7 STATICALLY DETERMINATE PLANE TRUSSES

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1 7 STATICALLY DETERMINATE PLANE TRUSSES OBJECTIVES: This chapter starts with the definition of a truss and briefly explains various types of plane truss. The determinancy and stability of a truss also will be discussed. The procedures of analyzing statically determinate trusses will be developed using the method of joints and the method of sections. For advance, the alternative computation using joint equilibrium method will be included in this topic. 6.1 Introduction A truss is defined as a structure composed of slender elements joined together at their end points. The members commonly used in construction consist of wooden struts, metal bars, angles or channels. Plane or planar truss composed of members that lie in the same plane and frequently used for bridge and roof support. Loads that cause the entire truss to bend are converted into tensile and compressive forces in the members. compression tension Loading causes bending of truss, which is develops compression in top members, tension in bottom members. Fig. 7-1 Mechanics of Materials 203

6.2 Types of trusses and its application 6.2.1 Roof trusses Often used as part of an industrial building frame as shown in Fig.

The roof truss along with its supporting columns called as a bent. To keep the bent rigid and capable of resisting horizontal wind forces, knee braces are sometimes used at the supporting columns.

2 6.2 Types of trusses and its application Roof trusses Often used as part of an industrial building frame as shown in Fig Fig. 7-2 The roof load is transmitted to the truss at the joints by series of purlins. Generally, the roof trusses are supported either by columns of wood, steel or reinforced concrete. The roof truss along with its supporting columns called as a bent. To keep the bent rigid and capable of resisting horizontal wind forces, knee braces are sometimes used at the supporting columns. The space between adjacent bents is called a bay. Bays are often tied together using diagonal bracing to maintain rigidity of the structure. The gusset plate is the connections which formed by bolting or welding at the ends of the members to a common plate. Fig. 7-3 : Gusset plate Trusses used to support roofs are selected on the basis of the span, the slope and the roof material. Some of common types of trusses are displayed in the Fig Mechanics of Materials 204

Fig. 7-4: Common types of plane trusses 1.2.2 Bridge trusses The main structural elements of a typical bridge truss are shown in Fig.

3 Fig. 7-4: Common types of plane trusses Bridge trusses The main structural elements of a typical bridge truss are shown in Fig Fig. 7-5 : The structure elements of typical bridge truss The load on the deck is first transmitted to stringers then to floor beam and finally to the joints of the two supporting side trusses. The top and bottom cords are connected by top and bottom lateral bracing which serves to resist the lateral forces caused by wind and the sidesway caused by moving vehicles. Additional stability is provided by the portal and sway bracing. Mechanics of Materials 205

4 The typical forms of bridge trusses currently used for single span are shown in Fig Fig. 7-6 : The forms of bridge trusses 6.3 Assumptions in analysis The assumptions are necessary to determine the force developed in each member when the truss is subjected to a given loading. C A to C, C to B, B to A the connection between them is called as truss member Fig. 7-7 : Truss member the assumptions are; i) All members are connected at both ends by smooth frictionless pins. Mechanics of Materials 206

5 ii) All loads are applied at joints (member weight is negligible). ASK STUDENTS: Why the analysis assume that the smooth pin is frictionless? Why the analysis assume that the member weight is negligible? Because of these two assumptions, each truss acts as an axial force member and the forces acting at the ends of the member must be directed along axis of the member. If the force tends to elongate the member, it is tensile force (T) as shown in Fig If the force tends to shorten the member, it is compressive force (C). In analysis it is important to state whether the force is tensile or compressive. T T The forces inside the member and outside should opposite C C Smooth pin Fig. 7-8 : The tensile and compressive force 6.4 Stability and Determinancy General The structures mechanics involves determination of unknown forces on the structures. Some of these structures can be completely analyzed by using the equations of equilibrium. ΣFx = 0 (F = F) ΣFy = 0 (F = F ) ΣMz =0 (M = M) The structures are known as statically determinate. On the other hand, if there exist extra redundant reaction components, then the structure is said to be statically indeterminate. Mechanics of Materials 207

6 6.4.2 Determinancy Criteria for Structures To be in a state of static equilibrium, a structure must meet the requirements of stability. A statically indeterminate structure is a structure that had more unknown forces. Eg; ( to determine whether the truss is determinate or indeterminate) For a pinned joint frame (trusses); i) m+r = 2j just stiff (statically determinate) ii) m+r < 2j under stiff (form a mechanism) unstable (it will collapse since there will be an insufficient number of reactions to constraint all the joints). iii) m+r > 2j over stiff (statically indeterminate) -A truss can be unstable if it is statically determinate or statically indeterminate. -A truss is externally unstable if all of its reaction is concurrent or parallel. Mechanics of Materials 208

