Lecture 3.pdf

Lecture 3.pdf
CENG6504: Finite Element Methods Truss Member Dr. Tesfaye Alemu 1
Learning Objectives • To derive the stiffness matrix for a bar element. • To illustrate how to solve a bar assemblage by the direct stiffness method. • To introduce guidelines for selecting displacement functions. • To describe the concept of transformation of vectors in two different coordinate systems in the plane. • To derive the stiffness matrix for a bar arbitrarily oriented in the plane. • To demonstrate how to compute stress for a bar in the plane. • To show how to solve a plane truss problem. • To develop the transformation matrix in three- dimensional space and show how to use it to derive the stiffness matrix for a bar arbitrarily oriented in space. • To demonstrate the solution of space trusses. 2
Having set forth the foundation on which the direct stiffness method is based, we will now derive the stiffness matrix for a linear-elastic bar (or truss) element using the general steps outlined in Chapter 2. We will include the introduction of both a local coordinate system, chosen with the element in mind, and a global or reference coordinate system, chosen to be convenient (for numerical purposes) with respect to the overall structure. We will also discuss the transformation of a vector from the local coordinate system to the global coordinate system, using the concept of transformation matrices to express the stiffness matrix of an arbitrarily oriented bar element in terms of the global system. Development of Truss Equations 3
Development of Truss Equations Next we will describe how to handle inclined, or skewed, supports. We will then extend the stiffness method to include space trusses. We will develop the transformation matrix in three-dimensional space and analyze a space truss. We will then use the principle of minimum potential energy and apply it to the bar element equations. Finally, we will apply Galerkin’s residual method to derive the bar element equations. 4
Development of Truss Equations 5
Stiffness Matrix for a Bar Element Consider the derivation of the stiffness matrix for the linear- elastic, constant cross-sectional area (prismatic) bar element show below. This application is directly applicable to the solution of pin- connected truss problems. 13
Stiffness Matrix for a Bar Element Consider the derivation of the stiffness matrix for the linear- elastic, constant cross-sectional area (prismatic) bar element show below. where T is the tensile force directed along the axis at nodes 1 and 2, x is the local coordinate system directed along the length of the bar. 14
Stiffness Matrix for a Bar Element Consider the derivation of the stiffness matrix for the linear- elastic, constant cross-sectional area (prismatic) bar element show below. The bar element has a constant cross-section A, an initial length L, and modulus of elasticity E. The nodal degrees of freedom are the local axial displacements u1 and u2 at the ends of the bar. 15
Stiffness Matrix for a Bar Element The strain-displacement relationship is: du E dx      From equilibrium of forces, assuming no distributed loads acting on the bar, we get: constant x A T    Combining the above equations gives: constant du AE T dx   Taking the derivative of the above equation with respect to the local coordinate x gives: 0 d du AE dx dx        16
Stiffness Matrix for a Bar Element The following assumptions are considered in deriving the bar element stiffness matrix: 1. The bar cannot sustain shear force: 1 2 0 y y f f   2. Any effect of transverse displacement is ignored. 3. Hooke’s law applies; stress is related to strain: x x E    17
Step 1 - Select Element Type We will consider the linear bar element shown below. Stiffness Matrix for a Bar Element The bar element has a constant cross-section A, an initial length L, and modulus of elasticity E. 18
Step 2 - Select a Displacement Function Stiffness Matrix for a Bar Element 1 2 u a a x   A linear displacement function u is assumed: The number of coefficients in the displacement function, ai, is equal to the total number of degrees of freedom associated with the element. Applying the boundary conditions and solving for the unknown coefficients gives: 2 1 1 u u u x u L          1 2 1 u x x u u L L                     19
Step 2 - Select a Displacement Function Stiffness Matrix for a Bar Element Or in another form: where N1 and N2 are the interpolation functions gives as:   1 1 2 2 u u N N u        1 2 1 x x N N L L    The linear displacement function plotted over the length of the bar element is shown here. u 20
Step 3 - Define the Strain/Displacement and Stress/Strain Relationships Stiffness Matrix for a Bar Element The stress-displacement relationship is: 2 1 x u u du dx L     Step 4 - Derive the Element Stiffness Matrix and Equations We can now derive the element stiffness matrix as follows: x T A  Substituting the stress-displacement relationship into the above equation gives: 2 1 u u T AE L         21
Stiffness Matrix for a Bar Element The nodal force sign convention, defined in element figure, is: Step 4 - Derive the Element Stiffness Matrix and Equations therefore, Writing the above equations in matrix form gives: 1 2 x x f T f T    1 2 2 1 1 2 x x u u u u f AE f AE L L                 1 1 2 2 1 1 1 1 x x f u AE f u L                      Notice that AE/L for a bar element is analogous to the spring constant k for a spring element. 22
Stiffness Matrix for a Bar Element The global stiffness matrix and the global force vector are assembled using the nodal force equilibrium equations, and force/deformation and compatibility equations. Step 5 - Assemble the Element Equations and Introduce Boundary Conditions     ( ) ( ) 1 1 n n e e e e K F         K k F f Where k and f are the element stiffness and force matrices expressed in global coordinates. 23
Stiffness Matrix for a Bar Element Solve the displacements by imposing the boundary conditions and solving the following set of equations: Step 6 - Solve for the Nodal Displacements  F Ku Step 7 - Solve for the Element Forces Once the displacements are found, the stress and strain in each element may be calculated from: 2 1 x x x u u du E dx L        24
Stiffness Matrix for a Bar Element Consider the following three-bar system shown below. Assume for elements 1 and 2: A = 1 in2 and E = 30 (106) psi and for element 3: A = 2 in2 and E = 15 (106) psi. Example 1 - Bar Problem Determine: (a) the global stiffness matrix, (b) the displacement of nodes 2 and 3, and (c) the reactions at nodes 1 and 4. 25
