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A B C D EF

GH

Influence Line for Girder

Forces Apply to the Girder

(b/L)P1 (a/L)P1 P2

P1

a b

P2

L

A B

P1

D

P2

A

B C D

F

GH

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A F

GH

Reaction

1

0.5 0.5

RGRH

RH

1

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A

F

GH

RG

1

0.5 0.5

RG

11

RH

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A

F

GH

Shear

L

a b b b

1

0.5 0.5

1

1

VC

VC

VCx

2b/L

-(a+b)/L

1

C

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A

F

GH

L

VCD

VCD

1

0.5 0.5

1

0.5 0.5

VCD x

-a/L

b/L

a b

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A

C

F

GH

Bending Moment

L

a b

ab/L

1

0.5 0.5

MC

x

1

1

MCMC

C

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A

B C D E

F

GH

L

MFx

1

0.5 0.5

1

0.5 0.5

MFMF

a b

ab/L

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Example 6-5

Draw the influence for

- the ReactionRGandRF- Shear VCD- the momentMCandMH

1.5 m 1.5 m

A

C D

H

F

3 m 3 m 3 m 3 m

G

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1

RG

x

RG

1.333

0.6670.333

1.5 m 1.5 m

A H

F

3 m 3 m 3 m 3 m

G

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RF

x

RF

10.667

0.333

-0.333

1.5 m 1.5 m

A H

F

3 m 3 m 3 m 3 m

G

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1.5 m 1.5 m

A H

F

3 m 3 m 3 m 3 m

G

VCD

x

VCD

VCD

1

0.5 0.5

4.5/9

-4.5/9

0.333

-0.333

0.333

C D

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1.5 m 1.5 m

A H

F

3 m 3 m 3 m 3 m

G

MCMC

1

1

MC

x

(3)(6)/9 = 2

-2

1

C

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MHMH

1

0.5 0.5

MH

x

(4.5)(4.5)/9 = 2.25

1.51.5

-1.5

1.5 m 1.5 m

A H

F

3 m 3 m 3 m 3 m

G

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Example 6-6

Draw the influence line diagrams of girder for

- the reaction at Cand G,

- shear atEandH,

- bending moment atH.