7 EXAMPLE 7.1 Determine number of redundant and state the determinancy criteria for the truss as shown below; a) m=10 r =3 j =6 m + r = 13 degree b) n= m + r 2j = (6) = 1..has a redundancy of 1th 2j = 12 =m+r > 2j.over stiff (statically indeterminate) EXAMPLE 7.2 Define whether the truss is statically determinate or indeterminate m = 7 m+r = 2j r = = 2(5) j = 5 10 = 10.statically determinate truss But this truss is unstable since the support reactions are parallel. Mechanics of Materials 209

8 m = 7 m+r = 2j r = = 2(5) j = 5 10 = 10.statically determinate truss But this truss unstable since the support reactions are concurrent. EXERCISE 7.1 (a) (c) (b) (d) 6.5 Method of Joints EXAMPLE 7.3 Calculate all member forces by using method of joints. H A V A V C Mechanics of Materials 210

9 Determine whether the truss is determinate or indeterminate. m = 3 r = 3 j = 3 m+r =2j 3+3 = 2(3) 6 = 6 the truss is determinate and stable. Calculate the support reactions. F = F H = H 0 kn = V A + V C 500 N = H A + M = 0 M A = V C ( 2) (2)=0 V C = - 500N ( ) V A -500 = 0 V A = 500N( ) -ve sign indicated that the original direction is upward V C = 500N ( ) change the direction; Calculate all member forces Joint A F AB ΣFx = ΣFx ΣFy = ΣFy F AB = 500N (T) Mechanics of Materials 211

Joint C 500 F CB 500 ΣFy = ΣFy 500 + F CB sin 45 = 0 = - 707 N (C) NOTES: -ve sign indicated the force tends to shorten the member, so the force is compression. How?

10 Joint C 500 F CB 500 ΣFy = ΣFy F CB sin 45 = 0 = N (C) NOTES: -ve sign indicated the force tends to shorten the member, so the force is compression. How? All loads are assumed moved outward from the joint. B The calculation shows the force is compression, so; C Change the direction; To stabilize inside the member; To stabilize inside and outside the member; The loads should be opposite at joint CHECKING : 500 (707) Joint B: ΣFx = ΣFx 500 =707 sin 45 0 = 0 (BALANCE) 500 Mechanics of Materials 212

11 Procedure for Analysis: The following procedures can be used as the guidance to solve the problems using the method of joints. Determine the truss is determinate or indeterminate by using the stability and determinancy equation. Calculate the support reactions if necessary. Draw the free-body diagram of a joint having at least one known force and at most two unknown forces. If the joint is located at the supports, it may be necessary to find the external reactions at the support. All loads are assumed moved outward from the joint. The x and y axes can be resolved into their x and y components. Two force equilibrium equations ΣFx =0 and ΣFy =0 are applied. Continue to analyze the member forces at the other joints where again it is necessary to choose a joint having at most two unknowns and at least one known force. Indicate whether the member in compression or tension. Mechanics of Materials 213

12 EXAMPLE 7.4 Determine the support reactions and member forces using method of joints. All members are connected at both ends by smooth frictionless pins. 6 kn C D 5m A H A F E B V A 8 kn 4m 5m 4m V B Solution; Determine the truss is determinate. Calculate the support reactions. [Ans: V A = 3.2 kn( ), V B = kn( ), H A = 6 kn( )] Calculate the member forces Joint A 6 θ F AC F AF 3.2 tan θ = 5/4 θ = ΣFy = ΣFy F AC sin = 0 F AC = kn (C) ΣFx = ΣFx F AF + F AC cos = 6 F AF = 8.6 kn (T) -4.1 Mechanics of Materials 214

13 Joint F 8.6 F FC F FE = 8.6 kn (T) F FC = 8 kn (T) Joint C 8 F FE 6 F CD ΣFy = ΣFy -4.1 θ 2 8 θ 1 F CE 0 = 8 + F CE sin 45 + (-4.1) cos F CE = - F CE = -6.8 kn (C) ΣFx = ΣFx tan θ 2 = 4/5 θ = tan θ 1 = 5/5 θ = F CD + F CE cos = (-4.1) sin F CD = F CD = kn (C) Joint B F BE F BD tan θ= 5/4 θ= θ ΣFy = ΣFy + F BD sin = 0 F BD = kn (C) ΣFx = ΣFx = F BE + F BD cos F BE = 3.84 kn (T) Joint D θ θ = = CHECKING: F DE ΣFy = ΣFy 0 = F DE + (-6.15) cos F DE = kn (T) Mechanics of Materials 215

14 6 kn C (3.75) D Checking: (6.8) (4.1) (6.15) Joint E ΣFx = ΣFx A F 8.6 E 3.84 B cos 45 = = 8.6 (BALANCE) 8 kn EXERCISE 7.2 H K V K 15 Determine the support reactions and member forces of GF, EF, GE, HG, HD, GD and HI. The method of joints is selected to solve the problem. H A V A 5 6 HD:? 15 = 6 5?? = 2m [V K = 0kN, V A = 7kN ( ), H K = 9.6kN ( ), H A = 9.6kN ( )] Mechanics of Materials 216

