Truss: Method of Joints and Sections Theory of Structure - I

Lecture Outlines Method of Joints Zero Force Members Method of Sections Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan

Method of Joints If a truss is in equilibrium, then each of its joints must also be in equilibrium The method of joints consists of satisfying the equilibrium conditions for the forces exerted “on the pin” at each joint of the truss Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan

Truss members are all straight two-force members lying in the same plane The force system acting at each pin is coplanar and concurrent (intersecting) Rotational or moment equilibrium is automatically satisfied at the joint, only need to satisfy ∑ Fx = 0, ∑ Fy = 0 Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan

Procedure for Analysis Draw the free-body diagram of a joint having at least one known force and at most two unknown forces (may need to first determine external reactions at the truss supports) Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan

Establish the sense of the unknown forces Always assume the unknown member forces acting on the joint’s free-body diagram to be in tension (pulling on the “pin”) Assume what is believed to be the correct sense of an unknown member force In both cases a negative value indicates that the sense chosen must be reversed Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan

Orient the x and y axes such that the forces can be easily resolved into their x and y components Apply ∑ Fx = 0 and ∑ Fy = 0 and solve for the unknown member forces and verify their correct sense Continue to analyze each of the other joints, choosing ones having at most two unknowns and at least one known force Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan

Members in compression “push” on the joint and members in tension “pull” on the joint Mechanics of Materials and building codes are used to size the members once the forces are known Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan

The Method of Joints 2 m 500 N 45o A B C Ax = 500 N Ay = 500 N Cy = 500 N 500 N B 45o Joint B SFx = 0: + x y 500 - FBCsin45o = 0 FBC = 707.11 N (C) FBA FBC SFy = 0: + - FBA + FBCcos45o = 0 FBA = 500 N (T) Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan

B 500 N 2 m 45o A C Ax = 500 N Ay = 500 N Cy = 500 N Joint A + SFx = 0: + 500 - FAC = 0 FAC = 500 N (T) FAC Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan

Zero Force Members Truss analysis using the method of joints is greatly simplified if one is able to determine those members which support no loading (zero-force members) These zero-force members are used to increase stability of the truss during construction and to provide support if the applied loading is changed Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan

If only two members form a truss joint and no external load or support reaction is applied to the joint, the members must be zero-force members. If three members form a truss for which two of the members are collinear, the third member is a zero-force member provided no external force or support reaction is applied. Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan

Zero-force members Frequently the analysis can be simplified by identifying members that carry no load two typical cases are found When only two members form a non-collinear joint and there is no external force or reaction at that joint, then both members must be zero-force If either TCB or TCD ≠ 0, then C cannot be in equilibrium, since there is no restoring force towards the right. Hence both BC and CD are zero-load members here. P A B C D E TCB TCD C Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan

When three members form a truss joint for which two members are collinear and the third is at an angle to these, then this third member must be zero-force in the absence of an external force or reaction from a support P Here, joint B has only one force in the vertical direction. Hence, this force must be zero or B would move (provided there are no external loads/reactions) Also TAB = TBC A B C D TBC TAB TBD B E Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan

While zero-force members can be removed in this configuration, care should be taken any change in the loading can lead to the member carrying a load the stability of the truss can be degraded by removing the zero-force member P You may think that we can remove AD and BD to make a triangle … This satisfies the statics requirements However, this leaves a long CE member to carry a compressive load. This long member is highly susceptible to failure by buckling. A B C D E Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan

P A B C D E Dx Ey Dy C FCD FCB SFx = 0: FCB = 0 + SFy = 0: FCD = 0 + FAE FAB SFy = 0: FABsinq = 0, FAB = 0 + A q SFx = 0: FAE + 0 = 0, FAE = 0 + Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan

Example 3-4 Using the method of joints, indicate all the members of the truss shown in the figure below that have zero force. P A B C D E F G H Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan

SOLUTION P A B C D E F G H Ax Gx Joint D SFy = 0: + x y FDC FDE Joint D SFy = 0: + x y FDC FDE FDCsinq = 0, FDC = 0 D q SFx = 0: + FDE + 0 = 0, FDE = 0 FEC FEF SFx = 0: + Joint E P E FEF = 0 Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan

P A B C D E F G H Ax Gx x y Joint H SFy = 0: + FHF FHA FHB H FHB = 0 Gx x y Joint H SFy = 0: + FHF FHA FHB H FHB = 0 FGF FGA Joint G SFy = 0: + G Gx FGA = 0 Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan

Method of Sections Based on the principle that if a body is in equilibrium, then any part of the body is also in equilibrium Procedure for analysis Section or “cut” the truss through the members where the forces are to be determined Before isolating the appropriate section, it may be necessary to determine the truss’s external reactions (then 3 equs. of equilibrium can be used to solve for unknown member forces in the section). Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan

Establish the sense of the unknown member forces Draw the free-body diagram of that part of the sectioned truss that has the least number of forces acting on it Establish the sense of the unknown member forces Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan

Apply 3 equations of equilibrium trying to avoid equations that need to be solved simultaneously Moments should be summed about a point that lies at the intersection of the lines of action of two unknown forces If two unknown forces are parallel – sum forces perpendicular to the direction of these unknowns Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan

The Method of Sections Ex Dx Dy a 100 N A B C D E F G 2 m + SMG = 0: FGF FGC FBC 100 N A B G 2 m 45o 100(2) - FBC(2) = 0 FBC = 100 N (T) C SFy = 0: + -100 + FGCsin45o = 0 FGC = 141.42 N (T) + SMC = 0: 100(4) - FGF(2) = 0 FGF = 200 N (C) Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan

Example 3-6 Determine the force in members GF and GD of the truss shown in the figure below. State whether the members are in tension or compression. The reactions at the supports have been calculated. 6 kN 8 kN 2 kN Ax = 0 Ay = 9 kN Ey = 7 kN A B C D E F G H 3 m 4.5 m Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan

6 kN 8 kN 2 kN Ax = 0 Ay = 9 kN Ey = 7 kN A B C D E F G H 3 m 4.5 m SOLUTION a 2 kN Ey = 7 kN D E F Section a-a 3 m FFG FDG FDC 26.6o 56.3o + SMD = 0: 3 m 26.6o O FFGsin26.6o(3.6) + 7(3) = 0, FFG = -17.83 kN (C) + SMO = 0: - 7(3) + 2(6) + FDGsin56.3o(6) = 0, FDG = 1.80 kN (C) Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan

Example 3-7 Determine the force in members BC and MC of the K-truss shown in the figure below. State whether the members are in tension or compression. The reactions at the supports have been calculated. A B C D E F G H I J K L M N O P 5.34 kN 6.67 kN 8 kN 7.11 kN 12.9 kN 3 m 4.6 m Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan

a SOLUTION A B C D E F G H I J K L M N O P 5.34 kN 6.67 kN 8 kN 3 m 4.6 m Section a-a A B L 5.34 kN 12.9 kN 4.6 m 6 m FLK FBM FLM FBC + SML = 0: FBC(6) - 12.9(4.6) = 0, FBC = 9.89 kN (T) Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan

-FCMcos33.1o(6) - 9.89(6) - 8(4.6) + 7.11(18.4) = 0 FCM = 6.90 kN (T) b A B C D E F G H I J K L M N O P 5.34 kN 6.67 kN 8 kN 7.11 kN 12.9 kN 3 m 4.6 m FKL FKM FCM C D E F G H I J K N O P 6.67 kN 8 kN 7.11 kN 3 m 4.6 m 9.89 kN 33.1o + SMK = 0: -FCMcos33.1o(6) - 9.89(6) - 8(4.6) + 7.11(18.4) = 0 FCM = 6.90 kN (T) Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan

Practice Problems Book Examples and exercise questions related to Structural Analysis by R.C. Hibbeler Chapter no. 03 Examples and exercise questions related to Joint Method Zero Force Members Section Method Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan

Thank You Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan