1. Structural Analysis II
Course Code: CIVL322
CH. 5
Dr. Aeid A. Abdulrazeg

2. Contents: Fundamentals of the stiffness method Member stiffness matrix
Displacement & force transformation matrix
Member global stiffness matrix
Truss stiffness matrix
Application of the stiffness method for truss analysis
Nodal coordinates
Trusses having thermal changes & fabrication errors

3. Matrix Analysis of Structures
Introduction
Two formulations are possible:
Force Method (Flexibility Method)
Displacement Method (Stiffness Method)
It is generally much easier to formulate the necessary matrices for the computer using the stiffness method.

4. Introduction
Nodes and Degrees of Freedom:
Nodes: are points which equilibrium will be in forces and displacements found.
Degrees of Freedom is equal to the number of possible displacements of the node.

5. Fundamentals of the Stiffness Method
Application of the stiffness method requires subdividing the structure into a series of discrete finite elements & identifying their end points as nodes
For truss analysis, the finite elements are represented by each of the members that compose the truss & the nodes represent the joints
The force-disp. properties of each element are determined & then related to one another using the force eqm eqn written at the nodes

6. Fundamentals of the Stiffness Method
These relationships for the entire structure are then grouped together into the structure stiffness matrix, K
The unknown disp. of the nodes can then be determined for any given loading on the structure
When these disp. are known, the external & internal forces in the structure can be calculated using the force-disp. relations for each member

7. Member stiffness matrix
To establish the stiffness matrix for a single truss member using local x’ and y’ coordinates as shown in Fig 1

8. When a +ve displacement dN is imposed on the near end of the member while the far end is held pinned, Fig (a)
The forces developed at the ends of the members are:

9. Likewise, a +ve displacement dF at the far end, keeping the near end pinned, Fig (b) results in member forces
By superposition, Fig (c), the resultant forces caused by both displacement are

10. These load-displacement eqn. may be written in matrix form as:
This matrix, k’ is called the member stiffness matrix

11. F N

12. Displacement & Force Transformation matrices
Since a truss is composed of many members, we will develop a method for transforming the member forces q and disp. d defined in local coordinates to global coordinates
Global coordinates convention: +ve x to the right and +ve y upward
x and y as shown in Fig

13. Displacement & Force Transformation matrices
The cosines of these angles will be used in the matrix analysis as follows
These will be identified as
For e.g. consider member NF of the truss as shown in Fig
The coordinates of N & F are (xN, yN ) and (xF, yF ) respectively

15. Displacement Transformation matrix
In global coordinates each end of the member can have 2 degrees of freedom or independent disp; namely joint N has DNx and DNy, Fig (a) and (b)

16. Joint F has DFx and DFy, Fig (c) and (d)
When the far end is held pinned & the near end is given a global disp, Fig (a), the corresponding displacement along member is DNxcosx

17. Displacement Transformation matrices
Disp Transformation matrix
A displacement Dny will cause the member to be displaced DNycosy along the x’ axis, Fig (b)

18. F Y N X

20. 2 1

21. Member global stiffness matrix
we can determine the member’s forces q in terms of the global disp. D at its end points

22. Performing the matrix operation yields:

23. Truss stiffness matrix
LOADS
RESPONSE ?
(Output)
(Input)
Structure
Displacement Vector
(System)
Load Vector Q
Support reaction
Support Displacement D
Internal Forces
Initial Deformation
Member Deformation

24. Truss stiffness matrix
Once all the member stiffness matrices are formed in the global coordinates, it becomes necessary to assemble them in the proper order so that the stiffness matrix K for the entire truss can be found
This is done by designating the rows & columns of the matrix by the 4 code numbers used to identify the 2 global degrees of freedom that can occur at each end of the member

25. The structure stiffness matrix will then have an order that will be equal to the highest code number assigned to the truss since this represent the total number. of degree of freedom for the structure
This method of assembling the member matrices to form the structure stiffness matrix will now be demonstrated by numerical e.g.
This process is somewhat tedious when performed by hand but is rather easy to program on computer

26. Example 1 (Matrix Analysis for Truss)
Determine the structure stiffness matrix for the 2 member truss shown in Fig 14.7(a)
AE is constant 3 1 2

