Bending of Beams

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Date: Aug 1, 2019

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This excerpt discusses the bending of straight as well as curved beams—that is, structural elements possessing one dimension significantly greater than the other two, usually loaded in a direction normal to the longitudinal axis.

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5.1 Introduction

In this chapter we are concerned with the bending of straight as well as curved beams—that is, structural elements possessing one dimension significantly greater than the other two, usually loaded in a direction normal to the longitudinal axis. We first examine the elasticity or “exact” solutions of beams that are straight and made of homogeneous, linearly elastic materials. Then, we consider solutions for straight beams using mechanics of materials or elementary theory, special cases involving members made of composite materials, and the shear center. The deflections and stresses in beams caused by pure bending as well as those due to lateral loading are discussed. We analyze stresses in curved beams using both exact and elementary methods, and compare the results of the various theories.

Except in the case of very simple shapes and loading systems, the theory of elasticity yields beam solutions only with considerable difficulty. Practical considerations often lead to assumptions about stress and deformation that result in mechanics of materials or elementary theory solutions. The theory of elasticity can sometimes be applied to test the validity of such assumptions. This theory has three roles in these problems: It can serve to place limitations on the use of the elementary theory, it can be used as the basis of approximate solutions through numerical analysis, and it can provide exact solutions for simple configurations of loading and shape.

Part A: Exact Solutions

5.2 Pure Bending of Beams of Symmetrical Cross Section

The simplest case of pure bending is that of a beam possessing a vertical axis of symmetry, subjected to equal and opposite end couples (Fig. 5.1a). The semi-inverse method is now applied to analyze this problem. The moment M z shown in Fig. 5.1a is defined as positive, because it acts on a positive (negative) face with its vector in the positive (negative) coordinate direction. This sign convention agrees with that of stress (Section 1.5). We will assume that the normal stress over the cross section varies linearly with y and that the remaining stress components are zero:


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Figure 5.1. (a) Beam of singly symmetric cross section in pure bending; (b) stress distribution across cross section of the beam.

Here k is a constant, and y = 0 contains the neutral surface—that is, the surface along which σx = 0 The intersection of the neutral surface and the cross section locates the neutral axis (abbreviated NA). Figure 5.1b shows the linear stress field in a section located an arbitrary distance a from the left end.

Since Eqs. (5.1) indicate that the lateral surfaces are free of stress, we need only be assured that the stresses are consistent with the boundary conditions at the ends. These conditions of equilibrium require that the resultant of the internal forces be zero and that the moments of the internal forces about the neutral axis equal the applied moment


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where A is the cross-sectional area. Note that the zero stress components τxy, τxz in Eqs. (5.1) satisfy the conditions that no y- and z-directed forces exist at the end faces. Moreover, because of the y symmetry of the section, σx = ky produces no moment about the y axis. The negative sign in the second expression implies that a positive moment mz is one that results in compressive (negative) stress at points of positive y. Substituting Eqs. (5.1) into Eqs. (5.2) yields


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Since k ≠ 0, Eq. (5.3a) indicates that the first moment of cross-sectional area about the neutral axis is zero. This requires that the neutral and centroidal axes of the cross section coincide. Neglecting body forces, it is clear that the equations of equilibrium (3.4), are satisfied by Eqs. (5.1). It can also readily be verified that Eqs. (5.1) together with Hooke's law fulfill the compatibility conditions, Eq. (2.12). Thus, Eqs. (5.1) represent an exact solution.

The integral in Eq. (5.3b) defines the moment of inertia Iz of the cross section about the z axis of the beam cross section (Appendix C); therefore,


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An expression for normal stress can now be written by combining Eqs. (5.1) and (a):


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This is the familiar elastic flexure formula applicable to straight beams.

Since, at a given section, M and I are constant, the maximum stress is obtained from Eq. (5.4) by taking |y|max = c:


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where S is the elastic section modulus. Equation (5.5) is widely employed in practice because of its simplicity. To facilitate its use, section moduli for numerous common sections are tabulated in various handbooks. A fictitious stress in extreme fibers, computed from Eq. (5.5) for the experimentally obtained ultimate bending moment (Section 12.7), is termed the modulus of rupture of the material in bending. This quantity, σmax=Mu/S, is frequently used as a measure of the bending strength of materials.

5.2.1 Kinematic Relationships

To gain further insight into the beam problem, we now consider the geometry of ­deformation—that is, beam kinematics. Fundamental to this discussion is the hypothesis that sections originally plane remain so subsequent to bending. For a beam of symmetrical cross section, Hooke′s law and Eq. (5.4) lead to


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where EIz is the flexural rigidity.

Let us examine the deflection of the beam axis, whose axial deformation is zero. Figure 5.2a shows an element of an initially straight beam, now in a deformed state. Because the beam is subjected to pure bending, uniform throughout, each element of infinitesimal length experiences identical deformation, with the result that the beam curvature is everywhere the same. The deflected axis of the beam or the deflection curve is thus shown deformed, with radius of curvature rx. The curvature of the beam axis in the xy plane in terms of the y deflection υ is


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Figure 5.2. (a) Segment of a bent beam; (b) geometry of deformation.

where the approximate form is valid for small deformations (dυ/dx≪1). The sign convention for curvature of the beam axis is such that this sign is positive when the beam is bent concave downward, as shown in the figure.

As shown by the geometry in Fig. 5.2b, the shaded sectors are similar. Hence, the radius of curvature and the strain are related as follows:


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where ds is the arc length mn along the longitudinal axis of the beam. For a small displacement, dsdx and θ represents the slope dυ/dx of the beam axis. Clearly, for the positive curvature shown in Fig. 5.2a, θ increases as we move from left to right along the beam axis. On the basis of Eqs. (5.6) and (5.8),


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Following a similar procedure and noting that εz ≈ -Vεx, we may also obtain the curvature in the yz plane as


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The basic equation of the deflection curve of a beam is obtained by combining Eqs. (5.7) and (5.9a) as follows:


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This expression, relating the beam curvature to the bending moment, is known as the Bernoulli—Euler law of elementary bending theory. It is observed from Fig. 5.2 and Eq. (5.10) that a positive moment produces a positive curvature. If the sign convention adopted in this section for either moment or deflection (and curvature) is reversed, the plus sign in Eq. (5.10) should likewise be reversed.

Reference to Fig. 5.2a reveals that the top and bottom lateral surfaces have been deformed into saddle-shaped or anticlastic surfaces of curvature 1/rz. The vertical sides have been simultaneously rotated as a result of bending. Examining Eq. (5.9b) suggests a method for determining Poisson’s ratio [Ref. 5.1]. For a given beam and bending moment, a measurement of 1/rz leads directly to v. The effect of anticlastic curvature is small when the beam depth is comparable to its width.

5.2.2 Timoshenko Beam Theory

The Timoshenko theory of beams, developed by S. P. Timoshenko at the beginning of the twentieth century, constitutes an improvement over the Euler—Bernoulli theory. In the static case, the difference between the two hypotheses is that the former includes the effect of shear stresses on the deformation by assuming a constant shear over the beam height, whereas the latter ignores the influence of transverse shear on beam deformation. The Timoshenko theory is also said to be an extension of the ordinary beam theory that allows for the effect of the transverse shear deformation while ­relaxing the assumption that plane sections remain plane and normal to the deformed beam axis.

The Timoshenko beam theory is well suited to describing the behavior of short beams and sandwich composite beams. In the dynamic case, the theory incorporates shear deformation as well as rotational inertia effects, and it will be more accurate for not very slender beams. By effectively taking into account the mechanism of deformation, Timoshenko′s theory lowers the stiffness of the beam, with the result being a larger deflection under static load and lower predicted fundamental frequencies of vibration for a prescribed set of boundary conditions.

