2.5 Theories of Straight Beams

Home , Bending moment, Deflection (engineering)

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2.5. THEORIES OF STRAIGHT BEAMS 71 2.5 Theories of Straight Beams 2.5.1 Introduction Most practical engineering structures, microscale or macroscale, consist of members that can be classified as beams, plates, and shells, called structural members. Beams are structural members that have a ratio of length-to-cross- sectional dimensions very large, say 10 to 100 or more, and subjected to forces both along and transverse to the length and moments that tend to rotate them about an axis perpendicular to their length [see Fig. 2.5.1(a)]. When all applied loads are along the length only, they are often called bars (i.e., bars experience only tensile or compressive strains and no bending deformation). Cables may be viewed as very flexible form of bars, and they can only take tension and not compression. Plates are two-dimensional versions of beams in the sense that one of the cross-sectional dimensions of a plate can be as large as the length. The smallest dimension of a plate is called its thickness. Thus, plates are thin bodies subjected to forces in the plane as well as in the direction normal to the plane and bending moments about either axis in the plane [see Fig. 2.5.1(b)]. A shell is a thin structure with curved geometry [see Fig. 2.5.1(c)] and can be subjected to distributed as well as point forces and moments. Because of their geometries and loads applied, beams, plates, and shells are stretched and bent from their original shapes. The difference between structural elements

Current

Cantilever beam

AFM (atomic micro scope) tip (a)

(b) (c)

Fig. 2.5.1: (a) An AFM cantilever beam. (b) A narrow plate strip subjected to a vertical force. (c) A hyperboloidal shell. The deformations shown for (b) and (c) are exaggerated.

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and three-dimensional solid bodies, such as solid blocks and spheres, that have no restrictions on their geometric make up, is that the latter may change their geometries but they may not show significant “bending” deformation. Although all solids can be analyzed for stress and deformation using the elasticity equations reviewed in the preceding sections of this chapter, their geometries allow us to develop theories that are simple and yet yield results that are accurate enough for engineering analysis and design. Structural mechanics is a branch of solid mechanics that deals with the study of beams (frames), plates, and shells using theories that are derived from three-dimensional elasticity theory by making certain simplifying assumptions concerning the deformation and stress states in these members. The present section is devoted to the development of structural theories of straight beams in bending for the static case. The equations governing beams will be referenced in the coming chapters and their development in this chapter will help the reader in understanding the material of the subsequent chapters. Plates will be studied in a chapter devoted entirely for them (see Chapter 7). The geometric description of shells is more involved and they are not covered in this book to keep the book size within reasonable limits (see, for additional information, Timoshenko and Woinowsky-Krieger [124] and Reddy [50, 51]). In this section we consider two most commonly used theories of straight beams. They differ from each other in the representation of displacement and strain fields. The first one is the Bernoulli–Euler beam theory3, and it is the theory that is covered in all undergraduate mechanics of materials books. In the Bernoulli–Euler beam theory, the transverse shear strain is neglected, making the beam infinitely rigid in the transverse direction. The second one is a refinement to the Bernoulli–Euler beam theory, known as the Timoshenko beam theory, which accounts for the transverse shear strain. These two beam theories will be developed assuming infinitesimal deformation. Therefore, we use σij to mean Sij, the second Piola–Kirchhoff stress tensor, and xi in place of Xi. We consider straight beams of length L and symmetric (about the x2-axis) cross section of area A. The x1-coordinate is taken along the centroidal axis of the beam with the x3-coordinate along the thickness (the height) and the x2-coordinate along the width of the beam (into the plane of the page), as shown in Fig. 2.5.2(a). In general, the cross-sectional area A can be a function of x1. Suppose that the beam is subjected to distributed axial force f(x1) and transverse load q(x1), and let (u1, u2, u3) denote the total displacements along the coordinates (x1 = x, x2 = y, x3 = z). For simplicity of the developments only stretching along the length of the beam and bending about the x2-axis are considered here. 3Jacob Bernoulli (1655–1705) was one of the many prominent Swiss mathematicians in the Bernoulli family. He, and along with his brother Johann Bernoulli, was one of the founders of the calculus of variations. Leonhard Euler (1707–1783) was a pioneering Swiss mathematician and physicist.

