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MANE 4240 & CIVL 4240 Introduction to Finite Elements Prof. Suvranu De Development of Truss Equations Reading assignment: Chapter 3: Sections 3.1-3.9 + Lecture notes Summary: • Stiffness matrix of a bar/truss element • Coordinate transformation • Stiffness matrix of a truss element in 2D space •Problems in 2D truss analysis (including multipoint constraints) •3D Truss element Trusses: Engineering structures that are composed only of two-force members. e.g., bridges, roof supports Actual trusses: Airy structures composed of slender members (I-beams, channels, angles, bars etc) joined together at their ends by welding, riveted connections or large bolts and pins A typical truss structure Gusset plate Ideal trusses: Assumptions • Ideal truss members are connected only at their ends. • Ideal truss members are connected by frictionless pins (no moments) • The truss structure is loaded only at the pins • Weights of the members are neglected A typical truss structure Frictionless pin These assumptions allow us to idealize each truss member as a two-force member (members loaded only at their extremities by equal opposite and collinear forces) member in compression member in tension Connecting pin FEM analysis scheme Step 1: Divide the truss into bar/truss elements connected to each other through special points (“nodes”) Step 2: Describe the behavior of each bar element (i.e. derive its stiffness matrix and load vector in local AND global coordinate system) Step 3: Describe the behavior of the entire truss by putting together the behavior of each of the bar elements (by assembling their stiffness matrices and load vectors) Step 4: Apply appropriate boundary conditions and solve Stiffness matrix of bar element E, A © 2002 Brooks/Cole Publishing / Thomson Learning™ L: Length of bar A: Cross sectional area of bar E: Elastic (Young’s) modulus of bar uˆ ( xˆ ) :displacement of bar as a function of local coordinate The strain in the bar at xˆ d uˆ ε( xˆ ) d xˆ The stress in the bar (Hooke’s law) ( xˆ ) E ε( xˆ ) xˆ of bar dˆ 2x Tension in the bar T( xˆ ) EA ε dˆ 1x xˆ Assume that the displacement uˆ ( xˆ ) xˆ L is varying linearly along the bar xˆ ˆ xˆ ˆ uˆ ( xˆ ) 1 d 1x d 2x L L Then, strain is constant along the bar: Stress is also constant along the bar: Tension is constant along the bar: xˆ xˆ uˆ ( xˆ ) 1 dˆ 1x dˆ 2x L L ε d uˆ d xˆ dˆ 2x dˆ 1x L E ˆ Eε d 2x dˆ 1x L EA ˆ T EA ε d 2x dˆ 1x L k The bar is acting like a spring with stiffness k EA L Recall the lecture on springs E, A © 2002 Brooks/Cole Publishing / Thomson Learning™ Two nodes: 1, 2 Nodal displacements: dˆ 1x dˆ 2x Nodal forces: fˆ1x fˆ 2x Spring constant: k EA L Element stiffness matrix in local coordinates Element force vector - k fˆ1x k dˆ 1x ˆf kˆ dˆ ˆ ˆ k - k d 2x 2x f Element nodal Element stiffness matrix displacement vector fˆ kˆ dˆ What if we have 2 bars? E1, A1 E2, A2 L2 