5.3 Pure Bending of Beams of Asymmetrical Cross Section

In this section, we extend the discussion in Section 5.2 to the more general case in which a beam of arbitrary cross section is subjected to end couples My and Mz about the y and z axes, respectively (Fig. 5.3). Following a procedure similar to that described in Section 5.2, plane sections are again taken to remain plane. Assume that the normal stress σx acting at a point within dA is a linear function of the y and z coordinates of the point; assume further that the remaining stresses are zero. Then the stress field is


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Figure 5.3. Pure bending of beams of asymmetrical cross section.

where c1, c2, c3 are constants to be evaluated.

The equilibrium conditions at the beam ends, as before, relate to the force and bending moment:


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Carrying σx, as given by Eq. (5.11), into Eqs. (a), (b), and (c) results in the following expressions:


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For the origin of the y and z axes to be coincident with the centroid of the section, it is required that


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Based on Eq. (d), we conclude that c1 = 0; based on Eqs. (5.11), we conclude that σx = 0 at the origin. Thus, the neutral axis passes through the centroid, as in the beam of symmetrical section. In addition, the field of stress described by Eqs. (5.11) satisfies the equations of equilibrium and compatibility and the lateral surfaces are free of stress. Now consider the defining relationships


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The quantities Iy and Iz are the moments of inertia about the y and z axes, respectively, and Iyz is the product of inertia about the y and z axes. From Eqs. (e) and (f), together with Eqs. (5.12), we obtain expressions for c2 and c3.

5.3.1 Stress Distribution

Substitution of the constants into Eqs. (5.11) results in the following generalized flexure formula:


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The equation of the neutral axis is found by equating this expression to zero:


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This is an inclined line through the centroid C. The angle ϕ between the neutral axis and the z axis is determined as follows:


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The angle ϕ (measured from the z axis) is positive in the clockwise direction, as shown in Fig. 5.3. The highest bending stress occurs at a point located farthest from the neutral axis.

There is a specific orientation of the y and z axes for which the product of inertia Iyz vanishes. Labeling the axes so oriented as y′ and z′, we have Iy′z′ = 0. The flexure formula under these circumstances becomes


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The y′ and z′ axes now coincide with the principal axes of inertia of the cross section, and we can find the stresses at any point by applying Eq. (5.13) or Eq. (5.16).

The kinematic relationships discussed in Section 5.2 are valid for beams of asymmetrical section provided that y and z represent the principal axes.

5.3.2 Transformation of Inertia Moments

Recall that the two-dimensional stress (or strain) and the moment of inertia of an area are second-order tensors (Section 1.17). Thus, the transformation equations for stress and moment of inertia are analogous (Section C.2.2). In turn, the Mohr′s circle analysis and all conclusions drawn for stress apply to the moment of inertia. With reference to the coordinate axes shown in Fig. 5.3, applying Eq. (C.12a), the moment of inertia about the y′ axis is


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From Eq. (C.13), the orientation of the principal axes is given by


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The principal moments of inertia, I1 and I2 from Eq. (C.14) are


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where the subscripts 1 and 2 refer to the maximum and minimum values, respectively.

Determination of the moments of inertia and stresses in an asymmetrical section is illustrated in Example 5.1.

5.4 Bending of A Cantilever of Narrow Section

Consider a narrow cantilever beam of rectangular cross section, loaded at its free end by a concentrated force of magnitude such that the beam weight may be neglected (Fig. 5.5). This situation may be regarded as a case of plane stress provided that the beam thickness t is small relative to the beam depth 2h. The distribution of stress in the beam, as we found in Example 3.1, is given by


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Figure 5.5. Deflections of an endloaded cantilever beam.

To derive expressions for the beam displacement, we must relate stress, described by Eq. (3.21), to strain. This is accomplished through the use of the strain-displacement relations and Hooke′s law:


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Integration of Eqs. (a) and (b) yields


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Differentiating Eqs. (d) and (e) with respect to y and x, respectively, and substituting into Eq. (c), we have


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In this expression, note that the left and right sides depend only on y and x, respectively. These variables are independent of each other, so we conclude that the equation can be valid only if each side is equal to the same constant:


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These are integrated to yield


in which a2 and a3 are constants of integration. The displacements may now be written


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The constants a1, a2, and a3 depend on known conditions. If, for example, the situation at the fixed end is such that


then, from Eqs. (5.20),


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The beam displacement is therefore


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On examining these equations, it becomes that u and υ do not obey a simple linear ­relationship with y and x. We conclude that plane sections do not, as assumed in elementary theory, remain plane subsequent to bending.

5.4.1 Comparison of the Results with the Elementary Theory Results

The vertical displacement of the beam axis is obtained by substituting y = 0 into Eq. (5.22):


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Introducing this relation into Eq. (5.7), the radius of curvature is given by


provided that /dx is a small quantity. Once again, we obtain Eq. (5.9a), the beam curvature—moment relationship of elementary bending theory.

It is also a simple matter to compare the total vertical deflection at the free end (x = 0) with the deflection derived in elementary theory. Substituting x = 0 into Eq. (5.23), the total deflection is


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where the deflection associated with shear is clearly Ph2L/2GI = 3PL/2GA. The ratio of the shear deflection to the bending deflection at x = 0 provides a measure of beam slenderness:


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If, for example, L = 10(2h), the preceding quotient is only inline254_1.jpg. For a slender beam, 2hL and the deflection is mainly due to bending. In contrast, in cases involving vibration at higher modes, and in wave propagation, the effect of shear is of great importance in slender as well as in other beams.

In the case of wide beams (t≫2h), Eq. (5.24) must be modified by replacing E and ν as indicated in Table 3.1.

5.5 Bending of a Simply Supported Narrow Beam

In this section, we consider the stress distribution in a narrow beam of thickness t and depth 2h subjected to a uniformly distributed loading (Fig. 5.6). The situation described here is one of plane stress, subject to the following boundary conditions, consistent with the origin of an x, y coordinate system located at midspan and midheight of the beam, as shown:


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Figure 5.6. Bending of a simply supported beam with a uniform load.

Since no longitudinal load is applied at the ends, it would appear reasonable to state that σx = 0 at x = ± L. However, this boundary condition leads to a complicated solution, and a less severe statement is used instead:


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The corresponding condition for bending couples at x = ±L is


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For y equilibrium, it is required that


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5.5.1 Use of Stress Functions

The problem is treated by superimposing the solutions Φ23, and Φ5 (Section 3.6) with


We then have


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The stresses are


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The conditions (a) are


and the solution is


The constant d3 is obtained from condition (c) as follows:




Expressions (e), together with the values obtained for the constants, also fulfill conditions (b) and (d).

The state of stress is thus represented by


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where inline256_43.jpg is the area moment of inertia taken about a line through the centroid, parallel to the z axis. Although the solutions given by Eqs. (5.25) satisfy the equations of elasticity and the boundary conditions, they are not exact. We can see this by substituting x = ±L into Eq. (5.25a) to obtain the following expression for the normal distributed forces per unit area at the ends:


This state cannot exist, as no forces act at the ends. From Saint-Venant′s principle, however, we may conclude that the solutions do predict the correct stresses throughout the beam, except near the supports.

5.5.2 Comparison of the Results with the Elementary Theory Results

Recall that the longitudinal normal stress derived from elementary beam theory is σx = -My/I; this is equivalent to the first term of Eq. (5.25a). The second term is then the difference between the longitudinal stress results given by the two approaches. To gauge the magnitude of the deviation, consider the ratio of the second term of Eq. (5.25a) to the result of elementary theory at x = 0. At this point, the bending moment is a maximum. Substituting y = h for the condition of maximum stress, we obtain


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For a beam of length 10 times its depth, this ratio is small, inline256_1.jpg. For beams of ordinary proportions, we can conclude that elementary theory provides a result of sufficient accuracy for σx. On the one hand, for σy, this stress is not found in the elementary theory. On the other hand, the result for τxy is the same as that yielded by elementary beam theory.

The displacement of the beam may be determined in a manner similar to that described for a cantilever beam (Section 5.4).