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2.5. THEORIES OF STRAIGHT BEAMS 73

2.5.2 The Bernoulli–Euler Beam Theory The Bernoulli–Euler beam theory is based on certain simplifying assumptions, known as the Bernoulli–Euler hypothesis, concerning the kinematics of bending deformation.Figure 2.5.2 The hypothesis states that straight lines perpendicular to the beam axis before deformation remain (a) straight, (b) inextensible, and (c) perpendicular to the tangent line to the beam axis after deformation.

dw q =- x dx z xz3 = qx() z • q • w x xx1 = z u x uz+ q f ()x x Undeformed edge Deformed edge (a) (b)

qx() qx()

Mx()+D Mx () Mx() M()x f(x) f(x) Nx() Nx()+D Nx () Nx() sxx f ()x f ()x Vx()+D Vx () s Vx() Dx Vx() xz (c) (d)

Fig. 2.5.2: (a) A typical beam with loads. (b) Kinematics of deformation of an Bernoulli– Euler beam theory. (c) Equilibrium of a beam element. (d) Deﬁnitions (or internal equilibrium) of stress resultants.

The Bernoulli–Euler hypothesis leads to the following displacement ﬁeld [see Fig. 2.5.2(b)] dw u (x, y, z) = u(x) − z , u = 0, u (x, y, z) = w(x), (2.5.1) 1 dx 2 3 where (u, v, w) are the displacements of a point on the x-axis in the x1 = x, x2 = y and x3 = z coordinate directions, respectively. The inﬁnitesimal strains are

2 ∂u1 du d w ∂u1 ∂u3 dw dw εxx = = − z 2 , 2εxz = + = − + = 0 (2.5.2) ∂x1 dx dx ∂x3 ∂x1 dx dx and all other strains are identically zero. A free-body-diagram of an element of length ∆x is shown, with all its forces, in Fig. 2.5.2(c). Summing the forces in the x- and z-directions and summing

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the moments at the right end of the beam element, we obtain the following equilibrium equations: X dN F = 0 : − = f(x), (2.5.3) x dx X dV F = 0 : − = q(x), (2.5.4) z dx X dM M = 0 : V − = 0. (2.5.5) y dx where N(x) is the net axial force, M(x) is the net bending moment about the y-axis, and V (x) is the net transverse shear force [see Fig. 2.5.2(d)] on the beam cross section: Z Z Z N(x) = σxx dA, M(x) = σxxz dA, V (x) = σxz dA. (2.5.6) A A A where dA denotes an area element of the cross section (dA = dydz). The set (N,M,V ) are known as the stress resultants (because they result from the stresses in the beam). The stresses in the beam can be computed using the linear elastic constitutive relation for an isotropic but possibly inhomogeneous material (i.e., the material properties may be functions of x and z), we have du d2w σ (x, z) = E(x, z)ε (x, z) = E − z , xx xx dx dx2 (2.5.7) σxz(x, z) = 2G(x, z)εxz(x, z) = 0. Although the transverse shear stress computed using constitutive equation is zero (hence, V = 0), it cannot be zero in a beam because the force equilibrium, Eq. (2.5.4) is violated. This is the inconsistency resulting from the Bernoulli– Euler beam hypothesis. Therefore, to respect the physical requirement of V 6= 0, we do not compute the shear stress from σxz = 2Gεxz; it is computed using the shear force V obtained from the equilibrium condition in Eq. (2.5.5), namely, V = dM/dx. The stress resultants (N,M) now can be related to the displacements u and w and back to the stress σxx, as discussed next. Using Eq. (2.5.6), we obtain Z du d2w N(x) = σ dA = A + B − , xx xx dx xx dx2 A (2.5.8) Z du d2w M(x) = zσxx dA = Bxx + Dxx − 2 , A dx dx

where Axx is the axial stiffness, Bxx is the bending-stretching stiffness, and Dxx is the bending stiffness Z Z Z 2 Axx = E(x, z) dA, Bxx = zE(x, z) dA, Dxx = z E(x, z) dA (2.5.9) A A A

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2.5. THEORIES OF STRAIGHT BEAMS 75

When E is a function of x only, Eq. (2.5.9) simpliﬁes to Z Z Axx = E(x) dA = E(x)A(x),Bxx = E(x) z dA = 0, A A Z Z (2.5.10) 2 2 Dxx = E(x) z dA = E(x)I(x),I = z dA, A A where we have used the fact that the x-axis coincides with the centroidal axis: Z z dA = 0. (2.5.11) A Then (N,M) of Eq. (2.5.8) reduce to du d2w N(x) = EA ,M(x) = −EI (2.5.12) dx dx2 Equations (2.5.8) and (2.5.9) can be inverted to express du/dx and d2w/dx2 in terms of N and M. We obtain 2 du DxxN − BxxM d w AxxM − BxxN = 2 , − 2 = 2 . (2.5.13) dx AxxDxx − Bxx dx AxxDxx − Bxx Substituting these relations for du/dx and −d2w/dx2 into the expression for σxx in Eq. (2.5.7), we obtain E σxx(x, z) = 2 [DxxN − BxxM + z (AxxM − BxxN)] . (2.5.14) AxxDxx − Bxx