Stiffness Matrix for a Bar Element For elements 1 and 2: Example 1 - Bar Problem For element 3:    6 (1) (2) 6 1 30 10 1 1 1 1 10 30 1 1 1 1 lb lb in in                     k k 1 2 node numbers for element 1    6 (3) 6 2 15 10 1 1 1 1 10 30 1 1 1 1 lb lb in in                    k 3 4 node numbers for element 3 As before, the numbers above the matrices indicate the displacements associated with the matrix. 2 3 node numbers for element 2 26
Assembling the global stiffness matrix by the direct stiffness methods gives: Relating global nodal forces related to global nodal displacements gives: 6 1 1 0 0 1 2 1 0 10 0 1 2 1 0 0 1 1                    K 1 1 2 2 6 3 3 4 4 1 1 0 0 1 2 1 0 10 0 1 2 1 0 0 1 1 x x x x F u F u F u F u                                                E1 E 2 E 3 Stiffness Matrix for a Bar Element Example 1 - Bar Problem 27
                                               1 2 2 6 3 3 4 1 1 0 0 0 1 2 1 0 10 0 1 2 1 0 0 1 1 0 x x x x F F u F u F Stiffness Matrix for a Bar Element The boundary conditions are: Example 1 - Bar Problem 1 4 0 u u   Applying the boundary conditions and the known forces (F2x = 3,000 lb) gives: 2 6 3 3,000 2 1 10 0 1 2 u u                      28
Stiffness Matrix for a Bar Element Example 1 - Bar Problem Solving for u2 and u3 gives:   2 3 0.002 0.001 u in u in The global nodal forces are calculated as: 1 2 6 3 4 1 1 0 0 0 2,000 1 2 1 0 0.002 3,000 10 0 1 2 1 0.001 0 0 0 1 1 0 1,000 x x x x F F lb F F                                                                 29
Stiffness Matrix for a Bar Element Consider the following guidelines, as they relate to the one- dimensional bar element, when selecting a displacement function. Selecting Approximation Functions for Displacements 1. Common approximation functions are usually polynomials. 2. The approximation function should be continuous within the bar element. 30
Stiffness Matrix for a Bar Element Selecting Approximation Functions for Displacements 3. The approximating function should provide interelement continuity for all degrees of freedom at each node for discrete line elements, and along common boundary lines and surfaces for two- and three-dimensional elements. Consider the following guidelines, as they relate to the one- dimensional bar element, when selecting a displacement function. 31
Stiffness Matrix for a Bar Element Selecting Approximation Functions for Displacements For the bar element, we must ensure that nodes common to two or more elements remain common to these elements upon deformation and thus prevent overlaps or voids between elements. The linear function is then called a conforming (or compatible) function for the bar element because it ensures both the satisfaction of continuity between adjacent elements and of continuity within the element. Consider the following guidelines, as they relate to the one- dimensional bar element, when selecting a displacement function. 32
Stiffness Matrix for a Bar Element Selecting Approximation Functions for Displacements 4. The approximation function should allow for rigid-body displacement and for a state of constant strain within the element. Completeness of a function is necessary for convergence to the exact answer, for instance, for displacements and stresses. Consider the following guidelines, as they relate to the one- dimensional bar element, when selecting a displacement function. 33
Stiffness Matrix for a Bar Element The interpolation function must allow for a rigid-body displacement, that means the function must be capable of yielding a constant value. Consider the follow situation: Selecting Approximation Functions for Displacements Therefore: 1 1 1 2 u a a u u      1 1 2 2 1 2 1 u N u N u N N a     Since u = a1 then: This means that: 1 2 1 N N     1 1 2 1 u a N N a    The displacement interpolation function must add to unity at every point within the element so the it will yield a constant value when a rigid-body displacement occurs. 34
Stiffness Matrix for a Bar Element In many problems it is convenient to introduce both local and global (or reference) coordinates. Local coordinates are always chosen to conveniently represent the individual element. Global coordinates are chosen to be convenient for the whole structure. Transformation of Vectors in Two Dimensions 35
Stiffness Matrix for a Bar Element Given the nodal displacement of an element, represented by the vector d in the figure below, we want to relate the components of this vector in one coordinate system to components in another. Transformation of Vectors in Two Dimensions 36
Stiffness Matrix for a Bar Element Let’s consider that d does not coincide with either the local or global axes. In this case, we want to relate global displacement components to local ones. In so doing, we will develop a transformation matrix that will subsequently be used to develop the global stiffness matrix for a bar element. Transformation of Vectors in Two Dimensions 37
Stiffness Matrix for a Bar Element We define the angle  to be positive when measured counterclockwise from x to x’. We can express vector displacement d in both global and local coordinates by: Transformation of Vectors in Two Dimensions 1 1 1 1 u v u v         d i j i j 38
Stiffness Matrix for a Bar Element Consider the following diagram: Transformation of Vectors in Two Dimensions Using vector addition:   a b i Using the law of cosines, we get: | | | | cos | | cos     a i a Similarly: | | | | sin | | sin     b i b 39
Stiffness Matrix for a Bar Element Consider the following diagram: Transformation of Vectors in Two Dimensions The vector a is in the direction and b is in the direction, therefore:  i  j        | | cos | | sin             a a i i b b j j 40
Stiffness Matrix for a Bar Element Consider the following diagram: Transformation of Vectors in Two Dimensions The vector i can be rewritten as: cos sin       i i j The vector j can be rewritten as: sin cos       j i j Therefore, the displacement vector is:     1 1 1 1 1 1 cos sin sin cos u v u v u v                    i j i j i j i j41
Stiffness Matrix for a Bar Element Consider the following diagram: Transformation of Vectors in Two Dimensions Combining like coefficients of the local unit vectors gives: 1 1 1 cos sin u v u      1 1 1 sin cos u v v       cos sin C S     1 1 1 1 u u C S v v S C                       42
Stiffness Matrix for a Bar Element Transformation of Vectors in Two Dimensions The matrix is called the transformation matrix: The previous equation relates the global displacement d to the d local displacements C S S C        The figure below shows u expressed in terms of the global coordinates x and y. u Cu Sv    43