15 EXERCISE 7.3 A pin-connected truss is loaded and supported as shown in figure; a) prove that the plane truss is a statically determinate structure. b) determine all member forces by using Method of Joint. 3m [Ans: F AC = 33.33kN (T), F AB = 6.66kN (C), F DC =26.67kN (T), F BC = 20kN (T)] 6.6 Method of Sections EXAMPLE 7.6 Determine the forces of member BC, CG and FG by using method of section. Prove the truss is statically determinate. State whether the members are in tension or compression. 2m Mechanics of Materials 217

16 Solution; Prove the truss is statically determinate. Calculate the support reaction if necessary. Cut the section of the truss through the members where forces are to be determined. Choose one section whether left or right sections. If choose right section, the external reactions at support have to calculate. F BC C F GC All member loads are assumed moved outward from the cut. F GF Calculate the member forces. + M G = 0 F BC (2) 100 (2) = 0 F BC = 100 N (T) ΣFy = ΣFy F GC sin 45 = 100 F GC = N (T) + M C = 0 -F GF (2) 100 (4) = 0 F GF = N (C) Checking: ΣFy = ΣFy F GC sin 45 = = 100 (BALANCE) Mechanics of Materials 218

17 Procedure for Analysis: The following procedure provides the guidance to determine the forces in the members of a truss by using the method of sections. Determine the truss is determinate or indeterminate by using the stability and determinancy equation. Calculate the support reactions if necessary. Make a decision as to how to cut the truss through the members where forces are to be determined. Draw the free-body diagram of the part of the section truss whether left or right section which has the least number of forces on it. Moments should be summed about a point that lies at the intersection of the lines of action of two unknown forces. The third unknown force is determined directly from the equation. Mechanics of Materials 219

18 EXAMPLE 7.7 Determine the force in members CA, CF and FE of the bridge truss by using method of section. State whether the members are in tension of compression. The support reactions have been calculated. 6 kn 5m C D A 4m F 4m E 4m 8 kn Solution: B 6 C D 6kN A 3.2kN B F E 8 kn Choose the left section F AC C + M C = 0 6 F FC -F FE (5) (4)+ 6(5) = 0 F FE = 8.6kN (T) A F FE + M A = 0 -F FC (4) + 8 (4) = 0 F FC = 8 kn (T) ΣFy = ΣFy F AC sin = 8 F AC = kn (C) Mechanics of Materials 220

19 EXAMPLE 7.8 Determine the member forces of HI, ID and CD by using the method of sections. tan θ = 3/1 θ =71.6º 15 = 6 = 9 5 HD IC HD = 2m, IC = 3m Solution: Choose right section + M I = 0 F DC sin 18.43(9) + 1(6) + 2(3)= 0 F DC = kn (C) I F DI F HI tan θ = 2/3 θ =33.7 ΣFy = ΣFy F DI cos 56.3 = F DC cos F DI = F DI = 3kN(T) F DC tan θ = 5/15 θ = M D = 0 -F HI (2) + 1(3) = 0 F HI = 1.5 kn (T) Mechanics of Materials 221

20 EXAMPLE 7.9 Determine the force in members HG, BC and BG of the truss. State whether the members are in tension or compression. 6 kn 8 kn 2 kn 3m 3m 3m 3m Solution; Determine the truss is statically determinate. Calculate the support reactions. [Ans: V A = 9 kn ( ),V E = 7 kn ( )] Cut the section of the truss through the members where forces are to be determined. 6 kn 8 kn 2 kn 3m 3m 3m 3m Mechanics of Materials 222

21 Choose one section whether left or right sections. F HG G θ 2 tan θ 1 = 4.5/3 = 56.3º θ 3 = θ 2 A B F BG F BC tan θ 2 = 1.5/3 = 26.6º 9kN 6kN θ 1 Calculate the member forces. + M B = 0 (9) 3 + F HG cos 26.6(3) = F HG = -27 F HG = kn (C) ΣFy = ΣFy 9 + F HG sin F BG sin 56.3 = = F BG F BG = 1.8 kn(t) + M G = 0 9(6) 6(3) F BC (4.5) = 0 F BC = 8 kn (T) Mechanics of Materials 223

EXERCISE 7.3 A pin-connected compound truss is loaded and supported shown in figure, calculate the member forces of HG, JC and BC by using the method of section. [Ans: F JC = 1.15kN (T), F BC = 2.

