Lecture 3: The Concept of Stress, Generalized Stresses and Equilibrium

Structural Mechanics 2.080 Lecture 3 Semester Yr Lecture 3: The Concept of Stress, Generalized Stresses and Equilibrium 3.1 Stress Tensor We start with the presentation of simple concepts in one and two dimensions before introducing a general concept of the stress tensor. Consider a prismatic bar of a square cross-section subjected to a tensile force F , F θ F ⇡ 2 0 - T1 T1 1 ✓ 2 3 Figure 3.1: A long bar with three different cuts at θ, θ = 0 and π/2 − θ. The force per unit area is called the surface traction T :   force F N T =σ= = area Ao mm2 (3.1) In the uniaxial case, the surface fraction is the only component of the stress tensor in the global coordinate system, commonly referred to as σ. How can one apply a force to the end section of a bar? This can be done in a number of different ways (see Fig. (3.2)). A pin connection can be glued (or welded) to the end section, or a hole can be drilled through the bar to attach a pin. Or an internal or external thread can be machined. Finally, axial force could be applied through frictional or mechanical grips. Except the welded or glued connector, a complex state of stress is created near the bar ends where the stress state is multi-axial. Such stress states is confined to a relatively short segment of the bar comparable with the height or diameter of the bar. Along this section a gradual transition takes place from the multi-axial state of stress to the uniaxial state, for which Eq.(3.1) holds. The above example an serve as a practical application of the Saint-Venant’s principle (1856). This principle named after the French elasticity theorist, Jean Claude Barre’ de Saint-Venant can be stated as: “the difference between the effects of two different but statically equivalent loads become very small at sufficiently large distances from load.” 3-1 Structural Mechanics 2.080 Lecture 3 Threaded Semester Yr Weld Hole for a pin Wedge grips Figure 3.2: How to apply tension to the end of a bar. Think what are the “two” equivalent loads that are applied to the bar ends? We usually think of a cross-section being  π cut perpendicular to the axis of the bar. Consider now two − θ to the normal direction. The planes are defined by the cuts at the angles θ and 2 unit normal vector n. Cut A-A F Cut B-B θ θ θ Figure 3.3: Normal and tangential forces acting on the slant section of the bar. From the free body diagram, the components of the normal and tangential forces: FN = F cos θ π Fn = F cos( − θ) 2 FT = F sin θ π Ft = F sin( − θ) 2 The slant cross-section A is larger and is related to the reference cross-section by π  Ao = AA cos θ, Ao = AB cos −θ 2 (3.2a) (3.2b) (3.2c) (3.2d) (3.3) Consider now a unit volume cubic element located at the intersections of cuts A-A and B-B, Fig.(3.4). 3-2 Structural Mechanics 2.080 Lecture 3 Semester Yr θ Figure 3.4: The volume element with surface traction acting on two adjacent facets. The surface traction (force per unit area) on the two perpendicular facets are Facet parallel to A-A: Tn = T cos2 θ (3.4a) (3.4b) Facet parallel to B-B: Tt = T sin θ cos θ π  Tn = T cos2 −θ  π2  π  Tt = T sin − θ cos −θ 2 2 (3.4c) (3.4d) It can be observed that the tangential components of the surface traction vector on A-A and B-B cuts are identical. The normalized plots of the above quantities versus the orientation angle of the cross-section are shown in Fig.(3.5). 1.0 Tn T AA B T n T B- 0.5 Tt T 0 ⇡ 4 ⇡ 2 Figure 3.5: Relative values of normal and shear components of the surface traction as a function of the orientation of the cut. It can be noted that the tangential component attains maximum at 45◦ . This means that if the material fails due to shear loading, the fracture surface will always be oriented at 45◦ . The above example teaches us that there are infinite combinations of normal and tangential components of surface tractions which are in equilibrium with the applied load. For each orientation of the cross-section there is a different pair of {Tn , Tt }. The orientation of the surface element is uniquely defined by the unit normal vector n{n1 , n2 , n3 }. At the 3-3 Structural Mechanics 2.080 Lecture 3 Semester Yr same time the components of the surface traction vector acting on the same element are T {T1 , T2 , T3 }. The components of the surface traction vector acting on this surface element are T {T1 , T2 , T3 }. For example, the orientation of facets of the unit material cube is shown in Fig.