Post on 30-Apr-2018
transcript
1 VukazichCE160TrussAnalysisMethodofSections[3]
CE 160 Notes: Truss Method of Sections Example
A truss is pin supported at point A, roller supported at point L and subjected to the point loads shown. Determine the axial force in truss member FH.
Draw a Free-Body Diagram (F.B.D.) of the entire truss to find the support reactions
5 kN 5 kN 5 kN
1 kN
1 kN
1 kN
1 kN
1 kN
8 m
5 m 5 m 5 m 5 m 5 m 5 m
A
B
D
F
H
J
L
C E G I K
5 kN 5 kN 5 kN
1 kN
1 kN
1 kN
1 kN
1 kN
8 m
5 m 5 m 5 m 5 m 5 m 5 m
A
B
D
F
H
J
L
C E G I K
Ay
Ax
Ly
2 VukazichCE160TrussAnalysisMethodofSections[3]
Notes:
β’ Senses for the reaction force components Ax , Ay and Ly are assumed; β’ All appropriate dimensions and directions of forces are shown β’ Three unknown forces: Ax , Ay , and Ly.
Apply Equations of Equilibrium to find support reactions
Usually we can isolate an unknown by taking moment equilibrium about a convenient point Choose point A
π! = 0
Counterclockwise moments about point A are positive
Note that unknown forces: Ax and Ay are concurrent at A and so their moment about A is zero.
β 1 ππ + 5 ππ 5 π β 1 ππ + 5 ππ 10 π β 1 ππ + 5 ππ 15 π β 1 ππ )( 20 π
β 1 ππ )(25 π + πΏ! )( 30 π = 0
π³π = π.π ππ΅ (Ly is positive so Ly acts upward as assumed)
Force Equilibrium
+β πΉ! = 0
forces positive to the right
π¨π = π
Force Equilibrium
+β πΉ! = 0
Upward forces positive
π΄! β 1 ππ β 5 ππ β 1π π β 5 ππ β 1 ππ β 5 ππ β 1 ππ β 1ππ + Ly 7.5 ππ = 0
π¨π = ππ.π ππ΅ (Ay is positive so Ay acts upward as assumed)
+
3 VukazichCE160TrussAnalysisMethodofSections[3]
F.B.D of entire truss in equilibrium with all support reaction forces shown
Draw a F.B.D. or a portion of the truss, cutting through the member whose force we are interested in finding F.B.D. of truss section to the right of cut a-a
Notes:
β’ Assume member forces in FH, GH, and GI are in tension (forces pulling away from joint);
β’ Moment equilibrium equation about point G will contain only unknown force FFH.
5 kN 5 kN 5 kN
1 kN
1 kN
1 kN
1 kN
1 kN
8 m
5 m 5 m 5 m 5 m 5 m 5 m
A
B
D
F
H
J
L
C E G I K
12.5 kN
7.5 kN
a
a
1 kN
1 kN
5 m 5 m
H
J
L
I K
7.5 kN
a
a
5 m
8 m
G
F FFH
FGH
FGI
4 VukazichCE160TrussAnalysisMethodofSections[3]
Principle of Transmissibility Advantageous to slide FFH to point F and break into components (note that it would also be advantageous to slide FFH to point L)
Moment Equilibrium of Truss Section about point G
π! = 0
Counterclockwise moments about point G are positive
Note that unknown forces FGH and FGI are concurrent at G and so their moment about G is zero.
πΉ!"1517 8 π β 1 ππ )( 5 π β 1 ππ)( 10 π + (7.5 ππ )( 15 π) = 0
πππ― = βππ.ππ ππ΅ (FFH is negative so FH is in compression)
+
5 m 5 m 5 m
1 kN
1 kN
H
J
L
I K
7.5 kN
a
a
8 m
G
F
FFH (8/17)
FGH
FGI
FFH (15/17)
17 8
15