Influence line

7/16/2019 Influence line

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1

! Comparison Between Indeterminate and

Determinate

! Influence line for Statically Indeterminate

Beams! Qualitative Influence Lines for Frames

INFLUENCE LINES FOR STATICALLY

INDETERMINATE BEAMS



2

A

B C E D

R A A

B C E D R A

Indeterminate Determinate

Comparison between Indeterminate and Determinate

11



3

A

B C E D

R A A B C E D

R A

A

B C E D

M E A

B C E D M E

A

B C E D

V D A B C E D

V D

1 1

1

Indeterminate Determinate

11

1



4

f 1j

f jj

∆´1 = f 1j

+

1

1 2 3 j4

Redundant R1 applied

1

1

=

f 11

f j1

× R1× R1

f 11

Influence Lines for Reaction

Compatibility equation:

011111 =∆=+ R f f j

)

1

( 1111 f f R j

−=

)(11

1

1 f

f R

j=



5

1

1 2 3 j4

1

=

+

× R2

Redundant R2 applied

f jj f 2j

1

f j2

f 22

Compatibility equation.

0' 22222 =∆=+∆ R f

022222 =∆=+ R f f j

)1

(22

22 f

f R j−=

)(22

2

2 f

f

R

j

=



6

1 2 3 j4

f j4

1

1

Influence Lines for Shear

4

44

)1

( j E f f

V =

f 44

1

1



7

4

44

)1

( j E f M α

=

1 2 3 j4

Influence Lines for Bending Moment

1 1

α 44

f j4

1 1



8

R3 )(33

3

3 f

f R

j=

R2 )(22

2

2 f

f R

j=

R1)(

11

1

1 f

f

R j

=

1 2 3 j4

1

1

• Influence line of Reaction

1

Using Equilibrium Condition for Shear and Bending Moment

111

11

= f

f 11

41

f

f

11

1

f

f j

122

22 =

f

f

22

2

f

f j

22

42

f

f

133

33 = f

f

33

3

f

f j

33

43

f f



9

1 2 3 j4

4V 4

V 4 = R1

R 1V4

M41

• Unit load to the right of 4

• Influence line of Shear

V 4 = R1

V 4 = R1 - 1

1

R2 R3 R1

1

1

1

R 1

1 x

V4

M41

• Unit load to the left of 4

V 4 = R1 - 1

01;0 41 =−−=Σ↑+ V R F y

1

R1

0;0 41 =−=Σ↑+ V R F y



10

1 2 3 j4

• Influence line of Bending moment

M 4 = - l + x + l R1

M 4 - 1 (l - x) - l R1 = 0+ Σ M 4 = 0:

R 1

1 x

V4

M4l 1

• Unit load to the left of 4

M 4 = l R1

R 1V4

M4

• Unit load to right of

l

4

1

1

R2 R3 R1

1

l 1 l 2

l

M 4

4

1

R1

1 1 M 4 = - l + x + l R1

M 4 - l R1 = 0+ Σ M 4 = 0:

M4 = lR1



11

Influence Line of V I

Maximum positive shear Maximum negative shear

Qualitative Influence Lines for Frames

I

1

1



12

Influence Line of M I

Maximum positive moment Maximum negative moment

I

11



13

A D G

15 m15 m

Ay Gy Dy

M A

Dy

1.0

Ay

1.0

Gy

1.0

Influence Line for MOF



14

M H

M A

1

1

A H G

15 m15 m

D



15

R A

1.0

R B

1.0

M B

1

M G

1

A E

G

B C D



16

V G1

V F 1

V H 1

A E

G

B C D



17

Example 1

Draw the influence line for

- the vertical reaction at A and B

- shear at C

- bending moment at A and C

EI is constant . Plot numerical values every 2 m.