(3.6). 3 {0,0,1} {0,1,0} {1,0,0} 2 1 Figure 3.6: Components of the unit normal vector on facets of a unit cube. The relation between the vectors of surface tractions, unit normal vector defining the surface element and the stress tensor are given by the famous Cauchy formula Ti = Tij nj (3.5) T1 = σ1j nj = σ11 n1 + σ12 n2 + σ13 n3 (3.6a) T2 = σ2j nj = σ21 n1 + σ22 n2 + σ23 n3 (3.6b) T3 = σ3j nj = σ31 n1 + σ32 n2 + σ33 n3 (3.6c) or in the expanded notation, To a large extent the Cauchy relation is analogous to the strain-displacement relation put in the form of Eq.(3.2). dui = Fij dxj (3.7) The displacement gradient Fij transforms the increment of the length element dxj into the increment of displacement dui . In the same way the stress tensor transforms the orientation of the surface element n into the surface traction acting on this element. In order to get a physical interpretation of the concept of the stress tensor, let us see how the Cauchy formula works in the case of one and two-dimensional problems of the axially loaded bar. Consider first the normal cut of the bar with the longitudinal axis as 3-4 Structural Mechanics 2.080 Lecture 3 Semester Yr 1-axis. The components of the surface tractions are given in Fig.(3.7). The corresponding F . components of the unit normal vector were defined in Fig. (3.6), where T = Ao 3 T{0,0,1} T{0,1,0} 2 T{1,0,0} 1 Figure 3.7: The unit volume element aligned with the axis of the bar. Substituting the values of the components of the two vectors into Eq.(3.6b) one gets the following expressions: Facet(1, 0, 0) Facet(0, 1, 0) Facet(0, 0, 1) T1 = σ11 0 = σ12 0 = σ13 0 = σ22 0 = σ23 0 = σ21 0 = σ31 0 = σ32 0 = σ33 (3.8) Therefore the components of the stress 3 × 3 matrix in the global coordinate system are σ= T 0 0 0 0 0 0 0 0 (3.9) This is the uniaxial state of stress. The two-dimensional example of the slant cut is much more interesting. This time a local coordinate system, rotated with respect to the 3-axis will be used. In this system the components n are the same as in the global system. The components of the surface traction vector on three facets, calculated in Eq.(3.4b) are defined in Fig.(3.8). 3-5 Structural Mechanics 2.080 Lecture 3 3 Semester Yr T{0,0,1} T{Tsin2θ,Tsinθcosθ,0} T{Tcos2θ,Tsinθcosθ,0} 2' 2 1 1' Figure 3.8: Components of the surface tractions on the rotated volume element. Substituting the above values into the Cauchy formula we obtain Facet(1, 0, 0) Facet(0, 1, 0) Facet(0, 0, 1) 2 T cos θ = σ11 T sin θ cos θ = σ12 0 = σ13 T sin2 θ = σ22 0 = σ23 T sin θ cos θ = σ21 0 = σ31 0 = σ32 0 = σ33 (3.10) The plane stress components of the stress tensor are σ= T cosθ T sin θ cos θ 0 T sin θ cos θ T sin2 θ 0 0 0 0 (3.11) It is interesting that the matrices Eq.(3.9) and Eq.(3.11) represent the same state of stress seen in two coordinate systems rotated with respect to one another. The transformation of the stress tensor from one coordinate system to the other is the subject Recitation 1 where the relation between Eq.(3.9) and Eq.(3.11) will be derived in a different way. Symmetry of the stress tensor It should also be noted from Eq.(3.11) that stress tensor is symmetric meaning that σ12 = σ21 . The symmetry of the stress tensor comes from the moment equilibrium equation of are infinitesimal volume element. In general σij = σji 3-6 (3.12) Structural Mechanics 2.080 Lecture 3 Semester Yr The symmetry of the stress tensor reduce the nine components of the 3 × 3 metric to only six independent components. The meaning of the two subscripts of the stress tensor is explained below σ?? The first subscript defines the plane on which the surface tractions are acting. For example “1” denotes the surface element perpendicular to the axis x1 . The second subscript indicates direction of a particular component of the surface tractions. This convention is explained in Fig.(3.9). 3 σ33 σ32 σ31 σ23 σ13 σ22 σ12 σ22 σ21 σ11 2 1 Figure 3.9: Components of the stress tensor on three facets of the infinitesimal surface element. Sign convention The Cauchy formula can also be consistently used to determine the sign of the components of the stress tensor. The point is that the sign of the components of the vectors is known from the chosen coordinate system. For illustration, let us orient the volume element along the x1 axis. With positive direction to the right. T1 n1 = -1 T1+ + 11 x2 n1 = 1 x1 Figure 3.10: Explanation of the sign convention of the stress tensor. 