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

10 VukazichCE160ConstructionofInfluenceLinesforTrusses[10]

Plot the influence lines for the member forces in DF, DG, and KJ

FDF

C E G I K L

-1.125

0 0 A

FDG

C E

G I K

L

0 0 A

0.625

-0.4167

FJK

C E G I K L

0 0 A

1.0