Plane Trusses (Initial notes are designed by Dr. Nazri Kamsah)

4-3 Plane Trusses A typical two-dimensional plane truss is shown. It comprises of two-force members, connected by frictionless joints. All loads and reaction forces are applied at the joints only. Note: There are two displacement components at a given node j, denoted by Q2j-1 and Q2j.

4-4 Local & Global Coordinate Systems Local and global coordinate systems are shown at left. In local coordinate (x’), every node has one degree of freedom, while in global coordinate (x, y), every node has two degrees of freedom. The nodal displacements, in the local coordinate system is, In the global coordinate system, Local and global coordinate systems.

4-5 Relation Between Coordinate Systems Consider a deformed truss member as shown. We can now establish a relationship between {q’} and {q} as follows:

4-6 Direction Cosines To eliminate the  terms from previous equations, we define direction cosines, such that and The relation between {q’} and {q} can now be written as which can be written in matrix form as

Or, in a condensed matrix form where [L] is a rectangular matrix called the transformation matrix, given by,

4-7 Formula for Evaluating l and m Using a trigonometry relation, we observe + Note: Coordinates (xi, yi) are based on local coordinate system.

4-8 Element Stiffness Matrix A truss element is a one-dimensional (bar) element, when it is viewed in local coordinate system. Thus, element stiffness matrix for a truss element in local coordinate, The internal strain energy in the truss element, in local coordinate system is, We need the expression for [k] when viewed in global coordinate… Substituting we get,

In the global coordinates system, Since internal strain energy is independent of coordinate system, Ue = U’e. Therefore, Simplifying,

Use: E = 180 GPa; d = 15 mm for all members. Exercise 4-1 For each element of a plane truss structure shown, Determine the transformation matrix, [L]e; Write the element stiffness matrix, [k]e; Assemble the global stiffness matrix, [K]. P = 50 kN 0.8 m 0.6 m A B C Use: E = 180 GPa; d = 15 mm for all members.

4-9 System of Linear Equations The system of linear equations for a single plane truss element in local coordinate system can be expressed as Where {q} is nodal displacement vector and {f} is nodal force vector, in the global coordinate direction. Substituting, we get Note: To assemble the global stiffness matrix, a local-global nodal connectivity will be required.

Exercise 4-2 Reconsider Exercise 4-1. a) Assemble global system of linear equations for the structure; b) Apply the boundary conditions, and c) Write the reduced system of linear equations. P = 50 kN 0.8 m 0.6 m A B C Boundary conditions: Q1 = Q2 = Q4 = 0 (homogeneous type) Use: E = 180 GPa; d = 15 mm for all members.

4-10 Stress Calculations Normal stress in a plane truss element, in local coordinate system is, In the global coordinate system, since Expanding the [B] and [L] matrices, Or

Exercise 4-3 Reconsider Exercise 4-2. a) Determine the unknown nodal displacements at B and C; b) Compute the stresses in the member AC and BC. P = 50 kN 0.8 m 0.6 m A B C Use: E = 180 GPa; d = 15 mm for all members.

Homework 4-1 For the truss structure shown, rod (1) has modulus E = 29 x 103 ksi and a yield point sY = 36 ksi. Rod (2) has modulus E = 10 x 103 ksi and a yield point sY = 60 ksi. If the horizontal force P = 10 kips, determine the cross-sectional areas of the two rods so that the yield stress in each rod is not exceeded. Note: This is a trial and error problem, so you may use Nastran simulation to help you solve the problem. Work in group.

Example 4-1 For a plane truss shown, determine: a) displacements at nodes 2 and 3; b) stresses in each element; c) reaction forces at the supports. Use: Ee = 29.5 x 106 psi and Ae = 1 in2 for all elements.

Solution 1. Establish nodal coordinate data and element connectivity information Nodal coordinate Node x y 1 2 40 3 30 4 Element connectivity Element Node 1 Node 2 1 2 3 4

2. Compute the direction cosines Element le l m 1 40 2 30 -1 3 50 0.8 0.6 4 3. Write the stiffness matrix for each element Element 1

For elements 2, 3, and 4

4. Assemble the global stiffness matrix [K] for the entire truss structure

5. Assemble the global system of linear equations.

Using elimination method, the above equations reduced to 6. Impose boundary conditions & write the reduced system of linear equations We have; Q1 = Q2 = Q4 = Q7 = Q8 = 0. Using elimination method, the above equations reduced to 7. Solve the reduced equations using the Gaussian elimination method, we get

8. Compute the stress in each element In the same manner,

9. Determine support reactions Using the 1st, 2nd, 4th, 7th, and the 8th equations, we get