When E is not a function of z, we obtain (Axx = EA and Dxx = EI) du N d2w M = , − = . (2.5.15) dx EA dx2 EI and N Mz σ = + . (2.5.16) xx A I We note that, when E is not a function of z, N is independent of w and M is independent of u. Therefore, Eq. (2.5.3) is independent of Eqs. (2.5.4) and (2.5.5). In that case the axial deformation of members can be determined by solving Eq. (2.5.3) independent of the bending deformation and vice versa. Also, Eqs. (2.5.4) and (2.5.5) can be combined to read d2M − = q(x), (2.5.17) dx2 which can be expressed in terms of the transverse deﬂection w using the second equation in Eq. (2.5.15) as d2 d2w EI = q(x). (2.5.18) dx2 dx2

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We now return to the computation of shear stress σxz in the Bernoulli–Euler beam theory. The shear stress is computed from equilibrium considerations. For beams with E = E(x), the expression for σxz is given by (see pages 256–261 of Fenner and Reddy [98] for a derivation),

V (x)Q(z) Z c1 σxz(x, z) = ,Q(z) = b(z)zdz, (2.5.19) I(x)b z where Q(z) denotes the ﬁrst moment of the hatched area [see Fig. 2.5.3] about the axis y of the entire cross section, and b is the width of the cross section at z where the longitudinal shear stress (σxz) is computed. The hatched area is only a portion of the total area of cross section that lies below the surface on which the shear force acts; Q is the maximum at the centroid (i.e., z = 0), and σxz is the maximum wherever Q/b is the maximum. For example, for a rectangular cross section beam of height 2h and width b, Q(z) takes the form

Z h b Q(z) = b zdz = h2 − z2 (2.5.20) z 2 and the shear stress becomes V (x)Q(z) 1 σ (x, z) = = V (x) h2 − z2 . (2.5.21) xz I(x)b 2I Thus, the shear stress varies quadratically through the beam height, vanishing at the top (z = −h) and bottom (z = h). The maximum shear stress is max 3 σFigurexz = σxz 2.5.3(x, 0) = 4bh V (x). Thus, the actual maximum occurs wherever V is the maximum.

z qx() z σ Dx σ xx(,)xz xx(,)xxz+D c 1 sxz z x c2 y sxz b

Fig. 2.5.3: Transverse shear stress σxz due to bending.

2.5.3 The Timoshenko Beam Theory The Timoshenko beam theory is based on a relaxation of the normality condition, namely, Part (c) of the Bernoulli–Euler hypothesis. In other words, the transverse normal has rotation φx that is not the same as the slope −dw/dx. The diﬀerence between these two quantities is the transverse shear strain. The relaxed Bernoulli–Euler hypothesis leads to the following displacement ﬁeld for the Timoshenko beam theory (see Fig. 2.5.4)

i i q0

M 0 i i

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Figure 2.5.4 2.5. THEORIES OF STRAIGHT BEAMS 77

dw q =- x dx

gxz z z • fx q • w x z u x

uz+ fx Undeformed edge Deformed edge

Fig. 2.5.4: Kinematics of deformation in the Timoshenko beam theory.

u1(x, y, z) = u(x) + zφx, u2 = 0, u3(x, y, z) = w(x), (2.5.22)

where φx is the rotation of a transverse normal line. The inﬁnitesimal strains are

∂u1 du dφx ∂u1 ∂u3 dw εxx = = + z , 2εxz = + = φx + (2.5.23) ∂x1 dx dx ∂x3 ∂x1 dx and all other strains are identically zero. The free-body-diagram of a typical element of the beam remains the same as depicted in Fig. 2.5.2(c). Therefore, the equilibrium equations given in Eqs. (2.5.3)–(2.5.5) are also valid for the Timoshenko beam theory, except that the stress resultants (N,M,V ) are related to (u, w,dw φx) in a diﬀerent way, as gfxz=+ x discussed in the following paragraphs. dx The stresses in the beam are fx dw x - du dφ dx σ (x, z) = E(x, z)ε (x, z) = E + z x , xx xx dx dx u (2.5.24) w dw dw σxz(x, z) = 2G(x, z)εxz(x, z) = G(x, z) φx + .- z dx dx

Note that the shear strain (hence, shear force) is not zero, removing the inconsistency of the Bernoulli–Euler beam theory. The stress resultants (N,M) in the Timoshenko beam theory have the form Z du dφx N(x) = σxx dA = Axx + Bxx , (2.5.25) A dx dx Z du dφx M(x) = zσxx dA = Bxx + Dxx , (2.5.26) A dx dx Z dw V (x) = Ksσxz dA = KsSxz φx + , (2.5.27) A dx

