In this chapter trusses, frames and machines will be examines as engineering structures.

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2 In the previous chapter we have employed the equations of equilibrium in order to determine the support / joint reactions acting on a single rigid body or a system of connected members treated as a single rigid body. Determination of the support / joint reactions constitutes only the first step of the analysis in engineering structures. From now on we will focus on the determination of the forces internal to a structure, that is, forces of action and reaction between the connected members. An engineering structure is any connected system of members built to support or transfer forces and to safely withstand the loads applied to it. To determine the forces internal to an engineering structure, we must dismember the structure and analyze separate free body diagrams of individual members or combination of members which are mostly connected each other using smooth pin connections. Determination of joint reactions is of great importance in the selection of connection members that hold the structure as a whole.

3 Joint reactions always occur in pairs that are equal in magnitude and opposite in direction. If not isolated from the rest of the structure or the environment by means of a free body diagram, joint forces are not included in the diagram since they will be internal forces. In order to determine the joint reactions, the structures must be separated into at least two or more parts. At these separation points the joint reactions become external forces and thus, are included in the equations of equilibrium. In this chapter trusses, frames and machines will be examines as engineering structures.

4 A framework composed of members joined at their ends to form a rigid structure is called a truss. Bridges, roof supports, derricks, grid line supports, motorway passages and other such structures are examples of trusses. Structural members commonly used are I-beams, channels, angles, bars and special shapes which are fastened together at their ends by welding, riveted connections, or large bolts or pins using large plates named as gusset plates. For bridges and similar structures, plane trusses are commonly utilized in pairs with one truss assembly placed on each side of the structure. The combined weight of the roadway and vehicles is transferred to either side.

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11 I-Beam (I-Kiriş) Channel Beam (U-Profil) A Gusset Plate (Bayrak) Angled Beam Bar (Çubuk)) (Köşebent L profil)

12 The basic element of a plane truss is the triangle. Three bars joined by pins at their ends constitute a rigid frame. In planar trusses all bars and external forces acting on the system lie in a single plane. P A B C

13 A Typical Roof Truss Joint (Düğüm) A Support Reactions G B F Member (Çubuk) External Force E D C Support Reaction A truss can be extended by additing extra tirangles to the system. Such trusses comprising only of trianges are called simple trusses basit kafes. In a simple truss it is possible to check the rigidity of the truss and whether the joint forces can be determined or not by using the following equation: m : number of members j : number of joints m=2j-3 should be satisfied for rigidity

14 1. In a truss system it is assumed that all bars are two forces members. The weights of members are neglected compared to the forces they are supporting. Therefore members work either in tension or compression. (Çeki) (Bası)

15 2. When welded or riveted connections are used to join structural members, we may usually assume that the connection is a pin joint if the centerline of the members are concurrent at the joint. In this case the joint does not support any moment since it allows for the rotation of the members. 3. It is assumed in the analysis of simple trusses that all external forces are applied at the pin connections. 4. Since bars used in trusses are long, slender elements they can support very little transverse loads or bending moments.

16 Determinaton of Zero-Force Members (Boş Çubukların Belirlenmesi) Determination of the zero-force members beforehand will generally facilitate the solution of the problem 1.Rule: When two collinear members are under compression, it is necessary to add a third member to maintain alignment of the two members and prevent buckling. We see from a force summation in the y direction that the force F 3 in the third member must be zero and from the x direction that F 1 =F 2. This conclusion holds regardless of the angle q and holds also if the collinear members are in tension. If an external force in y direction were applied to the joint, then F 3 would no longer be zero.

17 2. Rule : When two noncollinear members are joined as shown, then in the absence of an externally load at this joint, the forces in both members must be zero, as we can see from the two force summations.

18 Determine the Zero-Force Members in the plane truss.

19 Equal Force Members (Eşit Yük Taşıyan Elemanlar) When two pairs of collinear members are joined as shown, the forces in each pair must be equal and opposite. F 1 and F 2, F 3 and F 4 collinear

20 1 This method for finding the forces in the members of a truss consists of satisfying the conditions of equilibrium for the forces acting on the connecting pin of each joint. The method therefore deals with the equilibrium of concurrent forces, and only two independent equilibrium equations are involved (SF x =0, SF y =0).

21 Sign convention (İşaret anlaşması): It is initially assumed that all the members work in tension. Therefore, when the FBDs of pins are being constructed, members are shown directed away from the joint. After employing the equations of equilibrium, if the result yields a positive value (+), it means that the member actually works in tension (T) (çeki), if the result yields a negative value (-), it means that the member works in compression (C) (bası).

