareas times the distance to their individual centroids, divided by the sum of the subareas( i.e. the total area): � Ai y i y = �i i Ai For our example, this is 4 (d/2)(cd) + (d + b/2)(ab) cd + ab The moments of inertia of the individual parts of the compound area with respect to their own centroids are just ab3 /12 and cd3 /12. These moments can be referenced to the horizontal axis through the centroid of the compound area using the “parallel axis theorem” (see Prob. 3). This theorem states that the moment of inertia Iz� of an area A, relative to any arbitrary axis z � parallel to an axis through the centroid but a distance d from it, is the moment of inertia relative to the centroidal axis Iz plus the product of the area A and the square of the distance d: y= Iz� = Iz + Ad2 For our example, this is � �2 b ab3 + (ab) d + − y I = 12 2 � �2 cd3 d + (cd) −y I (2) = 12 2 (1) The moment of inertia of the entire compound area, relative to its centroid, is then the sum of these two contributions: I = I (1) + I (2) The maximum stress is then given by Eqn. 7 using this value of I and y = y/2 (the distance from the neutral axis to the outer fibers), along with the maximum bending moment Mmax . The result of these substitutions is � 2 � 3 d c + 6 abd + 3 ab2 wL2 σx = 2 c2 d4 + 8 abcd3 + 12 ab2 cd2 + 8 ab3 cd + 2 a2 b4 In practice, each step would likely be reduced to a numerical value rather than working toward an algebraic solution. In pure bending (only bending moments applied, no transverse or longitudinal forces), the only stress is σx as given by Eqn. 7. All other stresses are zero (σy = σz = τxy = τxz = τyz = 0). However, strains other than �x are present, due to the Poisson effect. This does not generate shear strain (γxy = γxz = γyz = 0), but the normal strains are �x = �y = 1 σx [σx − ν(σy + σz )] = E E 1 σx [σy − ν(σx + σz )] = −ν E E 1 σx [σz − ν(σx + σy )] = −ν E E The strains can also be written in terms of curvatures. From Eqn. 2, the curvature along the beam is �z = v,xx = − �x y This is accompanied by a curvature transverse to the beam axis given by 5 �z ν�x = = −νv,xx y y This transverse curvature, shown in Fig. 5, is known as anticlastic curvature; it can be seen by bending a “Pink Pearl” type eraser in the fingers. v,zz = − Figure 5: Anticlastic curvature. As with tension and torsion structures, bending problems can often be done more easily with energy methods. Knowing the stress from Eqn. 7, the strain energy due to bending stress Ub can be found by integrating the strain energy per unit volume U ∗ = σ 2 /2E over the specimen volume: � Ub = � � = Since � A y 2 dA L A 1 2E � V U ∗ dV = −M y I � � σx2 dA dL 2E L A �2 � dA dL = L M2 2EI 2 � A y 2 dA dL = I, this becomes � Ub = L M 2 dL 2EI (8) If the bending moment is constant along the beam (definitely not the usual case), this becomes U= M 2L 2EI This is another analog to the expression for uniaxial tension, U = P 2 L/2AE. Buckling Long slender columns placed in compression are prone to fail by buckling, in which the column develops a kink somewhere along its length and quickly collapses unless the load is relaxed. This is actually a bending phenomenon, driven by the bending moment that develops if and when when the beam undergoes a transverse deflection. Consider a beam loaded in axial compression and pinned at both ends as shown in Fig. 6. Now let the beam be made to deflect transversely by an amount v, perhaps by an adventitious sideward load or even an irregularity in the beam’s cross section. Positions along the beam will experience a moment given by 6 M (x) = P v(x) (9) The beam’s own stiffness will act to restore the deflection and recover a straight shape, but the effect of the bending moment is to deflect the beam more. It’s a battle over which influence wins out. If the tendency of the bending moment to increase the deflection dominates over the ability of the beam’s elastic stiffness to resist bending, the beam will become unstable, continuing to bend at an accelerating rate until it fails. Figure 6: Imminent buckling in a beam. The bending moment is related to the beam curvature by Eqn. 6, so combining this with Eqn. 9 gives P v (10) EI Of course, this governing equation is satisfied identically if v = 0, i.e. the beam is straight. We wish to look beyond this trivial solution, and ask if the beam could adopt a bent shape that would also satisfy the governing equation; this would imply that the stiffness is insufficient to restore the unbent shape, so that the beam is beginning to buckle. Equation 10 will be satisfied by functions that are proportional to their own second derivatives. Trigonometric functions have this property, so candidate solutions will be of the form v,xx = � v = c1 sin � P x + c2 cos EI P x EI It is obvious that c2 must be zero, since the deflection must go to zero at x = 0 and L. Further, the sine term must go to zero at these two positions as well, which requires that the length L be exactly equal to a multiple of the half wavelength of the sine function: � P L = nπ, EI n = 1, 2, 3, · · · The lowest value of P leading to the deformed shape corresponds to n = 1; the critical buckling load Pcr is then: Pcr = π 2 EI L2 (11) Note the dependency on L2 , so the buckling load drops with the square of the