3-7 + 11 Structural Mechanics 2.080 Lecture 3 Semester Yr From the Cauchy formula T1 = σ11 n1 (3.13) On the right facet both the surface traction and the unit normal vector is positive and so must be the normal component of the stress tensor σ11 . On the left facet both T1 and to the x1 axis. In order for Eq.(3.13) to hold the component σ11 must be positive, even if its visualization points out in the negative direction. in the above example the stress state is uniform along the x1 axis. This is the case of a bar under tension. In general there is a gradient of the components of the stress tensor so that stresses on both sides of the infinitesimal element differ by a small amount of dσ11 . The sign convention is opening the way for deriving the equations of equilibrium for the 3-D continuum. This topic is the subject of the next section. 3-D Equilibrium The equilibrium equation for an infinitesimal volume element are derived first using two methods. Referring to Fig.(3.9), indicated on Fig.(3.11) are only those components of the stress tensor that are directed along x2 axis. These are σ12 , σ22 and σ32 . 3 32 + @ 32 dx3 @x3 σ12 σ22 12 + @ 12 dx1 @x1 + @ 22 dx2 @x2 2 dx1 σ32 1 22 dx2 Figure 3.11: All components of the stress tensor contributing to the force equilibrium in x2 direction must be in equilibrium. According to Newton’s law, the sum of all forces (stress times the surface area) acting 3-8 Structural Mechanics 2.080 Lecture 3 Semester Yr along x2 must be zero     ∂σ12 ∂σ22 dx2 dx1 dx3 − σ2 dx1 dx3 + σ12 + dx1 dx2 dx3 − σ12 dx2 dx3 σ22 + ∂x2 ∂x1   ∂σ32 + σ32 + dx3 dx1 dx2 − σ32 dx1 dx2 + B2 dx1 dx2 dx3 = 0 ∂x3 (3.14) For generality, the body force (force per unit volume) was included as well. The body force represent for example gravity force B = ρg or d’Alambert inertia force B = ρü so that the derivation is valid both for static and dynamic problems. Summing up the forces one gets the first equilibrium equation ∂σ22 ∂σ12 ∂σ32 + + + B2 = 0 ∂x2 ∂x1 ∂x3 (3.15) ∂σj2 + B2 = 0 → σj2,j + B2 = 0 ∂xj (3.16) Invoking the index notation with the summation and coma convention. A similar procedure of summing-up forces can be repeated in the x1 and x3 direction, yielding two additional equations of equilibrium. One can immediately notice that by replacing the life subscripts “2” in Eq.(3.16) respectively by “1” and “3”, the final compact form of the equation of equilibrium reads σij,j + Bi = 0 or ∂σij + Bi = 0 ∂xj (3.17) In the expanded notation and replacing xi by (x, y, z), the familiar form of the equilibrium equation is ∂σxz ∂σxx ∂σxy + + + Bx = 0 ∂x ∂y ∂z ∂σyx ∂σyy ∂σyz + + + By = 0 ∂x ∂y ∂z ∂σzx ∂σzy ∂σzz + + + Bz = 0 ∂x ∂y ∂z (3.18a) (3.18b) (3.18c) The plane stress case, prevailing in thin plate and shells is defined by σ3j = 0 or σ31 = σ32 = σ33 = 0 (3.19) In other words all components of the stress tensor pointing out in the z-directions are zero, σzz = σzx = σzy = 0. The components of the plane stress tensor are highlighted by the framed area, thus σ is equal to 3-9 Structural Mechanics 2.080 Lecture 3  Semester Yr σxx σxy σxz     σyx σyy σyz   σzx σzy σzz      For plane stress, the subscripts run only over two dimensions and the Greek letters are commonly used, α, β = 1, 2. In the compact notation, the plane stress equilibrium equation reads σαβ,β + Bα = 0 (3.20) In the uniaxial case only one component of the equilibrium survives, giving dσxx +B =0 dx (3.21) With no body force, B = 0, Eq.(3.21) predicts a constant stress along the length of the bar. The addition of the d’Alambert inertia force will lead to the one-dimensional wave equation. 3-10 Structural Mechanics 2.080 Lecture 3 Semester Yr ADVANCED TOPIC 3.2 Local Equilibrium from the Principle of Virtual Work The derivation of the local equation of equilibrium from the global principle of virtual work is an elegant method in continuum and structural mechanics. This procedure also formulates static and kinematic boundary condition. There are two mathematical tools involved. One is the divergence theorem (Gauss-Green identity) and the other one is the concept of the calculus of variation. The Gauss theorem transforms the volume integral into a surface integral Z Z ∂Ai Ai,i dV = Ai ni dS, Ai,i = (3.22) ∂xi V S where Ai is a vector and ni is the unit normal vector of the surface element dS. In the simplest one-dimensional case, Eq.(3.22) is reduced to Z x2 x1 dA dx = A|xx21 = A(x2 ) − A(x1 ) dx (3.23) Starting from the definition of the infinitesimal strain given by Eq.(2.14), the increments of the strain tensor and displacement vector are also linearly related 1 δij = (δui,j + δuj,i ) 2 (3.24) There is a fine difference between the symbol δu and du, which is explained in Fig.(3.12). dx u u x1 x u δu(x) x2 du Figure 3.12: The local increment δu over the infinitesimal length dx and the global small (virtual) displacement from the equilibrium configuration satisfying kinematic boundary conditions. Both are linear operators and the rule for differentiations are the same. The principle of virtual work states that the incremental work of strains on the stresses over the volume of the body must be equal to the work of surface tractions in the incremental displacements over the surface of the body. Fig.(3.13) helps to visualize the notation. A part of the surface on which the displacement are zero δui = 0 is denoted by SU . The remainder of the surface S − SU is denoted by ST . Mathematically the principle of virtual work states Z Z σij δij dV = Ti δui dS (3.25) V S where δij are calculated from δui using Eq.(3.24). The one-dimensional graphical interpretation of the principle is shown in Fig.(3.14). 3-11 Structural Mechanics 2.080 Lecture 3 Ti Semester Yr δui S σij, δεij V Figure 3.13: The 3-D potato (body) subjected to stress and displacement boundary condition develops internal stresses and incremental displacements. σ T σ T ε δε ε u δu u Figure 3.14: Incremental internal and external energies. The integrand of the left hand side of Eq.(3.25) can be transformed to a simpler form using the symmetry property of the stress tensor σij = σji 1 1 σij δui,j + σji δuj,i = σij δui,j 2 2 (3.26) Recall an elementary rule of differentiation of the product of two functions (ab)0 = a0 b + ab0 (3.27) which in application to our problem reads σij (δui ),j = (σij δui ),j − (σij ),j δui Now, the left hand side of Eq.(3.25) is transformed to Z Z Z σij δij dV = (σij δui ),j dV − (σij ),j δui dV V V | {z } V (3.28) (3.29) Ai The first volume integral is now transformed to the surface integral according to Eq.(3.22). Substituting this result into the statement of virtual work one gets Z Z Z σij δui nj dS − σij,j δui dV = Ti δui dS (3.30) S V S 3-12 Structural Mechanics 2.080 Lecture 3 Combining the two surface integrals into one integral we finally arrive at Z Z σij,j δui dV = 0 (σij ni − Ti )δui dS − Semester Yr (3.31) V S The first integral vanishes when either σij nj − Ti = 0 on ST or δui = 0 on SU (3.32a) (3.32b) The above equations represent respectively the stress and displacement boundary condition. The meaning of second integral should be interpreted in the spirit of the first lemma of the calculus of variation. The increment of the displacement vector δui can not vanish over the whole volume of the body because this would mean rigid body motion. The point is that the second integral in Eq.(3.28) must be zero not for one particular shape of δui but for all possible variation of the displacement field, as shown in Fig.(3.12). Thus, the calculus of variation tell us that this is possible only when σij,j = 0 in V (3.33) The principle of virtual work is often called the weak (global) statement of equilibrium while Eq.(3.33) is the local equation of equilibrium but is not called strong. The weak statement of equilibrium is a starting point of developing most approximate methods in continuum and structural mechanics such as eigenvalue expansion, finite difference or finite element method. The critical assumption of the first lemma of the calculus of variation is that an infinity of different virtual velocities are considered. This is achieved by considering a large but finite degrees of freedom through many terms in the eigenvalue expansion or many discrete elements. Through this assumption the equivalence of the global and local formulation is achieved. An alternative form of the principle of virtual work, extensively in plasticity theory is the principle of virtual velocity. By observing that δu = du δt = u̇δt dt (3.34) Eq.(3.25) is transformed to Z Z σij ˙ij dV = V Ti u̇i dS (3.35) S where ˙ij is the instantaneous velocity field obtained from the incremental velocities u̇i through the linear geometric relation, Eq.(3.25). Generalized Stresses This concept is introduced in order to reduce the two-dimensional problem in x and z in beams to one-dimensional problem in x, governed by an ordinary differential equation. 