L1 This is equivalent to the following system of springs k1 E 1A 1 L1 k2 E 2A 2 L2 x Element 1 2 Element 23 1 d1x PROBLEM d2x d3x Problem 1: Find the stresses in the two-bar assembly loaded as shown below E, 2A E, A P 1 2 3 L L Solution: This is equivalent to the following system of springs k1 1 2E A L k2 EA L x Element 1 2 Element 23 d1x d2x d3x We will first compute the displacement at node 2 and then the stresses within each element The global set of equations can be generated using the technique developed in the lecture on “springs” k1 k 1 0 here k1 k1 k 2 k2 d1 x d 3 x 0 0 d 1x F1x k 2 d 2 x F2 x F k 2 d 3x 3x and F2 x P Hence, the above set of equations may be explicitly written as k 1 d 2 x F1 x (1) ( k1 k 2 ) d 2 x P (2 ) k 2 d 2 x F3 x (3) From equation (2) d 2 x P k1 k 2 PL 3EA To calculate the stresses: For element #1 first compute the element strain (1) d 2 x d1x d2x L L P 3EA and then the stress as (1) E P (1) (element in tension) 3A Similarly, in element # 2 (2) d3x d2x L (2) E (2) d2x L P 3A P 3EA (element in compression) © 2002 Brooks/Cole Publishing / Thomson Learning™ Inter-element continuity of a two-bar structure Bars in a truss have various orientations member in compression member in tension Connecting pin d 2y , f 2y y dˆ 2y , fˆ2y 0 yˆ dˆ 2x , fˆ 2x d 2x , f 2x d1y , f1y θ dˆ 1y , fˆ1y 0 dˆ 1x , fˆ1x xˆ d1x , f1x x At node 1: θ dˆ 1x d1x θ f1x dˆ 1y d1y fˆ1y 0 fˆ1 x f1y At node 2: θ dˆ 2 x f 2x d 2y fˆ 2y 0 d 2x θ dˆ 2y fˆ 2 x f 2y In the global coordinate system, the vector of nodal displacements and loads d 1x d 1y d ; d 2x d 2y f 1x f 1y f f 2x f 2y Our objective is to obtain a relation of the form f k d 4 1 4 4 4 1 Where k is the 4x4 element stiffness matrix in global coordinate system The key is to look at the local coordinates xˆ y dˆ 2y , fˆ2y 0 yˆ dˆ 2x , fˆ 2x dˆ 1y , fˆ1y 0 fˆ1x k ˆ f 2x - k θ - k dˆ 1x ˆ k d 2x dˆ 1x , fˆ1x k x Rewrite as fˆ 1x k 0 fˆ1y ˆ f 2x - k ˆ 0 f 2y 0 -k 0 0 0 k 0 0 EA L ˆ 0 d 1x ˆ 0 d 1y ˆ 0 d 2x 0 dˆ 2y fˆ kˆ dˆ NOTES 1. Assume that there is no stiffness in the local ^y direction. 2. If you consider the displacement at a point along the local x direction as a vector, then the components of that vector along the global x and y directions are the global x and y displacements. 3. The expanded stiffness matrix in the local coordinates is symmetric and singular. NOTES 5. In local coordinates we have fˆ kˆ dˆ 4 1 4 4 4 1 But or goal is to obtain the following relationship f k d 4 1 4 4 4 1 Hence, need a relationship between dˆ and d dˆ and between fˆ and f 1y ˆ d 1x d 1x ˆ d 1y ˆ d 1y d d ˆ d 2x d 2x d 2y ˆ d 2y θ dˆ 1x d1x θ d 2x d1y dˆ 2y dˆ 2 x d 2y Need to understand how the components of a vector change with coordinate transformation Transformation of a vector in two dimensions yˆ y v y cos θ vx ˆx v θ v vˆ y vy v x sin θ xˆ Angle q is measured positive in the counter clockwise direction from the +x axis) θ v x cos θ v y sin