Stiffness Matrix for a Bar Element Example 2 - Bar Element Problem Using the following expression we just derived, we get: The global nodal displacement at node 2 is u2 = 0.1 in and v2 = 0.2 in for the bar element shown below. Determine the local displacement. 2 cos60 (0.1) sin60 (0.2) o o u   u Cu Sv    0.223 in  44
Stiffness Matrix for a Bar Element Global Stiffness Matrix We will now use the transformation relationship developed above to obtain the global stiffness matrix for a bar element. 45
Stiffness Matrix for a Bar Element Global Stiffness Matrix We want to relate the global element forces f to the global displacements d for a bar element with an arbitrary orientation. We known that for a bar element in local coordinates we have: 1 1 2 2 1 1 1 1 x x f u AE f u L                          1 1 1 1 2 2 2 2 x y x y f u f v k f u f v                                  f k d f = kd 46
Stiffness Matrix for a Bar Element Global Stiffness Matrix Combining both expressions for the two local degrees-of- freedom, in matrix form, we get: Using the relationship between local and global components, we can develop the global stiffness matrix. We already know the transformation relationships: 1 1 1 2 2 2 cos sin cos sin u u v u u v           1 1 1 2 2 2 0 0 0 0 u u v C S u u C S v                               * d = T d * 0 0 0 0 C S C S        T 47
Stiffness Matrix for a Bar Element Global Stiffness Matrix Substituting the global force expression into element force equation gives: A similar expression for the force transformation can be developed. 1 1 1 * 2 2 2 0 0 0 0 x y x x x y f f f C S f f C S f                                f T f    f = k d    * * T f k T d     * T f k d  * d = T d Substituting the transformation between local and global displacements gives: 48
Stiffness Matrix for a Bar Element Global Stiffness Matrix The matrix T* is not a square matrix so we cannot invert it. Let’s expand the relationship between local and global displacement. where T is: 1 1 1 1 2 2 2 2 0 0 0 0 0 0 0 0 u u C S v v S C u u C S v v S C                                                0 0 0 0 0 0 0 0 C S S C C S S C                T  d = Td 49
Stiffness Matrix for a Bar Element Global Stiffness Matrix We can write a similar expression for the relationship between local and global forces. Therefore our original local coordinate force-displacement expression 1 1 1 1 2 2 2 2 0 0 0 0 0 0 0 0 x x y y x x y y f f C S f f S C f f C S f f S C                                                1 1 2 2 1 1 1 1 x x f u AE f u L                           f = Tf    f = k d 50
Stiffness Matrix for a Bar Element Global Stiffness Matrix May be expanded: The global force-displacement equations are: 1 1 1 1 2 2 2 2 1 0 1 0 0 0 0 0 1 0 1 0 0 0 0 0 x y x y f u f v AE f u L f v                                                        f k d Multiply both side by T -1 we get:   -1 f T k Td where T-1 is the inverse of T. It can be shown that:   1 T T T    Tf k Td 51
Stiffness Matrix for a Bar Element Global Stiffness Matrix The global force-displacement equations become: Where the global stiffness matrix k is: Expanding the above transformation gives: We can assemble the total stiffness matrix by using the above element stiffness matrix and the direct stiffness method. =  T f T k Td   T k T k T 2 2 2 2 2 2 2 2 CS CS C C S S CS CS AE L CS CS C C S S CS CS                      k     ( ) ( ) 1 1 n n e e e e K F         K k F f  F Kd 52
Stiffness Matrix for a Bar Element Global Stiffness Matrix Local forces can be computed as: 1 1 1 1 2 2 2 2 1 0 1 0 0 0 0 0 1 0 1 0 0 0 0 0 x y x y f u f v AE f u L f v                                                    1 1 1 2 2 1 2 1 1 2 2 2 0 0 x y x y f Cu Sv Cu Sv f AE f L Cu Sv Cu Sv f                                       1 1 2 2 1 0 1 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 u C S v S C AE u C S L v S C                                            53
Stiffness Matrix for a Bar Element Example 3 - Bar Element Problem Therefore: For the bar element shown below, evaluate the global stiffness matrix. Assume the cross-sectional area is 2 in2, the length is 60 in, and the E is 30 x 106 psi. 3 cos30 2 o C   2 2 2 2 2 2 2 2 CS CS C C S S CS CS AE L CS CS C C S S CS CS                      k 1 sin30 2 o S   54
Stiffness Matrix for a Bar Element Example 3 - Bar Element Problem Simplifying the global elemental stiffness matrix is: The global elemental stiffness matrix is:   2 6 3 3 3 3 4 4 4 4 3 3 1 1 (2 ) 30 10 4 4 4 4 60 3 3 3 3 4 4 4 4 3 3 1 1 4 4 4 4 in psi in                               k 6 0.750 0.433 0.750 0.433 0.433 0.250 0.433 0.250 10 0.750 0.433 0.750 0.433 0.433 0.250 0.433 0.250 lb in                      k 55
Stiffness Matrix for a Bar Element Computation of Stress for a Bar in the x-y Plane For a bar element the local forces are related to the local displacements by: 1 1 2 2 1 1 1 1 x x f u AE f u L                          The force-displacement equation for is: 2x f    1 2 2 1 1 x u AE f u L            The stress in terms of global displacement is:   1 1 2 2 0 0 1 1 0 0 u v C S E u L C S v                          1 1 2 2 E Cu Sv Cu Sv L      56
Stiffness Matrix for a Bar Element Example 4 - Bar Element Problem For the bar element shown below, determine the axial stress. Assume the cross-sectional area is 4 x 10-4 m2, the length is 2 m, and the E is 210 GPa. The global displacements are known as u1 = 0.25 mm, v1 = 0, u2 = 0.5 mm, and v2 = 0.75 mm.   1 1 2 2 E Cu Sv Cu Sv L       6 210 10 1 3 1 3 (0.25) (0) (0.5) (0.75) 2 2 4 2 4 KN m              2 3 81.32 10 kN    81.32 MPa  57
Stiffness Matrix for a Bar Element Solution of a Plane Truss We will now illustrate the use of equations developed above along with the direct stiffness method to solve the following plane truss example problems. A plane truss is a structure composed of bar elements all lying in a common plane that are connected together by frictionless pins. The plane truss also must have loads acting only in the common plane. 58
Stiffness Matrix for a Bar Element Example 5 - Plane Truss Problem The plane truss shown below is composed of three bars subjected to a downward force of 10 kips at node 1. Assume the cross-sectional area A = 2 in2 and E is 30 x 106 psi for all elements. Determine the x and y displacement at node 1 and stresses in each element. 59
Stiffness Matrix for a Bar Element Example 5 - Plane Truss Problem Element Node 1 Node 2  C S 1 1 2 90o 0 1 2 1 3 45o 0.707 0.707 3 1 4 0o 1 0 60