3-13 Structural Mechanics 2.080 Lecture 3 Semester Yr At an arbitrary cross-section in a beam one can distinguish a vector of bending moments {Mx , My , T } where Mx is bending the beam in the (x, z) plane, My and T is the torque. (Do not confuse torque with surface traction vector). The meaning of the moment vector is explained in Fig.(3.15). T N My Mx Vy Vx Figure 3.15: Imagine short shafts rotating a rigid slice of the beam. These are components of the bending moment vector. The components of the force vector acting at an arbitrary cross-section are {Vx , Vy , N } is the axial (membrane) force. In planar bending of a beam only three out of six components of the generalized stress resultants survival. They are defined by Z def Mx = M = σxx z dA [Nm] (3.36a) ZA def N = σxx dA [N] (3.36b) A Z def Vx = V = σxz dA [N] (3.36c) A The product σxx dA in Eq.(3.36a) is the incremental force. Multiplying this force by the “arm” z from the beam bending axis gives the incremental bounding moment dM = (σxx dA)z. The total bending moment is an integral of dM over the beam cross-section. The sign of the generalized quantities is decided by the sign of the stress. M M dz zz+ x σ+ z z Figure 3.16: The bending moment in a “smiling” beam is positive. Imagine that the beam is bent in the way shown is Fig.(3.16). On the tensile side of the beam the stress is positive, σ + and so is the distance from the beam axis. On the compressive side both the stress and force arms are negative σ − z − , but the product is positive. Therefore the tensile and compressive side of the beam contribute to the positive 3-14 Structural Mechanics 2.080 Lecture 3 Semester Yr bending moment. The beam (or its portion) where the bending moment is negative is called the “smiling beam”. Therefore looking at the deformed shape of the beam one can determine immediately the sign of the bending moment. The sign of the axial and shear force can be easily determined from Fig.(3.17). M+ M+ V+ N+ V+ Figure 3.17: Positive shear and normal stresses in a beam. END OF ADVANCED TOPIC 3-15 N+ Structural Mechanics 3.3 2.080 Lecture 3 Semester Yr Generalized Forces and Bending Moments in Plates In plates there are three in-plane components of the stress tensor σαβ {σxx , σyy , σxy }. Replacing σxx by σαβ or σzα in Eqs.(3.36a-3.36c) the generalized forces and couples are defined Z Mαβ = −h 2 Z Nαβ = h 2 −h 2 Z Vα = h 2 h 2 −h 2 σαβ z dz [Nm/m] = [N] (3.37a) σαβ dz [N/m] (3.37b) σzα dz [N/m] (3.37c) Note that in the plate theory the integration is performed over the thickness of the plate rather than the entire surface. Therefore the dimensions of the quantities defined by Eqs.(3.37a-3.37c) are “per unit length”. 3-16 Structural Mechanics 2.080 Lecture 3 Semester Yr ADVANCED TOPIC 3.4 Principle of Virtual Work for Beams This principle can be derived directly from the general 3-D principle, Eq.(3.24) assuming one-dimensional stress state and kinematic assumption of the elementary beam theory σij → σxx δij → δxx = δ◦ (x) + zδκ from Eq.(2.44) (3.38) dV = dAdx, 0 < x < l The left hand (LH) side of Eq.(3.25) becomes  Z Z l Z ◦ LH = σij δij dV = [σxx δ (x)dA + σxx zδκdA] dx V 0 (3.39) A Both δ◦ (x) and δκ(x) are extension and curvature of the beam axis and are constant with respect to integration over the area. The above equation can be further simplified  Z Z Z l ◦ δ (x) σxx dA + δκ(x) σxx zdA dx (3.40) LH = A 0 A Recalling the definition of the axial force, Eq.(3.36b) and the bending moment, Eq.(3.36a), the final expression for the virtual work inside the volume of the beam takes this simple form Z l LH = (N δ◦ + M δκ) dx (3.41) 0 where l is the length of the beam. Evaluation of the right hand side (RH) of Eq.(3.25) is more interesting. dx T1 = σxx P dx δw δux dl dS T3 = σxz δw Figure 3.18: The outer surface of the beam consists of two parts: the lateral surface SL on which the surface traction are acting and the end cuts A. Note that all points on a slice of the beam move downward with the virtual displacement δw. The end cuts translate and rotate, according to Eq.(2.35). Then the right hand side of Eq.(3.25) becomes Z l Z Z ◦ RH = qδw dx + σxx [δu − δθz] dA + σxz δw dA (3.42) 0 A A 3-17 Structural Mechanics 2.080 Lecture 3 Semester Yr where q is the integrated pressure over the circumference of a slice I q = Ti Vi ds (3.43) and Vi are direction cosine of the surface traction vector with respect to the z-axis. In the case of the rectangular section (h × b), Eq.