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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 PREREQUISITE KNOWLEDGE Units of measurement Trigonometry concepts Rectangular component concepts Equilibrium of forces in 2D 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 aa 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 aa? 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 bb and using ME = 0 B) By taking section bb, and using FX = 0 and FY = 0 C) By taking section aa and using MB = 0 D) By taking section aa 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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 Spring '08
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