Stiffness Matrix for a Bar Element Example 5 - Plane Truss Problem The global elemental stiffness matrix are: 0 1 C S   element 1: 1 0 C S   element 2: element 3: 2 2 2 2 C S   2 2 2 2 2 2 2 2 CS CS C C S S CS CS AE L CS CS C C S S CS CS                      k 1 1 2 2 2 6 (1) 0 0 0 0 0 1 0 1 (2 )(30 10 ) 0 0 0 0 120 0 1 0 1 lb in u v u v in psi in                  k 1 1 3 3 2 6 (2) 1 1 1 1 1 1 1 1 (2 )(30 10 ) 1 1 1 1 240 2 1 1 1 1 lb in u v u v in psi in                        k 1 1 4 4 2 6 (3) 1 0 1 0 0 0 0 0 (2 )(30 10 ) 1 0 1 0 120 0 0 0 0 lb in u v u v in psi in                k 61
Stiffness Matrix for a Bar Element Example 5 - Plane Truss Problem The total global stiffness matrix is: 5 1.354 0.354 0 0 0.354 0.354 1 0 0.354 1.354 0 1 0.354 0.354 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0.354 0.354 0 0 0.354 0.354 0 0 0.354 0.354 0 0 0.354 0.354 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 5 10 lb in                               K The total global force-displacement equations are: 5 0 1.354 0.354 0 0 0.354 0.354 1 0 10, 000 0.354 1.354 0 1 0.354 0.354 0 0 2 0 0 0 0 0 0 0 0 2 0 1 0 1 0 0 0 0 0.354 0.354 0 0 0.354 0.354 0 0 3 0.354 0.354 0 0 0.354 0.354 0 0 3 1 0 0 0 0 0 1 0 4 0 0 0 0 0 0 0 0 4 5 10 F x F y F x F y F x F y                                            1 1 0 0 0 0 0 0 u v                         element 1 element 2 element 3 62
5 0 1.354 0.354 0 0 0.354 0.354 1 0 10, 000 0.354 1.354 0 1 0.354 0.354 0 0 2 0 0 0 0 0 0 0 0 2 0 1 0 1 0 0 0 0 0.354 0.354 0 0 0.354 0.354 0 0 3 0.354 0.354 0 0 0.354 0.354 0 0 3 1 0 0 0 0 0 1 0 4 0 0 0 0 0 0 0 0 4 5 10 F x F y F x F y F x F y                                            1 1 0 0 0 0 0 0 u v                         Stiffness Matrix for a Bar Element Example 5 - Plane Truss Problem Applying the boundary conditions for the truss, the above equations reduce to: 63
Stiffness Matrix for a Bar Element Example 5 - Plane Truss Problem Applying the boundary conditions for the truss, the above equations reduce to: 1 5 1 0 1.354 0.354 5 10 10,000 0.354 1.354 u v                      Solving the equations gives: 2 1 2 1 0.414 10 1.59 10 u in v in        The stress in an element is: 1 1 2 2 E Cu Sv Cu Sv L           where is the local node number i 64
Stiffness Matrix for a Bar Element Example 5 - Plane Truss Problem Element Node 1 Node 2  C S 1 1 2 90o 0 1 2 1 3 45o 0.707 0.707 3 1 4 0o 1 0 element 1   6 (1) 1 30 10 3,965 120 v psi      1 1 2 2 E Cu Sv Cu Sv L           element 2   6 (2) 1 1 30 10 (0.707) (0.707) 1,471 120 u v psi       65
element 3   6 (3) 1 30 10 1,035 120 u psi       Stiffness Matrix for a Bar Element Example 5 - Plane Truss Problem Element Node 1 Node 2  C S 1 1 2 90o 0 1 2 1 3 45o 0.707 0.707 3 1 4 0o 1 0 1 1 2 2 E Cu Sv Cu Sv L           66
Stiffness Matrix for a Bar Element Example 5 - Plane Truss Problem Let’s check equilibrium at node 1: (2) 1 cos(45 ) x x F f    (2) 1 sin(45 ) y x F f    (3) 1x f  (1) 1x f  10,000lb  67
Stiffness Matrix for a Bar Element Example 5 - Plane Truss Problem Let’s check equilibrium at node 1: 2 2 (1,471 )(2 )(0.707) (1,035 )(2 ) 0 x F psi in psi in     2 2 (3,965 )(2 ) (1,471 )(2 )(0.707) 10,000 0 y F psi in psi in      68
Stiffness Matrix for a Bar Element Example 6 - Plane Truss Problem Develop the element stiffness matrices and system equations for the plane truss below. Assume the stiffness of each element is constant. Use the numbering scheme indicated. Solve the equations for the displacements and compute the member forces. All elements have a constant value of AE/L 69
Stiffness Matrix for a Bar Element Example 6 - Plane Truss Problem Develop the element stiffness matrices and system equations for the plane truss below. Member Node 1 Node 2 Elemental Stiffness  1 1 2 k 0 2 2 3 k 3/4 3 1 3 k /2 70
Stiffness Matrix for a Bar Element Example 6 - Plane Truss Problem Compute the elemental stiffness matrix for each element. The general form of the matrix is: Member Node 1 Node 2 Elemental Stiffness  1 1 2 k 0 2 2 3 k 3/4 3 1 3 k /2 2 2 2 2 2 2 2 2 CS CS C C S S CS CS AE k L CS CS C C S S CS CS                      71
Stiffness Matrix for a Bar Element Example 6 - Plane Truss Problem For element 1: Member Node 1 Node 2 Elemental Stiffness  1 1 2 k 0 2 2 3 k 3/4 3 1 3 k /2 1 1 2 2 1 1 (1) 2 2 1 0 1 0 0 0 0 0 1 0 1 0 0 0 0 0 u v u v u v k u v                k 72
2 2 3 3 2 2 (2) 3 3 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 u v u v u v k u v                      k Stiffness Matrix for a Bar Element Example 6 - Plane Truss Problem For element 2: Member Node 1 Node 2 Elemental Stiffness  1 1 2 k 0 2 2 3 k 3/4 3 1 3 k /2 73
1 1 3 3 1 1 (3) 3 3 0 0 0 0 0 1 0 1 0 0 0 0 0 1 0 1 u v u v u v k u v                k Stiffness Matrix for a Bar Element Example 6 - Plane Truss Problem For element 3: Member Node 1 Node 2 Elemental Stiffness  1 1 2 k 0 2 2 3 k 3/4 3 1 3 k /2 74
                               1 1 2 2 3 3 1 1 2 2 3 3 2 0 2 0 0 0 0 2 0 0 0 2 2 0 3 1 1 1 2 0 0 1 1 1 1 0 0 1 1 1 1 0 2 1 1 1 3 u v u v u v u v u k v u v K Stiffness Matrix for a Bar Element Example 6 - Plane Truss Problem Assemble the global stiffness matrix by superimposing the elemental global matrices. element 1 element 2 element 3 75
Stiffness Matrix for a Bar Element Example 6 - Plane Truss Problem The unconstrained (no boundary conditions satisfied) equations are: 1 1 1 1 1 2 2 2 3 3 3 3 2 0 2 0 0 0 0 2 0 0 0 2 2 0 3 1 1 1 2 0 0 1 1 1 1 0 0 1 1 1 1 0 2 1 1 1 3 x y x y F u F v P u k P v F u F v                                                                             The displacement at nodes 1 and 3 are zero in both directions. Applying these conditions to the system equations gives:                                                                         1 1 1 2 2 2 3 3 2 0 2 0 0 0 0 0 2 0 0 0 2 0 2 0 3 1 1 1 2 0 0 1 1 1 1 0 0 1 1 1 1 0 0 2 1 1 1 3 0 x y x y F F P u k P v F F 76
Stiffness Matrix for a Bar Element Example 6 - Plane Truss Problem Applying the boundary conditions to the system equations gives: Solving this set of equations is fairly easy. The solution is: 2 1 2 2 3 1 2 1 1 u P k v P                       1 2 2 P P u k   1 2 2 3 P P v k   77
Stiffness Matrix for a Bar Element Example 6 - Plane Truss Problem Member (element) 1: Using the force-displacement relationship the force in each member may be computed.   1 2 x f k Cu   1 1 1 2 2 1 2 1 1 2 2 2 0 0 x y x y f Cu Sv Cu Sv f k f Cu Sv Cu Sv f                                   1 0 C S   2 0 y f  1 0 y f  1 2 P P k k            1 2 P P      2 2 x f k Cu  1 2 P P k k         1 2 P P   78
Stiffness Matrix for a Bar Element Example 6 - Plane Truss Problem Member (element) 2: Using the force-displacement relationship the force in each member may be computed.   2 2 2 x f k Cu Sv   2 2 2 3 3 2 3 2 2 3 3 3 0 0 x y x y f Cu Sv Cu Sv f k f Cu Sv Cu Sv f                                   1 1 2 2 C S      3 2 2 x f k Cu Sv    1 2 1 2 3 1 1 2 2 P P P P k k k                                    1 2 1 2 3 1 1 2 2 P P P P k k k                                    2 2P   2 2P  79