θ x The vector v has components (vx, vy) in the global coordinate system and (v^x, v^y) in the local coordinate system. From geometry ˆ x v x cos θ v y sin θ v ˆ y v x sin θ v y cos θ v In matrix form ˆ x cos θ v ˆ v y sin θ sin θ v x v cos θ y Direction cosines Or ˆx l v ˆ v y m m vx l vy where l cos q m sin q Transformation matrix for a single vector in 2D l T m * where m l relates * ˆ T v v ˆx v vx ˆ and v v ˆ vy vy are components of the same vector in local and global coordinates, respectively. Relationship between dˆ and d for the truss element dˆ 1y At node 1 At node 2 d dˆ 1x * 1x ˆ T d1y d1y d dˆ 2x * 2x ˆ T d d 2y 2y Putting these together dˆ T d dˆ 1x θ dˆ 2y d1x θ dˆ 2 x d 2y d 2x dˆ 1x m 0 0 d 1x l m l 0 0 d 1y dˆ 1y ˆ d 0 0 l m 2x d 2x ˆ 0 0 m l d 2y d 2y d T dˆ d1y T * T 4 4 0 0 * T Relationship between fˆ and f for the truss element At node 1 At node 2 f fˆ1x * 1x ˆ T f1y f1y f fˆ2x * 2x ˆ T f 2y f 2y Putting these together fˆ T f fˆ1x m 0 0 f 1x l m l 0 0 f 1y fˆ1y ˆ 0 l m f 2x f 2x 0 ˆ 0 0 m l f 2y f 2y f T fˆ fˆ1 y θ f1y fˆ1 x fˆ 2 y f1x θ fˆ 2 x f 2y f 2x T * T 4 4 0 0 * T Important property of the transformation matrix T The transformation matrix is orthogonal, i.e. its inverse is its transpose T 1 T T Use the property that l2+m2=1 Putting all the pieces together xˆ y dˆ 2y , fˆ 2y yˆ dˆ T d dˆ 2x , fˆ 2x dˆ 1y , fˆ1y fˆ kˆ dˆ θ dˆ 1x , fˆ1x T f kˆ T d x The desired relationship is Where k T 4 4 fˆ T f T kˆ T 4 4 4 4 4 4 f k d 4 1 f T kˆ T d 1 k 4 4 4 1 is the element stiffness matrix in the global coordinate system l m T 0 0 m 0 l 0 0 l 0 m 0 0 m l l2 EA lm T ˆ k T kT L l2 lm k 0 ˆk - k 0 l lm m 2 lm 2 lm l m lm 2 2 0 -k 0 0 0 k 0 0 lm m lm m 2 2 0 0 0 0 Computation of the direction cosines l cos q x2 x1 L 2 (x2,y2) L m sin q y2 y1 θ 1 L (x1,y1) What happens if I reverse the node numbers? l ' cos q x1 x2 L l L m' sin q y1 y2 L m θ 2 (x ,y ) 2 2 Question: Does the stiffness matrix change? 1 (x1,y1) Example Bar element for stiffness matrix evaluation E 30 10 © 2002 Brooks/Cole Publishing / Thomson Learning™ A 2 in 6 psi 2 L 60 in q 30 l cos 30 3 2 3 4 3 6 30 10 2 4 k 60 3 4 3 4 3 4 1 4 3 3 4 3 4 3 4 1 4 3 4 4 3 4 1 4 lb in 3 4 1 4 m sin 30 1 2 Computation of element strains © 2002 Brooks/Cole Publishing / Thomson Learning™ Recall that the element strain is ε dˆ 2x dˆ 1x L 1 L 1 L 1 L 1 1 0 1 0 dˆ 1 0 1 0 T d 0 1 dˆ 1x dˆ 1y 0 ˆ d 2x ˆ d 2y ε 1 L 1 L 1 L 1 l l 0 l m 0 0 0 1 m m l l m 0 l 0 0 l 0 m m d d 1x d 1y m d 2x d 2y 0 d m l 0 Computation of element stresses stress and tension Recall that the element stress is Eε E ˆ E d 2x dˆ 1x l L L Recall that the element tension is T EAε EA L l m l md m l m d Steps in solving a problem Step 1: Write down the node-element connectivity table linking local and global nodes; also form the table of direction cosines (l, m) Step 2: Write down the stiffness matrix of each element in global coordinate system with global numbering Step 3: Assemble the element stiffness matrices to form the global stiffness matrix for the entire structure using the node element connectivity table Step 4: Incorporate appropriate boundary conditions Step 5: Solve resulting set of reduced equations for the unknown displacements Step 6: Compute the unknown nodal forces Node element connectivity table ELEMENT Node 1 Node 2 1 1 2 2 2 3 3 3 1 1 El 1 2 60 60 L El 3 60 El 2 θ 3 1 (x ,y ) 1 1 2 (x2,y2) Stiffness matrix of element 1 d1x d1y d2x d2y d1x d1y (1 ) k d2x d2y Stiffness matrix of element 3 d3x d3y d1x d1y d3x d3y (3) k d1x d1y Stiffness matrix of element 2 d2x d2y d3x d3y d2x d2y (2) k d3x d3y There are 4 degrees of freedom (dof) per element (2 per node) k (1 ) Global stiffness matrix d1x d1y d2x d2y d3x d3y d1x d1y d2x K d 2y d3x d3y 6 6 How do you incorporate boundary conditions? k k (2) (3) Example 2 The length of bars 12 and 23 are equal (L) E: Young’s modulus A: Cross sectional area of each bar Solve for P1 (1) d and d 2x 2y (2) Stresses in each bar y 3 El#2 P2 El#1 2 45o 1 x Solution Step 1: Node element connectivity table ELEMENT Node 1 Node 2 1 1 2 2 2 3 Table of nodal coordinates Node x y 1 0 0 2 Lcos45 Lsin45 3 0 2Lsin45 Table of direction cosines ELEMENT Length l x2 x1 length m y2 y1 length 1 L cos45 sin45 2 L -cos45 sin45 Step 2: Stiffness matrix of each element in global coordinates with global numbering Stiffness matrix of element 1 k (1) l2 E A lm L l 2 lm d1x 1 EA 1 2L 1 1 l lm m 2 lm 2 lm l m lm 2 2 d1y d2x 1 1 1 1 1 1 1 1 lm 2 m lm 2 m d2y 1 1 1 1 d1x d1y d2x d2y Stiffness matrix of element 2 k (2) d2x d2y d3x d3y 1 1 1 1 1 1 1 1 1 1 1 1 EA 1 2L 1 1 1 d2x d2y d3x d3y Step 3: Assemble the global stiffness matrix 1 1 EA 1 K 2L 1 0 0 1 1 1 0 1 1 1 0 1 2 0 1 1 0 2 1 0 1 1 1 0 1 1 1 0 0 1 1 1 1 The final set of equations is K d F Step 4: Incorporate boundary conditions 0 0 d2x d d2y 0 0 Hence reduced set of equations to solve for unknown displacements at node 2 EA 2 2 L 0 0 d2x P1 d 2 2y P2 Step 5: Solve for unknown displacements d2x d2y P1 L EA P2 L EA Step 6: Obtain stresses in the elements For element #1: (1) E 1 L 2 E 2L 1 2 (d 2 x d 2 y ) 1 2 P1 P2 A 2 d1x d 1 1y d 2 2x d2 y 0 0 For element #2: (2) E 1 L 2 E 2L 1 2 (d 2 x d 2 y ) 1 2 P1 P2 A 2 d2x d 1 2y d 2 3x d3y 0 0 Multi-point constraints © 2002 Brooks/Cole Publishing / Thomson Learning™ Figure 3-19 Plane truss with inclined boundary conditions at node 3 (see problem worked out in class) Problem 3: For the plane truss y P 3 El#2 2 El#1 P=1000 kN, L=length of elements 1 and 2 = 1m E=210 GPa A = 6×10-4m2 for elements 1 and 2 = 6 2 ×10-4 m2 for element 3 El#3 45o 1 x Determine the unknown displacements and reaction forces. Solution Step 1: Node element connectivity table ELEMENT Node 1 Node 2 1 2 3 1 2 1 2 3 3 Table of nodal coordinates Node x y 1 0 0 2 0 L 3 L L Table of direction cosines ELEMENT Length l x2 x1 m y2 y1 length length 1 L 0 1 2 L 1 0 3 L 2 