M Example Tangent to a Curve
Determine the unit vector tangent to the curve x t2 1, y 4t 3, z 2t2 6t for t 2. Solution dt t2 1 i 4t 3 j 2t2 6t k 2ti 4j 4t 6 k 2ti 4j 4t 6 k J 2t 2 4 2 4t 6 2 , - i i k for t 2 Mathematica solution is shown in Fig. 3.4 ParametricPlot3D t 2 1, 4 t - 3, 2 t 2 - 6 t , t, 0, 4 ParametricPlot3D t 2 1, 4 t - 3, 2 t 2 - 6 t , t, 0, 4 Figure 3.4 Mathematica Solution for the Tangent to a Curve in 3D
Introduction 1
1 We have thus far studied the stress tensors Cauchy, Piola Kirchoff , and several other tensors which describe strain at a point. In general, those tensors will vary from point to point and represent a tensor field. 2 We have also obtained only one differential equation, that was the compatibility equation. 3 In this chapter, we will derive additional differential equations governing the way stress and deformation vary at a point and with time. They will apply to any continuous medium, and yet...
Example Stress Transformation
Show that the transformation tensor of direction cosines previously determined transforms the original stress tensor into the diagonal principal axes stress tensor. Solution
Force Traction and Stress Vectors
1 There are two kinds of forces in continuum mechanics body forces act on the elements of volume or mass inside the body, e.g. gravity, electromagnetic fields. dF pbdVol. surface forces are contact forces acting on the free body at its bounding surface. Those will be defined in terms of force per unit area. 2 The surface force per unit area acting on an element dS is called traction or more accurately stress vector. I tdS if txdS j tydS k tzdS 2.1 Most authors limit the term traction to an...
MATHEMATICA ASSIGNMENT and SOLUTION
Connect to Mathematica using the following procedure 5. On the newly opened shell, enter your password first, and then type setenv DISPLAY xxx 0.0 where xxx is the workstation name which should appear on a small label on the workstation itself. and then solve the following problems 1. The state of stress through a continuum is given with respect to the cartesian axes 0x1x2x3 by Determine the stress vector at point P 1,1, a 3 of the plane that is tangent to the cylindrical surface x2 X3 4 at P....
Monotropic Material
33 A plane of elastic symmetry exists at a point where the elastic constants have the same values for every pair of coordinate systems which are the reflected images of one another with respect to the plane. The axes of such coordinate systems are referred to as equivalent elastic directions. 34 If we assume Xi xi, X2 X2 and X3 X3, then the transformation Xj aj Xj is defined through where the negative sign reflects the symmetry of the mirror image with respect to the x3 plane. 35 We next...
Load Shear Moment Relations
38 Let us derive the basic relations between load, shear and moment. Considering an infinitesimal length dx of a beam subjected to a positive load5 w x , Fig. 12.6. The infinitesimal section must also be in equilibrium. 39 There are no axial forces, thus we only have two equations of equilibrium to satisfy XFy 0 and XMz 0. 40 Since dx is infinitesimally small, the small variation in load along it can be neglected, therefore we assume w x to be constant along dx. 41 To denote that a small change...
Boundary Value Problem Formulation
9 Hence, the boundary value formulation is suumarized by and is illustrated by Fig. 9.3. This is now a well posed problem.
Principle of Virtual Work and Complementary Virtual Work
33 The principles of Virtual Work and Complementary Virtual Work relate force systems which satisfy the requirements of equilibrium,, and deformation systems which satisfy the requirement of compatibility. 1. In any application the force system could either be the actual set of external loads dp or some virtual force system which happens to satisfy the condition of equilibrium p. This set of external forces will induce internal actual forces da or internal hypothetical forces Sa compatible with...
Geometric Instability
33 The stability of a structure is determined not only by the number of reactions but also by their arrangement. 34 Geometric instability will occur if 1. All reactions are parallel and a non-parallel load is applied to the structure. 2. All reactions are concurrent, Fig. 12.4. Figure 12.4 Geometric Instability Caused by Concurrent Reactions Figure 12.4 Geometric Instability Caused by Concurrent Reactions 3. The number of reactions is smaller than the number of equations of equilibrium, that is...