Part B: Approximate Solutions

5.6 Elementary Theory of Bending

We may conclude, on the basis of the previous sections, that exact solutions are difficult to obtain. We also observed that for a slender beam, the results of the exact theory do not differ markedly from those found with the mechanics of materials or elementary approach provided that solutions close to the ends are not required. The bending deflection is very much larger than the shear deflection, so the stress associated with the former predominates. As a consequence, the normal strain εy resulting from transverse loading may be neglected.

Because it is more easily applied, the elementary approach is usually preferred in engineering practice. The exact and elementary theories should be regarded as complementary—rather than competitive—approaches, enabling the analyst to obtain the degree of accuracy required in the context of the specific problem at hand.

5.6.1 Assumptions of Elementary Theory

The basic presuppositions of the elementary theory [Ref. 5.2], for a slender beam whose cross section is symmetrical about the vertical plane of loading, are


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The first equation of Eqs. (5.26) is equivalent to the assertion υ = υ(x). Thus, all points in a beam at a given longitudinal location x experience identical deformation. The second equation of Eqs. (5.26), together with υ = υ(x), yields, after integration,


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The third equation of Eqs. (5.26) and Eqs. (5.27) imply that the beam is considered narrow, and we have a case of plane stress.

At y = 0 the bending deformation should vanish. Referring to Eq. (a), it is clear that u0(x) must represent axial deformation. The term /dx is the slope θ of the beam axis, as shown in Fig. 5.7a, and is very much smaller than unity. Therefore,


Figure 5.7. (a) Longitudinal displacements in a beam due to rotation of a plane section; (b) element between adjoining sections of a beam.

The slope is positive when clockwise, provided that the x and y axes have the directions shown. Since u is a linear function of y, this equation restates the kinematic hypothesis of the elementary theory of bending: Plane sections perpendicular to the longitudinal axis of the beam remain plane subsequent to bending. This assumption is confirmed by the exact theory only in the case of pure bending.

5.6.2 Method of Integration

In the next section, we obtain the stress distribution in a beam according to the elementary theory. We now derive some useful relations involving the shear force V, the bending moment M, the load per unit length p, the slope θ, and the deflection. Consider a beam element of length dx subjected to a distributed loading (Fig. 5.7b). Since dx is small, we omit the variation in the load per unit length p. In the free-body diagram, all the forces and the moments are positive. The shear force obeys the sign convention discussed in Section 1.4, while the bending moment is in agreement with the convention adopted in Section 5.2.

In general, the shear force and bending moment vary with the distance x, such that these quantities will have different values on each face of the element. The increments in shear force and bending moment are denoted by dV and dM, respectively. The equilibrium of forces in the vertical direction is governed by V − (V + dV) − p dx = 0 or


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That is, the rate of change of shear force with respect to x is equal to the algebraic value of the distributed loading. Equilibrium of the moments about a z axis through the left end of the element, neglecting the higher-order infinitesimals, leads to


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This relation states that the rate of change of bending moment is equal to the algebraic value of the shear force—a relation that is valid only if a distributed load or no load acts on the beam segment. Combining Eqs. (5.28) and (5.29), we have


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The basic equation of bending of a beam, Eq. (5.10), combined with Eq. (5.30), may now be written as


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For a beam of constant flexural rigidity EI, the beam equations derived here may be expressed as


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These relationships also apply to wide beams provided that we substitute E/(1 − V2) for E (Table 3.1).

In many problems of practical importance, the deflection due to transverse loading of a beam may be obtained through successive integration of the beam equation:


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Alternatively, we could begin with EIυ" = M(x) and integrate twice to obtain


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In either case, the constants c1, c2, c3, and c4, which correspond to the homogeneous solution of the differential equations, may be evaluated from the boundary conditions. The constants c1, c2, c3/EI, and c4/EI represent the values at the origin of V, M, θ, and υ, respectively. In the method of successive integration, there is no need to distinguish between statically determinate and statically indeterminate systems (Section 5.11), because the equilibrium equations represent only two of the boundary conditions (on the first two integrals), and because the total number of boundary conditions is always equal to the total number of unknowns.

5.7 Normal and Shear Stresses

When a beam is bent by transverse loads, usually both a bending moment M and a shear force V act on each cross section. The distribution of the normal stress associated with the bending moment is given by the flexure formula, Eq. (5.4):


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where M and I are taken with respect to the z axis (Fig. 5.7).

In accordance with the assumptions of elementary bending, Eqs. (5.26) and (5.27), we omit the contribution of the shear strains to beam deformation in these calculations. However, shear stresses do exist, and the shearing forces are the resultant of the stresses. The shearing stress τxy acting at section mn, which is assumed to be uniformly distributed over the area b·dx, can be determined on the basis of equilibrium of forces acting on the shaded part of the beam element (Fig. 5.9). Here b is the width of the beam a distance y1 from the neutral axis, and dx is the length of the element. The distribution of normal stresses produced by M and M + dM is indicated in the figure. The normal force distributed over the left face mr on the shaded area A* is equal to


Figure 5.9. (a) Beam segment for analyzing shear stress; (b) cross section of beam.

Similarly, an expression for the normal force on the right face ns may be written in terms of M + dM. The equilibrium of x-directed forces acting on the beam element is governed by


from which we have


After substituting in Eq. (5.29), we obtain the shear formula (also called the shear stress formula) for beams:


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The integral represented by Q is the first moment of the shaded area A* with respect to the neutral axis z:


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By definition, inline263_46.jpg is the distance from the neutral axis to the centroid of A*. In the case of sections of regular geometry, inline261_45.jpg provides a convenient means of calculating Q. The shear force acting across the width of the beam per unit length


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is called the shear flow.

5.7.1 Rectangular Cross Section

In the case of a rectangular cross section of width b and depth 2h, the shear stress at y1 is


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This equation shows that the shear stress varies parabolically with y1. It is zero when y1 = ±h, and has its maximum value at the neutral axis:


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where A = 2bh is the area of the rectangular cross section. Note that the maximum shear stress (either horizontal or vertical: τxy = τyx) is 1.5 times larger than the average shear stress V/A. As observed in Section 5.4, for a thin rectangular beam, Eq. (5.42) is the exact distribution of shear stress. More generally, for wide rectangular sections and for other sections, Eq. (5.39) yields only approximate values of the shearing stress.

5.7.2 Various Cross Sections

Because the shear formula for beams is based on the flexure formula, the limitations of the bending formula apply when it is used. Problems involving various types of cross sections can be solved by following procedures identical to that for rectangular sections. Table 5.1 shows some typical cases. Observe that shear stress can always be expressed as a constant times the average shear stress (P/A), where the constant is a function of the cross-sectional form. Nevertheless, the maximum shear stress does not always occur at the neutral axis. For instance, in the case of a cross section having nonparallel sides, such as a triangular section, the maximum value of Q/b (and thus τxy) occurs at midheight, h/2, while the neutral axis is located at a distance h/3 from the base.

Table 5.1. Maximum Shearing Stress for Some Typical Beam Cross-Sectional Forms

Cross Section

Maximum Shearing Stress










Hollow Circle







Halfway between top and bottom




At h/8 above and below the NA

Notes: A, cross-sectional area; V, transverse shear force; NA, the neutral axis.

The following examples illustrate the application of the normal and shear stress formulas.

5.7.3 Beam of Constant Strength

When a beam is stressed to a constant permissible stress, σall throughout, then clearly the beam material is used to its greatest capacity. For a given material, such a design is of minimum weight. At any cross section, the required section modulus S is defined as


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where M presents the bending moment on an arbitrary section. Tapered beams designed in this way are called beams of constant strength. Ultimately, the shear stress at those beam locations where the moment is small controls the design.

Examples of beams with uniform strength include leaf springs and certain cast machine elements. For a structural member, fabrication and design constraints make it impractical to produce a beam of constant stress. Hence, welded cover plates are often used for parts of prismatic beams where the moment is large—for instance, in a bridge girder. When the angle between the sides of a tapered beam is small, the flexure formula allows for little error. On the contrary, the results found by applying the shear stress formula may not be accurate enough for nonprismatic beams. Often, a modified form of this formula is used for design purposes. The exact distribution in a rectangular wedge is found by applying the theory of elasticity (Section 3.10).