length. This strong dependency on length shows why crossbracing is so important in preventing buckling. If a brace is added at the beam’s midpoint as shown in Fig. 7 to eliminate deflection 7 there, the buckling shape is forced to adopt a wavelength of L rather than 2L. This is equivalent to making the beam half as long, which increases the critical buckling load by a factor of four. Figure 7: Effect of lateral support and end conditions on beam buckling. Similar reasoning can be used to assess the result of having different support conditions. If for instance the beam is cantilevered at one end but unsupported at the other, its buckling shape will be a quarter sine wave. This is equivalent to making the beam twice as long as the case with both ends pinned, so the buckling load will go down by a factor of four. Cantilevering both ends forces a full-wave shape, with the same buckling load as the pinned beam with a midpoint support. Shear stresses Transverse loads bend beams by inducing axial tensile and compressive normal strains in the beam’s x-direction, as discussed above. In addition, they cause shear effects that tend to slide vertical planes tangentially to one another as depicted in Fig. 8, much like sliding playing cards past one another. The stresses τxy associated with this shearing effect add up to the vertical shear force we have been calling V , and we now seek to understand how these stresses are distributed over the beam’s cross section. The shear stress on vertical planes must be accompanied by an equal stress on horizontal planes since τxy = τyx , and these horizontal shearing stresses must become zero at the upper and lower surfaces of the beam unless a traction is applied there to balance them. Hence they must reach a maximum somewhere within the beam. The variation of this horizontal shear stress with vertical position y can be determined by examining a free body of width dx cut from the beam a distance y above neutral axis as shown in Fig. 9. The moment on the left vertical face is M (x), and on the right face it has increased to M + dM . Since the horizontal normal stresses are directly proportional to the moment (σx = M y/I), any increment in moment dM over the distance dx produces an imbalance in the horizontal force arising from the normal stresses. This imbalance must be compensated by a shear stress τxy on the horizontal plane at y. The horizontal force balance is written as � τxy b dx = A� 8 dM ξ dA� I Figure 8: Shearing displacements in beam bending. Figure 9: Shear and bending moment in a differential length of beam. where b is the width of the beam at y, ξ is a dummy height variable ranging from y to the outer surface of the beam, and A� is the cross-sectional area between the plane at y and the outer surface. Using dM = V dx from Eqn. 8 of Module 12, this becomes τxy = V Ib � A� ξ dA� = � VQ Ib (12) where here Q(y) = A� ξ dA� = ξA� is the first moment of the area above y about the neutral axis. The parameter Q(y) is notorious for confusing persons new to beam theory. To determine it for a given height y relative to the neutral axis, begin by sketching the beam cross section, and draw a horizontal line line at the position y at which Q is sought (Fig. 10 shows a rectangular beam of of constant width b and height h for illustration). Note the area A� between this line and the outer surface (indicated by cross-hatching in Fig. 10). Now compute the distance ξ from the neutral axis to the centroid of A� . The parameter Q(y) is the product of A� and ξ; this is the first moment of the area A� with respect to the centroidal axis. For the rectangular beam, it is 9 Figure 10: Section of a rectangular beam. � � h Q = A� ξ = b −y 2 �� � 1 y+ 2 � h −y 2 �� b = 2 � h2 − y2 4 � Note that Q(y), and therefore τxy (y) as well, is parabolic, being maximum at the neutral axis (y = 0) and zero at the outer surface (y = h/2). Using I = bh3 /12 for the rectangular beam, the maximum shear stress as given by Eqn. 12 is 3V 2bh (Keep in mind than the above two expressions for Q and τxy,max are for rectangular cross section only; sections of other shapes will have different results.) These shear stresses are most important in beams that are short relative to their height, since the bending moment usually increases with length and the shear force does not (see Prob. 11). One standard test for interlaminar shear strength2 is to place a short beam in bending and observe the load at which cracks develop along the midplane. τxy,max = τxy |y=0 = Example 2 Since the normal stress is maximum where the horizontal shear stress is zero (at the outer fibers), and the shear stress is maximum where the normal stress is zero (at the neutral axis), it is often possible to consider them one at a time. However, the juncture of the web and the flange in I and T beams is often a location of special interest, since here both stresses can take on substantial values. Consider the T beam seen previously in Example 1, and examine the location at point A shown in Fig. 11, in the web immediately below the flange. Here the width b in Eqn. 12 is the dimension labeled c; since the beam is thin here the shear stress τxy will tend to be large, but it will drop dramatically in the flange as the width jumps to the larger value