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where Ks is the shear correction coeﬃcient introduced to correct the energy loss 4 due to the constant state of transverse shear stress σxz and Sxz is the shear stiﬀness Z Sxz = G(x, z) dA (2.5.28) A Inverting Eqs. (2.5.25)–(2.5.27), we obtain

du DxxN − BxxM dφx AxxM − BxxN dw V = 2 , = 2 , φx + = . dx AxxDxx − Bxx dx AxxDxx − Bxx dx KsSxz (2.5.29) Substituting for du/dx, dφx/dx, and φx + dw/dx from Eq. (2.5.29) into the expressions for σxx and σxz in Eq. (2.5.24), we obtain E σxx(x, z) = 2 [DxxN − BxxM + z (AxxM − BxxN)] , AxxDxx − B xx (2.5.30) GV σxz(x, z) = . KsSxz

When E is not a function of z, we obtain (Axx = EA, Bxx = 0, and Dxx = EI) N Mz V σxx(x, z) = + , σxz(x) = . (2.5.31) A I KsA and the equations of equilibrium in Eqs. (2.5.3)–(2.5.5) in terms of the generalized displacements (u, w, φx) take the form d du − EA = f, (2.5.32) dx dx d dw − K GA + φ = q, (2.5.33) dx s dx x dw d dφ K GA + φ − EI x = 0. (2.5.34) s dx x dx dx

As in the case of the Bernoulli–Euler beam theory, the axial deformation of beams can be determined by solving Eq. (2.5.32) independent of the bending equations, Eqs. (2.5.33) and (2.5.34). Example 2.5.1 illustrates several ideas from this chapter.

Example 2.5.1

Consider the problem of an isotropic cantilever beam of rectangular cross section 2h×b (height 2h and width b), bent by an upward transverse load F0 applied at the free end, as shown in

4We note that the shear stress in the Timoshenko beam theory is not a function of z, implying that it has a constant value at any section of the beam, violating the condition that it be zero at the top and bottom of the beam.

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2.5. THEORIES OF STRAIGHT BEAMS 79

Fig. 2.5.5. Assume that the beam is isotropic, linearly elastic, and homogeneous. (a) Compute the stresses using Eqs. (2.5.16) and (2.5.19), (b) compute the strains using the uniaxial strain– Figurestress 2.5.5 relations, (c) show that the strains computed satisfy the compatibility condition in the x1x3-plane, and (d) determine the two-dimensional displacement ﬁeld (u1, u2) that satisﬁes the geometric boundary conditions. Solution: (a) The stresses in the beam are Mx F x x VQ F σ = 3 = − 0 1 3 , σ = = − 0 (h2 − x2). (1) 11 I I 13 Ib 2I 3

where I is the moment of inertia about the x2 axis, 2h is the height of the beam, and b is the width of the beam.

x3 x 3 F0 F0

x2 2h x1 MFx=- 01 b L VF=- 0 x1

Fig. 2.5.5: Cantilever beam bent by a point load, F0.

(b) The strains ε11 and ε13 are known from this section (except for a change of coordinates from Xi to xi); ε11 is the same from the Bernoulli–Euler or the Timoshenko beam theory, and we use the shear strain obtained from the equilibrium conditions: σ M(x )x F x x ε = 11 = 1 3 = − 0 1 3 , 11 E EI EI σ VQ (1 + ν)F ε = 13 = = − 0 (h2 − x2), (2) 13 2G 2IbG 2EI 3 νF x x ε = ε = −νε = 0 1 3 , 22 33 11 EI where ν is the Poisson ratio, E is Young’s modulus, and G is the shear modulus.

(c) Next, we determine if the strains computed are compatible. Substituting εij into the third equation in Eq. (2.3.17), we obtain 0 + 0 = 0. Thus, the strains satisfy the compatibility equations in two dimensions (i.e., in the x1x3-plane). Although the two-dimensional strains are compatible, the three-dimensional strains are not compatible. For example, using the additional strains, ε33 = ε22 = −νε11 and ε12 = ε23 = 0, one can show that all of the equations except the fifth equation in Eq. (2.3.17) are satisfied. We shall seek the two-dimensional displacement field (u1, u3) associated with the x1x3-plane. (d) Integrating the strain-displacement equations, we obtain

2 ∂u1 F0x1x3 F0x1x3 = ε11 = − or u1 = − + f(x3), (3) ∂x1 EI 2EI 2 ∂u3 νF0x1x3 νF0x1x3 = ε33 = or u3 = + g(x1), (4) ∂x3 EI 2EI

where f(x3) and g(x1) are functions of integration. Substituting u1 and u3 into the deﬁnition of 2ε13, we obtain 2 2 ∂u1 ∂u3 F0x1 df νF0x3 dg 2ε13 = + = − + + + . (5) ∂x3 ∂x1 2EI dx3 2EI dx1