5.8 Effect of Transverse Normal Stress

When a beam is subjected to a transverse load, a transverse normal stress is created. According to Eq. (5.26), this stress is not related to the normal strain εy, so it cannot be determined using Hooke′s law. However, an expression for the average transverse normal stress can be obtained from the equilibrium requirement of force balance along the axis of the beam. For this purpose, a procedure is used similar to that employed for determining the shear stress in Section 5.7.

Consider, for example, a rectangular cantilever beam of width b and depth 2h subject to a uniform load of intensity p (Fig. 5.14a). The free-body diagram of an isolated beam segment of length dx is shown in Fig. 5.14b. Passing a horizontal plane through this segment results in the free-body diagram of Fig. 5.14c, for which the condition of statics ΣFy = 0 yields


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Figure 5.14. (a) Uniformly loaded cantilever beam of rectangular cross section; (b) free-body diagram of a segment; (c) stresses in a beam element.

Here, the shear stress is defined by Eq. (5.41) as


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After substituting Eqs. (5.28) and (b) into Eq. (a), we have


Integration yields the transverse normal stress in the form


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This stress varies as a cubic parabola, ranging from −plb at the surface (y = −h) where the load acts, to zero at the opposite surface (y = h).

The distribution of the bending and the shear stresses in a uniformly loaded cantilever beam (Fig. 5.12a) is determined from Eqs. (5.38) and (b):


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The largest values of σx, τxy, and σy given by Eqs. (5.44) are


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To compare the magnitudes of the maximum stresses, consider the following ratios:


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Because L is much greater than h in most beams (L ≥ 20h), the shear and the transverse normal stresses will usually be orders of magnitude smaller than the bending stresses. This justification is the rationale for assuming γxy = 0 and ɛy = 0 in the technical theory of bending. Note that Eq. (e) results in even smaller values than Eq. (d). Therefore, in practice, it is reasonable to neglect σy.

The foregoing conclusion applies, in most cases, to beams of a variety of cross-­sectional shapes and under various load configurations. Clearly, the factor of proportionality in Eqs. (d) and (e) will differ for beams of different sectional forms and for different loadings of a given beam.

5.9 Composite Beams

Beams constructed of two or more materials having different moduli of elasticity are referred to as composite beams. Examples include multilayer beams made by bonding together multiple sheets, sandwich beams consisting of high-strength material faces separated by a relatively thick layer of low-strength material such as plastic foam, and reinforced concrete beams. The assumptions of the technical theory for a homogeneous beam (Section 5.6) are valid for a beam composed of more than one material.

5.9.1 Transformed Section Method

To analyze composite beams, we will use the common transformed-section method. In this technique, the cross sections of several materials are transformed into an equivalent cross section of one material on which the resisting forces and the neutral axis are the same as on the original section. The usual flexure formula is then applied to the new section. To illustrate this method, we will use a frequently encountered example: a beam with a symmetrical cross section built of two different materials (Fig. 5.15a).


Figure 5.15. Beam composed of two materials: (a) composite cross section; (b) strain distribution; (d) transformed cross section.

The cross sections of the beam remain plane during bending. Hence, the condition of geometric compatibility of deformation is satisfied. It follows that the normal strain ɛx varies linearly with the distance y from the neutral axis of the section; that is, ɛx = ky (Figs. 5.15a and b). The location of the neutral axis is yet to be determined. Both materials composing the beam are assumed to obey Hooke′s law, and their moduli of elasticity are designated as E1 and E2. Then, the stress—strain relation gives


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This result is sketched in Fig. 5.13c for the assumption that E2 > E1. We introduce the notation


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where n is called the modular ratio. Note that n >1 in Eq. (5.46). However, this choice is arbitrary; the technique applies as well for n > 1.

5.9.2 Equation of Neutral Axis

Referring to the cross section (Figs. 5.15a and c), the equilibrium equations ΣFx = 0 and ΣMz = 0 lead to


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where A1 and A2 denote the cross-sectional areas for materials 1 and 2, respectively. Substituting σx1, σx2 and n, as given by Eqs. (5.45) and (5.46), into Eq. (a) results in


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Using the top of the section as a reference (Fig. 5.15a), from Eq. (5.47) with


or, setting


we have


This expression yields an alternative form of Eq. (5.47):


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Equations (5.47) and (5.47') can be used to locate the neutral axis for a beam of two materials. These equations show that the transformed section will have the same neutral axis as the original beam, provided the width of area 2 is changed by a factor n and area 1 remains the same (Fig. 5.15d). Clearly, this widening must be effected in a direction parallel to the neutral axis, since the distance equ272_509a.jpg to the centroid of area 2 remains unchanged. The new section constructed in this way represents the cross section of a beam made of a homogeneous material with a modulus of elasticity E1 and with a neutral axis that passes through its centroid, as shown in Fig. 5.15d.

5.9.3 Stresses in the Transformed Beam

Similarly, condition (b) together with Eqs. (5.45) and (5.46) leads to




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where I1 and I2 are the moments of inertia about the neutral axis of the cross-sectional areas 1 and 2, respectively. Note that


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is the moment of inertia of the entire transformed area about the neutral axis. From Eq. (5.48), we have


The flexure formulas for a composite beam are obtained by introducing this relation into Eqs. (5.45):


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where σx1 and σx2 are the stresses in materials 1 and 2, respectively. Note that when E1 = E2 = E, Eqs. (5.50) reduce to the flexure formula for a beam of homogeneous material, as expected.

5.9.4 Composite Beams of Multi Materials

The preceding discussion may be extended to include composite beams consisting of more than two materials. It is readily shown that for m different materials, Eqs. (5.47′), (5.49), and (5.50) take the forms


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where i = 2, 3,..., m denotes the ith material.

The use of the formulas developed in this section is demonstrated in the solutions of the numerical problems in Examples 5.7 and 5.8.

5.10 Shear Center

Given any cross-sectional configuration, one point may be found in the plane of the cross section through which the resultant of the transverse shearing stresses passes. A transverse load applied on the beam must act through this point, called the shear center or flexural center, if no twisting is to occur [Ref. 5.3]. The center of shear is sometimes defined as the point in the end section of a cantilever beam at which an applied load results in bending only. When the load does not act through the shear center, in addition to bending, a twisting action results (Section 6.1).

The location of the shear center is independent of the direction and magnitude of the transverse forces. For singly symmetrical sections, the shear center lies on the axis of symmetry, while for a beam with two axes of symmetry, the shear center coincides with their point of intersection (also the centroid). It is not necessary, in general, for the shear center to lie on a principal axis, and it may be located outside the cross section of the beam.

5.10.1 Thin-Walled Open Cross Sections

For thin-walled sections, the shearing stresses are taken to be distributed uniformly over the thickness of the wall and directed so as to parallel the boundary of the cross section. If the shear center S for the typical section of Fig. 5.18a is required, we begin by calculating the shear stresses by means of Eq. (5.39). The moment Mx of these stresses about arbitrary point A is then obtained. Since the external moment attributable to Vy about A is Vye, the distance between A and the shear center is given by


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Figure 5.18. Shear centers.

If the force is parallel to the z axis rather than the y axis, the position of the line of action may be established in the manner discussed previously. If both Vy and Vz exist, the shear center is located at the intersection of the two lines of action.

The determination of Mx is simplified by propitious selection of point A, such as in Fig. 5.18b. There, the moment Mx of the shear forces about A is zero; point A is also the shear center. For all sections consisting of two intersecting rectangular elements, the same situation exists.

For thin-walled box beams (with boxlike cross section), the point or points in the wall where the shear flow q = 0 (or τxy = 0) is unknown. Here, shear flow is represented by the superposition of transverse and torsional flow (see Section 6.8). Hence, the unit angle of twist equation, Eq. (6.23), along with q = VQ/I, is required to find the shear flow for a cross section of a box beam. The analysis procedure is as follows: First, introduce a free edge by cutting the section open; second, close it again by obtaining the shear flow that makes the angle of twist in the beam zero [Refs. 5.4 through 5.6].