a. The normal stress at point A is computed from σx = M y/I, using y = d − y. This value will be almost as large as the outer-fiber stress if the flange thickness b is small compared with the web height d. The Mohr’s circle for the stress state at point A would then have appreciable contributions from both σx and τxy , and can result in a principal stress larger than at either the outer fibers or the neutral axis. This problem provides a good review of the governing relations for normal and shear stresses in beams, and is also a natural application for symbolic-manipulation computer methods. Using Maple software, we might begin by computing the location of the centroidal axis: 2 “Apparent Horizontal Shear Strength of Reinforced Plastics by Short Beam Method,” ASTM D2344, Amer­ ican Society for Testing and Materials. 10 Figure 11: Section of T beam. > ybar := ((d/2)*c*d) + ( (d+(b/2) )*a*b )/( c*d + a*b ); Here the “>” symbol is the Maple prompt, and the “;” is needed by Maple to end the command. The maximum shear force and bending moment (present at the wall) are defined in terms of the distributed load and the beam length as > V := w*L; > M := -(w*L)*(L/2); For plotting purposes, it will be convenient to have a height variable Y measured from the bottom of the section. The relations for normal stress, shear stress, and the first principal stress are functions of Y; these are defined using the Maple “procedure” command: > sigx := proc (Y) -M*(Y-ybar)/Iz end; > tauxy := proc (Y) V*Q(Y)/(Iz*B(Y) ) end; > sigp1 := proc (Y) (sigx(Y)/2) + sqrt( (sigx(Y)/2)^2 + (tauxy(Y))^2 ) end; The moment of inertia Iz is computed as > I1 := (a*b^3)/12 + a*b* (d+(b/2)-ybar)^2; > I2 := (c*d^3)/12 + c*d* ((d/2)-ybar)^2; > Iz := I1+I2; The beam width B is defined to take the appropriate value depending on whether the variable Y is in the web or the flange: > B:= proc (Y) if Y Q:= proc (Y) if Y int( (yy-ybar)*c,yy=Y..d) + int( (yy-ybar)*a,yy=d..(d+b) ) > else > int( (yy-ybar)*a,yy=Y.. (d+b) ) > fi end; Here “int” is the Maple command for integration, and yy is used as the dummy height variable. The numerical values of the various parameters are defined as > a:=3: b:=1/4: c:=1/4: d:=3-b: L:=8: w:=100: 11 Figure 12: Stresses at the web-flange junction in a short cantilevered T beam subjected to uniform loading. Finally, the stresses can be graphed using the Maple plot command > plot({sigx,tauxy,sigp1},Y=0..3,sigx=-500..2500); The resulting plot is shown in Fig. 12. Example 3 In the previous example, we were interested in the variation of stress as a function of height in a beam of irregular cross section. Another common design or analysis problem is that of the variation of stress not only as a function of height but also of distance along the span dimension of the beam. The shear and bending moments V (x) and M (x) vary along this dimension, and so naturally do the stresses σx (x, y) and τxy (x, y) that depend on them according to Eqns. 7 and 12. Figure 13: (a) Beam in four-point bending. (b) Free-body diagram. Consider a short beam of rectangular cross section subjected to four-point loading as seen in Fig. 13. The loading, shear, and bending moment functions are: 12 q(x) = P �x�−1 − P �x − a�−1 − P �x − 2a�−1 + P �x − 3a�−1 � V (x) = − � M (x) = − q(x) dx = −P �x�0 + P �x − a�0 + P �x − 2a�0 − P �x − 3a�0 V (x) dx = P �x�1 − P �x − a�1 − P �x − 2a�1 + P �x − 3a�1 The shear and normal stresses can be determined as functions of x and y directly from these functions, as well as such parameters as the principal stress. Since σy is zero everywhere, the principal stress is �� � σx σx 2 2 σp1 = + + τxy 2 2 One way to visualize the x-y variation of σp1 is by means of a 3D surface plot, which can be prepared easily by Maple. For the numerical values P = 100, a = h = 10, b = 3 , we could use the expressions (Maple responses removed for brevity): > > > > > > > > > > > > > > > > > > # use Heaviside for singularity functions readlib(Heaviside); sfn := proc(x,a,n) (x-a)^n * Heaviside(x-a) end; # define shear and bending moment functions V:=(x)-> -P*sfn(x,0,0)+P*sfn(x,a,0)+P*sfn(x,2*a,0)-P*sfn(x,3*a,0); M:=(x)-> P*sfn(x,0,1)-P*sfn(x,a,1)-P*sfn(x,2*a,1)+P*sfn(x,3*a,1); # define shear stress function tau:=V(x)*Q/(Iz*b); Q:=(b/2)*( (h^2/4) -y^2); Iz:=b*h^3/12; # define normal stress function sig:=M(x)*y/Iz; # define principal stress sigp:= (sig/2) + sqrt( (sig/2)^2 + tau^2 ); # define numerical parameters P:=100;a:=10;h:=10;b:=3; # make plot plot3d(sigp,x=0..3*a,y=-h/2 .. h/2); The resulting plot is shown in Fig. 14. The dominance of the parabolic shear stress is evident near the beam ends, since here the shear force is at its maximum value but the bending moment is small (plot the shear and bending moment diagrams to confirm this). In the central part of the beam, where a < x < 2a, the shear force vanishes and the principal stress is governed only by the normal stress σx , which varies linearly from the beam’s neutral axis. The first principal stress is zero in the compressive lower part of this section, since here the normal stress σx is negative and the right edge of the Mohr’s circle must pass through the zero value of the other normal stress σy . Working through the plot of Fig. 14 is a good review of the beam stress formulas. 13