Transversely Isotropic Material
38 A material is transversely isotropic if there is a preferential direction normal to all but one of the three axes. If this axis is x3, then rotation about it will require that cos 6 sin 6 0 sin 6 cos 6 0 0 0 1 substituting Eq. 7.33 into Eq. 7.41, using the above transformation matrix, we obtain C1111 cos4 6 c1111 cos2 6 sin2 6 2cm2 4cm2 sin4 6 c2222 C1122 cos2 6 sin2 6 cm1 cos4 6 cm2 4 cos2 6 sin2 6 cm2 sin4 6 c2211 sin2 6 cos2 6 c2222 cos2 6 c1133 sin2 6 c2233 sin4 6 cm1 cos2 6 sin2 6 2cm2...
Example Mohrs Circle in Plane Stress
An element in plane stress is subjected to stresses axx 15, ayy 5 and Txy 4. Using the Mohr's circle determine a the stresses acting on an element rotated through an angle 0 40 counterclockwise b the principal stresses and c the maximum shear stresses. Show all results on sketches of properly oriented elements. Solution 1. The center of the circle is located at 2 axx ayy 1 15 5 10. 2.63 2. The radius and the angle 2p are given by R y1 15 5 2 42 6.403 2.64-a 2 4 tan2 w 0.8 38.66 f3 19.33 2.64-b...
Summary
60 Summary of Virtual work methods, Table 13.2. Displacement strains Forces Stresses KAD Kinematically Admissible Dispacements SAS Statically Admissible Stresses Table 13.2 Comparison of Virtual Work and Complementary Virtual Work 61 A summary of the various methods introduced in this chapter is shown in Fig. 13.7. 62 The duality between the two variational principles is highlighted by Fig. 13.8, where beginning with kinematically admissible displacements, the principle of virtual work provides...
Divergence Vector
8 The divergence of a vector field of a body b with boundary Q, Fig. 3.5 is defined by considering that each point of the surface has a normal n, and that the body is surrounded by a vector field v x . The volume of the body is v b . Figure 3.5 Vector Field Crossing a Solid Region Figure 3.5 Vector Field Crossing a Solid Region The divergence of the vector field is thus defined as where v.n is often referred as the flux and represents the total volume of fluid that passes through dA in unit...
Draft Hqf
1S Beam Fixed at Both Ends Concentrated Load 1S Beam Fixed at Both Ends Concentrated Load 14 Cantilever Beam Triangular Unsymmetric Load 14 Cantilever Beam Triangular Unsymmetric Load 6b 3L - 3x - b P L 2 3b - L x 17 Cantilever Beam Point Load at Free End 18 Cantilever Beam Concentrated Force and Moment at Free End
Boundary Conditions
4 In describing the boundary conditions B.C. , we must note that 1. Either we know the displacement but not the traction, or we know the traction and not the corresponding displacement. We can never know both a priori. 2. Not all boundary conditions specifications are acceptable. For example we can not apply tractions to the entire surface of the body. Unless those tractions are specially prescribed, they may not necessarily satisfy equilibrium. 5 Properly specified boundary conditions result...
Principal Values and Directions of Symmetric Second Order Tensors
59 Since the two fundamental tensors in continuum mechanics are of the second order and symmetric stress and strain , we examine some important properties of these tensors. 60 For every symmetric tensor Tij defined at some point in space, there is associated with each direction specified by unit normal nj at that point, a vector given by the inner product If the direction is one for which vi is parallel to ni, the inner product may be expressed as and the direction ni is called principal...
Cauchys Reciprocal Theorem
21 If we consider t1 as the traction vector on a plane with normal n1, and t2 the stress vector at the same point on a plane with normal n2, then t1 n1 and t2 n2a ti L 1JM and 2 L -2JH If we postmultiply the first equation by n2 and the second one by n1, by virtue of the symmetry of 0 we have Figure 2.4 Cauchy's Reciprocal Theorem Figure 2.4 Cauchy's Reciprocal Theorem 22 In the special case of two opposite faces, this reduces to 23 We should note that this theorem is analogous to Newton's...
Introduction
1 A field is a function defined over a continuous region. This includes, Scalar Field g x , Vector Field v x , Fig. 3.1 or Tensor Field T x . 2 We first introduce the differential vector operator Nabla denoted by V 3 We also note that there are as many ways to differentiate a vector field as there are ways of multiplying vectors, the analogy being given by Table 3.1. Table 3.1 Similarities Between Multiplication and Differentiation Operators Table 3.1 Similarities Between Multiplication and...
Example Principal Stresses
The stress tensor is given at a point by determine the principal stress values and the corresponding directions. Solution Or upon expansion and simplification A 2 A 4 A 1 0, thus the roots are a l 4, a 2 1 and a 3 2. We also note that those are the three eigenvalues of the stress tensor. If we let xl axis be the one corresponding to the direction of a 3 and n3 be the direction cosines of this axis, then from Eq. 2.28 we have Similarly If we let x2 axis be the one corresponding to the direction...