A BC D

2 m 2 m 2 m



18

BB f BB f BB f 1= BB f

• Influence line of R B

1

A BC D

2 m 2 m 2 m

AB f CB f DB f

BB f



19

Real Beam A BC D

2 m 2 m 2 m

Conjugate Beam

• Find f xB by conjugate beam

11

6 kN•m

x

=−+ I

x

I I

x 1872

6

3

I

6

I

18

I

18

I

72

3

x

3

2 xV ́x

M ́x

x

I

72

I

18

I

x

2

2

I

x



20

x (m)

0

2

4

6

Point

B

D

C

A

x

f xB / f BB

1

0.518

0.148

0

A BC D

2 m 2 m 2 m

1

f BB

f xB

0

72/ EI

37.33/ EI 10.67/ EI

I

x

I I

x

M f x xB

1872

6'

3

−+==

I 72

I

33.37

I

67.10

0

/72 = 0.518

/72 = 0.148

1

37.33

10.67

0

Influence line of R B

172

72==

BB

BB

f

f

f xB



21

1 kN

Influence line of R A

A BC D

2 m 2 m 2 m

AA

CA

f

f

AA

DA

f

f

1= AA

AA

f f



22

Conjugate Beam

xV ́x

M ́x

I B y

18' =

I

x

2

2

I

x

3

x

3

2 x

A

B

C D

2 m 2 m 2 m

Real Beam

1 kN

• Find f xA by conjugate beam

1 kN

6 kN•m

I M A

72' =

x

I B y

18' =

=− I

x

I

x

6

18 3

I

18

I 6



23

Point

B

D

C

A

f xA / f AA

0

0.482

0.852

1.0

I

x

I

x

M f x xA 6

18

'

3

−==

x (m)

0

2

4

6

x

A BC D

2 m 2 m 2 m

1 kN

f AA

f xA

72/ EI 61.33 / EI 34.67 / EI

f xA

0

I

67.34

I

33.61

I

72

34.67

61.33

72

1 kN


/72 = 1.0

/72=0.852

/72 = 0.4821=

AA

AA

f

f



24

A B

B A

y

R R

R R

F

−=

=−+

=Σ↑+

1

01

;0

A BC D

2 m 2 m 2 m

1 x

R A

M A

R B

0.148.518

1 R B

1 R B

0.482

0.8521.0

1 kN

R A

Alternate Method: Use equilibrium conditions for the influence line of R A

R A = 1- R B



25

V C

1

R B

0.1480.518

1

A BC D

2 m 2 m 2 m R A

M A

R B

0.8520.482

-0.148

1 x

1

Using equilibrium conditions for the influence line of V C

V C = 1 - R B

• Unit load to the left of C

R B

1 x

VC

MC

01

;0

=+−+

=Σ↑+

BC

y

RV

F

R BVC

MC

V C = - R B

0

;0

=++=Σ↑+

C B

y

V R

F




26

B A

B A

A

R x M

R x M

M

66

06)6(1

;0

++−=

=+−−−

=Σ

A BC D

2 m 2 m 2 m R A

M A

R B

1 x

1

M A

1

R B0.148

0.518

1

-1.112 -0.892

Using equilibrium conditions for the influence line of M A

+



27

M C

C

1

R B

0.1480.518

1

A BC D

2 m 2 m 2 m

1 x

R A

M A

R B

0.074

1

Using equilibrium conditions for the influence line of M C

0.592 R B

1 x

VC

MC


R BVC

MC

4 m

M C = 4 R B

+

04

;0

=+−

=Σ

BC

C

R M

M


4 m

M C = -4 + x + 4 R B

04)4(1

;0

=+−−−

=Σ

BC

C

R x M

M +



28

Example 2

Draw the influence line and plot numerical values every 2 m for

- the vertical reaction at supports A, B and C

- Shear at G and E

- Bending moment at G and E

EI is constant.

A B C D E F

2@2=4 m 4@2 = 8 m

G



29


1

A B C D E F

2@2=4 m 4@2 = 8 m

G

1= AA

AA

f

f

AA

DA

f

f

AA

EA

f f

AA

FA

f

f



30

4 / EI

1

A B C D E F

4 m 6 m2 m


Real beam

0.51.5

0

18.67/ EI

64/ EI

4/ EI

4/ EI

Conjugate beam

I

33.5

I

67.10

0

I

67.10

I

M f A AA

64' ==



31

x1 x2

=− I

x

I

x 1

3

1 33.5

12

I I I

x 67.1864

6

3

2 −+=

V´ x1

M´ x1

x1 I

x

2

1

I 33.5

I

x

4

2

1

32 1 x

3

1 x

Conjugate beam

4/ EI

4 m 8 m

64/ EI

I

33.5

I

67.18

M´ x2

V´ x2

x2

I

x2

I

x

2

2

2

3

2 x

3

2 2 x

I

67.18

I

64

xx 3353



32

C to B for I

x

I

x M f x xA ,

33.5

12' 1 −==

I

28

I

64

Bto A for I I

x

I

x

M f x xA ,

6467.18

6'