1/ 2 1/ 2 Step 2: Stiffness matrix of each element in global coordinates with global numbering Stiffness matrix of element 1 k (1) l2 E A lm L l 2 lm lm l m lm lm 2 m lm 2 m d1x d1y d2x l lm m 2 lm 2 2 2 0 9 -4 (210 10 )(6 10 ) 0 0 1 0 0 0 1 0 0 0 1 0 d2y 1 0 1 0 d1x d1y d2x d2y Stiffness matrix of element 2 d2x d2y d3x d3y 1 9 -4 (210 10 )(6 10 ) 0 1 1 0 0 1 0 0 0 1 0 0 0 0 0 0 k (2) d2x d2y d3x d3y Stiffness matrix of element 3 k (3) d1x d1y d3x 0.5 9 -4 (210 10 )(6 2 10 ) 0.5 0.5 2 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 d3y 0.5 0.5 0.5 0.5 d1x d1y d3x d3y Step 3: Assemble the global stiffness matrix 0.5 0.5 0 5 K 1260 10 0 0.5 0.5 0.5 0 0 0.5 1.5 0 1 0.5 0 1 0 1 1 0 1 0 0.5 1 0 1.5 0.5 0 0 0.5 The final set of equations is K d F 0.5 0.5 0 0 0.5 0.5 Eq(1) N/m Step 4: Incorporate boundary conditions y 0 0 d2x d 0 d3x d3y Also, d 3y 0 P y x 3 El#2 2 El#1 El#3 45o 1 x in the local coordinate system of element 3 How do I convert this to a boundary condition in the global (x,y) coordinates? y F1 x F 1y P F F 2y F3 x F3 y Also, F 3x 0 x y P 3 El#2 2 El#1 El#3 45o 1 x in the local coordinate system of element 3 How do I convert this to a boundary condition in the global (x,y) coordinates? Using coordinate transformations d 3x d 3y l m d 3x d 3y m d3x d l m l 3y 1 2 1 2 d 3y 0 d 2 1 1 2 d3x 2 1 d3y 1 2 2 (Multi-point constraint) 3y 1 1 2 d d3 y d3x 0 3y d3x Eq (2) 0 d d 3x 3 y d 3 y d 3 x Similarly for the forces at node 3 F F 3x 3y F F l m 3x 3y m F3 x F l m n 3y 1 2 1 2 3x 2 1 1 2 F3 x 2 1 F3 y 1 2 2 F 3x 0 F 1 1 2 F F3 y F3 x 0 3y F3 x Eq (3) 0 F F 3x 3 y F3 y F3 x Therefore we need to solve the following equations simultaneously Kd F Eq(1) Eq(2) d3 y d3x 0 Eq(3) F3 y F3 x 0 Incorporate boundary conditions and reduce Eq(1) to 1 5 1260 10 1 0 1 1 .5 0 .5 0 0 .5 0 .5 d2x d3x d 3y P F3 x F 3y Write these equations out explicitly Eq(4) 1260 10 (d 2 x d 3x ) P 5 1 2 6 0 1 0 ( d 2 x 1 .5 d 3 x 0 .5 d 3 y ) F3 x Eq(5) 1 2 6 0 1 0 (0 .5 d 3 x 0 .5 d 3 y ) F3 y Eq(6) 5 5 Add Eq (5) and (6) 1 2 6 0 1 0 ( d 2 x 2 d 3 x d 3 y ) F3 x F3 y 0 using 5 1260 10 ( d 2 x 3d 3 x ) 0 5 using Eq(2) d 2 x 3 d 3 x Eq(7) 1 2 6 0 1 0 (3 d 3 x d 3 x ) P 5 Plug this into Eq(4) 2520 10 d3x 10 5 6 Eq(3) d 3 x 0 .0 0 3 9 6 8 m d 2 x 3 d 3 x 0 .0 1 1 9 m Compute the reaction forces F1 x F 1y F2 y F 3x F3 y 5 1 2 6 0 1 0 500 500 0 kN 500 500 0 0 0 1 0 0 .5 0 .5 0 1 .5 0 .5 0 .5 0 .5 0 0 .5 0 .5 d2x d3x d 3y Physical significance of the stiffness matrix In general, we will have a stiffness matrix of the form k 11 K k 21 k 31 k 12 k 22 k 32 k 13 k 23 k 33 And the finite element force-displacement relation k 11 k 21 k 31 k 12 k 22 k 32 k 13 d 1 F1 k 23 d 2 F2 k 33 d 3 F3 Physical significance of the stiffness matrix The first equation is k 11 d 1 k 12 d 2 k 13 d 3 F1 Force equilibrium equation at node 1 Columns of the global stiffness matrix What if d1=1, d2=0, d3=0 ? F1 k 11 F2 k 21 F3 k 31 While d.o.f 2 and 