Size Effect Griffith Theory
13 In his quest for an explanation of the size effect, Griffith came across Inglis's paper, and his strike of genius was to assume that strength is reduced due to the presence of internal flaws. Griffith postulated that the theoretical strength can only be reached at the point of highest stress concentration, and accordingly the far-field applied stress will be much smaller. 14 Hence, assuming an elliptical imperfection, and from equation 11.2 a is the stress at the tip of the ellipse which is...
Basic Kinematic Assumption Curvature
47 Fig.12.7 shows portion of an originally straight beam which has been bent to the radius p by end couples M. support conditions, Fig. 12.1. It is assumed that plane cross-sections normal to the Figure 12.7 Deformation of a Beam under Pure Bending length of the unbent beam remain plane after the beam is bent. Figure 12.7 Deformation of a Beam under Pure Bending length of the unbent beam remain plane after the beam is bent. 48 Except for the neutral surface all other longitudinal fibers either...
Tensor Product
45 Since a tensor primary objective is to operate on vectors, the tensor product of two vectors provides a fundamental building block of second-order tensors and will be examined next. 46 The Tensor Product of two vectors u and v is a second order tensor u v which in turn operates on an arbitrary vector w as follows In other words when the tensor product u v operates on w left hand side , the result right hand side is a vector that points along the direction of u, and has length equal to v-w u...
Isotropic Material
40 An isotropic material is symmetric with respect to every plane and every axis, that is the elastic properties are identical in all directions. 41 To mathematically characterize an isotropic material, we require coordinate transformation with rotation about x2 and xi axes in addition to all previous coordinate transformations. This process will enforce symmetry about all planes and all axes. 42 The rotation about the x2 axis is obtained through cos 6 0 sin 6 0 1 0 sin 6 0 cos 6 we follow a...
Draft Szr
max when a K h max when a gt h at x a when x K a at x a when x K a 7 Simple Beam Two Equally Concentrated Symmetric Loads when x K a Ax when a K x K L a Ax SEI 3La 3a x sei 3Lx 3x a 8 Simple Beam Two Equally Concentrated Unsymmetric Loads 10 Propped Cantilever, Concentrated Load at Center 10 Propped Cantilever, Concentrated Load at Center when x lt Ll Mx 5Pf 1 when L lt x Mx P L - 6 Jmax' at x .4472L Amax .009317E3 11 Propped Cantilever Concentrated Load 11 Propped Cantilever Concentrated Load...
Example Stress Vector normal to the Tangent of a Cylinder
The stress tensor throughout a continuum is given with respect to Cartesian axes as Determine the stress vector or traction at the point P 2, VS of the plane that is tangent to the cylindrical surface x2 x2 4 at P, Fig. 3.9. Figure 3.9 Radial Stress vector in a Cylinder Figure 3.9 Radial Stress vector in a Cylinder At point P, the stress tensor is given by The unit normal to the surface at P is given from Thus the traction vector will be determined from 20 We can also define the gradient of a...
Covariant Transformation
15 Similarly to Eq. 1.24, a covariant component transformation recognized by subscript will be defined as _ Vj apPvp and inversely vk bk vq We note that contrarily to the contravariant transformation, the covariant transformation uses the same transformation coefficients as the ones for the base vectors. 16 Finally transformation of tensors of order one and two is accomplished through
Operations
Addition of two vectors a b is geometrically achieved by connecting the tail of the vector b with the head of a, Fig. 1.2. Analytically the sum vector will have components a b a2 b2 a3 b3 J. Scalar multiplication aa will scale the vector into a new one with components aai aa2 aa3 J. Vector Multiplications of a and b comes in three varieties Dot Product or scalar product is a scalar quantity which relates not only to the lengths of the vector, but also to the angle between them. a-b a UK b cos9...
v
Finally, the displacement components can be obtained by integrating the above equations 10.2.2.3 Example Thick-Walled Cylinder 4i If we consider a circular cylinder with internal and external radii a and b respectively, subjected to internal and external pressures p and po respectively, Fig. 10.2, then the boundary conditions for the plane strain problem are Ttt Pi at r Oi Ttt -po at r b 42 These Boundary conditions can be easily shown to be satisfied by the following stress field These...
Internal Equilibrium Section Properties
55 Just as external forces acting on a structure must be in equilibrium, the internal forces must also satisfy the equilibrium equations. 56 The internal forces are determined by slicing the beam. The internal forces on the cut section must be in equilibrium with the external forces. 57 The first equation we consider is the summation of axial forces. 58 Since there are no external axial forces unlike a column or a beam-column , the internal axial forces must be in equilibrium. Fx 0 axdA 0 12.19...
