Q_-_Method_of_Sections

# Q_-_Method_of_Sections - STRUCTURAL ANALYSIS Method of...

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Unformatted text preview: STRUCTURAL ANALYSIS Method of Sections LEARNING OBJECTIVES Be able to determine the forces in members of a simple truss using the method of sections PRE-REQUISITE KNOWLEDGE Units of measurement Trigonometry concepts Rectangular component concepts Equilibrium of forces in 2-D METHOD OF SECTIONS In this method, a truss is divided into two sections. The equilibrium of both external and member forces at the division line is considered. The division line does not have to be linear but has to pass through the members where a solution of the forces is desired. EXAMPLE 1 Determine the forces in members BC, GC and GF SOLUTION 1 Use the method of sections through a-a to pass members BC, GC and GF : FY = 0; FGCsin 45o 1000 = 0, FGC = 1414 N (T) MG = 0; 1000(2) FBC(2) = 0, FX = 0; FBC + FGCcos 45o FGF = 0 FBC = 1000 N (T) FGF = 2000 N (C) EXAMPLE 2 Determine the forces in member EB of the roof truss. SOLUTION 2 MB = 0; 1000(4) + 3000(2) 4000(4) + FEDSin30(4) = 0 FED = 3000 N (C) SOLUTION 2 -continued FX = 0; FEFCos30 - 3000Cos30 = 0; F = 3000 N (C) EXAMPLE 3 Determine the forces in members HG, HC and BC EXAMPLE 3 Step 1 Determine the reactions at A and B EH A EV R R R ME = 0 12(9) + 14(6) + 18(3) RA(12) = 0 R = 20.5 kN upward EXAMPLE 3 Step 1 Determine the reactions at A and B EH A EV R R R FV = 0 12(9) + 14(6) + 18(3) RA(12) = 0 R = 20.5 kN upward SOLUTION 3 HG HC F F F A A BC R R SOLUTION 3 HG HC F F F A A BC R R ME = 0; 12(9) + 14(6) + 18(3) RA(12) = 0; R = 20.5 kN upward Where should the method of sections be applied? MH = 0; FBC(3) RA(3) = 0; F = 20.5 kN (T) EXAMPLE 4 Determine the forces in members GF, CF and CD Where should the method of sections be applied? GF CF SOLUTION 4 F FCD A A F ME = 0; 12(9) + 14(6) + 18(3) RA(12) = 0; R = 20.5 kN upward MC = 0; FGF(3) + 12(3) RA(6) = 0; F = 29kN (C) R R EXAMPLE 5 Determine the force in member GJ SOLUTION 5 Where should the method of sections be applied? GJ F CJ F CD` F EY EY MA = 0; REY(40)1000(30)1000(20)1000(10) = 0; R R MJ = 0; REY(10) FCD(10tan30o);F = 2598.1 lb FX = 0; FCD + FCJ cos 30o FGJ cos 30o = 0.0; FY = 0; REY 1000 FCJ cos 60o FGJ cos 60o = 0.0; F = 2000 lb (C); F = -500 lb (C) = 1500 lb R QUESTION In the method of sections, generally a "cut" passes through no more than _____ members in which the forces are unknown. A) 1 C) 3 B) 2 D) 4 QUESTION If a simple truss member carries a tensile force of T along its length, then the internal force in the member is ______ . A) B) C) D) tensile with magnitude of T/2 compressive with magnitude of T/2 compressive with magnitude of T tensile with magnitude of T QUESTION Can you determine the force in member ED by making the cut at section a-a? Explain your answer. A) No, there are 4 unknowns B) Yes, using MD = 0 C) Yes, using ME = 0 D) Yes, using MB = 0 QUESTION If you know FED, how would you determine FEB ? A) By taking section b-b and using ME = 0 B) By taking section b-b, and using FX = 0 and FY = 0 C) By taking section a-a and using MB = 0 D) By taking section a-a and using MD = 0 QUESTION A cut is made through members GH, BG and BC to determine the forces in them. Which section will you choose for analysis and why? A) Right, fewer calculations B) Left, fewer calculations C) Either right or left, same amount of work D) None of the above, too many unknowns QUESTION For determining the force in member GH which equation of equilibrium is best to use? A) MH = 0 B) MG = 0 C) MB = 0 D) MC = 0 QUESTION What is the force in member HG? Let L = 2 m. A) 450 N (C) B) 450 N (T) C) 800 N (C) D) 800 N (T) QUESTION P Truss ABC is changed by decreasing its height from H to 0.9 H while the width W and the load P are kept the same. Which one of the following statements is true for the revised truss as compared to the original one? A H B W C A) Force in all its members have decreased B) Force in all its members have increased C) Force in all its members have remained the same D) None of the above ...
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