Chapter 3

Chapter 3

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 (Youngs) modulus of bar u(x) :displacement of bar as a function of local coordinate x of bar The strain in the bar at x du (x) dx The stress in the bar (Hookes law)

(x) E (x) d 2x Tension in the bar T(x) EA x d1x x x x u(x) 1 d1x d 2x L L L Assume that the displacement u(x) is varying linearly along the bar x x u(x) 1 d1x d 2x L L du d 2x d1x Then, strain is constant along the bar: dx L E Stress is also constant along the bar: E d 2x d1x L

Tension is constant along the bar: EA T EA d 2x d 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: d1x d 2x Nodal forces: f1x f 2x EA Spring constant: k L Element stiffness matrix in local coordinates Element force vector f1x k - k d1x f k d

f 2x - k k d 2x Element nodal Element stiffness matrix displacement vector f k d What if we have 2 bars? E1, A1 E2, A2 L2 L1 This is equivalent to the following system of springs k1 E1A1 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 2EA k1 L k2 EA L x Element 1 2 Element 23 1 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 k1 k k k 1 1 2 0 k2 here 0 d1x F1x k2 d 2 x F2 x k2 d3 x F3 x d1x d 3 x 0 and F2 x P Hence, the above set of equations may be explicitly written as k1d 2 x F1x (1) (k1 k2 )d 2 x P (2) k2 d 2 x F3 x (3) P PL

From equation (2) d 2 x k1 k2 3EA To calculate the stresses: For element #1 first compute the element strain (1) d 2 x d1x d 2 x P L L 3EA and then the stress as (1) E (1) P 3A Similarly, in element # 2 (element in tension) d3 x d 2 x

d2 x P L L 3EA P (2) (2) E (element in compression) 3A (2) 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 y d 2y , f 2y d 2y , f2y 0 d 2x , f 2x y

d 2x , f 2x d1y , f1y d 1y , f1y 0 d1x , f1x x d1x , f1x x At node 1: d1x d1x f1x f1x d1y d1y f1y 0 f1y At node 2: d 2x d 2x f 2x f 2x d 2y d 2y f 2y 0

f 2y In the global coordinate system, the vector of nodal displacements and loads d1x d 1y d ; d 2x d 2y f1x f 1y f f 2x f 2y Our objective is to obtain a relation of the form f k d 41 44 41 Where k is the 4x4 element stiffness matrix in global coordinate system The key is to look at the local coordinates y x d 2y , f2y 0

y d 2x , f 2x d 1y , f1y 0 f1x k - k d1x f 2x - k k d 2x d1x , f1x x Rewrite as f 1x k f1y 0 f 2x - k f 0 2y k EA L 0 - k 0 d1x 0 0 0 d1y f k d 0 k 0 d 2x

0 0 0 d 2y 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 f k d 41 44 41 But or goal is to obtain the following relationship f k d 41 44 41 Hence, need a relationship between d and d d1y and between f and f d1x d d 1y d 2x d 2y

d1x d1y d d 2x d 2y d1x d1x d 2x d 2x d1y d 2y d 2y Need to understand how the components of a vector change with coordinate transformation Transformation of a vector in two dimensions y y v y cos vx v x v

v y vy v x cos v y sin v x sin x Angle is measured positive in the counter clockwise direction from the +x axis) x The vector v has components (vx, vy) in the global coordinate system ^ and (v^x, vy) in the local coordinate system. From geometry v x v x cos v y sin v y v x sin v y cos In matrix form v x cos sin v x v v sin cos y

y Or v x l m v x v y m l vy Direction cosines where l cos m sin Transformation matrix for a single vector in 2D * m l v T v relates T m l v x v x are components of the same where v and v v y v y vector in local and global * coordinates, respectively.

Relationship between d and d for the truss element d1y At node 1 At node 2 d1x * d1x T d1y d1y d 2x * d 2x T d 2y d 2y d1x d1x d m l 0 0 d 2y d 2x d 2y d 2x

Putting these together d Td d1x l d1y m d 2x 0 d 0 2y d1y 0 0 d1x 0 0 d1y d l m 2x d m l 2y T d T* T 44 0 0 * T

Relationship between f and f for the truss element At node 1 At node 2 f1y f1x * f1x T f1y f1y f 2x * f 2x T f 2y f 2y f m l 0 0 f 2y f1x f 2x Putting these together f Tf f1x l f1y m

f 2x 0 f 0 2y f1y f1x f 2y f 2x 0 0 f1x 0 0 f1y f l m 2x f m l 2y T f T* T 44 0 0 * T Important property of the transformation matrix T

The transformation matrix is orthogonal, i.e. its inverse is its transpose 1 T T T Use the property that l2+m2=1 Putting all the pieces together x y d 2y , f 2y y d T d d 2x , f 2x d1y , f1y f k d d1x , f1x Tf k T d x The desired relationship is Where f T f

f k d 41 1 f T k T d k 44 41 T k T k T is the element stiffness matrix in the 44 44 44 44 global coordinate system m 0 0 l m l 0 0 T 0 0 l m

0 m l 0 l2 lm 2 lm m EA T k T k T L l 2 lm 2 lm m k 0 k - k 0 0 - k 0 0 0 0 0 k 0 0 0 0 l 2 lm 2 lm m l2 lm 2

lm m Computation of the direction cosines x2 x1 L y2 y1 m sin L L l cos 2 (x2,y2) 1 (x ,y ) 1 1 What happens if I reverse the node numbers? x1 x2 l ' cos l L y y2 m' sin 1 m L L 2 (x ,y ) 2 2

Question: Does the stiffness matrix change? 1 (x1,y1) Example Bar element for stiffness matrix evaluation E 30 106 psi 2002 Brooks/Cole Publishing / Thomson Learning A 2 in 2 L 60 in 30 3 4 3 30 106 2 4 k 3 60 4 3 4 3 4 1

4 3 4 1 4 3 4 3 4 3 4 3 4 3 4 1 4 3 4 1 4 lb in

3 l cos 30 2 1 m sin 30 2 Computation of element strains 2002 Brooks/Cole Publishing / Thomson Learning Recall that the element strain is d 1x d 1y d 2x d 1x 1 1 0 1 0 L L d 2x d 2y 1 1 0 1 0 d L 1 1 0 1 0 T d L m 0 0 l m l

0 0 1 d 1 0 1 0 0 0 l m L 0 m l 0 1 l m l m d L d 1x d 1 l m l m 1y d L 2x d 2y Computation of element stresses stress and tension Recall that the element stress is E E E d 2x d 1x l L L

Recall that the element tension is EA T EA l L m l m d 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 3 1 (x ,y ) 1 1 2 (x2,y2)

Stiffness matrix of element 1 d1x d1y d2x d2y d1x k (1) d1y d2x d2y Stiffness matrix of element 3 d3x d3y d1x d1y d3x k ( 3)

d3y d1x d1y Stiffness matrix of element 2 d2x d2y d3x d3y d2x k ( 2) d2y 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

K d1x d1y d2x d2y d3x d3y 66 How do you incorporate boundary conditions? k k ( 2) ( 3) Example 2 y 3 The length of bars 12 and 23 are equal (L) E: Youngs modulus A: Cross sectional area of each bar Solve for

P1 (1) d and d 2x 2y (2) Stresses in each bar El#2 P2 El#1 45o 1 2 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 y y m 2 1 length 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 l2 lm 2 lm m EA (1) k L l 2 lm 2 lm m d1x l 2 lm 2 lm m l2 lm 2 lm

m d1y d2x d2y 1 1 1 1 1 1 1 1 EA 2L 1 1 1 1 1 1 1 1 d1x d1y d2x d2y Stiffness matrix of element 2 d2x d2y k (2) d3x d3y 1 1 1 1 d2x 1 1 1 1 EA d2y

2L 1 1 1 1 d3x d 1 1 1 1 3y Step 3: Assemble the global stiffness matrix 1 1 1 1 0 0 1 1 1 1 0 0 EA 1 1 2 0 1 1 K 2L 1 1 0 2 1 1 0 0 1 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 d2 x d d 2 y 0 0 Hence reduced set of equations to solve for unknown

displacements at node 2 EA 2 2 L 0 0 d 2 x P1 2 d 2 y P2 Step 5: Solve for unknown displacements P1L d2 x EA d 2 y P2 L EA Step 6: Obtain stresses in the elements For element #1: E 1 (1) L 2 1 2