Rotation of Axes
55 The rule for changing second order tensor components under rotation of axes goes as follow ajTjqvq From Eq. 1.39-a 1.71 But we also have Ui Tipvp again from Eq. 1.39-a in the barred system, equating these two expressions we obtain By extension, higher order tensors can be similarly transformed from one coordinate system to another. 56 If we consider the 2D case, From Eq. 1.38 2 sin 2aTxx sin 2aTyy 2 cos 2aTxy 2 sin 2aTxx sin 2aTyy 2 cos 2aTxy sin2 aTxx cos a cos aTyy 2 sin aTxy 0...
j Contravariant Transformation
12 The vector representation in both systems must be the same V Vqbq Vkbk Vk bkbq v - Vk b bq 0 1.23 since the base vectors bq are linearly independent, the coefficients of bq must all be zero hence showing that the forward change from components vk to vq used the coefficients bqk of the backward change from base bq to the original bk. This is why these components are called contravariant. 13 Generalizing, a Contravariant Tensor of order one recognized by the use of the superscript transforms a...
RayleighRitz Method
55 Continuous systems have infinite number of degrees of freedom, those are the displacements at every point within the structure. Their behavior can be described by the Euler Equation, or the partial differential equation of equilibrium. However, only the simplest problems have an exact solution which satisfies equilibrium, and the boundary conditions . 56 An approximate method of solution is the Rayleigh-Ritz method which is based on the principle of virtual displacements. In this method we...
Arch Elasticity
51 Fig. 2.10 illustrates the stresses acting on a differential element of a shell structure. The resulting forces in turn are shown in Fig. 2.11 and for simplification those acting per unit length of the middle surface are shown in Fig. 2.12. The net resultant forces are given by Figure 2.9 Mohr Circle for Stress in 3D Figure 2.10 Differential Shell Element, Stresses Figure 2.10 Differential Shell Element, Stresses Figure 2.11 Differential Shell Element, Forces Figure 2.11 Differential Shell...
Spherical and Deviatoric Stress Tensors
36 If we let o denote the mean normal stress p 0 -p 3 011 022 033 3on tr a then the stress tensor can be written as the sum of two tensors Hydrostatic stress in which each normal stress is equal to p and the shear stresses are zero. The hydrostatic stress produces volume change without change in shape in an isotropic medium. Deviatoric Stress which causes the change in shape.
Linear Momentum Principle Equation of Motion Momentum Principle
22 The momentum principle states that the time rate of change of the total momentum of a given set of particles equals the vector sum of all external forceps acting on the particles of the set, provided Newton's Third Law applies. The continuum form of this principle is a basic postulate of continuum mechanics. i tdS pbdV d f pvdV 6.22 Then we substitute ti Tijnj and apply the divergence theorm to obtain which is Cauchy's first equation of motion, or the linear momentum principle, or more...
Continuum Mechanics
LMC DMX EPFL Prof. Saouma Exam I Closed notes , March 27, 1998 3 Hours There are 19 problems worth a total of 63 points. Select any problems you want as long as the total number of corresponding points is equal to or larger than 50. 1. 2 pts Write in matrix form the following 3rd order tensor Dijk in R2 space. i,j,k range from 1 to 2. 2. 2 pts Solve for Eijai in indicial notation. 3. 4 pts if the stress tensor at point P is given by determine the traction or stress vector t on the plane passing...
Orthotropic Material
36 If the material possesses three mutually perpendicular planes of elastic symmetry, that is symmetric with respect to two planes x2 and X3 , then the transformation Xi aj Xj is defined through where the negative sign reflects the symmetry of the mirror image with respect to the x3 plane. Upon substitution in Eq.7.33 we now would have We note that in here all terms of cijkl with the indices 3 and 2 occuring an odd number of times are again set to zero. 37 Wood is usually considered an...
Cylindrical Coordinates
20 So far all equations have been written in either vector, indicial, or engineering notation. The last two were so far restricted to an othonormal cartesian coordinate system. 21 We now rewrite some of the fundamental relations in cylindrical coordinate system, Fig. 9.5, as this would enable us to analytically solve some simple problems of great practical usefulness torsion, pressurized cylinders, . This is most often achieved by reducing the dimensionality of the problem from 3 to 2 or even...
