Chapter 3Development of Truss Equation

3.1 Derivation of the Stiffness Matrix for a Bar Element in Local Coordinates Basic understanding of finite element bar• Local ( ) and global (reference) (x,y) coordinates• Node 1 and 2, nodal force and nodal displacement• Tension T apply to the bar• 1 dimen. only which is x direction (>2D later)• Cross area A, modulus of elasticity E, initial LFrom hook law

Force Equi.

Differentiate with Respect to

Assumption1) Ignore shear force and bending

moment2) Hook law applied3) No intermediate applied load or

distribute load

is the axial displacement function

Step how to solve the element forces for a given bar

Step 1: Labelling nodes and elements number

Step 2: Select a displacement function

Step 3: Define strain in term of displacement

The polynomial of the equation is depend on the DOF

For 2 DOF

Step 4: Derive element stiffness matrix and eq.

Step 5: Using boundary condition, solve displacement (using given value of force)

Step 6: Find out all the forces acting on the nodes (by individual equation)

Example

3.2 Selecting Approximation Functions for Displacement

• The simple linear function of is continues within the element

• Disp. at node 2(element1) = Disp. at node 2(element2)

• Function satisfied these criteria because a1 term allows for rigid body motion(const. motion w/o straining) and a2 term allows for constant strain because x = d/d = a2 is a const. (lower order term cannot be omitted)

• Should allow for rigid body displacement means the function must yield a constant value (say, a1)

3.3 Transformation of vectors in Two Dimensions

• Two dimension means move in x and y direction

• We want related vector d with both local and global axis

Similarly

Where C = cos , S = sin

Relationship between local and global displacements

If y=0, then

3.4 Global Stiffness Matrix

•Use the transformation relationship to obtain the global stiffness matrix K for a bar element.

•The global stiffness matrix of each element is needed to assemble the total global stiffness matrix of the structure.

• In local coordinate system

• In global coordinate system

• Transformation relationship for displacements

Expand it to square matrix

• Similarly for forces transformation

•Also, must be expanded to 4x4 matrix

• Substitute and , into the equation below

• yields

•Multiplying both sides by , and since

• Since , hence,

• Substituting T and into equation above, yields K in explicit form (Important!)

3.5 Computation of Stress for a Bar in the x-y Plane

• The local forces are related to the local displacement

•Axial tensile stress

• is the axial force that pull on the bar

• is equals

• Substitute into stress formulae yields

• Since,

•where,

• hence,

•Or simply as,

•Where

•After multiplying the matrices, we have

3.6 Solution of a plane truss

• Plane truss : A structure composed of a bar elements that

lie in a common plane and connected by frictionless pins.

have loads acting only in common plane All loads must be applied at node or joints

Boundary conditions•Homogeneous - completely prevented from movement - can solved using elimination of rows and columns•Non-homogeneous - finite non-zero values of displacement are specified - can solved by using partition with known displacement and simultaneous equation

Example Given E= 30x60psi

Area, A=2in2

1. Determine number of nodes, elements, and angle for each element

2. Construct table

3. Form global stiffness matrix

4. Combine the 3 individual element stiffness matrix to form a total stiffness matrix.

5. Accounting for the applied force at node 1 and the boundary constraints at nodes 2 to 4 which corresponding to zero-displacement.

6. By partition, rows and columns corresponding to zero-displacement was eliminate

7. Solve using simultaneous equation.

8. To determine stresses in each element.

where

For element 1,

The same method go for element2 and 3.

3.7 Transformation Matrix and Stiffness Matrix for a Bar in Three-Dimensional Space

Example:Given modulus of elasticity, E= 1.2x106 psi

1. First, compute the length and direction cosine for element 3 using formula.

2. Using formula below to compute the stiffness matrix for element 3

3. Substitution for Cx, Cy and Cz into .

4. Substitution for into to yield the stiffness matrix for element 3.

5. By summing individual stiffness matrix of element 1, 2 and 3 to yield total stiffness matrix.

6. Compute for global stiffness equation.

7. Solve for d1x and d1z using simultaneous equation

8. To determine stress for element 3,

3.8 Use of symmetry in structure

•Different type of symmetry: axial, mirror, skewed• Focus on mirror (reflective) symmetry • Size, shape, material properties and position are all correspondence• Purpose of symmetry: Reduce problems•Use symmetry method to solve displacement in truss system

Find , ,

•Boundary condition:

• Find , ,

3.9 Inclined or skewed support

• Solve for displacement and reaction force•Combination of global coordinate and local coordinate

•Boundary condition:

3.10 Potential Energy Approach to Derive Bar Element Equations• The total potential energy,

•where

Stress and Strain• Strain

• Stress

•Example

Element 1

Element 2

Final nodal force matrix

• The assembled global stiffness matrix

Element 1 Element 2

3.11 Comparison of Finite Element Solution to Exact Solution for Bar

Residual Methods and Application to 1-Dimensional Bar Problem

• Collocation• Subdomain• Least Squares• Galerkin’s Method

Residual Methods

• Substituting the trial function of an approximate solution into differential equation will results in some residuals or errors.• Each residual method requires the error to vanish over some chosen intervals or at some chosen points.• It is common practice to use the simple linear function in each elements used to model the rod yielding a closer and closer approximation to the actual displacement.

Residual Method

Residual MethodGoverning Equation

Collocation Method

Subdomain Method

Least Square Method

Least Square Method

Galerkin’s Method

Galerkin’s Method