1) The document discusses influence lines for statically indeterminate beams and frames. It provides examples of calculating the influence lines for reactions, shear, and bending moment at various points on indeterminate beams. 2) The examples show how to use the method of conjugate beams to determine the influence lines by considering equilibrium in the conjugate beam system. Numerical values for influence lines are plotted at regular intervals along the beam. 3) Qualitative influence lines for typical frames are also shown, indicating the maximum and minimum values for shear and bending moment.

1. 1
! Comparison Between Indeterminate and
Determinate
! Influence line for Statically Indeterminate
Beams
! Qualitative Influence Lines for Frames
INFLUENCE LINES FOR STATICALLY
INDETERMINATE BEAMS

2. 2
A
B C
E
D
RA
A
B C
E
D
RA
Indeterminate Determinate
Comparison between Indeterminate and Determinate
1
1

3. 3
A
B C
E
D
RA
A
B C
E
D
RA
A
B C
E
D
ME
A
B C
E
D
ME
A
B C
E
D
VD
A
B C
E
D
VD
1 1
1
Indeterminate Determinate
1
1
1

4. 4
f1j
fjj
∆´1 = f1j
+
1
1 2 3
j
4
Redundant R1 applied
1
1
=
f11
fj1
×R1
×R1
f11
Influence Lines for Reaction
Compatibility equation:
0
1
1
11
1 =
∆
=
+ R
f
f j
)
1
(
11
1
1
f
f
R j
−
=
)
(
11
1
1
f
f
R j
=

5. 5
1
1 2 3
j
4
1
=
+
×R2
Redundant R2 applied
fjj
f2j
1
fj2
f22
Compatibility equation.
0
' 2
2
22
2 =
∆
=
+
∆ R
f
0
2
2
22
2 =
∆
=
+ R
f
f j
)
1
(
22
2
2
f
f
R j
−
=
)
(
22
2
2
f
f
R
j
=

6. 6
1 2 3
j
4
fj4
1
1
Influence Lines for Shear
4
44
)
1
( j
E f
f
V =
f44
1
1

7. 7
4
44
)
1
( j
E f
M
α
=
1 2 3
j
4
Influence Lines for Bending Moment
1 1
α44
fj4
1 1

8. 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
j
4
1
1
• Influence line of Reaction
1
Using Equilibrium Condition for Shear and Bending Moment
1
11
11
=
f
f
11
41
f
f
11
1
f
f j
1
22
22
=
f
f
22
2
f
f j
22
42
f
f
1
33
33
=
f
f
33
3
f
f j
33
43
f
f

9. 9
1 2 3
j
4
4
V4
V4 = R1
R1
V4
M4
1
• Unit load to the right of 4
• Influence line of Shear
V4 = R1
V4 = R1 - 1
1
R2 R3
R1
1
1
1
R1
1
x
V4
M4
1
• Unit load to the left of 4
V4 = R1 - 1
0
1
;
0 4
1 =
−
−
=
Σ
↑
+ V
R
Fy
1
R1
0
;
0 4
1 =
−
=
Σ
↑
+ V
R
Fy

10. 10
1 2 3
j
4
• Influence line of Bending moment
M4 = - l + x + l R1
M4 - 1 (l-x) - l R1 = 0
+ Σ M4 = 0:
R1
1
x
V4
M4
l
1
• Unit load to the left of 4
M4 = l R1
R1
V4
M4
• Unit load to right of
l
4
1
1
R2 R3
R1
1
l1 l2
l
M4
4
1
R1
1 1
M4 = - l + x + l R1
M4 - l R1 = 0
+ Σ M4 = 0:
M4 = lR1

11. 11
Influence Line of VI
Maximum positive shear Maximum negative shear
Qualitative Influence Lines for Frames
I
1
1

12. 12
Influence Line of MI
Maximum positive moment Maximum negative moment
I
1
1

13. 13
A D G
15 m
15 m
Ay Gy
Dy
MA
Dy
1.0
Ay
1.0
Gy
1.0
Influence Line for MOF

14. 14
MH
MA
1
1
A H G
15 m
15 m
D

15. 15
RA
1.0
RB
1.0
MB
1
MG
1
A E
G
B C D

16. 16
VG
1
VF
1
VH
1
A E
G
B C D

17. 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 B
C D
2 m 2 m 2 m

18. 18
BB
f BB
f
BB
f 1
=
BB
f
• Influence line of RB
1
A B
C D
2 m 2 m 2 m
AB
f CB
f
DB
f
BB
f