1 2 E P P (d 2 x d 2 y ) 1 2 2L A 2 d1x d 1 1y 2 d2 x d 2 y 0 0 For element #2: E 1 1 (2) L 2 2 1 2 E P1 P2 (d 2 x d 2 y ) 2L

A 2 d2 x d 1 2y 2 d3 x d3 y 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 El#3 45 o 1

x P=1000 kN, L=length of elements 1 and 2 = 1m E=210 GPa A = 610-4m2 for elements 1 and 2 = 6 2 10-4 m2 for element 3 Determine the unknown displacements and reaction forces. Solution Step 1: Node element connectivity table ELEMENT Node 1 Node 2 1 1 2 2 2 3 3 1 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 y y m 2 1 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 l2 lm l 2 lm 2 2 lm m lm m EA (1) k L l 2 lm l 2 lm

2 2 lm m lm m d1x d1y d2x 0 0 9 -4 (210 10 )(6 10 ) 0 1 0 0 1 0 1 d2y 0 0 0 1 0 0 0 1

d1x d1y d2x d2y Stiffness matrix of element 2 d2x d2y k (2) 1 9 -4 (210 10 )(6 10 ) 0 1 1 0 Stiffness matrix of element 3 k (3) d1x d3x d3y 0 1 0 0 0 0 0 1 0 0 0 0

d1y d3x d3y d2x d2y d3x d3y 0.5 0.5 0.5 0.5 d1x (210 109 )(6 2 10-4 ) 0.5 0.5 0.5 0.5 d 1y 0.5 0.5 0.5 0.5 d 2 3x 0.5 0.5 0.5 0.5 d3y Step 3: Assemble the global stiffness matrix 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 5 K 1260 10 1 0 1

0 0 0 0.5 0.5 1 0 1.5 0.5 0.5 0.5 0 0 0.5 0.5 The final set of equations is K d F Eq(1) N/m y Step 4: Incorporate boundary conditions y 0 0 3 P El#2 d 2 x 2 d 0 El#1 El#3 d3 x

o d 45 3y x 1 Also, d 3 y 0 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 F1x F 1y P F F 2y F3 x F3 y

Also, F 3 x 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 3x d l d 3 y m

1 3x d 2 d 3 y 1 2 d 3 y 0 m d3 x 1 l m l d3 y 2 1 1 d d 3x 3y d

2 2 3 x 1 d3 y 1 d d 3 y 3 x 2 2 (Multi-point constraint) 1 d 3y d 3 y d 3 x 0 2 Eq (2) d 3 y d 3 x 0 Similarly for the forces at node 3

3x F l 3y m F 1 3x F 2 F 3y 1 2 m F3 x 1 l m n F3 y 2 1 1 F

F 3x 3y F 2 2 3 x 1 F3 y 1 F F 3 y 3 x 2 2 F 3 x 0 1 F 3x

F3 y F3 x 0 2 Eq (3) F3 y F3 x 0 Therefore we need to solve the following equations simultaneously K d F d 3 y d 3 x 0 Eq(1) Eq(2) Eq(3) F3 y F3 x 0 Incorporate boundary conditions and reduce Eq(1) to 1 1260 105 1 0 1 1.5 0.5 0 0.5 0.5 d 2 x P d F 3x 3x

d F 3y 3y Write these equations out explicitly Eq(4) 1260 105 ( d 2 x d 3 x ) P 1260 105 ( d 2 x 1.5d 3 x 0.5d 3 y ) F3 x Eq(5) 1260 105 (0.5d 3 x 0.5d 3 y ) F3 y Eq(6) Add Eq (5) and (6) 1260 105 ( d 2 x 2d 3 x d3 y ) F3 x F3 y 0 using Eq(3) 1260 105 ( d 2 x 3d 3 x ) 0 using Eq(2) d 2 x 3d 3 x Eq(7) 1260 105 (3d 3 x d 3 x ) P Plug this into Eq(4) 2520 105 d 3 x 106 d 3 x 0.003968m d 2 x 3d 3 x 0.0119m Compute the reaction forces F1x 0 F 0 1y 5 F2 y 1260 10 0 F 1 3x

