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