Chapter 3: Analysis of Statically Determinate Trusses PREPARED BY SHAMILAH

Transcription

1 Chapter 3: Analysis of Statically Determinate Trusses PREPARED BY SHAMILAH

2 INTRODUCTION Truss is a structure composed of slender members joined together at their end points The joint connections are usually formed by bolting or welding at the ends of the members to a common plate known as gusset plate.

3 Common Types of Trusses Roof Trusses Scissors short spans Howe and Pratt moderate span (18m to 30m) an and ink (>30m)

4 Common Types of Trusses (cont ) Bridge Trusses Pratt, Howe and Warren span <60m Parker span >60m

5 Assumptions for Design a Trusses 2 important assumptions need to be considered are:- The members are joined together by smooth pins All loadings are applied at the joints Due to the 2 assumptions, each truss member acts as an axial force member

6 Classification of Coplanar Trusses Trusses can be classify into 3 classes :- Simple Compound Complex Truss

7 Determinacy In order to ensure equilibrium force, equilibrium equation need to be satisfied:- x and 0 0 y By comparing the total unknowns with the total no. of available equilibrium eqn, we have: b r 2 j statically determinate b r 2 j statically indeterminate

8 Stability Stability can be categorized as internal or external stability. If b + r < 2j => collapse A truss can be unstable if it is statically determinate or statically indeterminate Stability will have to be determined either through inspection or by force analysis

9 Internal Stability Internal stability can be checked by careful inspection of the arrangement of its members If it can be determined that each joint is held fixed so that it cannot move in a rigid body sense with respect to the other joints, then the truss will be stable. A simple truss will always be internally stable If a truss is constructed so that it does not hold its joints in a fixed position, it will be unstable or have a critical form

10 Internal Stability (cont ) To determine the internal stability of a compound truss, it is necessary to identify the way in which the simple truss are connected together The truss shown is unstable since the inner simple truss ABC is connected to DE using 3 bars which are concurrent at point O.

11 External Stability A structure is externally unstable if all of its reactions are concurrent or parallel The trusses are externally unstable since the support reactions have lines of action that are either concurrent or parallel

12 Example 3.1 Classify each of the trusses as stable, unstable, statically determinate or statically indeterminate. The trusses are subjected to arbitrary external loadings that are assumed to be known & can act anywhere on the trusses.

13 solution Externally stable Reactions are not concurrent or parallel b = 19, r = 3, j = 11 b + r =2j = 22 Truss is statically determinate By inspection, the truss is internally stable Externally stable b = 15, r = 4, j = 9 b + r = 19 >2j Truss is statically indeterminate By inspection, the truss is internally stable

14 Solution (cont ) Externally stable b = 9, r = 3, j = 6 b + r = 12 = 2j Truss is statically determinate By inspection, the truss is internally stable Externally stable b = 12, r = 3, j = 8 b + r = 15 < 2j The truss is internally unstable

15 The Method of Joints In all cases, the joint analysis should start at joint having at least one known force and at most two unknown forces. and 0 x 0 y

16 Example 3.2 Determine the force in each member of the roof truss as shown. State whether the members are in tension or compression. The reactions at the supports are given as shown.

17 Solution Only the forces in half the members have to be determined as the truss is symmetric with respect to both loading & geometry, Joint A, 0; 4 AG AB 8kN( C) x y 0; AB 6.93kN( T ) AG sin 30 8cos

18 Joint G, GB G 0; 2.60kN( C) x y 0; 8 GB 6.50kN( C) 3sin 30 3cos G 0 0 Joint B, 0; B BC 2.60kN( T ) x y 0; BC B 4.33kN( T ) sin cos sin cos

19 EXAMPLE 3.3 REER PAGE 99 TEXT BOOK

20 Zero-orce Members Truss analysis using method of joints is greatly simplified if one is able to first determine those members that support no loading These zero-force members may be necessary for the stability of the truss during construction & to provide support if the applied loading is changed The zero-force members of a truss can generally be determined by inspection of the joints & they occur in 2 cases.