In the following examples, the shear center of an open, thin-walled section is determined for two typical situations. In the first, the section has only one axis of symmetry; in the second, there is an asymmetrical section.

5.10.2 Arbitrary Solid Cross Sections

The preceding considerations can be extended to beams of arbitrary solid cross section, in which the shearing stress varies with both cross-sectional coordinates y and z. For these sections, the exact theory can, in some cases, be successfully applied to locate the shear center. Examine the section of Fig. 5.18c subjected to the shear force Vz, which produces the stresses indicated. Denote y and z as the principal directions. The moment about the x axis is then


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Vz must be located a distance e from the z axis, where e = Mx/vz.

5.11 Statically Indeterminate Systems

A large class of problems of considerable practical interest relate to structural systems for which the equations of statics are not sufficient (but are necessary) for determination of the reactions or other unknown forces. Such systems are called statically indeterminate, and they require supplementary information for their solution. Additional equations usually describe certain geometric conditions associated with displacement or strain. These equations of compatibility state that the strain owing to deflection or rotation must preserve continuity.

With this additional information, the solution proceeds in essentially the same manner as for statically determinate systems. The number of reactions in excess of the number of equilibrium equations is called the degree of statical indeterminacy. Any reaction in excess of that which can be obtained by statics alone is said to be redundant. Thus, the number of redundants is the same as the degree of indeterminacy.

Several methods are available to analyze statically indeterminate structures. The principle of superposition, briefly discussed next, is an effective approach for many cases. In Section 5.6 and in Chapters 7 and 10, a number of commonly employed methods are discussed for the solution of the indeterminate beam, frame, and truss problems.

5.11.1 The Method of Superposition

In the event of complicated load configurations, the method of superposition may be used to good advantage to simplify the analysis. Consider, for example, the continuous beam shown in Fig. 5.21a, which is then replaced by the beams shown in Fig. 5.21b and c. At point A, the beam now experiences the deflections (υA)p and (υA)R due to P and R, respectively. Subject to the restrictions imposed by small deformation theory and a material obeying Hooke′s law, the deflections and stresses are linear functions of transverse loadings, and superposition is valid:


Figure 5.21. Superposition of displacements in a continuous beam.

This procedure may, in principle, be extended to situations involving any degree of indeterminacy.

5.12 Energy Method for Deflections

Strain energy methods are frequently employed to analyze the deflections of beams and other structural elements. Of the many approaches available, Castigliano′s second theorem is one of the most widely used. To apply this theory, the strain energy must be represented as a function of loading. Detailed discussions of energy techniques are found in Chapter 10. In this section, we limit ourselves to a simple example to illustrate how the strain energy in a beam is evaluated and how the deflection is obtained by the use of Castigliano′s theorem (Section 10.4).

The strain energy stored in a beam under bending stress σx only, substituting M = EI(d2υ/dx2 into Eq. (2.63), is expressed in the form


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where the integrations are carried out over the beam length. We next determine the strain energy stored in a beam that is only due to the shear loading V. As described in Section 5.7, this force produces shear stress τxy at every point in the beam. The strain energy density is, from Eq. (2.50), Uo = τxy/2G. Substituting τxy as expressed by Eq. (5.39), we have Uo = V2Q2/2GI2b2. Integrating this expression over the volume of the beam of cross-­sectional area A, we obtain


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5.12.1 Form Factor for Shear

Let us denote


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This value is termed the shape or form factor for shear. When it is substituted in Eq. (a), we obtain


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where the integration is carried over the beam length. The form factor is a dimensionless quantity specific to a given cross-section geometry.

For example, for a rectangular cross section of width b and height 2h, the first moment Q, from Eq. (5.41), is inline285_1.jpg. Because A/I2 = 9/2bh5, Eq. (5.63) provides the following result:


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In a like manner, the form factor for other cross sections can be determined. Table 5.2 lists several typical cases. Following the determination of α the strain energy is evaluated by applying Eq. (5.64).

Table 5.2. Form Factor for Shear for Various Beam Cross Sections

Cross Section

Form Factor α

A. Rectangle


B. I-section, box section, or channelsa


C. Circle


D. Thin-walled circular


aA = area of the entire section, Aweb = area of the web ht, where h is the beam depth and t is the web thickness.

For a linearly elastic beam, Castigliano′s theorem, from Eq. (10.3), is expressed by


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where P is a load acting on the beam and δ is the displacement of the point of application in the direction of P. Note that the strain energy U = Ub + Us is expressed as a function of the externally applied forces (or moments).

As an illustration, consider the bending of a cantilever beam of rectangular cross ­section and length L, subjected to a concentrated force P at the free end (Fig. 5.5). The bending moment at any section is M = Px and the shear force V is equal in magnitude to P. Substituting these together with inline286_47.jpg into Eqs. (5.62) and (5.64) and integrating, we find the strain energy stored in the cantilever to be


The displacement of the free end owing to bending and shear is, by application of Castigliano′s theorem, therefore


The exact solution is given by Eq. (5.24).

Part C: Curved Beams

5.13 Elasticity Theory

Our treatment of stresses and deflections caused by the bending has been restricted so far to straight members. In real-world applications, many members—such as crane hooks, chain links, C-lamps, and punch-press frames—are curved and loaded as beams. Part C deals with the stresses caused by the bending of bars that are initially curved.

A curved bar or beam is a structural element for which the locus of the centroids of the cross sections is a curved line. This section focuses on an application of the theory of elasticity to a bar characterized by a constant narrow rectangular cross section and a circular axis. The axis of symmetry of the cross section lies in a single plane throughout the length of the member.

5.13.1 Equations of Equilibrium and Compatibility

Consider a beam subjected to equal end couples M such that bending takes place in the plane of curvature, as shown in Fig. 5.24a. Given that the bending moment remains constant along the length of the bar, the stress distribution should be identical in any radial cross section. Stated differently, we seek a distribution of stress displaying θ independence. It is clear that the appropriate expression of equilibrium is Eq. (8.2):


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and that the condition of compatibility for plane stress, Eq. (3.41),


must also be satisfied. The latter is an equidimensional equation, which can be reduced to a second-order equation with constant coefficients by substituting r = et or t = In r. Direct integration then leads to σr +, σθ = c" + c′ 1n r which may be written in the form σr + σθ = c′'' + c′ In (r/a). Solving this expression together with Eq. (a) results in the following equations for the radial and tangential stress:


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Figure 5.24. Pure bending of a curved beam of rectangular cross section.

5.13.2 Boundary Conditions

To evaluate the constants of integration, the boundary conditions are applied as follows:

  1. No normal forces act along the curved boundaries at r = a and r = b, so that

  2. Because there is no force acting at the ends, the normal stresses acting at the straight edges of the bar must be distributed to yield a zero resultant:

    where t represents the beam thickness.

  3. The normal stresses at the ends must produce a couple M:

The conditions (c) and (d) apply not just at the ends; that is, because of σθ independence, they apply at any θ. In addition, shearing stresses are assumed to be zero throughout the beam, so τ = 0 is satisfied at the boundaries, where no tangential forces exist.

Combining the first equation of Eqs. (5.65) with the condition (b), we find that


These constants together with the second of Eqs. (5.65) satisfy condition (c). Thus, we have


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Finally, substitution of the second of Eqs. (5.65) and (e) into (d) provides


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5.13.3 Stress Distribution

When the expressions for constants c1, c2 and c3 are inserted into Eq. (5.65), the following equations are obtained for the radial stress and tangential stress:


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If the end moments are applied so that the force couples producing them are distributed in the manner indicated by Eq. (5.67), then these equations are applicable throughout the bar. If the distribution of applied stress (to produce M) differs from Eq. (5.67), the results may be regarded as valid in regions away from the ends, in accordance with Saint-Venant′s principle.