2

3

2

2+−==

f xA

0

0

I

16−

I

10−

I

14−

x (m)

0

2

4

6

Point

C

F

E

D

B

A

8

12

G 10

f xA / f AA

0

-0.1562

-0.25

-0.2188

0

1

0.4375

x1 x2

1

f AA f xA

A B C D E F G

4 m 6 m2 m

I 10−

I

16− I 14−

I

28 I

64

1


-0.219 -0.25 -0.156

0.438

1= AA

AA

f f



33

A B C D E F G

4 m 6 m2 m

Using equilibrium conditions for the influence line of R B

R B RC R A

x1

R A

1 -0.219 -0.25 -0.156

10.438

R B

1

0.4850.8751.078

10.59

R B

0

0.485

0.875

1.078

1

0.5939

0

x (m)

0

2

4

6

Point

C

F

E

D

B

A

8

12

G 10

R A

0

-0.1562

-0.25

-0.2188

0

0.4375

1

A B

A B

C

R x

R

R x R

M

8

12

8

0128

;0

−=

=−+−

=Σ+



34

A B C D E F G

4 m 6 m2 m

R A

1 -0.219 -0.25 -0.156

10.438

R B RC R A

x1

x (m)

0

2

4

6

Point

C

F

E

D

B

A

8

12

G 10

R A

0

-0.1562

-0.25

-0.2188

0

0.4375

1

RC

1

0.6719

0.375

0.1406

0

-0.0312

0 RC

1

10.672

0.3750.141

-0.0312

Using equilibrium conditions for the influence line of RC

18

5.0

08)8(14

;0

+−=

=−−−

=Σ+

x R R

R x R

M

AC

C A

B



35

A B C D E F G

4 m 6 m2 m

• Check Σ F y = 0

R B RC R A

x1

R A

1 -0.219 -0.25 -0.156

1

0.438

R B

1

0.490.875

1.0810.59

RC 1

10.672

0.3750.141

-0.0312

1

;0

=++

=Σ↑+

C B A

y

R R R

F

RC

1

0.6719

0.375

0.1406

0

-0.0312

0

Point

C

F

E

D

B

A

G

R A

0

-0.1562

-0.25

-0.2188

0

0.4375

1

R B

0

0.485

0.875

1.078

1

0.5939

0

Σ R

1

1

1

1

1

1

1

U i ilib i di i f h i fl li f V



36

V G

R A

1 -0.219 -0.25 -0.156

1

0.438

1

x

V G

= R A

R A

A

VG

MG

• Unit load to the right of G

0.438

-0.219 -0.25-0.156-0.562

1

A B C D E F G

4 m 6 m2 m

Using equilibrium conditions for the influence line of V G

V G = R A - 1

R A

1 x

A

VG

MG

• Unit load to the left of G

01;0 =−−=Σ↑+ G A y V R F

U i ilib i diti f th i fl li f V



37

V E

RC

1

10.672

0.3750.141

-0.0312

1

x

0.625 0.3280.0312

-0.141-0.375

1

A B C D E F G

4 m 6 m2 m

Using equilibrium conditions for the influence line of V E

R C V E

M E

• Unit load to the left of E

V E = - RC

0;0 =+=Σ↑+ E C y V R F

V E = 1 - RC

• Unit load to the right of E

R C

1 x

V E

M E

01;0 =+−=Σ↑+ C E y RV F

Using eq ilibri m conditions for the infl ence line of M



38

M G

R A

1 -0.219 -0.25 -0.156

1

0.438

1

x

-0.438 -0.5-0.312

1

A B C D E F G

4 m 6 m2 m

Using equilibrium conditions for the influence line of M G

0.876

M G = -2 + x + 2 R A

• Unit load to the left of G

R A

1 x

A

VG

MG

2 m

02)2(1

;0

=−−+

=Σ

AG

G

R x M

M +

M G = 2 R A

• Unit load to the right of G

R A

A

VG

MG2 m

02

;0

=−

=Σ

AG

G

R M

M +

Using equilibrium conditions for the influence line of M



39

RC

1

10.672

0.3750.141

-0.0312

M E

1

x

1.5

0.6881

4 m 6 m2 m

A B C D E F G

Using equilibrium conditions for the influence line of M E

-0.125

0.564

R C V E

M E

4 m• Unit load to the left of E

M E = 4 RC

+ ΣM E = 0;

