An introduction to the module is given, including forces, moments, and the important concepts of free-body diagrams and static equilibrium. These concepts will then be used to solve static framework (truss) problems using two methods: the method of joints and the method of sections.

1. Section 1 Statics © Loughborough University 2010. This work is licensed under a Creative Commons Attribution 2.0 Licence . An introduction to the module is given, including forces, moments, and the important concepts of free-body diagrams and static equilibrium. These concepts will then be used to solve static framework (truss) problems using two methods: the method of joints and the method of sections.

17. (i) Check Determinacy of the Truss R AX R AY R DY A B C D P 1 P 2 x y l  

18. (ii) Moment Equilibrium Equation with Respect to Joint A Treat 2d truss as a free body. A B C D P 1 P 2 R AX R AY R DY x y l  

19. (iii) Determine Support Reactions Notes: We could have taken moments about any point e.g. point B or C or D Treat 2d truss as a free body. There are 3 reactions so need 3 eqns. to find 3 unknowns. A B C D P 1 P 2 R AX R AY R DY x y l  

20. (iv) Force Equilibrium Equations for Joint B Free body diagram for joint B FBD shows ALL forces acting on free body. Here 3 members are connected to joint B. (These members have an internal force) And an external force P1. B P 1   P BD P BC P BA x y A B C D P 1 P 2 R AX R AY R DY l  

21. Section 1 Statics – Plane Truss

25. Finding Reactions

27. Joint A  x y

28. Joint B B P 1 P BA P BE P BD P BC  Note that force P AB = P BA due to Newton’s 3 rd Law. Note that angle DBE =  .

29. Joint C C P CE P CB P CA

30. Joint D  D P DF P DB P DE

31. Joint E E P 2 P EB P EF P ED P EC 

32. Joint F  F R FY P FE P FD

34. Section 1 Statics – Computer Methods

35. Computer Methods – Method of Joints A B C E D F 8m 6m 8m 6m 6m 6m 4 kN 4 kN

36. Computer Methods – Method of Joints In this method we do not calculate the reactions first. We write down each equilibrium equation for each joint in turn i.e. 2j equations (12 in this case): We have 12 equations and 12 unknowns so what do we do next?

37. Computer Methods – Method of Joints We can rewrite each equation in turn and show all the unknown forces in each of these equations. I wont do that here for all junctions so lets just look at joint A only: Notice that including all the other forces has not changed the equations because they all have a zero coefficient. Note the order of forces has changed and where the coefficient is ‘1’ then this is included in the written equation.

38. Computer Methods – Method of Joints Instead of writing down all twelve equations I can write a matrix of the coefficients only multiplied by vector of unknown forces and add in the known external loads. (Remember the equations all add up to zero). A The coefficient matrix of the unknown forces (square matrix) (A-1 is inverse) x Column vector of UNKNOWN forces (member forces + reactions) y Column vector of KNOWN external forces. ( A) (x) (y)

39. Computer Methods – Method of Joints The matrix is square because the truss is statically determinate. The number of rows is 2 x the number of joints, i.e. 12; and the number of columns is equal to the number of unknown forces i.e. 12, that is 9 member forces and 3 reactions. Because the matrix is square we can use a numerical procedure to determine the unknown forces. First I will summarise the approach and then utilise it to find the unknown forces. A The coefficient matrix of the unknown forces (square matrix) (A -1 is inverse) x Column vector of UNKNOWN forces (member forces + reactions) y Column vector of KNOWN external forces.

40. Computer Methods – Solution

41. Computer Methods – Solution

42. Computer Methods – Solution

43. Section 1 Statics – Method of Sections

55. Worked Example - Methods of Sections Find all the member forces. A B C E D F 8m 6m 8m 6m 6m 6m 4 kN 4 kN

58. Section 1 – P AB and P AC x y P AB P AC 4 kN

60. Section 2 – P BC and P CE -5 KN P CE 4kN 4kN P CB x y

62. Section 3 – P BD and P DE A B C B P BD P BE 3 KN 4 KN 4 KN x y

64. Section 4 – P DE and P FE A B C E A B C -3 KN P ED P EF 4 KN 4 KN 4 KN x y

66. Section 5 – P DF We have now determined all the internal forces. F F P FD 3 KN 4 KN x y

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