These results, when applied to a beam with radius a, large relative to its depth h, yield an interesting comparison between straight and curved beam theory. For slender beams with h << a, the radial stress σr in Eq. (5.67) becomes negligible, and the tangential stress σθ is approximately the same as that obtained from My/I. Note that the radial stresses developed in nonslender curved beams made of isotropic materials are small enough that they can be neglected in analysis and design.

The bending moment is taken as positive when it tends to decrease the radius of ­curvature of the beam, as in Fig. 5.24a. Employing this sign convention, σr as determined from Eq. (5.67) is always negative, indicating that this stress is compressive. Similarly, when σθ is positive, the stress is tensile; otherwise, it is compressive. Figure 5.24b plots the stresses at section mn. Note that the maximum stress magnitude is found at the extreme fiber of the concave side.

5.13.4 Deflections

Substitution of σr and σθ from Eq. (5.67) into Hooke′s law provides expressions for the strains ɛθ, ɛr, and γ. The displacements u and υ then follow, upon integration, from the strain-displacement relationships, described by Eqs. (3.33). The resulting displacements indicate that plane sections of the curved beam subjected to pure bending remain plane subsequent to bending. Castigliano′s theorem (Section 5.12) is a particularly attractive method for determining the deflection of curved members.

For beams in which the depth of the member is small relative to the radius of curvature or, as is usually assumed, inline289_48.jpg, the initial curvature may be neglected in evaluating the strain energy. In such a case, inline289_49.jpg represents the radius to the centroid, and c is the distance from the centroid to the extreme fiber on the concave side. Thus, the strain energy due to the bending of a straight beam [Eq. (5.62)] is also a good approximation for curved, slender beams.

5.14 Curved Beam Formula

The approach to curved beams explored in this section was developed by E. Winkler (1835—1888). As an extension of the elementary theory of straight beams, Winkler′s theory assumes that all conditions required to make the straight-beam formula applicable are satisfied except that the beam is initially curved.

Consider the pure bending of a curved beam as illustrated in Fig. 5.25a. The distance from the center of curvature to the centroidal axis is inline291_51.jpg. The positive y coordinate is measured toward the center of curvature O from the neutral axis (Fig. 5.25b). The outer and inner fibers are at distances of ro and ri from the center of curvature, respectively.


Figure 5.25. (a) Curved beam in pure bending with a cross-sectional vertical (y) axis of symmetry; (b) cross section; (c) stress distributions over the cross section.

5.14.1 Basic Assumptions

Derivation of the stress in the beam is again based on the three principles of solid mechanics and the familiar presuppositions:

  1. All cross sections possess a vertical axis of symmetry lying in the plane of the centroidal axis passing through C.

  2. The beam is subjected to end couples M. The bending moment vector is everywhere normal to the plane of symmetry of the beam.

  3. Sections originally plane and perpendicular to the centroidal beam axis remain so subsequent to bending. (The influence of transverse shear on beam deformation is not taken into account.)

Referring to assumption (3), note the relationship in Fig. 5.25a between lines bc and ef representing the plane sections before and after the bending of an initially curved beam. Note also that the initial length of a beam fiber such as gh depends on the distance r from the center of curvature O. On the basis of plane sections remaining plane, we can state that the total deformation of a beam fiber obeys a linear law, as the beam element rotates through small angle.

5.14.2 Location of the Neutral Axis

In Fig. 5.25a, it is clear that the initial length of any arbitrary fiber gh of the beam depends on the distance r from the center of curvature O. Thus, the total deformation of a beam fiber obeys a linear law, as the beam element rotates through a small angle . Conversely, the normal or tangential strain ɛθ does not follow a linear relationship. The contraction of fiber gh equals -(R - r)dθ where R is the distance from O to the neutral axis (yet to be determined) and its initial length is . So, the normal strain of this fiber is given by ɛθ = - (R - r)dθ/rθ. For convenience, we denote λ = dθ/θ which is constant for any element.

The tangential normal stress, acting on an area dA of the cross section, can now be obtained through the use of Hooke8217;s law, σθ = θ. It follows that


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The equations of equilibrium, ΣFx = 0 and ΣMz = 0 are, respectively,


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When the tangential stress of Eq. (a) is inserted into Eq. (b), we obtain


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Since and R are constants, they may be moved outside the integral sign, as follows:


The radius of the neutral axis R is then written in the form


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where A is the cross-sectional area of the beam. The integral in Eq. (5.68) may be evaluated for various cross-sectional shapes (see Example 5.13 and Probs. 5.46 through 5.48). For reference, Table 5.3 lists explicit formulas for R and A for some commonly used cases.

Table 5.3. Properties for Various Cross-Sectional Shapes

Cross Section

Radius of Neutral Surface R

A. Rectangle



B. Circle



C. Ellipse



D. Triangle



E. Trapezoid


The distance e between the centroidal axis and the neutral axis (y = 0) of the cross section of a curved beam (Fig. 5.25b) is equal to


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Thus, in a curved member, the neutral axis does not coincide with the centroidal axis. This differs from the case involving straight elastic beams.

5.14.3 Tangential Stress

Once we know the location of the neutral axis, we can obtain the equation for the stress distribution by introducing Eq. (a) into Eq. (c). Therefore,


Expanding this equation, we have


Here, the first integral is equivalent to A/R, as determined by Eq. (5.68), and the second integral equals the cross-sectional area A. The third integral, by definition, represents inline291_50.jpg, where inline291_51.jpg is the radius of the centroidal axis. Therefore, inline291_52.jpg.

We now introduce E from Eq. (a) into the discussion and solve for σθ from the resulting expression. Then, the tangential stress in a curved beam, subject to pure bending at a distance r from the center of curvature, is expressed in the following form:


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where e is defined by Eq. (5.69). Alternatively, substituting y = R − r or r = R − y (Fig. 5.25a) into Eq. (5.70) yields


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5.14.4 Winkler’s Formula

Equations (5.70) and (5.71) represent two forms of the curved-beam formula. Another alternative form of these equations is often referred to as Winkler’s formula. The variation of stress over the cross section is hyperbolic, as sketched in Fig. 5.25c. The sign convention applied to bending moment is the same as that used in Section 5.13—namely, the bending moment is positive when directed toward the concave side of the beam, as shown in the figure. If Eq. (5.70) or Eq. (5.71) results in a positive value, a tensile stress is present.

5.15 Comparison of the Results of Various Theories

We now examine the solutions obtained in Sections 5.13 and 5.14 with results determined using the flexure formula for straight beams. To do so, we consider a curved beam of rectangular cross section and unit thickness experiencing pure bending. The tangential stress predicted by the elementary theory (based on a linear distribution of stress) is My/I. The Winkler approach, which leads to a hyperbolic distribution, is given by Eq. (5.70) or Eq. (5.71), while the exact theory results in Eqs. (5.67). In each case, the maximum and minimum values of stress are expressed by


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Table 5.4 lists values of m as a function of b/a for the four cases cited [Ref. 5.1], in which b = ro and a = rj (see Figs. 5.24 and 5.25). As can be seen, there is good agreement between the exact and Winkler results. On this basis, as well as from more extensive comparisons, we may conclude that the Winkler approach is adequate for practical applications. Its advantage lies in the relative ease with which it may be applied to any symmetric section.

Table 5.4. The Values of m for Typical Ratios of Outer Radius b to Inner Radius a



Curved Beam Formula

Elasticity Theory


Flexure Formula

r = a

r = b

r = a

r = b

























The agreement between the Winkler approach and the exact analyses is not as good in situations involved combined loading as it is for the case of pure bending. As might be expected, for beams of only slight curvature, the simple flexure formula provides good results while requiring only simple computation. The linear and hyperbolic stress distributions are approximately the same for b/a = 1.1. As the curvature of the beam increases, (b/a > 1.3) the stress on the concave side rapidly increases over the one given by the flexure formula.