M E = - 4 + x+ 4 RC

• Unit load to the right of E

R C

1 x

V E

M E

4 m

04)4(1

;0

=+−−−

=Σ+

C E

E

R x M

M



40

Example 3

For the beam shown

(a) Draw quantitative influence lines for the reaction at supports A and B, and

bending moment at B.

(b) Determine all the reactions at supports, and also draw its quantitative shear,

bending moment diagrams, and qualitative deflected curve for - Only 10 kN downward at 6 m from A

- Both 10 kN downward at 6 m from A and 20 kN downward at 4 m

from A

10 kN

A

C

B

4 m

2 EI 3 EI

2 m 2 m

20 kN

Influence line of R



411

f CA

f AA

f EA

f DA

1

/ f AA / f AA / f AA

/ f AA

f CA

f AA

f EA

f DA

A

C

B

4 m

2 EI 3 EI

2 m 2 m





42Conjugate Beam

A

C

B

4 m

2 EI 3 EI

2 m 2 m1

Real Beam

1 kN

8 kN•m

x (m)V (kN)1 1+

x (m)M

(kN•m)

8

4+

2 EI

f AA = M ́ A = 60.44/ EI

60.44/ EI

12/ EI

1.33 EI I 67.2

• Quantitative influence line of R A



43

I

67.22 EI

60.44/ EI

Conjugate Beam12/ EI

1.33 EI

37.11/ EI 17.77/ EI

4.88/ EI

60.44/ EI

0

R A = f xA /f AA

0.6140.294

0.081

1

0

f xA

A C B

Using equilibrium conditions for the influence line of R B and MB



44

A B

4 m 2 m 2 m

g q B B

R B = 1 - R A

10.386

0.7060.919

0

M B = 8 R A - (8- x)(1)0

-1.352

-1.088 -1.648

0

R A R B

M B

R A

0.6140.294

0.081

1

0

1 x

Using equilibrium conditions for the influence line of VB



45

A B

4 m 2 m 2 m

g q B

R A

0.6140.294

0.081

1

0

-1-0.386

-0.706-0.919

0 V B = R A -1

R A

1 x

V B = R A - 1

Using equilibrium conditions for the influence line of V C and M C



46

A B

4 m 2 m 2 m

C

g q C C

R A

R A

0.614 0.294 0.081

1

0

1 x

R B

M B

M C = 4 R A M C = 4 R A - (4- x)(1)

M C

V C = R AV C = R A - 1

V C 1

-0.386-0.706

0.3240.4561.176

0.2940.081 0

The quantitative shear and bending moment diagram and qualitative deflected curve



47

A B

4 m 2 m 2 m

10 kN

R A

0.6140.294

0.081

1

0

10(0.081)=0.81 kN

M A (kN•m)+

-

-13.53

4.86

V (kN)

-9.19

0.81 0.81

-

R B=9.19 kN

M B= 13.53 kN•m

The quantitative shear and bending moment diagram and qualitative deflected curve



48

A B

4 m 2 m 2 m

10 kN20 kN

20(.294) +1(0.081)

= 6.69 kN

V (kN)

-23.31

6.696.69

-

-13.31 -23.31

M A (kN•m)+

-

-46.48

26.760.14

R A

0.6140.294 0.081

1

0

R B=23.31 kN

M B=46.48 kN•m

APPENDIX



49

Example 1

Draw the influence line for

- the vertical reaction at B

A BC D

2 m 2 m 2 m

• Muller-Breslau for the influence line of reaction, shear and moment

• Influence lines for MDOF beams




50

1 kN

A BC D

2 m 2 m 2 m

1= AA

AA

f

f AA

CA

f

f

AA

DA

f

f




51

Conjugate Beam

xV ́x

M ́x

I B y

18' =

I

x

2

2

I

x

3

x

3

2 x

A BC D

2 m 2 m 2 m

Real Beam

1 kN

1 kN

6 kN•m

I M A

72' =

x

I B y

18' =

6 /EI

I

18

=− I

x

I

x

6

18 3

x



52

Point

B

D

C

A

f xA / f AA

0

0.482

0.852

1.0

x (m)