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But this must be equal to the shear strain known from Eq. (2):

F0 2 df νF0 2 dg (1 + ν) 2 2 − x1 + + x3 + = − F0(h − x3). (6) 2EI dx3 2EI dx1 EI

Separating the terms that depend only on x1 and those depend only on x3 (the constant term can go with either one), we obtain

2 dg F0 2 (1 + ν)F0h df (2 + ν)F0 2 − + x1 − = − x3. (7) dx1 2EI EI dx3 2EI

Since the left side depends only on x1 and the right side depends only on x3, and yet the equality must hold, it follows that both sides should be equal to a constant, say c0:

2 df (2 + ν)F0 2 dg F0 2 (1 + ν)F0h − x3 = c0, − + x1 − = c0. dx3 2EI dx1 2EI EI Integrating the expressions for f and g, we obtain (2 + ν)F f(x ) = 0 x3 + c x + c , 3 6EI 3 0 3 1 (8) F (1 + ν)F h2 g(x ) = 0 x3 − 0 x − c x + c , 1 6EI 1 EI 1 0 1 2

where c1 and c2 are constants of integration. Thus, the most general form of displacement ﬁeld (u1, u3) that corresponds to the strains in Eq. (2) is given by F (2 + ν)F u (x , x ) = − 0 x2x + 0 x3 + c x + c , 1 1 3 2EI 1 3 6EI 3 0 3 1 (9) (1 + ν)F h2 νF F u (x , x ) = − 0 x + 0 x x2 + 0 x3 − c x + c . 3 1 3 EI 1 2EI 1 3 6EI 1 0 1 2

The constants c0, c1, and c2 are determined using suitable boundary conditions. We impose the following boundary conditions that eliminate rigid-body displacements (that is, rigid-body translation and rigid-body rotation):

1 ∂u ∂u 3 1 u1(L, 0) = 0, u3(L, 0) = 0, Ω13 = − = 0. (10) x1=L,x3=0 2 ∂x1 ∂x3 x1=L,x3=0 Imposing the boundary conditions from Eq. (10) on the displacement ﬁeld in Eq. (9), we obtain u1(L, 0) = 0 ⇒ c1 = 0, (1 + ν)F h2L F L3 u (L, 0) = 0 ⇒ c L − c = − 0 + 0 , 3 0 2 EI 6EI (11) 2 2 ∂u3 ∂u1 F0L (1 + ν)F0h − = 0, ⇒ c0 = − ∂x1 ∂x3 x1=L,x3=0 2EI 2EI Thus, we have

F L2 (1 + ν)F h2 F L3 (1 + ν)F h2L c = 0 − 0 , c = 0, c = 0 + 0 . (12) 0 2EI 2EI 1 2 3EI 2EI Then the ﬁnal displacement ﬁeld in Eq. (9) becomes

F L2x x2 x2 h2 u (x , x ) = 0 3 3 1 − 1 + (2 + ν) 3 − 3(1 + ν) , 1 1 3 6EI L2 L2 L2 (13) F L3 x x2 x3 h2 x u (x , x ) = 0 2 − 3 1 1 − ν 3 + 1 + 3(1 + ν) 1 − 1 . 3 1 3 6EI L L2 L3 L2 L

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In the Euler–Bernoulli beam theory (EBT), where one assumes that L >> 2h and ν = 0, we have u1 = 0, and u3 is given by

F L3 x x3 uEBT(x , x ) = 0 2 − 3 1 + 1 , (14) 3 1 3 6EI L L3

2 while in the Timoshenko beam theory (TBT) we have u1 = 0 [E = 2(1 + ν)G, I = Ah /3, and A = 2bh], and u3 is given by

3 3 TBT F0L x1 x1 F0L x1 u3 (x1, x3) = 2 − 3 + 3 + 1 − . (15) 6EI L L KsGA L

Here Ks denotes the shear correction factor. Thus, the Timoshenko beam theory with shear correction factor of Ks = 4/3 predicts the same maximum deﬂection, u3(0, 0), as the two- dimensional elasticity theory. Both beam theory solutions, in general, are in error compared to the plane elasticity solution (primarily because of the Poisson eﬀect).

