ce 160 notes: construction of influence lines for beams 160 constructing infl... · 1 vukazich ce...
TRANSCRIPT
1 VukazichCE160ConstructionofInfluenceLinesforBeams[8]
CE 160 Notes: Construction of Influence Lines for Beams
Construct the influence line for the vertical reaction at the pin support for the beam (point A).
Place unit dimensionless load at x = 0 (point A) and draw F.B.D.
Use moment equilibrium to find support reaction at A
Choose point B
𝑀! = 0
Counterclockwise moments about point B are positive
1 )( 12 𝑚 − 𝐴! )(12 𝑚 = 0
𝑨𝒚 = 𝟏 (Ay is positive so Ay acts upward as assumed)
12 m
A
B
C
3 m
12 m
C
3 m
1
Ax
Ay
By
x
+
2 VukazichCE160ConstructionofInfluenceLinesforBeams[8]
Place unit dimensionless load at x = 6 m and draw F.B.D.
Use moment equilibrium to find support reaction at A
Choose point B
𝑀! = 0
Counterclockwise moments about point B are positive
1 )( 6 𝑚 − 𝐴! )( 12 𝑚 = 0
𝑨𝒚 = 𝟎.𝟓 (Ay is positive so Ay acts upward as assumed)
Place unit dimensionless load at x = 12 m (point B) and draw F.B.D.
Use moment equilibrium to find support reaction at A
6 m
C
3 m
1
Ax
Ay
By
x
6 m
+
12 m
C
3 m
1
Ax
Ay
By
x
3 VukazichCE160ConstructionofInfluenceLinesforBeams[8]
Choose point B
𝑀! = 0
Counterclockwise moments about point B are positive
− 𝐴! )( 12 𝑚 = 0
𝑨𝒚 = 𝟎
Place unit dimensionless load at x = 15 m and draw F.B.D.
Use moment equilibrium to find support reaction at A
Choose point B
𝑀! = 0
Counterclockwise moments about point B are positive
− 1 )( 3 𝑚 − 𝐴! )( 12 𝑚 = 0
𝑨𝒚 = −𝟎.𝟐𝟓 (Ay is negative so Ay acts downward)
+
12 m
C
3 m
1
Ax
Ay
By
x
+
4 VukazichCE160ConstructionofInfluenceLinesforBeams[8]
Table to keep track of results
Unit load at x = Ay 0 m 1 6 m 0.5 12 m 0 15 m -0.25
The plot of the value of the support reaction versus unit load is the influence line
1
0.5
-0.25
0
Ay
A B
C
5 VukazichCE160ConstructionofInfluenceLinesforBeams[8]
Construct the influence line for the internal shear and internal bending moment at point D between the pin and roller supports for the same beam.
Place unit dimensionless load at x = 0 (point A) and draw F.B.D. to find the support reactions
Note that from moment equilibrium and force equilibrium in the vertical and horizontal directions we find that Ay = 1; By = 0; and Ax = 0
F.B.D. of beam cut at point D
A
B
C
3 m 6 m 6 m
D
6 m
C
3 m
1
1 x
6 m
D A B
6 VukazichCE160ConstructionofInfluenceLinesforBeams[8]
Use moment equilibrium to find MD
𝑀! = 0
Counterclockwise moments about point D are positive
− 1 )( 6 𝑚 + 1 )( 6 𝑚 +𝑀! = 0
𝑴𝑫 = 𝟎 (note no internal forces will be developed in the beam)
Use force equilibrium to find VD
+↑ 𝐹! = 0
Upward forces positive
−1+ 1− 𝑉! = 0
𝑽𝑫 = 𝟎
Place unit dimensionless load at x = 6-- m (just to the left of point D, x = 5.99999… m) and draw F.B.D. to find the support reactions
6 m
