A) Determine The Force In Members BC, HC, And HG. After The Truss Is Sectioned Use A Single Equation Of Equilibrium For The Calculation Of Each Force. State If These Members Are In Tension Or Compression. B) Determine The Force In Members CD, CF, And CG And State If These Members Are In Tension Or Compression.

A) Determine The Force In Members BC, HC, And HG. After The Truss Is 
Sectioned Use A Single Equation Of Equilibrium For The Calculation Of Each 
Force. State If These Members Are In Tension Or Compression. B) Determine 
The Force In Members CD, CF, And CG And State If These Members Are In 
Tension Or Compression.

Transcribed Text:
Question: A) Determine the force in members BC, HC, and HG. After the tr…
Stru
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a) Determine the force in members BC, HC, and HG. After the truss is sectioned use a single equation of equilibrium for the calculation of each force. State if these members are in tension or compression. b) Determine the force in members CD, CF, and CG and state if these members are in tension or compression.
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3 KN
3 m
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Probs. 6836
Expert Answer
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General guidance
Concepts and reason The concept used to solve this problem is truss analysis. First, calculate the reactions at the support using force equilibrium equation or by taking moment about a particular point. Then use method of sections to calculate the force in each member. Finally take a particular section and assume the forces in that section as tension then use equilibrium conditions for forces and moments to form equations to calculate the force in each member. The type of force is decided based on the sign obtained in the final answer. If the answer has negative sign it is said to be in compression instead if it has a positive sign it is said to be in tension.
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Step-by-step
FIRST STEP ALL STEPS ANSWER ONLY
Step 1 of 7 ^
Draw the free body diagram of the entire truss as shown in Figure (4). 2 kN 4 kN
4 kN
5 kN
3 KN
3 m
elle
! b
5 m 15 m 15 m 15 m 1
Figure 4 Here, reaction at A along horizontal direction is Ac, reaction at A along vertical direction is Ay, and reaction at E along vertical direction is Ey. Apply the force equilibrium condition along horizontal direction.
E =0 A =0 Take the moment about point A.
MA=0
(-4kN x 5m) – (4kN x 10m) – (5kN x 15m) – (3kN x 20m) + (Ey x 20m) = 0
-20 – 40 – 75 – 60+ (E, *20) = 0
Ey = 1,25
Ey = 9.75kN Apply the force equilibrium condition along vertical direction.
Fy = 0
Ay + Ey – 2kN – 4kN – 4kN – 5kN – 3 kN = 0
Ay + Ey = (2+4+4+5+3) Substitute 9.75 kN for Ey.
Ay + (9.75kN) = 18kN
Ay = 8.25kN
Explanation Common mistakes Hint for next step From Figure (4), the sum of forces acting along the horizontal direction is calculated to determine the horizontal reaction force at joint A. From Figure (4), the sum of moments about the joint A is calculated to determine the vertical reaction force at joint E. From Figure (4), the sum of forces acting along the vertical direction is calculated to determine the vertical reaction force at joint A.
Step 2 of 7 ^
The free body diagram of left part of the section a-a as shown in Figure (5).
2 kN
4 kN
=>FBC
FHC
VLAS—–
=> FHG
5 m 15 m
Figure 5
Here, force in the member HG is FHG , force in the member HC is FHC, and force in the member BC is FBC. From Figure (5), calculate the angle (@). tan = DF Substitute 3m for DF and 5m for BC.
tan 0 = 3m
0 = tan-? (im)
0 = 30.96° From Figure (5), calculate the angle (0) tan o = GCDF Substitute 5m for GC, 3m for DF, and 5m for BC.
tan o = (m) o = tan-1 (cm)
6 = 21.8° Take the moment about point C.
Mc=0 (2kN x 10m) + (4kN x 5m) -(Ay x 10) + ((FHG cos 0) < 3m) + (FHG sin o) x 5m) = 0 Substitute 21.8o for o and 8.25kN for Ay. (2 x 10) + (4 x 5) -(8.25 kN x 10) + ((FHG cos 21.8°) ~ 3) + ((FHG sin 21.8°) ~ 5) = 0 20+20 - (82.5) +2.78FHG +1.86FHG = 0 -42.5+ (4.64FHG) = 0 4.64FHG = 42.5 FHG = 4261 FHG = 9.159kN (Tension) Part a. 3 The force in the member HG (FHG) is 9.159kN (Tension). Explanation Common mistakes Hint for next step The tangential angle between D F and BC is calculated by the ratio of opposite side to the adjacent side. Substitute the values of DF and BC in tangential ratio to get the value of angle 6). Similarly calculate the angle (o) for the triangle AGJF The free body diagram of left part of the section a - a is drawn. The force in the member HG is calculated by applying the moment law of equilibrium about point C to form an equation and then substitute the reaction of A along y-direction in that equation. Step 