22 1. Determine the force in each member of the loaded truss.

23 2) METHOD OF SECTIONS (Kesim Yöntemi) When analyzing plane trusses by the method of joints, we need only two of the three equilibrium equations because the procedures involve concurrent forces at each joint. We can take advantage of the third or moment equation of equilibrium by selecting an entire section of the truss for the free body in equilibrium under the action of nonconcurrent system of forces. The Method of Sections is often employed when forces in limited number of members are asked for and is based on the two dimensional equilibrium of rigid bodies (SF x =0, SF y =0, SM=0). This method has the basic advantage that the force in almost any desired member may be found directly from an analysis of a section which has cut that member. Thus, it is not necessary to proceed with the calculation from joint to joint until the member in question has been reached. In choosing a section of the truss, in general, not more than three members whose forces are unknown should be cut, since there are only three available independent equilibrium relations.

24 Once a truss is cut into two parts, one of the parts is taken into consideration and all the internal forces now become external from where the cut was passed. The forces are initially assumed as working in tension, so, they are shown directed away from the FBD. After employing the equations of equilibrium, if the result yields a positive value (+), it means that the member actually works in tension (T) (çeki), if the result yields a negative value (-), it means that the member works in compression (C) (bası). Before starting to solve with method, if necessary the support reactions can be determined from the FBD of the whole truss and also zero-force members can be identified. It is very important to recognize that, only the forces acting on the part are considered, the forces acting on the other part, which is not considered, should not be included. The moment center can be any point on or out of the part in consideration.

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26 1. Determine the forces in members FG, CG, BC, and EF for the loaded crane truss.

27 2. Determine the forces in members BC and FG.

28 Cut F BC F CJ F FJ FG

29 3. Determine the forces in members CD, CJ and DJ, state whether they work in tension (T) or compression (C).

30 I. Cut 3 m T F JI F DJ F CD A x A y

31 T F KJ F CJ F CD A x A y II. Cut

32 4. If it is known that the center pin A supports one-half of the vertical loading shown, determine the force in member BF.

33 AB AF Joint A Ay=26 kn I. Cut DE BF DF AF

34 5. a) By inspection, identify the zero-force members in the truss. b) Find the forces in members GI, GJ and GH. H m m D F J L 0.8 m B E G I M m m C 450 N K 0.25 m A N 0.6 m 0.3 m 0.3 m 0.3 m 0.3 m 0.6 m

35 I. Cut m m D F H HJ J L GJ 0.8 m B E G GI I M m m A C 900 N K N 0.25 m 0.6 m 0.3 m 0.3 m 0.3 m 0.3 m 0.6 m A y =450 N N y =450 N

36 6. The truss shown consists of 45 triangles. The cross members in the two center panels that do not touch each other are slender bars which are incapable of carrying compressive loads. Determine the forces in members GM and FL.

37 I. Cut A x A y B y

38 7. The hinged frames ACE and DFB are connected by two hinged bars, AB and CD, which cross without being connected. Compute the force in AB.

39 8. Determine the force acting in member DK.

40 U x U y V y

41 I. Cut III. Cut II. Cut U y =15 kn V y =20 kn

42 9. Determine the forces in members ME, NE and QG.

43 I. Cut II. Cut III. Cut F DE F EK F EK F ME F LB F MB F NF F AF F FQ F FG

44 10. In the truss system shown determine the forces in members EK, LF, FK and CN, state whether they work in tension (T) or compression (C). Crossed members do not touch each other and are slender bars that can only support tensile loads. 4 kn 6 kn 10 kn E F H 2 m B C D G 2 m N J 2 m L K 20 kn A P M 3 m 3 m 4 m 4 m 4 m 4 m Radii of pulleys H, F and K 400 mm

45 4 kn B y 6 kn 10 kn E F 10 kn 10 kn H 2 m B C D 10 kn G B x 2 m N 10 kn J 20 kn 2 m L K 10 kn A x A P M 3 m 3 m 4 m 4 m 4 m 4 m Radii of pulleys H, F and K 400 mm

46 1 st cut 4 kn B y 6 kn 10 kn E F EF F 10 kn 10 kn H 2 m B C D F FL 10 kn G B x 2 m N F EK 10 kn J 20 kn 2 m L F KL K 10 kn A x A P M 3 m 3 m 4 m 4 m 4 m 4 m Radii of pulleys H, F and K 400 mm