0

2

4

6

A BC D

2 m 2 m 2 m

1 kN

f AA f

xA

72/ EI 61.33 / EI 34.67 / EI

I

x

I

x M f x xA

6

18'

3

−==

f xA

0

I

67.34

I

33.61

I

72

34.67

61.33

72

1 kN


/72 = 1.0

/72=0.852

/72 = 0.4821=

AA

AA

f

f

• Influence line of R B



53

1

A BC D

2 m 2 m 2 m

1= BB

BB

f

f

BB

AB

f

f

BB

CB

f

f BB

DB

f

f

• Find f xB by conjugate beam



54

Real Beam

A BC D

2 m 2 m 2 m

Conjugate Beam

11

6 kN•m

x

=−+ I x

I I x 18726

3

I

6 I

18

I 18

I

72

3

x

3

2 xV ́x

M ́x

x

I 72

I

18

I

x

2

2

I

x

x



55

x (m)

0

2

4

6

Point

B

D

C

A

f xB / f BB

1

0.518

0.148

0

A BC D

2 m 2 m 2 m

1

f BB

f xB

0

72/ EI

37.33/ EI 10.67/ EI

/72 = 0.518

/72 = 0.148

/72 = 1

1

37.33

10.670

f BB

Influence line of R B

f BB

= 1

72

I

x

I I

x M f x xB

1872

6'

3

−+==

f xB

0

I 72

I

33.37

I

67.10

Example 2



56

Example 2

For the beam shown

(a) Draw the influence line for the shear at D for the beam

(b) Draw the influence line for the bending moment at D for the beam

EI is constant.Plot numerical values every 2 m.

A B C D E

2 m 2 m 2 m 2 m

The influence line for the shear at D



57

A B C D E

D

2 m 2 m 2 m 2 m

1 kN

1 kN

DV D

1 kN

1 kN DD

ED

f

f

1= DD

DD

f

f

• Using conjugate beam for find f xD



58

A B C D E

2 m 2 m 2 m 2 m

2 kN•m

1 kN

1 kN2 k

1 kN

1 kN

2 kN•m

1 kN2 kN1 kN

1 kN

1 kN



59

A B C D E

2 m 2 m 2 m 2 m

1 kN1 kN

Real beam

V ( kN)

x (m)

1

-1

M

(kN •m) x (m)

4

Conjugate beam

4/ EI

M ́D

1 kN

1 kN

2kN

4/EI

• Determine M ́D at D



60

2 m 2 m 2 m 2 m

M ́D

Conjugate beam

4/ EI

A B

C D E

4/EI

0

8/ EI

m3

8

I 3

16

I 3

8

128/3 EI

4/EI

8/ EI

m

3

8

I 3

16

I 3

40

4/ EI



61

=−=− )2)(3

8()

3

2)(

2(

4

I I I

I I I I 3

52)2)(

3

40(

3

128)

3

2)(

2( =−+=

I I I 3

76)2)(

3

40()

3

2)(

2( −=−=

128/3 EI

2 m 2 m 2 m 2 m

Conjugate beam A

BC

D E

I 3

8 I 3

40

V ́DL

M ́DL

2/ EI

2/ EI

3

2

I 3

40

I 3

40V ́DR

M ́DR

2/ EI

3

2

128/3 EI

2/ EI 2/ EI

V ́E

M ́E

3

2

I 3

8

Conjugate beam

4/ EI

A



62

128/3 EI = M ́ D = f DD

V ´ x (m) θ

f xD = M ´ x (m) ∆

I 3

76−

I 3

40−

I 3

34−

I 3

2

I 3

16−

I 3

8

I 3

52

I

4

−

Influence line of V D = f xD/ f DD

76

52

4

0.406 =128

/128 = -0.594 /(128/3) = -0.094

Conjugate beam

I 3

8128/3 EI

2 m 2 m 2 m 2 m

A

B

C

D E

I 3

40

The influence line for the bending moment at D

αDD



63

A B C D E

2 m 2 m 2 m 2 m

1 kN •m1 kN •m

M D

DD

ED f

α

DD

DD f

α

• Using conjugate beam for find f xD



64

1 kN•m

0.5 kN

1 kN•m

0.5 kN1 kN

0.5 kN

2 m 2 m 2 m 2 m

0.5 kN

0.5 kN

0.5 kN1 k

A B C D E

1 kN •m1 kN •m



65

2 m 2 m 2 m 2 m

Real beam A B C D E

0.5 kN1 kN0.5 kN

1 kN •m1 kN •m

V (kN)

x (m)