27. The direction cosines & the stiffness matrix for each member can now be determined1 1 2

28. L = 3m
Member 2
L = 5m
dy2 3
dx2
dy1 1
dx1

29. structure stiffness matrix1 2 3 4 5 6 1 2 3 4 5 6

31. 1 1

32. 1 2 1

34. Application of the stiffness method for truss analysis
The member forces can be determined using equation

35. Example 2 (Matrix Analysis for Truss)
Determine the force in each member of the 2-member truss shown in Fig (a)
AE is constant

36. The origin of x , y and the numbering of the joints & members are shown in Figure
By inspection, it is seen that the known external displacement are D3=D4=D5=D6=0
Also, the known external loads are Q1=0, Q2=-2kN
Hence,

37. Using the same notation as used here, this matrix has been developed in example 1
Q = KD for the truss we have
We can now identify K11 and thereby determine Du

38. By matrix multiplication,
Solving, we get

39. The force in each member is found from eqn.

40. Example 3 (Truss with Support Settlement)
Determine the force in member 2 of the truss shown in Fig (a), if the support at joint A settles downward 25 mm.
AE is 8(103) kN. C B A

41. 1 4 2 3
Member #1 L= 3 m Ɵ = 90o

42. Member #2 L= 5 m Ɵ = o
Member #3 L= 4 m Ɵ = 0o

43. By assembling these matrices, the global stiffness matrix becomes:

45. Member #2 L= 5 m Ɵ = 216.86o AE= 8(103) kN.

46. Trusses Having Thermal Changes and Fabrication Errors
If some of the members of the truss are subjected to an increase or decrease in length due to thermal changes or fabrication errors, then it is necessary to use the method of superposition to obtain the solution.
This requires 3 steps

47. Trusses Having Thermal Changes and Fabrication Errors
First, the fixed end forces necessary to prevent movement of the nodes as caused by temperature or fabrication are calculated
Second, equal but opposite forces are placed on the truss at the nodes & the displacement of the nodes are calculated using the matrix analysis
Third, the actual forces in the members & the reactions on the truss are determined by superposing these 2 results

48. Trusses Having Thermal Changes and Fabrication Errors
This force will hold the nodes of the member fixed as shown in figure

49. Trusses Having Thermal Changes and Fabrication Errors
This procedure is only necessary if the truss is statically indeterminate
If a truss member of length L is subjected to a temperature increase T, the member will undergo an increase in length of
L =  TL
A compressive force q0 applied to the member will cause a decrease in the member’s length of
L’ = q0L/AE
If we equate these 2 displacement q0 = AE T

50. Trusses Having Thermal Changes and Fabrication Errors
This force will hold the nodes of the member fixed as shown in the previous figure
If a temperature decrease occurs then T becomes negative & these forces reverse direction to hold the member in equilibrium

51. Trusses Having Thermal Changes and Fabrication Errors
If a truss member is made too long by an amount L before it is fitted into a truss, the force q0 needed to keep the member at its design length L is q0 = AEL /L

52. Trusses Having Thermal Changes and Fabrication Errors
If the member is too short, then L becomes negative & these forces will reverse
In global coordinates, these forces are:

53. Trusses Having Thermal Changes and Fabrication Errors
With the truss subjected to applied forces, temperature changes and fabrication errors, the initial force-displacement relationship for the truss then becomes:
Qo is the column matrix for the entire truss of the initial fixed-end forces caused by temperature changes & fabrication errors of the member defined as:

54. Trusses Having Thermal Changes and Fabrication Errors
Carrying out the multiplication, we obtain:
According to the superposition procedure described above, the unknown displacement are determined from the first equation by subtracting K12Dk and (Qk)0 from both sides & then solving for Du

55. Trusses Having Thermal Changes and Fabrication Errors
Once these nodal displacement are obtained, the member forces are determined by superposition:
If this equation is expanded to determine the force at the far end of the member, we obtain:
here we have the additional term which represents the initial fixed-end member force due to temperature changes and/or fabrication error as defined previously. Realize that if the computed result from this equation is negative, the member will be in compression.

56. Example 4 (Truss with fabrication error )
Determine the force in members 1 and 2 of the pin-connected assembly if member 2 was made 0.01 m too short before it was fitted into place.
AE is 8(103) kN. 1 4 2 3

57. Example 4 (Truss with fabrication error )
Since the member is short, then L = -0.01m

58. Example 4 (Truss with fabrication error )
Partitioning the matrices as shown and carrying out the multiplication to obtain the equations for the unknown displacements yields

59. Example 4 (Truss with fabrication error )

60. Example 4 (Truss with fabrication error )