3 are held fixed Force along d.o.f 1 due to unit displacement at d.o.f 1 Force along d.o.f 2 due to unit displacement at d.o.f 1 Force along d.o.f 3 due to unit displacement at d.o.f 1 Similarly we obtain the physical significance of the other entries of the global stiffness matrix In general k ij = Force at d.o.f ‘i’ due to unit displacement at d.o.f ‘j’ keeping all the other d.o.fs fixed Example The length of bars 12 and 23 are equal (L) E: Young’s modulus A: Cross sectional area of each bar Solve for d2x and d2y using the “physical P1 interpretation” approach y 3 El#2 P2 El#1 2 45o 1 x Solution Notice that the final set of equations will be of the form k 11 k 21 k 12 d 2 x P1 d k 22 2 y P2 Where k11, k12, k21 and k22 will be determined using the “physical interpretation” approach To obtain the first column 2 1.cos(45) y 3 F2y=k21 El#2 2 2 x 1 1.cos(45) d2x=1 Force equilibrium F F x k11 T1 cos(45) T 2 cos(45) 0 y d2x 1 d2y 0 F2y=k21 T2 F2x=k11 2’ El#1 1 1 k1 1 apply k 21 y k 21 T1 sin(45) T 2 sin(45) 0 F2x=k11 T1 2 x 1 2 Force-deformation relations T1 EA T2 EA L L 1 2 Combining force equilibrium and force-deformation relations k11 k 21 T1 T2 2 EA 2L T1 T2 2 EA 2L 1 2 1 2 Now use the geometric (compatibility) conditions (see figure) 1 1.cos(45) 1 2 1.cos(45) 1 2 2 Finally k 11 EA k 21 EA 2L 2L 1 2 EA 1 2 0 2L ( 2 2 ) EA L k To obtain the second column 1 2 apply d 2 x 0 y d2y 1 k 2 2 y 3 1 1.cos(45) F2y=k22 2 2’ T2 El#2 d2y=1 F2x=k12 2 El#1 T1 2 x 1 x 1.cos(45) 1 2 1 Force equilibrium F F x k12 T1 cos(45) T 2 cos(45) 0 y k 22 T1 sin(45) T 2 sin(45) 0 2 Force-deformation relations T1 EA T2 EA L L 1 2 Combining force equilibrium and force-deformation relations k12 k 22 T1 T2 EA 2 T1 T2 2L EA 2 2L 1 2 1 2 Now use the geometric (compatibility) conditions (see figure) 1 1.cos(45) 1 2 2 1.cos(45) 1 This negative is due to compression 2 Finally k 12 EA k 22 EA 2L 2L 1 2 0 1 2 EA 2L ( 2 2 ) EA L © 2002 Brooks/Cole Publishing / Thomson Learning™ 3D Truss (space truss) fˆ kˆ dˆ In local coordinate system fˆ1x fˆ1y fˆ1z ˆ f 2x ˆ f 2y ˆ f 2z k 0 0 k 0 0 0 0 k 0 0 0 0 0 0 0 0 0 0 0 k 0 0 0 0 0 0 0 0 0 dˆ 1x 0 dˆ 0 1y 0 dˆ 1z ˆ 0 d 2x 0 dˆ 2y 0 ˆ d 2z The transformation matrix for a single vector in 3D * dˆ T d l1 * T l2 l 3 m1 m2 m3 n1 n2 n3 © 2002 Brooks/Cole Publishing / Thomson Learning™ l1, m1 and n1 are the direction cosines of x^ l1 cos q x m 1 cos q y n 1 cos q z Transformation matrix T relating the local and global displacement and load vectors of the truss element dˆ T d T * T 6 6 0 fˆ T f 0 * T Element stiffness matrix in global coordinates k T 6 6 T kˆ T 6 6 6 6 6 6 l1 2 l1 m 1 EA l1 n 1 T ˆ k T kT 2 L l1 l m 1 1 l1 n 1 l1 m 1 m1 2 m 1 n1 l1 m 1 m1 2 m 1 n1 2 l1 n 1 l1 m 1 n1 l1 m 1 m1 l1 n 1 m 1 n1 n1 2 l1 n 1 m 1 n1 n1 2 l1 2 l1 m 1 2 l1 m 1 2 l1 m 1 m1 l1 n 1 m 1 n1 l1 n 1 m 1 n1 2 n1 l1 n 1 m 1 n1 2 n 1 Notice that the direction cosines of only the local ^x axis enter the k matrix