2.5.4 The von KármánTheory of Beams 2.5.4.1 Preliminary discussion In the preceding sections on beam theories, we made an assumption of infinitesimal strains to write the linear strain–displacement relations [see Eqs. (2.5.2) and (2.5.23)]. When one assumes the strain are small but the rotation of the transverse normal line is moderate, the strain fields of the Bernoulli–Euler and Timoshenko beam theories will have additional term, which is nonlinear in the transverse deflection w, in the extensional part of the strain. The nonlinear strain is often referred to as the von Kármánstrain. To see how this nonlinear term enters the calculation, we first make certain assumptions concerning the magnitude of the various terms in the Green strain tensor components and compute the nonzero strains for beams. We begin with the assumption that the axial strain (du /dx) and the cur- √1 vature (d2w/dx2) are of order , and (dw/dx) is of order , where << 1 is a small parameter. Then the nonzero Green strain tensor components from Eq. (2.3.11) are (with Xi ≈ xi) 2 ∂u1 1 ∂u3 εxx = + , (2.5.35) ∂x1 2 ∂x1 1 ∂u1 ∂u3 εxz = + . (2.5.36) 2 ∂x3 ∂x1 Thus, the underlined term in Eq. (2.5.35) is new and it is nonlinear. The beam theory resulting from the inclusion of this nonlinear term is termed the von Kármánbeam theory. Here we present the complete development for each of the beam theories considered before. Before we embark on the development of the beam theories, we must consider suitable equations of equilibrium for this nonlinear case. We adopt the

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equations of equilibrium in terms of the second Piola–Kirchhoﬀ stress tensor S. The vector equation of equilibrium in terms of S is [see Eqs. (2.2.14), (2.2.9), and (2.3.7)] T ∇0 · S · (I + ∇0u) + ρ0f = 0, which in component form is ∂ ∂uI δKI + SKJ + ρ0fI = 0, (2.5.37) ∂XJ ∂XK

where fI are body forces measured per unit mass. In the present case, the equilibrium equations associated with balance of forces in the x- and z-directions are ∂ ∂u ∂u ∂ ∂u S + 1 S + 1 S + S + 1 S + f¯ = 0, (2.5.38) ∂x xx ∂x xx ∂z xz ∂z xz ∂x xz x ∂ ∂u ∂ ∂u S + 3 S + 3 S + f¯ = 0, (2.5.39) ∂x xz ∂x xx ∂z ∂x xz z

where f¯x = ρ0f1 and f¯z = ρ0f3 are the body forces measured per unit volume. They are assumed, in the present case, to be only functions of x.

2.5.4.2 The Bernoulli–Euler beam theory The displacement ﬁeld of the Bernoulli–Euler beam theory is given in Eq. (2.5.1). Integration of Eqs. (2.5.38) and (2.5.39) over the beam area of cross section yields

d du d2w dw d2w N + N − M − V − V + f = 0, (2.5.40) dx dx dx2 dx dx2 d dw V + N + q = 0, (2.5.41) dx dx

where f and q are the axially and transversely distributed forces per unit length. We note that Z Z ∂ ∂u1 ∂u1 z Sxz + Sxz dA = − Sxz + Sxz dA, (2.5.42) A ∂z ∂x A ∂x

where we have used the assumption that Sxz = 0 on the surface of the beam. Next, multiply Eq. (2.5.38) with z, integrate over the beam area of cross section, and use the identity in Eq. (2.5.42) to obtain

d du d2w dw du d2w M + M − P − R − V − V + R = 0, (2.5.43) dx dx dx2 x dx x dx dx2 x

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2.5. THEORIES OF STRAIGHT BEAMS 83

where (Px,Rx) are higher-order stress resultants Z Z 2 Px = σxxz dA, Rx = σxzz dA. (2.5.44) A A

Note that area integrals of zfx and zfz vanish because fx and fz are only functions of x and the x-axis coincides with the geometric centroidal axis. Using the order-of-magnitude assumption of various quantities, Eqs. (2.5.40), (2.5.41), and (2.5.43) can be simplified as dN − = f, (2.5.45) dx dV d dw − − N = q, (2.5.46) dx dx dx dM − + V = 0, (2.5.47) dx where the underlined expression is the only new term due to the von Kármán nonlinearity when compared to Eqs. (2.5.3)–(2.5.5). Of course, the stress re- sultant N will also contain nonlinear term when it is expressed in terms of the displacements. Since in the Bernoulli–Euler beam theory V cannot be determined independently, we use V = dM/dx from Eq. (2.5.47) and substitute for V into Eq. (2.5.46) the following equations of equilibrium for the Bernoulli–Euler beam theory: dN − = f (2.5.48) dx d2M d dw − − N = q. (2.5.49) dx2 dx dx In view of the displacement field in Eq. (2.5.1), we obtain the following simplified Green strain tensor component du d2w 1 dw 2 E ≈ ε = − z + ≡ ε(0) + zε(1), γ = 0, (2.5.50) xx xx dx dx2 2 dx xx xx xz where du 1 dw 2 d2w ε(0) = + , ε(1) = − , (2.5.51) xx dx 2 dx xx dx2 and the stress resultants (N,M) can be expressed in terms of the displacements (u, w) (when linear stress–strain relations are used) as [see Eqs. (2.5.8) and(2.5.9)] " # Z du 1 dw 2 d2w N(x) = σxx dA = Axx + + Bxx − 2 , (2.5.52) A dx 2 dx dx " # Z du 1 dw 2 d2w M(x) = zσxx dA = Bxx + + Dxx − 2 . (2.5.53) A dx 2 dx dx