5.15.1 Correction of σθ for Beams with Thin-Walled Cross Sections

Where I-beams, T-beams, or thin-walled tubular curved beams are involved, the approaches developed in this chapter will not accurately predict the stresses in the system. The error in such cases is attributable to the high stresses existing in certain sections such as the flanges, which cause significant beam distortion. A modified Winkler’s equation can be applied in such situations if more accurate results are required [Ref. 5.6]. The distortion, and thus the error in σθ, is reduced when the flange thickness is increased. Given that material yielding is highly localized, its effect is not of concern unless the curved beam is under fatigue loading.

5.16 Combined Tangential and Normal Stresses

Curved beams are often loaded so that there is both an axial force and a moment on the cross section. The tangential stress given by Eq. (5.70) may then be algebraically added to the stress due to an axial force P acting through the centroid of cross-sectional area A. For this simple case of superposition, the total stress at a point located at distance r from the center of curvature O may be expressed as follows:


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As before, a negative sign would be associated with a compressive load P.

Of course, the theory developed in this section applies only to the elastic stress distribution in curved beams. Stresses in straight members under various combined loads are discussed in detail throughout this text.

The following problems illustrate the application of the formulas developed to statically determinate and statically indeterminate beams under combined loadings. In the latter case, the energy method (Section 10.4) facilitates the determination of the unknown, redundant moment in the member.


5.1. TIMOSHENKO, S. P., and GOODIER, J. N. Theory of Elasticity, 3rd ed. New York: McGraw-Hill, 1970.

5.2. UGURAL, A. C. Mechanics of Materials. Hoboken, NJ: Wiley, 2008, Chapter 8.

5.3. SOKOLNIKOFF, I. S. Mathematical Theory of Elasticity, 2nd ed. Melbourne, FL: Krieger, 1986.

5.4. COOK, R. D., and YOUNG, W. C. Advanced Mechanics of Materials. New York: Macmillan, 1985.

5.5. BUDYNAS, R. G. Advanced Strength and Applied Stress Analysis, 2nd ed. New York: McGraw-Hill, 1999.

5.6. BORESI, A. P., and SCHMIDT, R. J. Advanced Mechanics of Materials, 6th ed. Hoboken, NJ: Wiley, 2003.


Sections 5.1 through 5.5

5.1. A simply supported beam constructed of a 0.15 × 0.015 m angle is loaded by concentrated force P = 22.5 KN at its midspan (Fig. P5.1). Calculate stress σx at A and the orientation of the neutral axis. Neglect the effect of shear in bending and assume that beam twisting is prevented.


Figure P5.1.

5.2. A wood cantilever beam with cross section as shown in Fig. P5.2 is subjected to an inclined load P at its free end. Determine (a) the orientation of the neutral axis and (b) the maximum bending stress. Given: P = 1 kN, α = 30°, b = 80 mm, h = 150 mm and length L = 1.2 m.


Figure P5.2.

5.3. A moment Mo is applied to a beam of the cross section shown in Fig. P5.3 with its vector forming an angle α. Use b = 100 mm, h = 40 mm, Mo = 800N·m, and α = 25°. Calculate (a) the orientation of the neutral axis and (b) the maximum bending stress.


Figure P5.3.

5.4. Couples My = Mo and Mz = 1.5Mo are applied to a beam of cross section shown in Fig. P5.4. Determine the largest allowable value of Mo for the maximum stress not to exceed 80 MPa. All dimensions are in millimeters.


Figure P5.4.

5.5. For the simply supported beam shown in Fig. P5.5, determine the bending stress at points D and E. The cross section is a 0.15 × 0.15 × 0.02 angle (Fig. 5.4).


Figure P5.5.

5.6. A concentrated load P acts on a cantilever, as shown in Fig. P5.6. The beam is constructed of a 2024-T4 aluminum alloy having a yield strength σyp = 290 MPa, L = 1.5 m, t = 20 mm, c = 60 mm, and b = 80 mm. Based on a factor of safety n = 1.2 against initiation of yielding, calculate the magnitude of P for (a) α = 0° and (b) α = 15°. Neglect the effect of shear in bending and assume that beam twisting is prevented.


Figure P5.6.

5.7. Re-solve Prob. 5.6 for α = 30°. Assume the remaining data are unchanged.

5.8. A cantilever beam has a Z section of uniform thickness for which inline302_58.jpg,inline302_59.jpg and Iyz = - th3. Determine the maximum bending stress in the beam subjected to a load P at its free end (Fig. P5.8).


Figure P5.8.

5.9. A beam with cross section as shown in Fig. P5.9 is acted on by a moment Mo 3KN·m, with its vector forming an angle α = 20°. Determine (a) the orientation of the neutral axis and (b) the maximum bending stress.

5.10 and 5.11. As shown in the cross section in Figs. P5.10 and P5.11, a beam carries a moment M, with its vector forming an angle α with the horizontal axis. Find the stresses at points A, B, and D.


Figure P5.9.


Figure P5.10.


Figure P5.11.

5.12. For the thin cantilever shown in Fig. P5.12, the stress function is given by


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  1. Determine the stresses σxy, and τxy by using the elasticity method.

  2. Determine the stress σx by using the elementary method.

  3. Compare the values of maximum stress obtained by the preceding approaches for L = 10h.


Figure P5.12.

5.13. Consider a cantilever beam of constant unit thickness subjected to a uniform load of p = 2000 KN per unit length (Fig. P5.13). Determine the maximum stress in the beam:

  1. Based on a stress function

  2. Based on the elementary theory. Compare the results of (a) and (b).


Figure P5.13.

Sections 5.6 through 5.11

5.14. A bending moment acting about the z axis is applied to a T-beam, as shown in Fig. P5.14. Take the thickness t = 15 mm and depth h = 90 mm. Determine the width b of the flange needed so that the stresses at the bottom and top of the beam will be in the ratio 3:1, respectively.


Figure P5.14.

5.15. A wooden, simply supported beam of length L is subjected to a uniform load p. Determine the beam length and the loading necessary to develop simultaneously σmax = 8.4 MPa and τmax = 0.7 MPa Take thickness t = 0.05 m and depth h = 0.15 m.

5.16. A box beam supports the loading shown in Fig. P5.16. Determine the maximum value of P such that a flexural stress σ = 7 MPa or a shearing stress τ = 0.7 will not be exceeded.


Figure P5.16.

5.17. Design a rectangular cantilever beam of constant strength and width b, to carry a uniformly distributed load of intensity w (Fig. P5.17). Assumption: Only the normal stresses due to the bending need be taken into account; the permissible stress equals σall.


Figure P5.17.

5.18. Design a simply supported rectangular beam of constant strength and width b, supporting a uniformly distributed load of intensity w (Fig. P5.18). Assumption: Only the normal stresses due to the bending need be taken into account; the allowable stress is σall.


Figure P5.18.

5.19. A steel beam of the tubular cross section seen in Fig. P5.19 is subjected to the bending moment M about the z axis. Determine (a) the bending moment M and (b) the radius of curvature rx of the beam. Given: σall = 150 MPa, E = 70 GPa, b = 120 mm, h = 170 mm, E = 70 GPa, b = 120 mm, h = 170 mm, and t = 10 mm.


Figure P5.19.

5.20. An aluminum alloy beam of hollow circular cross section is subjected to a bending moment M about the z axis (Fig. P5.20). Determine (a) the normal stress at point A, (b) the normal stress at point B, and (c) the radius of curvature rx of the beam of a transverse cross section. Given: M = 600 N·m, D = 60 mm, d = 40 mm, E = 70 GPa, and v = 0.29.


Figure P5.20.

5.21. A simply supported beam AB of the channel cross section carries a concentrated load P at midpoint (Fig. P5.21). Find the maximum allowable load P based on an allowable normal stress of σall = 60 MPa in the beam.


Figure P5.21.

5.22. A uniformly loaded, simply supported rectangular beam has two 15-mm deep vertical grooves opposite each other on the edges at midspan, as illustrated in Fig. P5.22. Find the smallest permissible radius of the grooves for the case in which the normal stress is limited to σmax = 95 MPa. Given: p = 12 KN/m, L = 3, b = 80 mm, and h = 120 mm.


Figure P5.22.