0.5

1

M

(kN •m) x (m)

2

2/ EI

Conjugate beam

2/ EI



66

2/EI

4/ EI

m

3

8

I 3

8

2/EI

0

4/ EI

m3

8

2 m 2 m 2 m 2 m

Conjugate beam

I 3

4

I 3

8

I 3

32

I 3

4

2/ EI



67

I I I 3

26

)2)(

4

()3

2

)(

1

(=+=

=−=− )2)(3

4()

3

2)(

1(

2

I I I

2 m 2 m 2 m 2 m

Conjugate beam

I 3

32 I 3

4

I

4

M ́D

V ́D

1/ EI 1/ EI

3

2

I

4

1/ EI

1/ EI

V ́E

M ́E

3

2 I 3

4

2/ EI

Conjugate beam



68/(32/3) = -0.188-2

Influence line of M D

813.032

26=

xD

xD f

α

αDD = 32/3 EI

2 m 2 m 2 m 2 m

j g

4

3 EI 4

EI 32

3 EI

x (m)V ´ θ

I

4

I 3

8−

I 3

1

I 3

4

I 3

17−

I 5

f xD = M ´

x (m) ∆

-2/ EI

θ D = 0.469 + 0.531 = 1 rad

θ DL = 5/(32/3) = 0.469 rad.

θ DR = -17/32 = -0.531 rad.

26/3 EI

Example 3



69

Draw the influence line for the reactions at supports for the beam shown in thefigure below. EI is constant.

A D B C

5 m 5 m 5 m 5 m5 m 5 m

G E F

Influence line for R D

A D B C G E F



70

5 m 5 m 5 m 5 m5 m 5 m

1

f BD

f CD f DD f ED f FD

f XD

1

f XD /f DD = Influence

line for R D

DD

BD

f

f 1= DD

DD

f

f

DD

CD

f

f

DD

ED

f

f

DD

FD

f

f

• Use the consistency deformation method

A G



71

Conjugate beam

15 + (2/3)(15)

I

5.112

I

15

1

+x RG

- Use conjugate beam for find ∆´G and f GG

∆´G + f GG RG = 0 ------(1)

1

Real beam

15 m 15 m

A G

1

15

1

Real beam

30 m

A G

1

30

1

A

3@5 =15 m 3@5 =15 m

f GG

∆´G

1

=

112.5/EI

I M C C

5.2812'' ==∆

Conjugate beam

20 m

I

15 I

450

I M f GGG

9000''' ==

I 450

090005.2812

=+ G R

Substitute ∆´G and f GG in (1) :



72

x RG = -0.3125 kN

0G

I I ↓−= ,3125.0 kN RG

1

A

G5.625

0.6875 0.3125

11

15

=

1

+

1

30

f BD

f CD f DD f ED f FD

A G

Real beam

5.625

• Use the conjugate beam for find f XD



73

8.182 m

6.818 m

I

16.35

I

98.15

I

01.23

)3

(2

3125.0 2

2

2 x x−

2

2

'13.28

x M x I

=+

)2

(6875.0625.5 1

2

11 x

I

x x −=

)

