Power point on Trusses

MANE 4240 & CIVL 4240Introduction to Finite ElementsDevelopment of Truss EquationsProf. Suvranu De

Reading assignment:

Chapter 3: Sections 3.1-3.9 + Lecture notesSummary:

Stiffness matrix of a bar/truss element Coordinate transformation Stiffness matrix of a truss element in 2D spaceProblems 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 supportsActual 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

Gusset plateA typical truss structure

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 neglectedA typical truss structureFrictionless 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 compressionmember in tensionConnecting 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

L: Length of barA: Cross sectional area of barE: Elastic (Youngs) modulus of bar :displacement of bar as a function of local coordinate of barThe strain in the bar at Stiffness matrix of bar element

E, AThe stress in the bar (Hookes law)

Tension in the barAssume that the displacement is varying linearly along the barThen, strain is constant along the bar: LStress is also constant along the bar: Tension is constant along the bar: The bar is acting like a spring with stiffness

Recall the lecture on springsTwo nodes: 1, 2 Nodal displacements:Nodal forces:Spring constant:Element force vectorElement nodal displacement vectorElement stiffness matrixElement stiffness matrix in local coordinates

What if we have 2 bars?E1, A1E2, A2L1L2This is equivalent to the following system of springsPROBLEM

Problem 1: Find the stresses in the two-bar assembly loaded as shown belowE, 2AE, ALLSolution: This is equivalent to the following system of springs123PWe 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 springshereHence, the above set of equations may be explicitly written as From equation (2)

To calculate the stresses:For element #1 first compute the element strainand then the stress asSimilarly, in element # 2 (element in tension)(element in compression)

Inter-element continuity of a two-bar structure

Bars in a truss have various orientationsmember in compressionmember in tensionConnecting pin

xyAt node 1:At node 2:

In the global coordinate system, the vector of nodal displacements and loadsOur objective is to obtain a relation of the formWhere k is the 4x4 element stiffness matrix in global coordinate system

The key is to look at the local coordinatesRewrite as

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.^

NOTES5. In local coordinates we have

But or goal is to obtain the following relationship

Hence, need a relationship between and and between and Need to understand how the components of a vector change with coordinate transformation

Transformation of a vector in two dimensionsAngle q is measured positive in the counter clockwise direction from the +x axis)

In matrix formOrwhereTransformation matrix for a single vector in 2Drelateswhere are components of the same vector in local and global coordinates, respectively.Direction cosines

Relationship between and for the truss element At node 1At node 2Putting these together

Important property of the transformation matrix T The transformation matrix is orthogonal, i.e. its inverse is its transposeUse the property that l2+m2=1

Putting all the pieces togetherThe desired relationship isWhere is the element stiffness matrix in the global coordinate system

Computation of the direction cosinesWhat happens if I reverse the node numbers?Question: Does the stiffness matrix change?

Example Bar element for stiffness matrix evaluation

Computation of element strainsRecall that the element strain is

Computation of element stresses stress and tensionRecall that the element stress isRecall that the element tension is

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 displacementsStep 6: Compute the unknown nodal forces

Node element connectivity table123El 1El 2El 3606060

ELEMENTNode 1Node 2112223331

Stiffness matrix of element 1d1xd2xd2xd1xd1yd2yd1yd2yStiffness matrix of element 3d3xd1xd1xd3yd1yd3yd1yd3xThere are 4 degrees of freedom (dof) per element (2 per node)

Global stiffness matrixd2xd3xd3xd2xd1xd1xd3yd2yd1yd1yd2yd3yHow do you incorporate boundary conditions?