Stiffness Matrix for a Bar Element Example 6 - Plane Truss Problem Member (element) 3: Using the force-displacement relationship the force in each member may be computed. 1 1 3 3 0 0 0 0 x y x y f f f f     The solution to this simple problem can be readily checked by using simple static equilibrium equations. 1 1 1 3 3 1 3 1 1 3 3 3 0 0 x y x y f Cu Sv Cu Sv f k f Cu Sv Cu Sv f                                   0 1 C S   80
Stiffness Matrix for a Bar Element Example 7 - Plane Truss Problem Consider the two bar truss shown below. Determine the displacement in the y direction of node 1 and the axial force in each element. Assume E = 210 GPa and A = 6 x 10-4 m2 81
Stiffness Matrix for a Bar Element Example 7 - Plane Truss Problem The global elemental stiffness matrix for element 1 is: (1) 3 cos 0.6 5    6 4 (1) 0.36 0.48 0.36 0.48 0.48 0.64 0.48 0.64 210 10 (6 10 ) 5 0.36 0.48 0.36 0.48 0.48 0.64 0.48 0.64                         k Simplifying the above expression gives: 1 1 2 2 (1) 0.36 0.48 0.36 0.48 0.48 0.64 0.48 0.64 25,200 0.36 0.48 0.36 0.48 0.48 0.64 0.48 0.64 u v u v                      k (1) 4 sin 0.8 5    82
Stiffness Matrix for a Bar Element Example 7 - Plane Truss Problem The global elemental stiffness matrix for element 2 is: Simplifying the above expression gives: (2) cos 0   6 4 (2) 0 0 0 0 0 1 0 1 (210 10 )(6 10 ) 4 0 0 0 0 0 1 0 1                    k 1 1 3 3 (2) 0 0 0 0 0 1.25 0 1.25 25,200 0 0 0 0 0 1.25 0 1.25 u v u v                k (2) sin 1   83
Stiffness Matrix for a Bar Element Example 7 - Plane Truss Problem The total global equations are: The displacement boundary conditions are: 1 1 1 1 2 2 2 2 3 3 3 3 0.36 0.48 0.36 0.48 0 0 0.48 1.89 0.48 0.64 0 1.25 0.36 0.48 0.36 0.48 0 0 25,200 0.48 0.64 0.48 0.64 0 0 0 0 0 0 0 0 0 1.25 0 0 0 1.25 x y x y x y F u F v F u F v F u F v                                                                          1 2 2 3 3 0 u u v u v       element 1 element 2 84
                                                                     1 1 1 1 2 2 3 3 0.36 0.48 0.36 0.48 0 0 0.48 1.89 0.48 0.64 0 1.25 0.36 0.48 0.36 0.48 0 0 0 25,200 0.48 0.64 0.48 0.64 0 0 0 0 0 0 0 0 0 0 0 1.25 0 0 0 1.25 0 x y x y x y F u F v F F F F Stiffness Matrix for a Bar Element Example 7 - Plane Truss Problem The total global equations are: By applying the boundary conditions the force-displacement equations reduce to: 1 25,200(0.48 1.89 ) P v     1 2 2 3 3 x x y x y F P F F F F                       85
Stiffness Matrix for a Bar Element Example 7 - Plane Truss Problem Solving the equation gives: By substituting P = 1,000 kN and  = -0.05 m in the above equation gives: The local element forces for element 1 are: 5 1 (2.1 10 ) 0.25 v P      1 0.0337 v m  1 1 1 2 2 2 0.05 0.0337 1 1 0.6 0.8 0 0 25,200 1 1 0 0 0.6 0.8 x x u f v f u v                                       The element forces are: 1 2 76.6 76.7 x x f kN f kN    Tension 86
Stiffness Matrix for a Bar Element Example 7 - Plane Truss Problem The local element forces for element 2 are: The element forces are: 1 1 1 3 3 3 0.05 0.0337 1 1 0 1 0 0 31,500 1 1 0 0 0 1 x x u f v f u v                                       1 3 1,061 1,061 x x f kN f kN    Compression 87
Stiffness Matrix for a Bar Element Transformation Matrix and Stiffness Matrix for a Bar in Three-Dimensional Space Let’s derive the transformation matrix for the stiffness matrix for a bar element in three-dimensional space as shown below: 88
Stiffness Matrix for a Bar Element Transformation Matrix and Stiffness Matrix for a Bar in Three-Dimensional Space The coordinates at node 1 are x1, y1, and z1, and the coordinates of node 2 are x2, y2, and z2. Also, let x, y, and z be the angles measured from the global x, y, and z axes, respectively, to the local axis. 89
Stiffness Matrix for a Bar Element Transformation Matrix and Stiffness Matrix for a Bar in Three-Dimensional Space The three-dimensional vector representing the bar element is gives as: u v w u v w             d i j k i j k 90
Stiffness Matrix for a Bar Element Transformation Matrix and Stiffness Matrix for a Bar in Three-Dimensional Space Taking the dot product of the above equation with gives:  i ( ) ( ) ( ) u v w u           i i j i k i By the definition of the dot product we get: 2 1 x x x C L      i i where 2 2 2 2 1 2 1 2 1 ( ) ( ) ( ) L x x y y z z       cos x x C   where Cx, Cy, and Cz are projections of on to i, j, and k, respectively.  i 2 1 y y y C L      j i 2 1 z z z C L      k i cos y y C   cos z z C   91
Stiffness Matrix for a Bar Element Transformation Matrix and Stiffness Matrix for a Bar in Three-Dimensional Space Therefore: x y z u C u C v C w     The transformation between local and global displacements is:   * d T d 1 1 1 1 2 2 2 2 0 0 0 0 0 0 x y z y x z u v u w C C C C u C C u v w                                  0 0 0 0 0 0 x y z y x z C C C C C C        * T 92
Stiffness Matrix for a Bar Element Transformation Matrix and Stiffness Matrix for a Bar in Three-Dimensional Space The transformation from the local to the global stiffness matrix is:   T k T k T 0 0 0 0 0 0 1 1 0 1 1 0 0 0 0 0 x y z x y z y x x z y z C C C C C C AE C C C C L C C                                  k 2 2 2 2 2 2 2 2 2 2 2 2 x y x y x z x z x x y y y z y z x y x y y z y z z z x z x z x y x y x z x z x x y y y z y z x y x y y z y z z z x z x z C C C C C C C C C C C C C C C C C C C C C C C C C C C C C C AE C C C C C C C C L C C C C C C C C C C C C C C C C C C C C C C                                        k 93
Stiffness Matrix for a Bar Element Transformation Matrix and Stiffness Matrix for a Bar in Three-Dimensional Space The global stiffness matrix can be written in a more convenient form as: AE L              k 2 2 2 x x y x z x y y y z x z y z z C C C C C C C C C C C C C C C             94
Stiffness Matrix for a Bar Element Inclined, or Skewed Supports If a support is inclined, or skewed, at some angle  for the global x axis, as shown below, the boundary conditions on the displacements are not in the global x-y directions but in the x’-y’ directions. 95
Stiffness Matrix for a Bar Element Inclined, or Skewed, Supports We must transform the local boundary condition of v’3 = 0 (in local coordinates) into the global x-y system. 96
Stiffness Matrix for a Bar Element Inclined, or Skewed, Supports Therefore, the relationship between of the components of the displacement in the local and the global coordinate systems at node 3 is: 3 3 3 3 ' cos sin ' sin cos u u v v                         We can rewrite the above expression as:     3 3 3 ' [ ] d t d  We can apply this sort of transformation to the entire displacement vector as:         1 1 ' [ ] [ ] ' T d T d or d T d     3 cos sin sin cos t             97