3 of 7 ^ For Figure (5), apply the force law of equilibrium along the y-direction. Fy=0 Ay + (FHC sin ) - (FHG sin 6) = 6 Substitute 21.8° for ,30.96 for 0,9.159kN for FHG and 8.25kN for Ay. (8.25kN) + (Fhcsin 30.96o) - ((9.159kN) sin 21.8°) = 6 (8.25) + FHC (0.514) - (9.159) x 0.371 = 6 FHC (0.514) + 4.852 = 6 FHC (0.514) = 6 – 4.852 FHC = 1.518 FHC = 2.23kN (Tension) Part a.2 The force in the member HC (FHC) is 2.23kN (Tension). Explanation Common mistakes Hint for next step Substitute the reaction force at A along vertical direction and the force in the member HG in the equation obtained by applying the force law of equilibrium along vertical direction and calculate the force in the member HC. Step 4 of 7 ^ For Figure (5), apply the force law of equilibrium along the horizontal direction. F = 0 FBC + (Fhc cos ) + (FHG cos 0) = 0 Substitute 21.8 for , 30.96 for 0,2.23kN for FHC, and 9.159kN for FHG. FBC + ((2.23 kN) cos 30.96°) + ((9.159kN) cos 21.8°) = 0 FBC +(2.23) x 0.858) + ((9.159) < 0.928) = 0 FBC = -10.413kN FBC = 10.413kN (Compression) Part a.1 The force in the member BC (FBC) is 10.413kN (Compression). Explanation Common mistakes Hint for next step The force in the member BC is calculated by substituting the force in the member HG and HC in the equation obtained by applying the force law of equilibrium along x-direction. Step 5 of 7 ^ (b) The free body diagram of the right part of the section b-bis shown as in Figure (6). 5 kN 3 kN FCDA FCERES 3 m FFG D5 m 1 Figure 6 Here, force in the member CD is FCD, force in the member CF is FCF, and force in the member FG is FFG. Take the moment about point F. ME=0 (FCD X 3m) - (3kN x 5m) + (Ey x 5m) = 0 (FCD X 3) - (15) + (Ey x 5) = 0 Substitute 9.75kN for Ey. FCD x 3 = 15 – ((9.75kN) * 5) FcD = _ 33,75 Fcp = -11.25KN Fcp = 11.25kN (Compression) Part b.1 The force in the member CD (FCD) is 11.25kN (Compression). Explanation Hint for next step The force in the member CD is determined by substituting the value of the reaction at E along y-axis in the equation obtained by taking moment about point F. Step 6 of 7 A For Figure (6), apply the force law of equilibrium along horizontal direction. Fr=0 - FCD - (FCF cos ) - (FFG cos 6) = 0 Substitute 11.25 kN for FCD, 21.8° for , and 30.96° for . -(-11.25kN) - (FCF cos 30.96 ) - (FFG cos 21.8°) = 0 11.25 – (0.8575FCF) - (0.928FFG) = 0 0.8575FCF = 11.25 -0.928FFG FCF = 13.12 - 1.08FFG ...... (1) Apply the force law of equilibrium along vertical direction. -5-3+ Ey + (FCF sin ) - (FFG sin o) = 0 Substitute 13.12 – 1.08FFG for FCF, 9.75 kN for Ey 21.89 for o, and 30.96° for . -8+ (9.75kN) + (13.12 - 1.08FFG) sin 30.96o - (FFG sin 21.80°) = 0 8.50 = FFG (0.555 +0.371) FFG = 8.926 FFG = 9.18kN Find the value of (FCF) from the Equation (1), FCF = 13.12 - 1.08FFG Substitute 9.18 kN for FFG. FCF = 13.12 - (1.08 x (9.18kN)) FCF = 3.2056kN (Tension) Part b.2 The force in the member CF (FCF) is 3.2056kN (Tension). Explanation | Hint for next step From the Figure (6), first apply force equilibrium condition along horizontal direction to frame the equation in terms of CF and FG. Then apply force equilibrium condition along vertical direction to get the force in the member FG. Substitute the force in the member FG in equation (1) to calculate the force in the member of CF. Step 7 of 7 ^ The free body diagram of the forces acting at point G as shown in Figure (7). FCG FGHA = FEG Figure 7 Apply the force law of equilibrium along the horizontal direction. F = 0 -FGH COS 0 + FFG COS O = 0 Here, force in the member GH is FGH. - FGH COS O = -FFG COS O FGH = FFG Substitute 9.18KN for FFG. FGH = 9.18kN Apply the force law of equilibrium along the vertical direction. Fy = 0 FGH sin o + FFG sin o + Fco = 0 Here, force in the member CG is FCG. Substitute 9.18KN for FGH , 21.80° for o, and 9.18kN for FFG. ((9.18kN) x sin 21.80°) + ((9.18kN) x sin 21.80°) + FCG = 0 6.82 +FCG = 0 FCG = -6.82KN FCG = 6.82kN (Compression) Part b.3 The force in the member CG (FCG) is 6.82kN (Compression). Explanation From Figure (7), apply horizontal force equilibrium condition, obtain the force in the member GH, then apply vertical force equilibrium condition and substitute the force in the member FGH and FFG to calculate the force in the member CG. Answer Part a.3 The force in the member HG (FHG) is 9.159kN (Tension). Part a.2 The force in the member HC (FHC) is 2.23kN (Tension). Part a.1 The force in the member BC (FBC) is 10.413kN (Compression). Part b.1 The force in the member CD (FCD) is 11.25kN (Compression). Part 6.2 The force in the member CF (FCF) is 3.2056kN (Tension). Part b.3 The force in the member CG (FCG) is 6.82kN (Compression)

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