1
1
A
MD VD
+
7 VukazichCE160ConstructionofInfluenceLinesforBeams[8]
Use moment equilibrium to find support reaction at A
Choose point B
𝑀! = 0
Counterclockwise moments about point B are positive
1 )( 6! 𝑚 − 𝐴! )( 12 𝑚 = 0
𝑨𝒚 = 𝟎.𝟓! (Ay approaches 0.5 in the limit)
F.B.D. of beam cut at point D
Use moment equilibrium to find MD
𝑀! = 0
6 m
C
3 m
1
Ax
Ay By
5.999…
6 m
D
+
6 m
0.5
A
MD
VD
1
8 VukazichCE160ConstructionofInfluenceLinesforBeams[8]
Counterclockwise moments about point D are positive
− 0.5 )( 6 𝑚 +𝑀! = 0
𝑴𝑪 = 𝟑 𝒎 (units of the moment arm are important to include)
Use force equilibrium to find VD
+↑ 𝐹! = 0
Upward forces positive
0.5 − 1− 𝑉! = 0
𝑽𝑫 = −𝟎.𝟓
Place unit dimensionless load at x = 6+ m (just to the right of point D, x = 6.00000…0001 m) and draw F.B.D. to find the support reactions
Use moment equilibrium to find support reaction at A
Choose point B
𝑀! = 0
Counterclockwise moments about point B are positive
1 )( 6! 𝑚 − 𝐴! )( 12 𝑚 = 0
+
6 m
C
3 m
1
Ax
Ay By
6.00…01
6 m
D
+
9 VukazichCE160ConstructionofInfluenceLinesforBeams[8]
𝑨𝒚 = 𝟎.𝟓! (Ay approaches 0.5 in the limit)
F.B.D. of beam cut at point D
6 m
0.5
A
MD
VD
10 VukazichCE160ConstructionofInfluenceLinesforBeams[8]
Use moment equilibrium to find MD
𝑀! = 0
Counterclockwise moments about point D are positive
− 0.5 )( 6 𝑚 +𝑀! = 0
𝑴𝑫 = 𝟑 𝒎
Use force equilibrium to find VD
+↑ 𝐹! = 0
Upward forces positive
0.5 − 𝑉! = 0
𝑽𝑫 = 𝟎.𝟓
Place unit dimensionless load at x = 12 m (point B) and draw F.B.D. to find the support reactions
Note that from moment equilibrium and force equilibrium in the vertical and horizontal directions we find that Ay = 0; By = 1; and Ax = 0 Similar to the case when the unit load was at point A, no internal forces are developed in the beam so both VD and MD are zero.
+
11 VukazichCE160ConstructionofInfluenceLinesforBeams[8]
Place unit dimensionless load at x = 15 m and draw F.B.D. to find the support reactions
Note that from moment equilibrium and force equilibrium in the vertical and horizontal directions we find that Ay = -0.25; By = 1.25; and Ax = 0
F.B.D. of beam cut at point D
Use moment equilibrium to find MD
𝑀! = 0
Counterclockwise moments about point D are positive
0.25 )(6 𝑚 +𝑀! = 0
𝑴𝑫 = −𝟏.𝟓 𝒎
Use force equilibrium to find VD
6 m
C
3 m
1
Ax
Ay By
15 m
6 m
D
6 m
0.25
A
MD
VD
+
12 VukazichCE160ConstructionofInfluenceLinesforBeams[8]
+↑ 𝐹! = 0
Upward forces positive
−0.25 − 𝑉! = 0
𝑽𝑫 = −𝟎.𝟐𝟓
Table to keep track of results
Unit load at x = VD MD 0 mt 0 0 6 m− -0.5 3 m 6 m+ 0.5 3 m 12 m 0 0 15 m -0.25 -1.5 m
Plot values of internal force versus position of unit load is the influence line
Note that the influence lines of statically determinate structures will always consist of straight line segments
-0.5
0.5
-0.25 0
VD
A C
B D 0
3 m
-1.5 m 0
MD
A C
B D 0