ce 160 notes: construction of influence lines for … 160 truss influence... · 2 vukazich ce 160...
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1 VukazichCE160ConstructionofInfluenceLinesforTrusses[10]
CE 160 Notes: Construction of Influence Lines for Trusses
The truss shown is pin supported at point A, roller supported at point L. Construct the influence lines for:
1. The vertical support reaction at the pin at point A; 2. The axial force in member DF; 3. The axial force in member DG; 4. The axial force in member KJ.
For a load moving across the bridge deck from points A to L
Free-Body Diagram (F.B.D.) of the entire truss
A
C E G I K L
B D F H J
6 ft 6 ft 6 ft 6 ft 6 ft 6 ft
8 ft
Bridge Deck
A
C E G I K
L
B D F H J
6 ft 6 ft 6 ft 6 ft 6 ft 6 ft
8 ft
Ly Ay
Ax
2 VukazichCE160ConstructionofInfluenceLinesforTrusses[10]
1. Influence Line for support reaction Ay
Unit Load at C
𝑀! = 0
Counterclockwise moments about point L are positive
− 𝐴!)(36 𝑓𝑡 + 1 )(30 𝑓𝑡 = 0
𝑨𝒚 =𝟓𝟔 = 𝟎.𝟖𝟑𝟑𝟑
By moving the unit load to the other points and performing the analysis, one can verify that:
Unit load at: Ay A 1 C 0.833 E 0.667 G 0.5 I 0.333 K 0.167 L 0
Plot the influence line for the vertical support reaction at A
A
C E G I K
L
B D F H J
6 ft 6 ft 6 ft 6 ft 6 ft 6 ft
8 ft
Ly Ay
Ax
1
+
3 VukazichCE160ConstructionofInfluenceLinesforTrusses[10]
2.-4. Influence Lines for the axial force in members DF, DG, and KJ
Unit Load at C
F.B.D of entire truss and application of equations of equilibrium show that the support reactions are:
Ay = 5/6 (0.83333) Ly = 1/6 (0.16667)
F.B.D. of truss section to the left of cut a-a
Ay
C E G I K L 0 0
1.0
A
C E G I K
L
B D F H J
6 ft 6 ft 6 ft 6 ft 6 ft 6 ft
8 ft
Ly Ay
Ax
1
A
C E G I K
L
B D F H J
1
6 ft 6 ft 6 ft 6 ft 6 ft 6 ft
8 ft
0.1667 0.8333
a
a
4 VukazichCE160ConstructionofInfluenceLinesforTrusses[10]
Moment Equilibrium of Truss Section about point G
𝑀! = 0
Counterclockwise moments about point G are positive
− 𝐹!")(8 𝑓𝑡 + 1 12 𝑓𝑡 − 0.833 18 𝑓𝑡 = 0
𝑭𝑫𝑭 = −𝟑𝟖 = −𝟎.𝟑𝟕𝟓
Force Equilibrium
+↑ 𝐹! = 0
Upward forces positive
0.8333− 1−45𝐹!" = 0
𝑭𝑫𝑮 = −𝟓𝟐𝟒 = −𝟎.𝟐𝟎𝟖𝟑
Note that member KJ is a zero-force member when the load is applied at any joint other than joint K
𝑭𝑲𝑱 = 𝟎
A
C E G
B D
1
6 ft 6 ft 6 ft
0.8333
a
a
8 ft
FDF
FDG
FEG
+
5 VukazichCE160ConstructionofInfluenceLinesforTrusses[10]
Unit Load at E
F.B.D of entire truss can show that the support reactions are: Ay = 2/3 (0.6667) Ly = 1/3 (0.3333)
F.B.D. of truss section to the left of cut a-a
Moment Equilibrium of Truss Section about point G
𝑀! = 0
Counterclockwise moments about point G are positive
− 𝐹!")(8 𝑓𝑡 + 1 6 𝑓𝑡 − 0.6667 18 𝑓𝑡 = 0
𝑭𝑫𝑭 = −𝟑𝟒 = −𝟎.𝟕𝟓
A
C E G I K
L
B D F H J
1
6 ft 6 ft 6 ft 6 ft 6 ft 6 ft
8 ft
0.3333 0.6667
a
a
A
C E G
B D
1
6 ft 6 ft 6 ft
0.6667
a
a
8 ft
FDF
FDG
FEG
+
6 VukazichCE160ConstructionofInfluenceLinesforTrusses[10]
Force Equilibrium
+↑ 𝐹! = 0
Upward forces positive
0.6667− 1−45𝐹!" = 0
𝑭𝑫𝑮 = −𝟓𝟏𝟐 = −𝟎.𝟒𝟏𝟔𝟕
Note that member KJ is a zero-force member when the load is applied at any joint other than joint K
𝑭𝑲𝑱 = 𝟎
Unit Load at G
F.B.D of entire truss can show that the support reactions are: Ay = 1/2 (0.5) Ly = 1/2 (0.5)
F.B.D. of truss section to the left of cut a-a
A
C E G I K
L
B D F H J
1
6 ft 6 ft 6 ft 6 ft 6 ft 6 ft
8 ft
0.5 0.5
a
a
7 VukazichCE160ConstructionofInfluenceLinesforTrusses[10]
Moment Equilibrium of Truss Section about point G
𝑀! = 0
Counterclockwise moments about point G are positive
− 𝐹!")(8 𝑓𝑡 − 0.5 18 𝑓𝑡 = 0
𝑭𝑫𝑭 = −𝟗𝟖 = −𝟏.𝟏𝟐𝟓
Force Equilibrium
+↑ 𝐹! = 0
Upward forces positive
0.5−45𝐹!" = 0
𝑭𝑫𝑮 =𝟓𝟖 = 𝟎.𝟔𝟐𝟓
Note that member KJ is a zero-force member when the load is applied at any joint other than joint K
𝑭𝑲𝑱 = 𝟎
A
C E G
B D
6 ft 6 ft 6 ft
0.5
a
a
8 ft
FDF
FDG
FEG
+
8 VukazichCE160ConstructionofInfluenceLinesforTrusses[10]
Unit Load at K
F.B.D. of truss section to the left of cut a-a
Moment Equilibrium of Truss Section about point G
𝑀! = 0
Counterclockwise moments about point G are positive
− 𝐹!")(8 𝑓𝑡 − 0.1667 18 𝑓𝑡 = 0
𝑭𝑫𝑭 = −𝟑𝟖 = −𝟎.𝟑𝟕𝟓
Force Equilibrium
+↑ 𝐹! = 0
A
C E G I K L
B D F H J
1
6 ft 6 ft 6 ft 6 ft 6 ft 6 ft
8 ft
0.8333 0.1667
a
a
A
C E G
B D
6 ft 6 ft 6 ft
0.1667
a
a
8 ft
FDF
FDG
FEG
+
9 VukazichCE160ConstructionofInfluenceLinesforTrusses[10]
Upward forces positive
0.1667−45𝐹!" = 0
𝑭𝑫𝑮 =𝟓𝟐𝟒 = 𝟎.𝟐𝟎𝟖𝟑
F.B.D. of joint K
𝑭𝑲𝑱 = 𝟏
Summary of Analyses
Unit load at: FDF FDG FKJ A 0 0 0 C -0.375 -0.2083 0 E -0.75 -0.4167 0 G -1.125 0.6250 0 I -0.75 0.4167 0 K -0.375 0.2083 1.0 L 0 0 0
1
K FKL
FKJ = 1
FIK
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