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Relations in Eq. (2.5.13) take the form 2 2 du 1 dw DxxN − BxxM d w AxxM − BxxN + = 2 , − 2 = 2 . (2.5.54) dx 2 dx AxxDxx − Bxx dx AxxDxx − Bxx Consequently, Eq. (2.5.14) and (2.5.16) remain unchanged.

2.5.4.3 The Timoshenko beam theory Using the displacement field in Eq. (2.5.22) and following the procedure as in Eqs. (2.5.40)–(2.5.43), we obtain d du dφ dφ N + N + x M + φ V + x V + f = 0, (2.5.55) dx dx dx x dx d dw V + N + q = 0, (2.5.56) dx dx d du dφ du dφ M + M + x P + φ R − V − V − x R = 0. (2.5.57) dx dx dx x x x dx dx x Using the order-of-magnitude assumption of various quantities, Eqs. (2.5.55)– (2.5.57) are simplified to those listed in Eqs. (2.5.45)–(2.5.47). The simplified Green strain tensor components of the Timoshenko beam theory are du dφ 1 dw 2 dw E ≈ ε = + z x + ≡ ε(0) + zε(1), γ = φ + (2.5.58) xx xx dx dx 2 dx xx xx xz x dx where du 1 dw 2 dφ ε(0) = + , ε(1) = x . (2.5.59) xx dx 2 dx xx dx The stress resultants (N,M,V ) are known in terms of the generalized displacements (u, w, φx) as Z " 2# du 1 dw dφx N(x) = σxx dA = Axx + + Bxx , (2.5.60) A dx 2 dx dx Z " 2# du 1 dw dφx M(x) = zσxx dA = Bxx + + Dxx , (2.5.61) A dx 2 dx dx Z dw V (x) = Ks σxz dA = KsSxz φx + . (2.5.62) A dx Relations in Eq. (2.5.29) take the form 2 du 1 dw DxxN − BxxM + = 2 , dx 2 dx AxxDxx − B xx (2.5.63) dφx AxxM − BxxN dw V = 2 , φx + = . dx AxxDxx − Bxx dx KsSxz

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2.6. SUMMARY 85

Therefore, Eqs. (2.5.30) and (2.5.31) remain unchanged. This completes the development of the Bernoulli–Euler and Timoshenko beam theories with the von Kármánnonlinear strain. The equations governing bars can be obtained as a special case from the equations of beams by setting bending related quantities to zero (i.e., w = 0, φx = 0, M = 0, V = 0, and q = 0). Thus, the kinematic, constitutive, and equilibrium equations of bars (written in terms of forces as well as displacements) are du du ε = , σ = Eε ,N = EA xx dx xx xx dx (2.5.64) dN d du − + f = 0 ⇒ − EA + f = 0 , 0 < x < L dx dx dx Here u denotes the axial displacement and f is the distributed axial force. The governing equations for cables (also termed rods, ropes, strings, and so on) in a plane are the same as those for bars, with the exception that axial stiffness EA is replaced with the tension T (x) in the cable. Thus, we have

d du − T + f = 0 , 0 < x < L, (2.5.65) dx dx

where u is the transverse displacement and f(x) is the distributed force transverse to the cable.

2.6 Summary In this chapter a summary of the equations of elasticity are presented. These in- clude the equations of motion, strain–displacement equations, and constitutive (or stress–strain) relations. The stress and strain transformations and compatibility conditions for strains are also discussed. Also, two beam theories, namely the Bernoulli–Euler and Timoshenko beam theories, are also fully developed in this chapter. Finally, the von K´arm´annonlinear beam theories using the Bernoulli–Euler and Timoshenko kinematics are developed. The equations governing the two beam theories will be used extensively in the coming chapters. The main equations of this chapter for linearized elasticity and bending of beams are summarized here. Equations of motion [Eqs. (2.2.7) and (2.2.8)]