TENSORS order 1
Cauchy's deformation tensor N.D. Green's deformation tensor metric tensor, right Cauchy-Green deformation tensor Rate of deformation tensor Stretching tensor Lagrangian or Green's finite strain tensor Eulerian or Almansi finite strain tensor Strain deviator Material deformation gradient Spatial deformation gradient Idendity matrix Material displacement gradient Thermal conductivity Spatial displacement gradient Spatial gradient of the velocity Orthogonal rotation tensor First Piola-Kirchoff...
Elastic Potential or Strain Energy Function
16 Green defined an elastic material as one for which a strain-energy function exists. Such a material is called Green-elastic or hyperelastic if there exists an elastic potential function W or strain energy function, a scalar function of one of the strain or deformation tensors, whose derivative with respect to a strain component determines the corresponding stress component. 17 For the fully recoverable case of isothermal deformation with reversible heat conduction we have hence W p0 is an...
Mohrs Circle for Plane Stress Conditions
41 The Mohr circle will provide a graphical mean to contain the transformed state of stress axx,ayy, xy at an arbitrary plane inclined by a in terms of the original one axx, ayy, axy . cos2 a 1 c s2a sin2 a -c s2 cos 2a cos2 a - sin2 a sin 2a 2 sin a cos a into Eq. 2.49 and after some algebraic manipulation we obtain 7 axx Vyy T. Jxx - yy cos2a aXy sin2a 2.57-a TXy axy cos 2a - ctxx - yy sin2a 2.57-b 43 Points axx,axy , axx, 0 , ayy, 0 and axx ayy 2, 0 are plotted in the stress representation...
NavierCauchy Equations
11 One such approach is to substitute the displacement-strain relation into Hooke's law resulting in stresses in terms of the gradient of the displacement , and the resulting equation into the equation of motion to obtain three second-order partial differential equations for the three displacement components known as Navier's Equation Figure 9.3 Fundamental Equations in Solid Mechanics
Cartesian Coordinate System
17 If we consider two different sets of cartesian orthonormal coordinate systems ei, e2, e3 and e, e2, e3 , any vector v can be expressed in one system or the other is To determine the relationship between the two sets of components, we consider the dot product of v with one any of the base vectors 19 We can thus define the nine scalar values which arise from the dot products of base vectors as the direction cosines. Since we have an orthonormal system, those values are nothing else than the...
SHEAR MOMENT and DEFLECTION DIAGRAMS for BEAMS
Adapted from 1 Simple Beam uniform Load 2 Simple Beam Unsymmetric Triangular Load 3 Simple Beam Symmetric Triangular Load 3 Simple Beam Symmetric Triangular Load 4 Simple Beam Uniform Load Partially Distributed 4 Simple Beam Uniform Load Partially Distributed 5 Simple Beam Concentrated Load at Center at x 2 when x lt 2 whenx lt 2 at x L 6 Simple Beam Concentrated Load at Any Point
Strain Decomposition
72 In this section we first seek to express the relative displacement vector as the sum of the linear Lagrangian or Eulerian strain tensor and the linear Lagrangian or Eulerian rotation tensor. This is restricted to small strains. 73 For finite strains, the former additive decomposition is no longer valid, instead we shall consider the strain tensor as a product of a rotation tensor and a stretch tensor. 4.3.1 jLinear Strain and Rotation Tensors 74 Strain components are quantitative measures of...
C Euler Equation
1 The fundamental problem of the calculus of variation1 is to find a function u x such that 2 We define u x to be a function of x in the interval a, b , and F to be a known function such as the energy density . 3 We define the domain of a functional as the collection of admissible functions belonging to a class of functions in function space rather than a region in coordinate space as is the case for a function . 4 We seek the function u x which extremizes n. 5 Letting u to be a family of...
Principle of Virtual Work
35 Derivation of the principle of virtual work starts with the assumption of that forces are in equilibrium and satisfaction of the static boundary conditions. The Equation of equilibrium Eq. 6.26 which is rewritten as where b representing the body force. In matrix form, this can be rewritten as Note that this equation can be generalized to 3D. 37 The surface r of the solid can be decomposed into two parts r and T where tractions and displacements are respectively specified. t t on rt Natural...
Compatibility Equation
93 If ij 1 ui,j uj,i then we have six differential equations in 3D the strain tensor has a total of 9 terms, but due to symmetry, there are 6 independent ones for determining upon integration three unknowns displacements ui. Hence the system is overdetermined, and there must be some linear relations between the strains. 94 It can be shown through appropriate successive differentiation of the strain expression that the compatibility relation for strain reduces to There are 81 equations in all,...