0 F3 y 500 500 0 kN 500 500 0.5 0.5 0 1.5 0.5 0.5 0.5 0 0.5 0.5 d 2 x d 3x d 3y

Physical significance of the stiffness matrix In general, we will have a stiffness matrix of the form k11 K k 21 k 31 k12 k 22 k 32 k13 k 23 k 33 And the finite element force-displacement relation k11 k12 k13 d1 F1 k d F k k 2 22 23 2 21 k 31 k 32 k 33 d 3 F3 Physical significance of the stiffness matrix The first equation is k11d1 k12 d 2 k13d 3 F1 Force equilibrium equation at node 1

Columns of the global stiffness matrix What if d1=1, d2=0, d3=0 ? F1 k11 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 Force at d.o.f i due to unit displacement at d.o.f j k ij = keeping all the other d.o.fs fixed Example y 3 The length of bars 12 and 23 are equal (L) E: Youngs modulus A: Cross sectional area of each bar Solve for d2x and d2y using the physical P1 interpretation approach El#2 P2 El#1 45o 1 2 x

Solution Notice that the final set of equations will be of the form k11 k 21 k12 d 2 x P1 k22 d 2 y P2 Where k11, k12, k21 and k22 will be determined using the physical interpretation approach k11 d 1 2 x apply 1 1.cos(45) k d 2 y 0 21 2 y To obtain the first column y

2 3 F2y=k21 El#2 El#1 1 2 x 1 1.cos(45) d2x=1 Force equilibrium F F T2 F2x=k11 2 x k11 T1 cos(45) T2 cos(45) 0 y k21 T1 sin(45) T2 sin(45) 0 T1 1 2

F2y=k21 F2x=k11 2 x Force-deformation relations EA T1 1 L EA T2 2 L Combining force equilibrium and force-deformation relations k11 T1 T2 EA k21 2 T1 T2 2 2L 1 2 EA 1 2 2L Now use the geometric (compatibility) conditions (see figure)

1 2 1 2 1.cos(45) 2 1 1.cos(45) Finally EA EA 2 EA k11 1 2 ( ) L 2L 2L 2 EA k21 1 2 0 2L k12 d 0 2 x apply d 2 y 1 k22 y To obtain the second column y 3 El#2 El#1

1 2 1.cos(45) 2 2 d2y=1 T2 T1 x 1 1.cos(45) 1 2 Force equilibrium F F 1 2 x k12 T1 cos(45) T2 cos(45) 0 y k22 T1 sin(45) T2 sin(45) 0 F2y=k22 F2x=k12 2

x Force-deformation relations EA T1 1 L EA T2 2 L Combining force equilibrium and force-deformation relations k12 T1 T2 EA k22 2 T1 T2 2 2L 1 2 EA 1 2 2L Now use the geometric (compatibility) conditions (see figure) 1 1.cos(45) 1

2 2 1.cos(45) 1 2 This negative is due to compression Finally EA k12 1 2 0 2L EA EA 2 EA k22 ( ) 1 2 L 2L 2L 2 2002 Brooks/Cole Publishing / Thomson Learning 3D Truss (space truss) In local coordinate system f 1x k

f 1y 0 f 1z 0 f 2x k f 0 2y f 0 2z f k d 0 0 k 0 0 0 0 0 0 0 0 0 0 0 0 0 k 0 0 0 0 0 0 0 d 1x 0 0 d 1y

0 d 1z 0 d 2x 0 d 2y 0 d 2z The transformation matrix for a single vector in 3D * d T d l1 * T l 2 l3 m1 m2 m3 n1 n2 n3 2002 Brooks/Cole Publishing / Thomson Learning l1, m1 and n1 are the direction cosines of x^ l1 cos x m1 cos y n1 cos z Transformation matrix T relating the local and global displacement and load vectors of the truss element d T d

T* T 66 0 f T f 0 * T Element stiffness matrix in global coordinates T k T k T 66 66 66 66 2 l1 2 l1m1 l1n1 l1 l1 m1 l1 n1 2 2 m1 m1 n1 l1m1 m1 m1n1 l1 m1 2 2 m1 n1 n1 l1 n1

m1 n1 n1 EA l1 n1 T k T k T 2 2 L l1 l1 m1 l1 n1 l1 l1m1 l1 n1 2 l m m 2 m n l m m m n 1 1 1 1 1 1 1 1 1 1 2 2 l1 n1 m1 n1 n1 l1 n1 m1 n1 n1 Notice that the direction cosines of only the local ^x axis enter the k matrix

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