(3.43) reduces to q = pb (3.44) where p is the distributed pressure on the upper side of the beam and q is called the line load. The second term in Eq.(3.42) can be simplified using the definitions Eqs.(3.36a-3.36c) Z M= σxx z dA (3.45a) σxx dA (3.45b) σxz dz (3.45c) ZAend N= Aend Z V = Aend where the bar over the symbol indicates that this is the value at the beam end. The final expression for the principle of virtual work for a beam takes the form Z l ◦ Z (N δ + M δκ) dx = 0 0 l q(x)δw dx + N δu◦ − M δθ + V δw (3.46) The above principle will be used to derive approximate solutions of the beam problems and also to obtain the equations of equilibrium and boundary conditions. END OF ADVANCED TOPIC 3-18 Structural Mechanics 3.5 2.080 Lecture 3 Semester Yr Derivation of Equation of Equilibrium for Beams from the Principle of Virtual Work The needed mathematical apparatus is the integration by parts. The starting point in Eq.(3.27) which is put in an alternative form da d db b= (ab) − a dx dx dx (3.47) Integrating both sides of the above equation on gets Z da b dx = ab|ends − dx Z a db dx dx (3.48) To simplify the notation the “prime” convention will be used throughout d[] d2 [] = []0 ; = []00 dx dx2 (3.49) We turn now the left hand side of the principle of virtual work, Eq.(3.14) and recall the definition of beam curvature and axial strain κ = −w00 (3.50a) ◦ = u0 (3.50b) The virtual increments are δκ = −δw00 = (δw0 )0 ◦ (3.51a) 0 δ = δu (3.51b) Substituting Eq.(3.51b) into the LH side of Eq.(3.14) and integrating twice by parts we get Z l Z l M (δw ) dx + N δu0 dx 0 0     Z l Z l l 0 l 0 0 0 = − M δw 0 − M δw dx + N δu 0 − N δu dx LH = − 0 0 0 = l −M δw0 0 0 + M δw (3.52) 0 l 0 − l Z 0 00 l 0 M δw dx + N δu − Z l N 0 δu dx 0 The second term represents the work increment at the beam end on downward virtual displacement. Therefore the corresponding generalized force must be the shear force V V = M0 Introducing Eq.(3.52) into Eq.(3.46) and grouping the terms yields Z l Z l l l 00 (M + q)δw dx + N 0 δu dx + (M − M )δw0 0 − (N − N )δu 0 − (V − V )δw 0 0 3-19 (3.53) l 0 = 0 (3.54) Structural Mechanics 2.080 Lecture 3 Semester Yr The above equation should hold not for one specific incremental displacement but for arbitrary variations (δw, δw0 , δu), independent inside 0 < x < l and on the boundaries. Therefore by the first lemma of the calculus of variation, the local (strong) form of the beam equilibrium follows M 00 + q = 0 (3.55a) N0 = 0 (3.55b) along with the boundary conditions (M − M )δw0 = 0 (3.56a) (N − N )δu = 0 (3.56b) (V − V )δw = 0 (3.56c) In order to satisfy the boundary condition either M = M or δw0 = 0 (3.57a) either V = V or δw = 0 (3.57b) either N = N or δu = 0 (3.57c) The quantities with a bar denotes the quantities prescribed at the ends of a beam. In particular M , V , and N could be equal to zero. The first column in Eq.(3.57) represents the static boundary conditions while the second column the kinematic boundary conditions. There are also mixed boundary conditions. The following combinations satisfy all boundary conditions, Fig. (3.19). Free (static B.C.) Simply supported (mixed B.C.) M=0 M=0 V=0 δw = 0 Clamped (kinematic B.C.) Sliding (mixed B.C.) δw = 0 V=0 δw' = 0 δw' = 0 Figure 3.19: Boundary conditions in the axial direction. In addition the beam could freely slide at the end along x-axis or can be restrained from sliding, Fig. (3.20). In the case of symmetric loading of the beam, it suffices to consider only one half of the beam with the symmetry boundary condition. The symmetry B.C. is identical to the sliding boundary conditions, as explained in Fig.(3.21). 3-20 Structural Mechanics 2.080 Lecture 3 Sliding Semester Yr Axially restrained N=0 N=0 δu = 0 δu = 0 Figure 3.20: Shear force V and rotation angle δw0 vanishes at the symmetry plane. V=0 δw' = 0 l l/2 Figure 3.21: Shear force V and rotation angle δw0 vanishes at the symmetry plane. 3-21 Structural Mechanics 2.080 Lecture 3 Semester Yr ADVANCED TOPIC 3.6 Mathematical Theory of Beams The equations of equilibrium of a beam with a rectangular cross-section can be derived in an elegant way from the 3-D equilibrium equation. With zero body forces, the equation of equilibrium in the compact index notation is σij,j = 0 (3.58) i = 1, σ1j,j = 0 → σ11,1 + σ12,2 + σ13,3 = 0 (3.59a) or the expanded notation i = 2, σ2j,j = 0 → σ21,1 + σ22,2 + σ23,3 = 0 i = 3, σ3j,j = 0 → σ31,1 + σ32,2 + σ33,3 = 0 (3.59b) (3.59c) In the engineering notation, the full set of equilibrium equation, already given by Eq.