Example 2 P1P2123xyEl#1El#2The length of bars 12 and 23 are equal (L)E: Youngs modulusA: Cross sectional area of each barSolve for d2x and d2yStresses in each barSolutionStep 1: Node element connectivity table45o


Table of nodal coordinatesTable of direction cosines

Nodexy1002Lcos45Lsin45302Lsin45

ELEMENTLength1Lcos45sin452L-cos45sin45

Step 2: Stiffness matrix of each element in global coordinates with global numberingStiffness matrix of element 1d1xd2xd2xd1xd1yd2yd1yd2y

Stiffness matrix of element 2

Step 3: Assemble the global stiffness matrixThe final set of equations is

Step 4: Incorporate boundary conditionsHence reduced set of equations to solve for unknown displacements at node 2

Step 5: Solve for unknown displacementsStep 6: Obtain stresses in the elementsFor element #1:00

For element #2:00

Figure 3-19 Plane truss with inclined boundary conditions at node 3 (see problem worked out in class)Multi-point constraints

Problem 3: For the plane trussP=1000 kN, L=length of elements 1 and 2 = 1mE=210 GPaA = 610-4m2 for elements 1 and 2 = 6 10-4 m2 for element 3

Determine the unknown displacements and reaction forces.SolutionStep 1: Node element connectivity table


Table of nodal coordinatesTable of direction cosines

Nodexy10020L3LL

ELEMENTLength1L012L103L

Step 2: Stiffness matrix of each element in global coordinates with global numberingStiffness matrix of element 1d1xd2xd2xd1xd1yd2yd1yd2y

Stiffness matrix of element 2d2xd3xd3xd3yd2yd3yd2xd2yStiffness matrix of element 3d1xd3xd3xd3yd1yd3yd1xd1y

Step 3: Assemble the global stiffness matrixThe final set of equations isN/mEq(1)

Step 4: Incorporate boundary conditionsP123xyEl#1El#245oEl#3Also, How do I convert this to a boundary condition in the global (x,y) coordinates?in the local coordinate system of element 3

P123xyEl#1El#245oEl#3Also, How do I convert this to a boundary condition in the global (x,y) coordinates?in the local coordinate system of element 3

Using coordinate transformations Eq (2)(Multi-point constraint)

Similarly for the forces at node 3Eq (3)

Therefore we need to solve the following equations simultaneouslyEq(1)Eq(2)Eq(3)Incorporate boundary conditions and reduce Eq(1) to

Write these equations out explicitlyEq(4)Eq(5)Eq(6)Add Eq (5) and (6)using Eq(3)using Eq(2)Eq(7)Plug this into Eq(4)

Compute the reaction forces

Physical significance of the stiffness matrixIn general, we will have a stiffness matrix of the form

And the finite element force-displacement relation

Physical significance of the stiffness matrixThe first equation isForce equilibrium equation at node 1What if d1=1, d2=0, d3=0 ?Force along d.o.f 1 due to unit displacement at d.o.f 1Force along d.o.f 2 due to unit displacement at d.o.f 1Force along d.o.f 3 due to unit displacement at d.o.f 1While d.o.f 2 and 3 are held fixedSimilarly we obtain the physical significance of the other entries of the global stiffness matrixColumns of the global stiffness matrix

= Force at d.o.f i due to unit displacement at d.o.f j keeping all the other d.o.fs fixedIn general

Example P1P2123xyEl#1El#2The length of bars 12 and 23 are equal (L)E: Youngs modulusA: Cross sectional area of each barSolve for d2x and d2y using the physical interpretation approachSolutionNotice that the final set of equations will be of the formWhere k11, k12, k21 and k22 will be determined using the physical interpretation approach45o

F2x=k11F2y=k21123xyEl#1El#2To obtain the first columnapply 2xT1y2T2F2x=k11F2y=k21Force equilibriumForce-deformation relationsd2x=1

Combining force equilibrium and force-deformation relationsNow use the geometric (compatibility) conditions (see figure)Finally

123xyEl#1El#2To obtain the second columnapply 2xT1y2T2F2x=k12F2y=k22Force equilibriumForce-deformation relationsd2y=1

Combining force equilibrium and force-deformation relationsNow use the geometric (compatibility) conditions (see figure)FinallyThis negative is due to compression

3D Truss (space truss)

In local coordinate system

The transformation matrix for a single vector in 3Dl1, m1 and n1 are the direction cosines of x^

Transformation matrix T relating the local and global displacement and load vectors of the truss elementElement stiffness matrix in global coordinates

Notice that the direction cosines of only the local x axis enter the k matrix^

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Power point on Trusses