21 Zero-orce Members (cont ) Case 1 If only two non-collinear members form a truss joint and no external load or support reaction is applied to the joint, the members must be zeroforce members.

22 Case 2 Zero-orce Members (cont ) Zero-force members also occur at joints having a geometry as joint D Three members form a truss joint for which two of the members are collinear, the third member is zero-force member provided no external load or support reaction at the joint.

23 Zero-orce Members (cont ) Case 2 No external load acts on the joint, so a force summation in the y-direction which is perpendicular to the 2 collinear members requires that D = 0 Using this result, C is also a zero-force member, as indicated by the force analysis of joint

24 Example 3.4 Using the method of joints, indicate all the members of the truss that have zero force.

25 Solution Joint D, 0 DC DE x 0 y 0; 0; DE DC sin 0 0 0

26 Solution (cont ) Joint E, x 0; E 0 Joint H, y 0; HB 0 Joint G, y 0; GA 0

27 The Method of Sections If the forces in only a few members of a truss are to be found, the method of sections generally provide the most direct means of obtaining these forces This method consists of passing an imaginary section through the truss, thus cutting it into 2 parts Provided the entire truss is in equilibrium, each of the 2 parts must also be in equilibrium

28 The Method of Sections (cont ) The 3 eqns of equilibrium may be applied to either one of these 2 parts to determine the member forces at the cut section A decision must be made as to how to cut the truss In general, the section should pass through not more than 3 members in which the forces are unknown

29 The Method of Sections (cont ) If the force in GC is to be determined, section aa will be appropriate Also, the member forces acting on one part of the truss are equal but opposite The 3 unknown member forces, BC, GC & G can be obtained by applying the 3 equilibrium eqns

30 Example 3.5 Determine the force in members GJ and CO of the roof truss. State whether the members are in tension or compression. The reactions at the supports have been calculated. SCAN PIC IN PAE 105

31 A direct solution for sin kN( C) (2) 1200(1.155) 0 Solution ApplyingPrincipal of transmissibility, is slide to point G for simplicity. can be obtained by applying M With anti - clockwise moments as ve, M GJ GJ GJ o GJ I 0 I KJ GJ GH 0 2m 1200N 1.155m 600N I 4637N Momentswill be summed about point A in order to eliminate unknowns With anti- clockwise moments as ve, M 693kN( T ) and 1200(1.155) CO OP CO CD (2) 0 A 0 the

32 Determine the force in members C and GC of the roof truss. State whether the members are in tension or compression. The reactions at the supports have been calculated.

33 Solution The free-body diagram of member C can be obtained by considering the section aa, A direct solution for can be obtained by ApplyingPrincipal of transmissibility, C applying M E 0 is slide to point C for simplicity. With anti - clockwise moments as ve, C C C sin 30 o 1.73kN( C) (4) 1.50(2.31) 0 M E 0

34 Example 3.6 Determine the force in member G and GD of the truss. State whether the members are in tension or compression. The reactions at the supports have been calculated.

35 Solution The distance EO can be determined by proportional triangles or realizing that member G drops vertically = 1.5m in 3m. Hence, to drop 4.5m from G the distance from C to O must be 9m

36 Solution (cont ) The angles GD and G make with the horizontal are tan -1 (4.5/3) = 56.3 o tan -1 (4.5/9) = 26.6 o The force in Gcan be determined directly by applying M 0 is slide to point O. With anti - clockwise moments as ve, M G G G D sin 26.6 o 7.83kN( C) (6) 7(3) 0 D 0

37 Solution (cont ) The force in GD can be determined directly by applying M 0 is slide to point D. With anti - clockwise moments as ve, M 7(3) 2(6) GD GD O GD 1.80kN( C) sin 56.3 o (6) 0 O 0

38 REER PAGE 107 Example 3.7

39 THANK YOU VERY MUCH OR YOUR ATTENTION