(3.18), is  ∂σxx ∂σxy ∂σxz   ∂x ∂y ∂z    ∂σ  yx ∂σyy ∂σyz   ∂x ∂y ∂z    ∂σzx ∂σzy ∂σzz  ∂x ∂y ∂z              (3.60) which of the components of the stress tensor will survive beam assumption. Consider a rectangular cross-section beam (h × b) undergoing planar bending, Fig.(3.22). h The beam is subjected to pressure loading p at the plane z = − . In the case of 2 planar bending there must be no gradient of stresses in the y-direction. Therefore the term ∂σxy = 0. The surviving components lie outside the shaded box in Eq.(3.60), and are: ∂y ∂σxx ∂σxz + =0 ∂x ∂z Equation in the y-direction is satisfied identically ∂σzx ∂σzz + =0 ∂x ∂z 3-22 (3.61a) (3.61b) (3.61c) Structural Mechanics 2.080 Lecture 3 P Semester Yr x σyx b h σyz y σyy z Figure 3.22: Vanishing components of the stress vector. Boundary Conditions b Boundary conditions are specified by the Cauchy formula. The lateral surfaces y = ± as 2 h h well as the lower shelf z = are stress free. Consider only the upper shelf z = − , defined 2 2 by the unit normal vector n[0, 0, −1]. Assume that no shear loading is applied, so that the distributed load is directed along z-axis. The components of surface traction on the upper shelf are T [0, 0, p]. From the Cauchy formula, calculate the z-component of the surface traction T3 = σ31 n1 + σ32 n2 + σ33 n3 . Only the last term, for which n3 = −1 remains and so T3 = p = −σ33 = −σzz (3.62) It was straightforward to see that the pressure p must be equilibrated by the σzz component. However, for a precise determination of the sign, the Cauchy formula happened to be useful. All other components of the stress tensor on the lateral surface of the beam are zero. The derivation consists of three steps. First Eq.(3.61a) is integrated with respect to z and multiplied by the beam width b Z b h 2 −h 2 ∂σxx dz + b ∂x Z h 2 −h 2 ∂σxz dz = 0 ∂z (3.63) Next, for a definite integral, the differentiation of the integrant in the first is equivalent to the differentiation of the integral. The second term can be integrated to give d dx Z h 2 −h 2 σxx b dz + σxz h 2 −h 2 =0 (3.64) Noting that b dz = dA, the first integral represents the axial force N , according to the definition, Eq.(3.36b). The shear stress σxz is non-zero inside the beam height but vanishes h h at the lower and upper shelves, σxz = 0, at z = and z = − . Only the first term survives, 2 2 3-23 Structural Mechanics 2.080 Lecture 3 Semester Yr which is the force equilibrium in the x-direction dN = 0 or N 0 = 0 dx (3.65) In the second step both sides of Eq.(3.61a) are multiplied by bz and integrated again with respect to z Z h Z h 2 ∂σxx 2 ∂σxz z(bdz) + z(bdz) = 0 (3.66) − h ∂x − h ∂z 2 2 The second term is now integrated by parts d dx Z h 2 −h 2 σxx z dA + {σxz bz h 2 −h 2 − Z h 2 −h 2 σxz dA} = 0 (3.67) The second term vanishes, as before. The first integral is the bending moment M while the second one-the shear force V (see the definitions, Eqs. (3.36a-3.36c). So, the above equation represent the moment equilibrium dM −V =0 dx (3.68) In the third, final step, Eq.(3.61c) is integrated with respect to z after being multiplied by b. Z h Z h 2 ∂σxz 2 ∂σzz (bdz) + (bdz) = 0 (3.69) − h ∂x − h ∂z 2 2 After integration one gets d dx Z h 2 −h 2 h σxz dA + b σzz Recalling the boundary condition σzz h 2 h 2 − σzz = 0 and σzz −h 2 i −h 2 =0 (3.70) = −p, Eq.(3.70) yields dV + b(−1)(−p) = 0 dx or using bp = q (3.71) dV +q =0 (3.72) dx Eliminating the shear force between Eq.(3.68) and Eq.(3.72), the beam equilibrium equation is obtained. d2 M + q(x) = 0 (3.73) dx2 This equation is identical to the one derived from the principle of virtual work. END OF ADVANCED TOPIC 3-24 Structural Mechanics 3.7 2.080 Lecture 3 Semester Yr Equilibrium in the Theory of Moderately Large Deflections of Beams In Lecture 2, it was shown that finite rotation of the beam element introduced the additional 1 term θ2 in the expression for the axial strain. Let’s see if consideration of finite slope would 2 require modification of the equation of the equilibrium. In Fig.(3.21) the beam element is shown in the theory of small deflections (infinitesimal rotation) and moderately large deflections (finite rotation). N N V N dw dx V Figure 3.23: In finite rotation the axial force contributes to the total shear force. The so-called effective shear force V ∗ is a sum of the cross-sectional shear V and projection of the axial force into the vertical direction. Thus, V∗ =V +N dw dx (3.74) Note that this result is valid as long as cos θ ≈ 1 and sin θ ≈ tan θ ≈ θ. Will the derivation of the force equilibrium change? The answer is no. V* V V + dV dx dx V* + dV* Figure 3.24: Direction of forces to equilibrate the infinitesimal beam element. To ensure vertical equilibrium or (V ∗ + dV ∗ ) − V ∗ + qdx = 0 (3.75) dV ∗ +q =0 dx (3.76) 3-25 Structural Mechanics 2.080 Lecture 3 Semester Yr where V ∗ is defined by Eq.