Stiffness Matrix for a Bar Element Inclined, or Skewed, Supports Where the matrix [T1]T is: Both the identity matrix [I] and the matrix [t3] are 2 x 2 matrices. The force vector can be transformed by using the same transformation. T 1 3 [ ] [0] [0] [ ] [0] [ ] [0] [0] [0] [ ] I T I t                1 ' [ ] f T f  In global coordinates, the force-displacement equations are:     [ ] f K d  98
Stiffness Matrix for a Bar Element Inclined, or Skewed, Supports Applying the skewed support transformation to both sides of the equation gives: By using the relationship between the local and the global displacements, the force-displacement equations become: Therefore the global equations become:     1 1 [ ] [ ][ ] T f T K d       1 1 ' [ ][ ][ ] ' T f T K T d   1 1 1 1 2 2 1 1 2 2 3 3 3 3 [ ][ ][ ] ' ' ' ' x y x T y x y F u F v F u T K T F v F u F v                                              [ ] f K d      1 [ ] ' T d T d  99
Stiffness Matrix for a Bar Element Example 9 – Space Truss Problem Determine the stiffness matrix for each element. Consider the plane truss shown below. Assume E = 210 GPa, A = 6 x 10-4 m2 for element 1 and 2, and A = (6 x 10-4)m2 for element 3. 2 2 2 2 2 2 2 2 2 CS CS C C S S CS CS AE k L CS CS C C S S CS CS                      100
Stiffness Matrix for a Bar Element Example 9 – Space Truss Problem The global elemental stiffness matrix for element 1 is: (1) cos 0   1 1 2 2 6 2 4 2 (1) 0 0 0 0 0 1 0 1 (210 10 / )(6 10 ) 0 0 0 0 1 0 1 0 1 u v u v kN m m m                    k 2 2 2 2 2 2 2 2 CS CS C C S S CS CS AE k L CS CS C C S S CS CS                      (1) sin 1   101
(2) cos 1   Stiffness Matrix for a Bar Element Example 9 – Space Truss Problem The global elemental stiffness matrix for element 2 is: 2 2 3 3 6 2 4 2 (2) 1 0 1 0 0 0 0 0 (210 10 / )(6 10 ) 1 0 1 0 1 0 0 0 0 u v u v kN m m m                   k 2 2 2 2 2 2 2 2 CS CS C C S S CS CS AE k L CS CS C C S S CS CS                      (2) sin 0   102
(3) 2 cos 2   Stiffness Matrix for a Bar Element Example 9 – Space Truss Problem The global elemental stiffness matrix for element 3 is: 1 1 3 3 6 2 4 2 (3) 1 1 1 1 1 1 1 1 (210 10 / )(6 2 10 ) 1 1 1 1 2 2 1 1 1 1 u v u v kN m m m                         k 2 2 2 2 2 2 2 2 CS CS C C S S CS CS AE k L CS CS C C S S CS CS                      (3) 2 sin 2   103
Stiffness Matrix for a Bar Element Example 9 – Space Truss Problem Using the direct stiffness method, the global stiffness matrix is:                                 5 . 0 5 . 0 0 0 5 . 0 5 . 0 5 . 0 5 . 1 0 1 5 . 0 5 . 0 0 0 1 0 1 0 0 1 0 1 0 0 5 . 0 5 . 0 1 0 5 . 1 5 . 0 5 . 0 5 . 0 0 0 5 . 0 5 . 0 10 260 , 1 5 m N K We must transform the global displacements into local coordinates. Therefore the transformation [T1] is: 2 2 2 2 2 2 2 2 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0                     1 3 [ ] [0] [0] [ ] [0] [ ] [0] [0] [0] [ ] I T I t            104
Stiffness Matrix for a Bar Element Example 9 – Space Truss Problem The first step in the matrix transformation to find the product of [T1][K].    1 T K    5 0.5 0.5 0 0 0.5 0.5 0.5 1.5 0 1 0.5 0.5 0 0 1 0 1 0 0 1 0 1 0 0 0.707 0.707 0.707 0 1.414 0.707 0 0 0.707 0 0.707 0 1,260 10 1 N T K m                          2 2 2 2 2 2 2 2 5 1 0 0 0 0 0 0.5 0.5 0 0 0.5 0.5 0 1 0 0 0 0 0.5 1.5 0 1 0.5 0.5 0 0 1 0 0 0 0 0 1 0 1 0 1, 0 1 0 1 0 0 0 0 0 1 0 0 0.707 0.707 0.707 0 1.414 0.707 0 0 0 0 0 0 0.707 0 0.707 0 0 0 0 0 260 10 N m                                                   1 T   K 105
Stiffness Matrix for a Bar Element Example 9 – Space Truss Problem The next step in the matrix transformation to find the product of [T1][K][T1]T.     2 2 2 2 2 2 2 2 T 5 1 0 0 0 0 0 0.5 0.5 0 0 0.5 0.5 0 1 0 0 0 0 0.5 1.5 0 1 0.5 0.5 0 0 1 0 0 0 0 0 1 0 1 0 0 1 0 1 0 0 0 0 0 1 0 0 0.707 0.707 0.707 0 1.414 0.707 0 0 0 0 0 0 0.707 0 0.707 0 0 0 0 0 1,260 10 1 1 N T K m T                                                     T 0.5 0.5 0 0 0.707 0 0.5 1.5 0 1 0.707 0 0 0 1 0 0.707 0.707 0 1 0 1 0 0 0.707 0.707 0.707 0 1.5 0.5 0 0 0.707 0 0.5 0.5 5 1 ,260 10 1 1 T K N T m                           T 1 T    1 T K 106
Stiffness Matrix for a Bar Element Example 9 – Space Truss Problem The displacement boundary conditions are: 1 1 2 3 ' 0 u v v v     5 1 1 2 2 3 3 1 1 2 2 3 3 1 ,260 10 0.5 0.5 0 0 0.707 0 0.5 1.5 0 1 0.707 0 0 0 1 0 0.707 0.707 0 1 0 1 0 0 0.707 0.707 0.707 0 1.5 0.5 0 0 0.707 0 0.5 0.5 ' ' ' ' x y x y x y N m F F F F F F u v u v u v                                                                               107
Stiffness Matrix for a Bar Element Example 9 – Space Truss Problem By applying the boundary conditions the global force- displacement equations are: 2 5 3 1 0.707 1,260 10 ' 0.707 1.5 u N m u                Solving the equation gives: 2 11.91 u mm  2 3 1,000 ' 0 x x F kN F          3 ' 5.61 u mm  108
Stiffness Matrix for a Bar Element Example 9 – Space Truss Problem Therefore: The global nodal forces are calculated as: 1 1 2 2 2 3 3 0.5 0.5 0 0 0.707 0 0 0.5 1.5 0 1 0.707 0 0 0 0 1 0 0.707 0.707 11.91 0 1 0 1 0 0 0 0.707 0.707 0.707 0 1.5 0.5 5.61 0 0 0.707 0 0.5 0.5 0 1,260 10 ' ' x y x N mm y x y F F F F F F                                                               1 1 500 500 x y F kN F kN     2 3 0 ' 707 y y F F kN   1 1 1 1 2 2 1 1 2 2 3 3 3 3 [ ][ ][ ] ' ' ' ' x y x T y x y F u F v F u T K T F v F u F v                                          109
Stiffness Matrix for a Bar Element Potential Energy Approach to Derive Bar Element Equations The differential internal work (strain energy) dU in a one- dimensional bar element is: Let’s derive the equations for a bar element using the principle of minimum potential energy. The total potential energy, p, is defined as the sum of the internal strain energy U and the potential energy of the external forces : p U     ( )( )( ) x x dU y z x d       110
Stiffness Matrix for a Bar Element Potential Energy Approach to Derive Bar Element Equations Summing the differential energy over the whole bar gives: If we let the volume of the element approach zero, then: x x dU d dV    0 x x x V U d dV                 For a linear-elastic material (Hooke’s law) as shown below: x x E    0 x x x V E d dV                 2 1 2 x V E dV    1 2 x x V U dV     111
Stiffness Matrix for a Bar Element Potential Energy Approach to Derive Bar Element Equations The potential energy of the external forces is: The internal strain energy statement becomes 1 2 x x V U dV     s V S X u dV T u dS f u          M b x ix i i 1 where Xb is the body force (force per unit volume), Tx is the traction (force per unit area), and fix is the nodal concentrated force. All of these forces are considered to act in the local x direction. 112