2 2 ∂ u ∂σji ∂ ui ∇ · σ + ρf = ρ 2 + ρfi = ρ 2 . (2.6.1) ∂t ∂xj ∂t Strain–displacement relations [Eq. (2.3.12)] 1 h Ti 1 ∂ui ∂uj ε = 2 (∇u) + (∇u) εij = 2 + . (2.6.2) ∂xj ∂xi

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Strain compatibility equations [Eqs. (2.3.18) and (2.3.19)]

∂2ε ∂2ε ∂2ε ∂2ε ∇ × (∇ × ε)T = 0 ij + k` = `j + ki . (2.6.3) ∂xk∂x` ∂xi∂xj ∂xk∂xi ∂x`∂xj Stress–strain relations [modiﬁed Eqs. (2.4.11) and (2.4.12) to account for thermal strains; see Problem 2.39]

E νE Eα σ = 1+ν ε + (1+ν)(1−2ν) tr(ε) I − 1−2ν (T − T0) I (2.6.4) E νE Eα σij = 1+ν εij + (1+ν)(1−2ν) εkk δij − 1−2ν (T − T0)δij .

Equilibrium equations of the Bernoulli–Euler beam theory [Eqs. (2.5.48) and (2.5.49); set the nonlinear terms (underlined) to obtain the linear equations]

dN d2M d dw − = f, − − N = q. (2.6.5) dx dx2 dx dx

Kinematic relations of the Bernoulli–Euler beam theory [Eq. (2.5.50)]

du 1 dw 2 d2w ε = + − z , γ = 0. (2.6.6) xx dx 2 dx dx2 xz

Constitutive relations of the homogeneous Bernoulli–Euler beam theory [Eqs. (2.5.52) and (2.5.53)] " # du 1 dw 2 d2w N = EA + ,M = −EI . (2.6.7) dx 2 dx dx2

Equilibrium equations of the Timoshenko beam theory [Eqs. (2.5.45)– (2.5.47); set the nonlinear terms (underlined) to obtain the linear equations]

dN dV d dw dM − = f, − − N = q, − + V = 0. (2.6.8) dx dx dx dx dx

Kinematic relations of the Timoshenko beam theory [Eqs. (2.5.58) and (2.5.59)]

du 1 dw 2 dφ dw ε = + + z x , γ = φ + . (2.6.9) xx dx 2 dx dx xz x dx

Constitutive relations of the homogeneous Timoshenko beam theory [Eqs. (2.5.60)– (2.5.62)] " # du 1 dw 2 dφ dw N = EA + ,M = EI x ,V = GAK φ + . dx 2 dx dx s x dx (2.6.10)

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2.6. SUMMARY 87

Table 2.6.1 contains a summary of governing equations of homogeneous, isotropic, and linear elastic bodies in terms of displacements (i.e., equilibrium equations, kinematics relations, and constitutive equations are combined to express the governing equations in terms of displacements). The nonlinear equilibrium equations of beams with the von K´arm´annonlinearity can be expressed in terms of displacements with the help of equations presented in Section 2.5. In the later chapters of this book we shall make reference to the equations developed in this chapter and summarized in this section. Some of the equations developed here may be re-derived using the energy approach in Chapters 3 through 5.

Table 2.6.1: Summary of equations of elasticity, membranes, cables, bars, and beams.

1. Linearized Elasticity (µ and λ are Lam´econstants, u is the displacement vector, and f is the body force vector measured per unit volume)

µ∇2u + (λ + µ)∇(∇ · u) + f = 0 in Ω. (2.6.11)

2. Membranes (a11 and a22 are tensions in the x- and y-coordinate directions, respectively, u is the transverse deﬂection, and f is the distributed transverse force measured per unit area)

∂ ∂u ∂ ∂u − a − a = f in Ω. (2.6.12) ∂x 11 ∂x ∂y 22 ∂y

3. Cables (T is the tension in the cable, u is the transverse deﬂection, and f is the distributed transverse load) d du − T = f, 0 < x < L. (2.6.13) dx dx

4. Bars (E is Young’s modulus, A is the area of cross section, u is the axial displacement, and f is the distributed axial force)

d du − EA = f, 0 < x < L. (2.6.14) dx dx

5. Bernoulli–Euler beam theory (E is Young’s modulus; I is the moment of inertia; w is the transverse displacement, and q is the distributed load)

d2 d2w EI = q, 0 < x < L (2.6.15) dx2 dx2

6. Timoshenko beam theory (E is Young’s modulus, G is the shear modulus, A is the area of cross section, Ks shear correction coeﬃcients, I is the moment of inertia; w is the transverse displacement, φx rotation of a transverse normal line, and q is the distributed load)

d dφ dw − EI x + GAK φ + = 0 dx dx s x dx (2.6.16) d dw − GAK φ + = q dx s x dx

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