3

2(

2

6875. 1

2

1 x

I

x+

13@5 =15 m 3@5 =15 m0.6875 0.3125

28.13 EI x1

x2

x1 = 5 m -----> f BD = M ́1= 56/ EI

x1 = 10 m -----> f CD = M ́1= 166.7/ EI

x1 = 15 m -----> f DD = M ́1= 246.1/ EI

x2 = 5 m -----> f FD = M ́2= 134.1/ EI

x2 = 10 m -----> f ED = M ́2= 229.1/ EI

x2 = 15 m -----> f DD = M ́2= 246.1/ EI

A G Conjugate beam I

625.5

I

688.4−

A

x1

(5.625-0.6875 x1)/ EI

V´1

M´1

I

625.5 I

x x2

11 6875.0625.5 − I

x

2

6875.02

1

G

x2

0.3125 x2

V´2

M´2

I

13.28

I

x

2

3125.02

2

• Influence Line for R D

I

56 I

7.166 I

1.246

I

2.229

I

1.134



74

Influence Line for R D

0.228 0.677 1.0 0.931 0.545

1

f XD

I I

1

f XD /f DD

1.246

561.246

7.166

1.246

1.246

1.246

2.229

1.246

1.134

Influence line for RG

A D B C G E F



75

5 m 5 m 5 m 5 m5 m 5 m

f XG

1 f BG f CG

f GG f

EG

f FG

f XG /f GG

1GG

BG

f

f

GG

CG

f

f

GG

EG

f

f

GG

FG

f

f 1=

GG

GG

f

f

• Use consistency deformations

f XG



76

Conjugate beam

20 m

I 450

I

30

1

=

∆ ́D + f DD R D = 0 ------(2)

- Use conjugate beam for find ∆ ́D and f DD

1

+

1

Real beam

30 m

A G

1

30

1

Real beam

15 m 15 m

A G

1

15

X R D

f XG

13@5 =15 m 3@5 =15 m

∆ ́D

f DD

Conjugate beam

15 + (2/3)(15)

I

15 I

5.112

450/EI

I

9000

M´

I

5.112

I

5.1 I

9000

I

15 I

5.112



77

I I I I M D

5.2812)15(

4509000)

3

15(

5.112'' =−+==∆

I I M f DD

1125)15

3

2(

5.112'' =×==

↓=−==+ ,5.25.2,011255.2812

kN kN R R I I

D DSubstitute ∆ ́D and f DD in (2) :

x R D = -2.5 kN

=

1

11

30

+

11

15

1.5

7.5

2.5

15 mV´ 450/EI

M´´

V´´

I

• Use the conjugate beam for find f XG

Real beam

f GG f EG

f FG7.5



78

I I f BG

5.62)5

3

2(

75.18−=×−==

1 f BG f CG

3@5 =15 m 3@5 =15 m1.5 2.5

f GG = M´G = 1968.56/EI

168.75/EI

I I f CG

06.125)67.6(

75.18−=−==

A G Conjugate beam

I

5.7−

I

15

10 m

15 + (10/3) = 18.33 m

25 + (2/3)(5) = 28.33 m

I

5.112

I

75

I 75.18

A5 m

V´1

M´1

I 5.7−

I

75.18

A

6.67 m

V´2

M´2

I 75.18

I

5.7−

I

75.18−

3

2

3 75.16856.1968 x x

f GG = M´G = 1968.56/EI

x

2

2 x



79

=−+3

33

)3(2 x I I

x = 5 m -----> f FG = M´= 1145.64/ EI

x = 10 m -----> f EG = M´ = 447.73/ EI


-0.064-0.032

0.227 0.5821.0

M´ G x

168.75/EIV´2

1

f XG

I

5.62−

I

125−

I

73.447 I

64.1145 I

56.1968

56.19685.62

−

56.1968125

−

56.1968

73.44756.1968

64.114556.1968

56.1968

1

f XG /f GG

Using equilibrium condition for the influence line for A y

A D B C G E F

1 x



80

5 m 5 m 5 m 5 m5 m 5 mMA

Ay R D R G

Unit load

1 1

Influence Line for R D

0.228 0.678 1.0 0.929 0.542


-0.064-0.032

0.2270.582

1.0

0.386 Influence line for A y

0.804

-0.156 -0.124

1.0

C D A y R R R F −−==Σ↑+ 1:0

Using equilibrium condition for the influence line for M A

A D B C G E F

1 x



81

5 m 5 m 5 m 5 m5 m 5 mMA

Ay R D R G

x 15

x 30

R D

0.228 0.678 1.0 0.929 0.542

1x

5 10 15 20 25 30

RG

-0.064-0.032

0.2270.582

1.0

Influence line for M A

2.541.75

-0.745 -0.59

C D A R R x M 30151:0 −−=Σ+

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Influence line

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