Stiffness Matrix for a Bar Element Potential Energy Approach to Derive Bar Element Equations 1. Formulate an expression for the total potential energy. 2. Assume a displacement pattern. 3. Obtain a set of simultaneous equations minimizing the total potential energy with respect to the displacement parameters. Apply the following steps when using the principle of minimum potential energy to derive the finite element equations. 113
Stiffness Matrix for a Bar Element Potential Energy Approach to Derive Bar Element Equations We can approximate the axial displacement as: Consider the following bar element, as shown below: 2 2 0 L p x x 1x 1 x s V S A dx f u f u 2 X u dV T u dS            b x   1 1 2 2 u u N N u        1 1 x N L   2 x N L  114
Stiffness Matrix for a Bar Element Potential Energy Approach to Derive Bar Element Equations where N1 and N2 are the interpolation functions gives as: Using the stress-strain relationships, the axial strain is: x du dx   1 2 1 1 x u u L L                1 1 B L L         The axial stress-strain relationship is:     [ ] x x D      [ ]{ } x B d   1 1 2 2 u dN dN u dx dx              115
Stiffness Matrix for a Bar Element Potential Energy Approach to Derive Bar Element Equations The total potential energy expressed in matrix form is: For the one-dimensional stress-strain relationship [D] = [E] where E is the modulus of elasticity. Therefore, stress can be related to nodal displacements as: where {P} represented the concentrated nodal loads.     [ ][ ] x D B d                   0 L T T T T p x x V S A dx d P u X dV u T dS 2           b x 116
Stiffness Matrix for a Bar Element Potential Energy Approach to Derive Bar Element Equations If we substitute the relationship between and into the energy equations we get: û d̂                          0 L T T T T p T T T T s V S A d B D B d dx d P 2 d N X dV d N T dS         b x In the above expression for potential energy p is a function of the d, that is: p = p( ). 1, u 2 u However, [B] and [D] and the nodal displacements u are not a function of x. x      x      117
Stiffness Matrix for a Bar Element Potential Energy Approach to Derive Bar Element Equations where Integration the energy expression with respect to x gives:         [ ] [ ] [ ] 2 T T T T p AL d B D B d d f            [ ] [ ] V S f P N X dV N X dS      T T b b We can define the surface tractions and body-force matrices as:     [ ] x S f N T dS   T s     [ ] V f N X dV   T b b 118
Stiffness Matrix for a Bar Element Potential Energy Approach to Derive Bar Element Equations Minimization of p with respect to each nodal displacement requires that: For convenience, let’s define the following 1 2 0 0 p p u u               * [ ] [ ] [ ] T T T U d B D B d      1 * 1 2 2 1 1 1 [ ] 1 u L U u u E u L L L                              119
Stiffness Matrix for a Bar Element Potential Energy Approach to Derive Bar Element Equations Simplifying the above expression gives: The loading on a bar element is given as:   * 2 2 1 1 2 2 2 2 E U u u u u L        1 1 2 2 T x x d f u f u f   Therefore, the minimum potential energy is:   1 2 1 1 2 2 0 2 p x AE u u f u L          1 2 2 2 2 2 0 2 p x AE u u f u L         120
Stiffness Matrix for a Bar Element Potential Energy Approach to Derive Bar Element Equations The above equations can be written in matrix form as: The stiffness matrix for a bar element is: This form of the stiffness matrix obtained from the principle of minimum potential energy is identical to the stiffness matrix derived from the equilibrium equations.   1 1 2 2 1 1 0 1 1 p x x u f AE u f d L                            1 1 1 1 AE k L          121
Stiffness Matrix for a Bar Element Example 10 - Bar Problem Consider the bar shown below: The energy equivalent nodal forces due to the distributed load are:     0 [ ] x S f N T dS   T     0 0 L x 1 f L f Cx dx f x L                         1x 2x 122
Stiffness Matrix for a Bar Element Example 10 - Bar Problem The total load is the area under the distributed load curve, or: The equivalent nodal forces for a linearly varying load are: f f       1x 2x 2 1 ( )( ) 2 2 CL F L CL   1 1 of the total load 3 3 x F f   2 2 2 of the total load 3 3 x F f     0 L x 1 L Cx dx x L                  0 0 L 2 3 L 3 Cx Cx 2 3L Cx 3L                 2 2 CL 6 CL 3                123
Stiffness Matrix for a Bar Element Example 11 - Bar Problem Consider the axially loaded bar shown below. Determine the axial displacement and axial stress. Let E = 30 x 106 psi, A = 2 in2, and L = 60 in. Use (a) one and (b) two elements in the finite element solutions. 124
Stiffness Matrix for a Bar Element Example 11 - Bar Problem The one-element solution: The distributed load can be converted into equivalent nodal forces using:     0 [ ] x S F N T dS   T   0 1x 2x F F F          0 10 L x 1 L x dx x L                   125
Stiffness Matrix for a Bar Element Example 11 - Bar Problem The one-element solution:   0 10 L 1x 2x x 1 F L x dx F x L                         6,000 12,000 1x 2x F lb F lb                2 2 2 10 10 - + 2 3 10 - 3 L L L                2 2 10 - 6 10 - 3 L L                126
Stiffness Matrix for a Bar Element Example 11 - Bar Problem The one-element solution: The element equations are: (1) 6 1 1 10 1 1          k 1 6 2 6,000 1 1 10 12,000 1 1 0 x u R                        1 0.006 u in   The second equation gives: 6 1 2 10 ( ) 12,000 x u R    2 18,000 x R lb   127
Stiffness Matrix for a Bar Element Example 11 - Bar Problem The one-element solution: The axial stress-strain relationship is: { } [ ]{ } x x D      { } [ ] x E B d   1 2 1 1 u E u L L               6 0 0.006 30 10 60          2 1 u u E L         3,000 ( ) psi T  128
Stiffness Matrix for a Bar Element Example 11 - Bar Problem The two-element solution: The distributed load can be converted into equivalent nodal forces. For element 1, the total force of the triangular-shaped distributed load is: 1 (30 .)(300 ) 4,500 2 lb in in lb   129
Stiffness Matrix for a Bar Element Example 11 - Bar Problem The two-element solution: Based on equations developed for the equivalent nodal force of a triangular distributed load, develop in the one-element problem, the nodal forces are: (1) 1 (1) 2 1 (4,500) 3 2 (4,500) 3 x x f f                        1,500 3,000 lb lb          130