2 m

A

GB

6 @ 4 m = 24 m

C D E FH

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SOLUTION

RC

RC

10.75

0.500.25

1.25

A

GB

6 @ 4 m = 24 m

C D E FH

2 m

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A

GB

6 @ 4 m = 24 m

C D E FH

2 m

RG

RG

1

1

10.25

0.500.75

-0.25

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A

GB

6 @ 4 m = 24 m

C D E FH

2 m

VE

1

1

1

1

VE

VE

8/16 = 0.5

-8/16 = 0.5-0.25

0.250.25

8 m 8 m

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A

GB

6 @ 4 m = 24 m

C D E FH

2 m

6/16 = 0.375

-10/16 = 0.625

VH

1

1

VH

VH

1

0.5 0.5

-0.25

0.250.25

-0.50

10 m 6 m

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A

GB

6 @ 4 m = 24 m

C D E FH

2 m

MH

1

1

1

0.5 0.5

MHMH

(10)(6)/16 = 3.75

1.502.50

-1.50

3

10 m 6 m

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79

Draw the influence for

- the ReactionRGandRH- Shear VCand VCD- the momentMC,MD, andMF

and determine the maximum for

- the Reaction (RG)maxand (RH)max- Shear (VCD)max

due to

- a uniform dead load 2 kN/m

- a uniform live load 5 kN/m

- a concentrated live load 50 kN

Example 6-7

2 m 2 m 4 m 4 m 4 m

2 m 2 m

A

B C D

F

GH

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2 m 2 m 4 m 4 m 4 m

2 m 2 m

A

B C D

F

GH

RG

1

0.5 0.5

10/14

6/14

2/141/14

RG

11

RH

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2 m 2 m 4 m 4 m 4 m

2 m 2 m

A

B C D

E

F

GH

1

0.5 0.5

RGRH

RH

1 12/14

8/144/146/14

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2 m 2 m 4 m 4 m 4 m

2 m 2 m

A

B C D

F

GH

VC

VC

1

0.5 0.5

VCx

8/14

-6/14

1

1

1

-2/14

4/14

-1/14

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-8/14

6/14

2 m 2 m 4 m 4 m 4 m

2 m 2 m

A

B C D

F

GH

VF

VF

1

0.5 0.5

1

0.5 0.5

VCD x

-1/14 -2/14 -6/14

4/14

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2 m 2 m 4 m 4 m 4 m

2 m 2 m

A

B C D

F

GH

(6)(8)/14 = 3.43

1

0.5 0.5

MCx

(4/8)(3.43)(2/6)(3.43)

1

1

MCMC

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2 m 2 m 4 m 4 m 4 m

2 m 2 m

A

B C D

E

F

GH

1

0.5 0.5

1

1

MD x

(10)(4)/14 = 2.86(6/10)(2.86)(2/10)(2.86)

MDMD

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(8)(6)/14=3.428

2 m 2 m 4 m 4 m 4 m

2 m 2 m

A

B C D

F

GH

MFx

1

0.5 0.5

1

0.5 0.5

2.2852.571

0.8570.429

MFMF

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[0.54 (2/14)] + [(0.5)(2/14 + 1)(12) = 7.143

RG

110/14

6/14

2/141/14

Maximum Reaction

2 kN/m

5 kN/m

50 kN

(RG)max = (2)(7.143) + (5)(7.143) + (50)(1)

= 100 kN

2 m 2 m 4 m 4 m 4 m

2 m 2 m

A

B C D E

F

GH

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2 m 2 m 4 m 4 m 4 m

2 m 2 m

A

B C D E

F

GH

(0.5)(16)(12/14) = 6.857

RH

12/14

8/144/146/14

Maximum Reaction

2 kN/m

5 kN/m

50 kN

(RH)max = (2)(6.857) + (5)(6.857) + (50)(12/14)

= 90.86 kN

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2 m 2 m 4 m 4 m 4 m

2 m 2 m

A

B C D E

F

GH

+ (0.5)(5.6)(4/14) = + 0.8

-[0.5(4)(2/14) + 0.5(2/14) + (6/14)(4) + 0.5(2.4)(6/14)] = -1.943

2.4 m

VCD

-1/14 -2/14-6/14

4/14

Maximum Shear

2 kN/m

5 kN/m

50 kN

(VCD)max = (2)(-1.943 + 0.8) (5)(-1.943) + (50)(-6/14)

= -33.43 kN

C D

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SP 6-1

Draw the influence line for

-the reaction at supportAandD

-the shear and moment atE

- the moment atB andC .

2 m 2 m 2 m 2 m 2 m

A B E C D

I fl Li f T

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Influence Line for Trusses

RA

x

1

RE

x

1

(3/4) (2/4)(1/4)

(2/4)(3/4)

(1/4)

A B C

E

GH

D

F

4 @xm

h

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A B C

E

GH

D

F

4 @xm

h

1 kN

Consider the section to the right

FCB

RE

FGB

FGH

FBC x

(2x/h)(1/4)

+ !MG= 0:

0)()2( #$ hFxR BCE

EBC Rh

xF

2#

RE

x

1

(2/4) (3/4)(1/4)

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A B C

E

GH

D

F

4 @xm

h

1 kN

Consider section to the left

FHG

FBC

RA

FBG

+ !MG= 0:

0)2()( #$ xRhF ABC

ABC Rh

xF

2#

FBC x

(2x/h)(1/4)

RA

x

1

(3/4) (2/4)(1/4)

(2x/h)(1/4)

(2x/h)(2/4)

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Determine the maximum forcedeveloped

in memberBC , BG, andCG of the truss

due to the wheel loads of the car. Assume the

loads are applied directly to the truss.