(3.74). Equilibrium of the horizontal (axial) forces stays the same as before since cos θ ≈ 1. dN =0 (3.77) dx Eliminating V ∗ between Eq.(3.74) and Eq.(3.76) gives   d dw dV dN dw d2 w dV + N +q = + +N 2 +q =0 (3.78) dx dx dx dx dx dx dx The second term vanishes on account of Eq.(3.77). The modified force equilibrium equation becomes dV d2 w +N 2 +q =0 (3.79) dx dx The new nonlinear term vanishes if (i) the axial force is zero or (ii) for small deflections and rotation. The moment equilibrium equation, Eq.(3.68) is not affected by moderately large rotations. Together with Eq.(3.79) we arrive at the governing equation of the theory of moderately large deflection of beams d2 M d2 w + N +q =0 (3.80) dx2 dx2 On closing this section, two important remarks should be made. All equations of equilibrium for infinitesimal deformations of 3-D bodies and small deflections of beams involved only static quantities and their gradients (M, V, N ). In the theory of moderately large deflections there is coupling between static and kinematic quantities through the second nonlinear terms. Secondly, Eq.(3.80) includes leading in the in-plane direction (through N ) and outof-plane direction through q. Therefore, it is after referred as the equation describing a beam/columns. 3.8 Equilibrium of Rectangular Plates A step-by-step derivation of the equation of equilibrium and boundary conditions for rectangular plates is presented in the lecture notes of the course 2.081 Plates and Shells. This equation takes the following form in the tensor notation Mαβ,αβ + p = 0 (3.81) ∂ 2 Myy ∂ 2 Mxy ∂ 2 Mxx +2 + +p=0 2 ∂x ∂x∂y ∂y 2 (3.82) and in the extended notation Recall that the dimensions of the bending moments in plates are [Nm/m] = [N ]. In the case of cylindrical bending the twist Mxy and Myy vanish. Multiplying Eq.(3.82) by the width b, one gets d2 [bMxx ] + q = 0 (3.83) dx2 3-26 Structural Mechanics 2.080 Lecture 3 Semester Yr which is identical to the previously derived equation equilibrium of a beam, Eq.(3.73). Therefore wide beams are a special class of rectangular plates. The boundary conditions for plates are similar to those for beams in the local coordinate system at the edges, (n, t), Fig.(3.25). Mt n t Nt Mn Nn Vn Figure 3.25: Local coordinate system at the plate edge with applied generalized forces. Therefore Eq.(3.56) for beams should now read (Mn − M n )δw0 = 0 (3.84a) (Vn − V n )δw = 0 (3.84b) (Nt − N t )δut = 0 (3.84d) (Nn − N n )δun = 0 3.9 (3.84c) Circular Plates It is relatively easy to derive the equation of equilibrium of a circular plate from the principle of virtual work. Bending and in-plane responses is considered separately Z R2 Z R2 (Mr δκr + Mθ δκθ )r dr = R1 pδwr dr + rM r δw0 R1 R2 R1 + rV r δw R2 R1 (3.85) where the radial and circumferential curvatures and their variations are defined (without proof) by ∂2w ∂ 2 (δw) , δκ = r ∂r2 ∂r2 1 ∂(δw) 1 ∂w κθ = , δκθ = r ∂r r ∂r κr = (3.86a) (3.86b) Integrating the left hand side of Eq.(3.85) by parts and using similar arguments as in the case of a beam, one gets equilibrium:   d dMr dMr dMθ r + − = pr (3.87) dr dr dr dr 3-27 Structural Mechanics 2.080 Lecture 3 Semester Yr and boundary conditions (Mr − M r )δw0 = 0 (3.88a) (Vr − V r )δw = 0 (3.88b) where d (rMr ) (3.89) dr When R1 = 0, we have a circular plate. Otherwise the plate is annular with the inner and outer radius R1 and R2 , respectively. When the circular plate is loaded in the in-plane direction only, it remains flat and the components of the mid-surface extensions and their variations are Vr = dur d , δ◦r = (δur ) dr dr ur δur ◦θ = , δ◦θ = r r ◦r = (3.90a) (3.90b) The principle of virtual work can be easily established in the form Z R2 (Nr δ◦r + Nθ δ◦θ )rdr = rNr δur R1 R2 R1 (3.91) The equation of equilibrium in the in-plane direction are easily derived by integrating by parts d (rNr ) − Nθ = 0 (3.92) dr subject to the boundary condition (Nr − N r )δur R2 R1 =0 (3.93) Note that N θ is zero at the boundaries, ensuring that there will be no in-plane shearing force Nrθ and the radial and hoop membrane forces are principal forces. 3-28