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Find the axial forces at CG, DH, GH, BC, and BG in the truss shown below: 2m 2m 2m 2m C B D 1m E 1m A F G H 15 KN 20 KN 15 KN

Find the axial forces at CG, DH, GH, BC, and BG in the truss shown below: 2m 2m 2m 2m C B D 1m E 1m A F G H 15 KN 20 KN 15 KN

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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*Draw and label the parts of a typical truss.

 

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Find the axial forces at CG, DH, GH, BC, and BG in the truss shown below: 2m 2m 2m 2m D 1m B E 1m A F G H 15 KN 20 KN 15 KN
Transcribed Image Text:Find the axial forces at CG, DH, GH, BC, and BG in the truss shown below: 2m 2m 2m 2m D 1m B E 1m A F G H 15 KN 20 KN 15 KN
IENTIA HENTIA Examples: 1. Find the axial forces in the members of the truss shown. Given: EDIFICAL The FBD of joint A is constructed and the forces in members AB and AC are obtained. (Always assume the unknown member forces acting on the joint's FBD to be in tension). +1.2 m + 1.2 m→→+1.2 m → 1.2 m AB C +t ΣV 0 VER |0.9 m В D e) (AB) = 0 10.5 + E [0.9 m A AF negative sign means that the member AB = 17.5 kN (C) АВ — —17.5 kN A F G H is in compression. 000 +→ EH = 0 6 kN 9 kN 6 kN Av = 10.5 kN AF + (E) (AB) = 0 substitute the computed value of AB. Required: Force in each member of the truss Solution: AF + E)(-17.5) = 0 AF = 14 kN (T) Solve for the reactions by constructing a free diagram of the truss and using the equations of equilibrium. AF = 14 kN BF Joint F is isolated next. (It is not advisable to isolate joint B at this stage since there are three unknown forces acting on B.) +1.2 m →+ 1.2 m→|+ 1.2 m →+ 1.2 m→ C В D |0.9 m FG +1 ΣV - 0 F E 0.9 m AF = 14 kN BF – 6 = 0 Ан BF = 6 kN BF = 6 kN (T) EDIFICA F G H V 6 KN +-ΣΗ 0 FG – 14 = 0 6 kN 9 kN 6 kN CHY Av RE FG = 14 kN FG = 14 kN (T) Joint B can now be isolated and only two unknowns will appear in the FBD since the forces AB and BF are already known. ВС %3D Since tan-1 = 36.87°, then angle 0 = 2(36.87°) = 73.74º AĦ = 0 B. + ΣΥ= 0 +u £ Mg -BG sin 73.74° – 6 cos 36.87° = 0 A, (4.8) – 6(3.6) – 9(2.4) – 6(1.2) = 0 4 - AB=17.5 kN BG BG = -5 kN BG = 5 kN (C) Ay = 10.5 kN ↑ BF=6 kN +7 EX = 0 +u £MA 17.5 + BC – 6 sin 36.87° + BG cos 73.74° = 0 -RE (4.8) + 6(3.6) + 9(2.4) + 6(1.2) = 0 17.5 + BC – 6 sin 36.87° + (-5) cos 73.74° = 0 RE = 10.5 kN ↑ ВС %3D —12.5 kN BC = 12.5 kN (C) SITY
Transcribed Image Text:IENTIA HENTIA Examples: 1. Find the axial forces in the members of the truss shown. Given: EDIFICAL The FBD of joint A is constructed and the forces in members AB and AC are obtained. (Always assume the unknown member forces acting on the joint's FBD to be in tension). +1.2 m + 1.2 m→→+1.2 m → 1.2 m AB C +t ΣV 0 VER |0.9 m В D e) (AB) = 0 10.5 + E [0.9 m A AF negative sign means that the member AB = 17.5 kN (C) АВ — —17.5 kN A F G H is in compression. 000 +→ EH = 0 6 kN 9 kN 6 kN Av = 10.5 kN AF + (E) (AB) = 0 substitute the computed value of AB. Required: Force in each member of the truss Solution: AF + E)(-17.5) = 0 AF = 14 kN (T) Solve for the reactions by constructing a free diagram of the truss and using the equations of equilibrium. AF = 14 kN BF Joint F is isolated next. (It is not advisable to isolate joint B at this stage since there are three unknown forces acting on B.) +1.2 m →+ 1.2 m→|+ 1.2 m →+ 1.2 m→ C В D |0.9 m FG +1 ΣV - 0 F E 0.9 m AF = 14 kN BF – 6 = 0 Ан BF = 6 kN BF = 6 kN (T) EDIFICA F G H V 6 KN +-ΣΗ 0 FG – 14 = 0 6 kN 9 kN 6 kN CHY Av RE FG = 14 kN FG = 14 kN (T) Joint B can now be isolated and only two unknowns will appear in the FBD since the forces AB and BF are already known. ВС %3D Since tan-1 = 36.87°, then angle 0 = 2(36.87°) = 73.74º AĦ = 0 B. + ΣΥ= 0 +u £ Mg -BG sin 73.74° – 6 cos 36.87° = 0 A, (4.8) – 6(3.6) – 9(2.4) – 6(1.2) = 0 4 - AB=17.5 kN BG BG = -5 kN BG = 5 kN (C) Ay = 10.5 kN ↑ BF=6 kN +7 EX = 0 +u £MA 17.5 + BC – 6 sin 36.87° + BG cos 73.74° = 0 -RE (4.8) + 6(3.6) + 9(2.4) + 6(1.2) = 0 17.5 + BC – 6 sin 36.87° + (-5) cos 73.74° = 0 RE = 10.5 kN ↑ ВС %3D —12.5 kN BC = 12.5 kN (C) SITY
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