Stiffness Matrix for a Bar Element Example 11 - Bar Problem The two-element solution: For element 2, the applied force is in two parts: a triangular- shaped distributed load and a uniform load. The uniform load is: (30 )(300 / ) 9,000 in lb in lb   (2) 2 (2) 3 1 1 (9,000) (4,500) 2 3 1 2 (9,000) (4,500) 2 3 x x f f                                     The nodal forces for element 2 are: 6,000 7,500 lb lb          131
Stiffness Matrix for a Bar Element Example 11 - Bar Problem The two-element solution: The final nodal force vector is: The element stiffness matrices are: (1) 1 1 (1) (2) 2 2 2 (2) 3 3 x x x x x x x F f F f f F f                       (1) (2) 1 1 2 1 1 AE L           k k 1 2 2 3 3 1,500 9,000 7,500 x R               132
Stiffness Matrix for a Bar Element Example 11 - Bar Problem The two-element solution: The assembled global stiffness matrix is: The assembled global force-displacement equations are: 6 1 1 0 2 10 1 2 1 0 1 1                 K 1 6 2 3 1 1 0 1,500 2 10 1 2 1 9,000 0 1 1 0 7,500 x u u R                                        element 1 element 2 133
Stiffness Matrix for a Bar Element Example 11 - Bar Problem The two-element solution: After the eliminating the row and column associated with u3x, we get: Solving the equation gives: 1 6 2 1 1 1,500 2 10 1 2 9,000 u u                         1 2 0.006 0.00525 u in u in     6 2 3 2 10 7,500 x u R     3 18,000 x R   Solving the last equation gives: 134
Stiffness Matrix for a Bar Element Example 11 - Bar Problem The two-element solution: The axial stress-strain relationship is: 1 (1) 2 1 1 x x x d E d L L                0.006 1 1 30 30 0.00525 E                 750 ( ) psi T  135
Stiffness Matrix for a Bar Element Example 11 - Bar Problem The two-element solution: The axial stress-strain relationship is: 2 (2) 3 1 1 x x x d E d L L                0.00525 1 1 30 30 0 E                5,250 ( ) psi T  136
Stiffness Matrix for a Bar Element Comparison of Finite Element Solution to Exact Solution In order to be able to judge the accuracy of our finite element models, we will develop an exact solution for the bar element problem. The exact solution for the displacement may be obtained by: where the force P is shown on the following free-body diagram. 0 1 ( ) L u P x dx AE     1 ( ) (10 ) 2 P x x x  2 5x  137
Stiffness Matrix for a Bar Element Comparison of Finite Element Solution to Exact Solution Therefore: Applying the boundary conditions: 0 1 ( ) L u P x dx AE   2 1 5 x o u x dx AE   3 1 5 ( ) 0 3 x u L C AE    The exact solution for axial displacement is:   3 3 5 ( ) 3 u L x L AE   ( ) ( ) P x x A   3 1 5 3 x C AE   3 1 5 3 L C AE    2 5x A  138
Stiffness Matrix for a Bar Element Comparison of Finite Element Solution to Exact Solution A plot of the exact solution for displacement as compared to several different finite element solutions is shown below. One element Two elements 139
Stiffness Matrix for a Bar Element Comparison of Finite Element Solution to Exact Solution A plot of the exact solution for axial stress as compared to several different finite element solutions is shown below. One element Two elements 140
Stiffness Matrix for a Bar Element Comparison of Finite Element Solution to Exact Solution A plot of the exact solution for axial stress at the fixed end (x = L) as compared to several different finite element solutions is shown below. 141
Stiffness Matrix for a Bar Element Galerkin’s Residual Method and Its Application to a One-Dimensional Bar There are a number of weighted residual methods. However, the Galerkin’s method is more well-known and will be the only weighted residual method discussed in this course. In weighted residual methods, a trial or approximate function is chosen to approximate the independent variable (in our case, displacement) in a problem defined by a differential equation. The trial function will not, in general, satisfy the governing differential equation. Therefore, the substitution of the trial function in the differential equation will create a residual over the entire domain of the problem. 142
Stiffness Matrix for a Bar Element Galerkin’s Residual Method and Its Application to a One-Dimensional Bar Therefore, the substitution of the trial function in the differential equation will create a residual over the entire domain of the problem. minimum V RdV   In the residual methods, we require that a weighted value of the residual be a minimum over the entire domain of the problem. The weighting function W allows the weighted integral of the residuals to go to zero. 0 V RW dV   143
Stiffness Matrix for a Bar Element Galerkin’s Residual Method and Its Application to a One-Dimensional Bar Using Galerkin’s weighted residual method, we require the weighting functions to be the interpolation functions Ni. Therefore: 0 1, 2, , i V RN dV i n     144
Stiffness Matrix for a Bar Element Example 12 - Bar Element Formulation Let’s derive the bar element formulation using Galerkin’s method. The governing differential equation is: Applying Galerkin’s method we get: 0 d du AE dx dx        0 0 1, 2, , L i d du AE N dx i n dx dx           We now apply integration by parts using the following general formula: rds rs sdr     145
Stiffness Matrix for a Bar Element Example 12 - Bar Element Formulation If we assume the following: then integration by parts gives: i r N  d du ds AE dx dx dx        0 0 0 L L i i dN du du N AE AE dx dx dx dx    0 L i d du AE N dx dx dx         rds rs sdr     i dN dr dx dx  du s AE dx  146
Stiffness Matrix for a Bar Element Example 12 - Bar Element Formulation Recall that: 1 2 1 2 dN dN du u u dx dx dx   1 2 1 1 u du u dx L L               Our original weighted residual expression, with the approximation for u becomes: 1 2 0 1 1 L i u dN AE dx u dx L L               0 L i du N AE dx  147
Stiffness Matrix for a Bar Element Example 12 - Bar Element Formulation Substituting N1 for the weighting function Ni gives: 1 1 2 0 1 1 L u dN AE dx u dx L L               1 2 0 1 1 1 L u AE dx u L L L                        1 2 1x AE u u f L    1 0 L du N AE dx 0 x du AE dx   0 x x AE   1x f  0 x x A     1 2 2 AEL u u L   1 0 L du N AE dx  148
Stiffness Matrix for a Bar Element Example 12 - Bar Element Formulation Substituting N2 for the weighting function Ni gives: 1 2 2 0 1 1 L u dN AE dx u dx L L               1 2 0 1 1 1 L u AE dx u L L L                       1 2 2x AE u u f L     2 0 L du N AE dx x L du AE dx   x x L AE   2x f  x x L A     1 2 2 AEL u u L    2 0 L du N AE dx  149
Stiffness Matrix for a Bar Element Example 12 - Bar Element Formulation Writing the last two equations in matrix form gives: 1 1 2 2 1 1 1 1 x x u f AE u f L                      This element formulation is identical to that developed from equilibrium and the minimum potential energy approach. 150

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