47 2 nd cut 4 kn B y 6 kn 10 kn E F EF F 10 kn 10 kn H 2 m B C D F FL FFK 10 kn G B x 2 m N 10 kn F JK J 20 kn 2 m L K 10 kn A x A P M 3 m 3 m 4 m 4 m 4 m 4 m Radii of pulleys H, F and K 400 mm

48 4 kn B y 3 rd cut 6 kn 10 kn E F 10 kn 10 kn H 2 m B C F CD D 10 kn G B x 2 m 2 m F DN N F MN L K 10 kn 10 kn J 20 kn A x A P F PM M 3 m 3 m 4 m 4 m 4 m 4 m Radii of pulleys H, F and K 400 mm

49 4 kn B y 6 kn 10 kn E F 10 kn 10 kn H 2 m B C 4 th cut F CD D 10 kn G B x 2 m 2 m F CN F PN N L K 10 kn 10 kn J 20 kn A x A P F PM M 3 m 3 m 4 m 4 m 4 m 4 m Radii of pulleys H, F and K 400 mm

50 11. Determine the forces in members ON, NL and DL.

51 A x A y I y From equilibrium of whole truss; SF SM SF x y 0 A 0 0 A A A x y y 6 kn (18) 6(2) - 2(15) - 4(9) - 2(6) - 2(3) I y I y 6 kn 0 A y 4 kn

52 F ON F OC F BC I.cut I.cut ) ( (3) (3) 6(2) (6) Compression kn F F F A M ON ON ON kn y C - - S

53 Joint M 4 kn F ML F MN ) ( C kn F F F F F F F F F ML MN MN y ML MN ML MN x S - S

54 II.cut SM SF y D 0 F 0 NL Ay (9) - 2(6) 6(2) F 3 F 4 MN 2 2 MN kn -4.5 kn A y - 2 F ( C) MN F DL F DL F NL 0 ( Zero - force (4) 0 member ) II.cut F MN F NL F DL F DE

55 20 kn 12. Determine the forces in members HG and IG.

56 II.cut I.cut 20 kn 20 kn 20 kn 20 kn 20 kn 20 kn 20 kn

57 II.cut I.cut F CD F CD F HG F BA F HI F HG F GI F GJ 20 kn 20 kn 20 kn 20 kn 20 kn 20 kn 20 kn I.cut SM G =0 F CD =54.14 kn (T) II.cut SM A =0 F HG = kn (C) I.cut SF x =0 F GI =18.29 kn (T)

58 1 kn C 2 kn 2 kn 2 kn 5 kn 3 D E F G 4 4 m B N M O H 4 m A L K J 2 kn 2 kn 2 kn 3 m 3 m 3 m 3 m I 13. Determine the forces in members EF, NK and LK.

59 2 kn 2 kn 2 kn 3 kn I. Cut Top Part 1 kn B C F BN D E F G N F MN M F MO O F HO 4 kn H 4 m From the equilibrium of whole truss A x, A y and I y are determined A x F BA A L K J F HI I 4 m I. Cut SM H =0 F AB is determined A y 2 kn 2 kn 2 kn 3 m 3 m 3 m 3 m I y

60 2 kn 2 kn 2 kn 3 kn C D E F EF F G 4 kn 1 kn II. Cut Top Part B N M F MF O H 4 m II. Cut SM M =0 F BA F BN F MN F MO 4 m F EF and F MF are determined A L K J 2 kn 2 kn 2 kn 3 m 3 m 3 m 3 m I

61 2 kn 2 kn 2 kn 3 kn C F D E EF F G 4 kn 1 kn F MF 4 m B N M F MO O H III. Cut A F NK L K J F LK 2 kn 2 kn 2 kn I 4 m SM N =0 F LK and F NK are determined 3 m 3 m 3 m 3 m III. Cut Left Side

62 G 10 2 kn I H P F E 1 m 10 2 kn 10 2 kn 1 m J M K L N C O D 1 m 25 2 kn 20 2 kn 2 m A B 2 m 1 m 1 m 2 m 14. Determine the forces in members KN and FC.

63 I III. Cut H G P 10 2 F kn I. Cut E 1 m 10 2 kn 10 2 kn 1 m M N O 1 m 25 J 2 kn II. Cut K L C 20 2 D kn 2 m A x A B 2 m 1 m 1 m 2 m Ay B y