Figure P5.23.

5.23. A simple wooden beam is under a uniform load of intensity p, as illustrated in Fig. P5.23. (a) Find the ratio of the maximum shearing stress to the largest bending stress in terms of the depth h and length L of the beam. (b) Using σall = 9 MPa,τall = 1.4 MPa, b = 50 mm and h = 160 mm, calculate the maximum permissible length L and the largest permissible distributed load of intensity p.

5.24. A composite cantilever beam 140 mm wide, 300 mm deep, and 3 m long is fabricated by fastening two timber planks, 60 mm × 300 mm, to the sides of a steel plate (Es = 200 GPa), 20 mm wide by 300 mm deep. Note that the 300-mm dimension is vertical. The allowable stresses in bending for timber and steel are 7 and 120 MPa, respectively. Calculate the maximum vertical load P that the beam can carry at its free end.

5.25. A simple beam pan length 3 m supports a uniformly distributed load of 40 kN/m. Find the required thickness t of the steel plates. Given: The cross section of the beam is a hollow box with wood flanges (Ew = 10.5 GPa) and steel (Es = 210 GPa), as seen in Fig. P5.25. Let a = 62.5 mm, b = 75 mm, and h = 225 mm. Assumptions: The permissible stresses are 140 MPa for the steel and 10 MPa for the wood.


Figure P5.25.

5.26. A 180-mm-wide by 300-mm-deep wood beam (Ew = 10 GPa), 4 m long, is reinforced with 180-mm-wide and 10-mm-deep aluminum plates (Ea = 70 GPa) on the top and bottom faces. The beam is simply supported and subject to a uniform load of intensity 25 kN/m over its entire length. Calculate the maximum stresses in each material.

5.27. Referring to the reinforced concrete beam of Fig. 5.17a, assume b = 300, d = 450 mm, As = 1200 m2 and n = 10. Given allowable stresses in steel and concrete of 150 and 12 MPa, respectively, calculate the maximum bending moment that the section can carry.

5.28. Referring to the reinforced concrete beam of Fig. 5.17a, assume b = 300 mm, d = 500 mm, and n = 8. Given the actual maximum stresses developed to be σs = 80 MPa and σc = 5 MPa, calculate the applied bending moment and the steel area required.

5.29. A channel section of uniform thickness is loaded as shown in Fig. P5.29. Find (a) the distance e to the shear center, (b) the shearing stress at D, and (c) the maximum shearing stress. Given: b = 100 mm, h = 90 mm, t = 4 mm, Vy = 5 kN.


Figure P5.29.

5.30. A beam is constructed of half a hollow tube of mean radius R and wall thickness t (Fig. P5.30). Assuming tR, locate the shear center S. The moment of inertia of the section about the z axis is Iz = πR3tl2.


Figure P5.30.

5.31. An H-section beam with unequal flanges is subjected to a vertical load P (Fig. P5.31). The following assumptions are applicable:

  1. The total resisting shear occurs in the flanges.

  2. The rotation of a plane section during bending occurs about the symmetry axis so that the of curvature of both flanges are equal.


Figure P5.31.

Find the location of the shear center S.


Figure P5.32.

5.32. Determine the shear center S of the section shown in Fig. P5.32. All dimensions are in millimeters.

5.33. A cantilever beam AB supports a triangularly distributed load of maximum intensity po (Fig. P5.33). Determine (a) the equation of the deflection curve, (b) the deflection at the free end, and (c) the slope at the free end.


Figure P5.33.

5.34. The slope at the wall of a built-in beam (Fig. P5.34a) is as shown in Fig. P5.34b and is given by PL3/96EI. Determine the force acting at the simple support, expressed in terms of p and L.


Figure P5.34.

5.35. A fixed-ended beam of length L is subjected to a concentrated force P at a distance c away from the left end. Derive the equations of the elastic curve.

5.36. A propped cantilever beam AB is subjected to a couple Mo acting at support B, as shown in Fig. P5.36. Derive the equation of the deflection curve and determine the reaction at the roller support.


Figure P5.36.

5.37. A clamped-ended beam AB carries a symmetric triangular load of maximum intensity p0 (Fig. P5.37). Find all reactions, the equation of the elastic curve, and the maximum deflection, using the second-order differential equation of the deflection.


Figure P5.37.

5.38. A welded bimetallic strip (Fig. P5.38) is initially straight. A temperature increment ΔT causes the element to curve. The coefficients of thermal expansion of the constituent metals are α1 and α2. Assuming elastic deformation and α2 > α determine (a) the radius of curvature to which the strip bends, (b) the maximum stress occurring at the interface, and (c) the temperature increase that would result in the simultaneous yielding of both elements.


Figure P5.38.

Sections 5.12 through 5.16

5.39. Verify the values of α for cases B, C, and D of Table 5.2.

5.40. Consider a curved bar subjected to pure bending (Fig. 5.24). Assume the stress function


Re-derive the stress field in the bar given by Eqs. (5.67).

5.41. The allowable stress in tension and compression for the clamp body shown in Fig. P5.41 is 80 MPa. Calculate the maximum permissible load that the member can resist. Dimensions are in millimeters.


Figure P5.41.

5.42. A curved frame of rectangular cross section is loaded as shown in Fig. P5.42. Determine the maximum tangential stress (a) by using the second of Eqs. (5.67) together with the method of superposition and (b) by applying Eq. (5.73). Given: h = 100 mm, inline311_60.jpg, and P = 70 kN.


Figure P5.42.

5.43. A curved frame having a channel-shaped cross section is subjected to bending by end moments M, as illustrated in Fig. P5.43. Determine the dimension b required if the tangential stresses at points A and B of the beam are equal in magnitude.


Figure P5.43.

5.44. A curved beam of a circular cross section of diameter d is fixed at one end and subjected to a concentrated load P at the free end (Fig. P5.44). Calculate (a) the tangential stress at point A and (b) the tangential stress at point B. Given: P = 800 N, d = 20 mm, a = 25 mm, and b = 15 mm.


Figure P5.44.


Figure P5.45.

5.45. A circular steel frame has a cross section approximated by the trapezoidal form shown in Fig. P5.45. Calculate (a) the tangential stress at point A and (b) the tangential stress at point B. Given: ri = 100 mm, b = 75 mm, b = 50 mm, and P = 50 kN.

5.46. The triangular cross section of a curved beam is shown in Fig. P5.46. Derive the expression for the radius R along the neutral axis. Compare the result with that given for Fig. D in Table 5.3.


Figure P5.46.

5.47. The circular cross section of a curved beam is illustrated in Fig. P5.47. Derive the expression for the radius R along the neutral axis. Compare the result with that given for Fig. B in Table 5.3.


Figure P5.47.

5.48. The trapezoidal cross section of a curved beam is depicted in Fig. P5.48. Derive the expression for the radius R along the neutral axis. Compare the result with that given for Fig. E in Table 5.3.


Figure P5.48.

5.49. A machine component of channel cross-sectional area is loaded as shown in Fig. P5.49. Calculate the tangential stress at points A and B. All dimensions are in millimeters.


Figure P5.49.

5.50. A load P is applied to an eye bar with rigid insert for the purpose of pulling (Fig. P5.50). Determine the tangential stress at points A and B (a) by the elasticity theory, (b) by Winkler’s theory, and (c) by the elementary theory. Compare the results obtained in each case.


Figure P5.50.

5.51. A ring of mean radius inline313_61.jpg and constant rectangular section is subjected to a concentrated load (Fig. P5.51). You may omit the effect of shear in bending. Derive the following general expression for the tangential stress at any section of the ring:


Click to view larger image


Figure P5.51.



Use Castigliano’s theorem.

5.52. The ring shown in Fig. P5.51 has the following dimensions: inline314_62.jpg, t = 50 mm, and h = 100 mm. Taking inline314_63.jpg, determine (a) the tangential stress on the inner fiber at θ = π/4 and (b) the deflection along the line of action of the load P, considering the effects of the normal and shear forces, as well as bending moment (Section 10.4).

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