22 EXERCISE 7.3 A pin-connected compound truss is loaded and supported shown in figure, calculate the member forces of HG, JC and BC by using the method of section. [Ans: F JC = 1.15kN (T), F BC = 2.9kN (T), F HG = 3.47kN (C)] 6.7 Alternative Computation using Joint Equilibrium Method An alternative method can be applied to determine the member forces. The purpose is to reduce the time calculation. If the problems stated need to solve by using the method of joint or method of section, please follow the instruction. EXAMPLE 7.10 Calculate all member forces. 6 kn D 5m A H A E B V A 8 kn 4m 5m 4m V B Mechanics of Materials 224

23 Solution; The calculation can be started at joint A or B. 6 A C D 6 kn 2 nd 3.84 F E kn 2 F BD = = 6.15 kn 1 st DE = x x EB x = 5 4 x = F BD B Opposite direction 5 th 6 kn C D 3 rd DE = x FE x th x = 5 5 x = A 8 kn F F CE = = 6.8 kn F CE E B Mechanics of Materials 225

24 7 th C D 6 kn F EF = F FA =8.6kN 8 th A F E 3.84 B 6 th 3.84 = (to balance) F EF = 8.6kN F CF =8 kn 8 kn 6 kn 10 th C F CD D 10 th = (to balance) F CD =3.8 kn 9 th = 8.6 so = A 9 th y 2.6 AF = 8.6 y CF 2.6 y = 4 5 y = kn E 2 F AC = = 4.16 kn 2 B Mechanics of Materials 226

25 TUTORIAL 6 1) The pin-jointed truss shown in figure is subjected to a uniformly distributed load of 40kN/m along member CD and a vertical point load of 90kN at D. a) Determine the reactions at A and F. b) Determine the forces in member BC, BE and EF by using the method of joint. c) Calculate the forces in members CD and DE using the method of section. 40kN/m 90kN [Ans: V A = 98.89kN ( ),V F = 379kN ( ), F DE = 330 kn (C), F EF = 435 kn (C)] 2) The truss shown in figure is loaded with concentrated loads at B, D, E and G. The truss is pinned to the foundation at A and supported on rollers at F. Determine the internal forces in members LK, LC and BC using the method of joint and checks your answers using the method of section. State if the forces are tensile or compressive. 30kN 30kN 30kN 30kN [Ans: F BC = 36 kn (T)] Mechanics of Materials 227

26 3) Figure shows a pin-jointed plane truss pinned at A and supported on rollers at B. Determine the forces in each member of the truss by using the method of joint. Recalculate the force in member BC by using the method of section. 60kN 30kN [Ans: F AC = kN (C), F AB = 38.14kN (T), F DE = kn (T), F BD = 88.8kN (T)] 4) A pin-jointed is supported on roller support and pinned support at A and B respectively as shown in figure. Determine the forces in each member of the truss by using the method of joint. Recalculate the force in members AB and DE by using the method of section. 20kN 30kN [Ans: F AB = 5.3kN (C), F ED = 20kN (C), F EB = kn (C), F DC = 20kN (C)] Mechanics of Materials 228

27 5) (a) truss. (b) List two assumptions of analysis to determine the member s force of the Identify each of the trusses in Figure Q5(a) to Q5(d) as a determinate or statically indeterminate and stable or unstable. c) In Figure Q5(e), a statically determinate pin-jointed plane truss is pinned at A and supported on rollers at B. It carries two loads of 5000kg at joints F and G. (i) Prove that the plane truss is a statically determinate structure. (ii) Using the method of joint determine all member forces. (iii) Using the method of section, determine the member forces of CD and FG. L= 5m for each member. Figure Q5(a) Figure Q5(b) Figure Q5(c) Figure Q5(d) Mechanics of Materials 229

28 5000kg 5000kg Figure Q5(e) (FINAL EXAM SEM I 2007/2008 UTHM) 6) The pin-jointed plane frame shown in figure carries a vertical load 10 kn at D. The frame is supported by a hinge at support A and by a roller support at B. The roller support at B is designed to ensure that the raction at B is in the direction BC. (a) Determine the reaction at support A and B. (b) Determine the internal force in all member. C 10kN 4m A B D 2m 3m 2m 8m (FINAL EXAM SEM II 2007/2008 UTHM) Mechanics of Materials 230

29 REFERENCES 1) R.C.Hibbeler, Mechanics of Materials, Sixth Edition, ) Ferdinand P.Beer, Johnston Jr, T.Dewolf, Mechanics of Materials, Fourth Edition, 2006 Mechanics of Materials 231

CHAPTER 5 Statically Determinate Plane Trusses

CHAPTER 5 Statically Determinate Plane Trusses TYPES OF ROOF TRUSS TYPES OF ROOF TRUSS ROOF TRUSS SETUP ROOF TRUSS SETUP OBJECTIVES To determine the STABILITY and DETERMINACY of plane trusses To analyse