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

2. 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.

3. 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. 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:

5. 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

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

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

8. 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,

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

10. 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.

11. 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.

12. 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.

13. 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

14. 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.

15. 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.

16. 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.

17. 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

18. 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

19. For elements 2, 3, and 4

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

21. 5. Assemble the global system of linear equations.

22. 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

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

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