the wheel loads of the car

4 kN 1.5 kN2 m

Example 6-8-1

AB C

E

GH

4 m

D

F

3 m 3 m 3 m 3 m

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RA

x

10.75

0.5 0.25

RE

x

0.50.75

0.25

1

SOLUTION

AB C

E

GH

4 m

D

F

3 m 3 m 3 m 3 m

Consider the section to the right

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AB C

E

GH

4 m

D

F

3 m 3 m 3 m 3 m

Influence Line for FBC

1 kN

Consider the section to the right

FBC= (6/4)RE = 1.5RE

FCB

RE

FGB

FGH

RE

x

0.50.75

0.25

1

FBC x

1.5(0.25) = 0.375

+ !MG= 0:

Consider the section to the left

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AB C

E

GH

4 m

D

F

3 m 3 m 3 m 3 m

0.375

1.5(0.5) = 0.75

1 kN

FHG

FBC

RA

FBG

RA

x

10.75

0.5 0.25

FBC= 1.5RA

FBC

x

0.375

+ !MG

= 0:

Consider the section to the right

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AB C

E

GH

4 m

D

F

3 m 3 m 3 m 3 m

RE

x

0.50.75

0.25

1

Influence Line for FBG

1 kN

Consider the section to the right

!Fy= 0 ;

FBG cos % =RE

FGH

FGB

FCB

RE

%

FBG= 1.25RE

FBG

(4/5) =RE1.25(0.25) = 0.3125

FBG x

Consider the left handF

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AB C

E

GH

4 m

D

F

3 m 3 m 3 m 3 m

1 kN

!Fy= 0 ;

FBG cos % +RA = 0

RA

x

10.75

0.50.25

FHG

FBC

RA

FBG%

0.3125

FBG x

FBG= -1.25RA

FBG(4/5) = -RA

-0.3125

-1.25(0.5) = -0.625

Hf i f

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AB C

E

GH

4 m

D

F

3 m 3 m 3 m 3 m

FCG x

Influence Line for FCG

1 kN

0

1 kN

0

H

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AB C

E

GH

4 m

D

F

3 m 3 m 3 m 3 m

FCG x

1 kN

1

1

(FBC

) by Loads

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AB C

E

GH

4 m

D

F

3 m 3 m 3 m 3 m

(FBC

)max by Loads

(FBC)max= (4)(0.75) + (1.5)(0.5)

= 3.75 kN (T)

0.375

0.75FBC

x

0.375

4 kN 1.5 kN2 m

0.5

(FBG

)max by Loads

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AB C

E

GH

4 m

D

F

3 m 3 m 3 m 3 m

(BG

)max by o ds

(FBG)max= (4)(-0.625) + (1.5)(-0.417) = -3.126 kN (C)

FBG x

-0.3125

0.3125

-0.625

4 kN 1.5 kN2 m

-0.417

(FCG

)max by Loads

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AB C

E

GH

4 m

D

F

3 m 3 m 3 m 3 m

(CG

)max y

FCG x

1

0.333

(FCG)max= (4)(1) + (1.5)(0.333) = 4.50 kN (T)

4 kN 1.5 kN2 m

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105

Draw the influence line for FBC, FBH,andFCH.

Example 6-8-2

AB C

E

HI

4 m

D

F

3 m 3 m 3 m 3 m

GJ

SOLUTION HI FGJ

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RA

x

10.75

0.5 0.25

RE

x

0.50.75

0.25

1

SOLUTION

AB C

E

HI

4 m

D

F

3 m 3 m 3 m 3 m

GJ

HI G FJ

Influence Line for FBC

Consider the section to the rightF

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AB C

E

HI

4 m

D

G

3 m 3 m 3 m 3 m

FJInfluence Line for FBC

1 kN

FBC= (6/4)RE = 1.5RE

FCB

RE

FHB

FHI

RE

x

0.50.75

0.25

1

FBC x

1.5(0.25) = 0.375

+ !MH= 0:

HI G FJConsider the section to the left

FHI

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AB C

E

H

4 m

D

G

3 m 3 m 3 m 3 m

F

0.375

1.5(0.5) = 0.75

1 kN

FHI

FBC

RA

FBH

RA

x

10.75

0.5 0.25

FBC= 1.5RA

FBC

x

0.375

+ !MH= 0:

HI G FJ

Influence Line for FBH

Consider the section to the rightFHI

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AB C

E

H

4 m

D

G

3 m 3 m 3 m 3 m

F

RE

x

0.50.75

0.25

1

BH

1 kN

!Fy= 0 ;

FBH cos % =RE

HI

FBH

FCB

RE

%

FBH= 1.25RE

FBH(4/5) =RE1.25(0.25) = 0.3125

FBH x

HI G FJConsider the section to the left

FHI

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AB C

E

H

4 m

D

G

3 m 3 m 3 m 3 m

F

1 kN

!Fy= 0 ;

FBH cos % +RA = 0

RA

x

10.75

0.50.25

FBC

RA

FBH%

0.3125

FBH x

FBH= -1.25RA

FBH(4/5) = -RA

-0.3125

-1.25(0.5) = -0.625

HI G FJ

Influence Line for FCH

FGH

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AB C

E

H

4 m

D

G

3 m 3 m 3 m 3 m

F

FCH x

CH

1 kN

FCH

FCB

RE

x

0.50.75

0.25

1

RE!Fy= 0 ;

FCH= -RE

-0.25

HI G FJ FGH

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AB C

E

4 m

D

3 m 3 m 3 m 3 m

FCH x

1 kN

FCH

FCB

RA!Fy= 0 ;

FCH=RA

-0.25

RA

x

10.75

0.50.25

0.50.25

E l 6 8 3

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113

Draw the influence line forFBC, FBG,andFHG.

Example 6-8-3

AB C

E

G

H

3 m

D

F

3 m 3 m 3 m 3 m

1.5 m

SOLUTION G

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RA

x

1

0.750.5

0.25

RE

x

0.50.75

0.25

1

AB C

E

H

3 m

D

F

3 m 3 m 3 m 3 m

1.5 m

GInfluence Line for FBCConsider the section to the right

FHG

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AB C

E

H

3 m

D

F

3 m 3 m 3 m 3 m

1.5 m

1 kN

FBC= (6/4.5)RE = 1.33RE

RE

x

0.50.75

0.25

1

FBC x

1.33(0.25) = 0.333

+ !MG = 0:

RE

FBC

FBG

FHG

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GInfluence Line for FHGConsider the section to the right

FHG

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AB C

E

H

3 m

D

F

3 m 3 m 3 m 3 m

1.5 m

1 kN

RE

x

0.50.75

0.25

1

FHG x

+ !MB = 0:

RE

FBC

FBG

HG

26.57o

(FHGcos 26.57)(3) = -9REFHG= -3.35RE

-3.35(0.25) = -0.84

GConsider the section to the left

F

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AB C

E

H

3 m

D

F

3 m 3 m 3 m 3 m

1.5 m

1 kN

RA

RA

x

1

0.75 0.50.25

FBC

FBG

FHG

FHG x

-0.84

+ !MB = 0:

(FHGcos 26.57)(3) = -3RAFHG= -1.12RA

-1.12(0.5) = -0.56-1.12(0.25) = -0.28

GInfluence Line for FBGConsider the section to the right

FHG

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AB C

E

H

3 m

D

F

3 m 3 m 3 m 3 m

1.5 m

1 kN

RE

x

0.50.75

0.25

1

FBG x

RE

FBC

FBG26.57o

FHG x

-0.84-0.56 -0.28

!Fy= 0:+

-FBGcos 33.69-FHGsin26.57 +RE = 0

FHG= +1.2RE- 0.54FHG

33.69o

1.2(0.25)-0.54(-0.84) = 0.75

G

F

Consider the section to the left

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FHG x

-0.84-0.56

-0.28

!Fy= 0:+

RA+FBGcos33.69+FHGsin 26.57 = 0

FBG = -1.2RA- 0.54FHG

AB C

E

H

3 m

D

F

3 m 3 m 3 m 3 m

1.5 m

1 kN

RA

FBC

FBG

FHG

RA

x

10.75

0.5 0.25

FBG x

0.75

-1.2(0. 5)-0.54(-0.56) = -0.30 -1.2(0. 25)-0.54(-0.28) = -0.15