19. 19
Real Beam
A B
C D
2 m 2 m 2 m
Conjugate Beam
• Find fxB by conjugate beam
1
1
6 kN•m
x
=
−
+
EI
x
EI
EI
x 18
72
6
3
EI
6
EI
18
EI
18
EI
72
3
x
3
2x
V´x
M´x
x
EI
72
EI
18
EI
x
2
2
EI
x

20. 20
x (m)
0
2
4
6
Point
B
D
C
A
x
fxB / fBB
1
0.518
0.148
0
A B
C D
2 m 2 m 2 m
1
fBB
fxB
0
72/EI
37.33/EI
10.67/EI
EI
x
EI
EI
x
M
f x
xB
18
72
6
'
3
−
+
=
=
EI
72
EI
33
.
37
EI
67
.
10
0
/72 = 0.518
/72 = 0.148
1
37.33
10.67
0
Influence line of RB
1
72
72
=
=
BB
BB
f
f
fxB

21. 21
1 kN
Influence line of RA
A B
C D
2 m 2 m 2 m
AA
CA
f
f
AA
DA
f
f
1
=
AA
AA
f
f

22. 22
Conjugate Beam
x
V´x
M´x
EI
B y
18
' =
EI
x
2
2
EI
x
3
x
3
2x
A B
C D
2 m 2 m 2 m
Real Beam
1 kN
• Find fxA by conjugate beam
1 kN
6 kN•m
EI
M A
72
' =
x
EI
B y
18
' =
=
−
EI
x
EI
x
6
18 3
EI
18
EI
6

23. 23
Point
B
D
C
A
fxA / fAA
0
0.482
0.852
1.0
EI
x
EI
x
M
f x
xA
6
18
'
3
−
=
=
x (m)
0
2
4
6
x
A B
C D
2 m 2 m 2 m
1 kN
fAA
fxA
72/EI 61.33 /EI
34.67 /EI
fxA
0
EI
67
.
34
EI
33
.
61
EI
72
34.67
61.33
72
1 kN
Influence line of RA
/72 = 1.0
/72=0.852
/72 = 0.482
1
=
AA
AA
f
f

24. 24
A
B
B
A
y
R
R
R
R
F
−
=
=
−
+
=
Σ
↑
+
1
0
1
;
0
A B
C D
2 m 2 m 2 m
1
x
RA
MA
RB
0.148
.518
1
RB
1
RB
0.482
0.852
1.0
1 kN
RA
Alternate Method: Use equilibrium conditions for the influence line of RA
RA = 1- RB

25. 25
VC
1
RB
0.148
0.518
1
A B
C D
2 m 2 m 2 m
RA
MA
RB
0.852
0.482
-0.148
1 x
1
Using equilibrium conditions for the influence line of VC
VC = 1 - RB
• Unit load to the left of C
RB
1 x
VC
MC
0
1
;
0
=
+
−
+
=
Σ
↑
+
B
C
y
R
V
F
RB
VC
MC
VC = - RB
0
;
0
=
+
+
=
Σ
↑
+
C
B
y
V
R
F
• Unit load to the left of C

26. 26
B
A
B
A
A
R
x
M
R
x
M
M
6
6
0
6
)
6
(
1
;
0
+
+
−
=
=
+
−
−
−
=
Σ
A B
C D
2 m 2 m 2 m
RA
MA
RB
1 x
1
MA
1
RB
0.148
0.518
1
-1.112 -0.892
Using equilibrium conditions for the influence line of MA
+

27. 27
MC
C
1
RB
0.148
0.518
1
A B
C D
2 m 2 m 2 m
1 x
RA
MA
RB
0.074
1
Using equilibrium conditions for the influence line of MC
0.592 RB
1 x
VC
MC
• Unit load to the left of C
RB
VC
MC
4 m
MC = 4RB
+
0
4
;
0
=
+
−
=
Σ
B
C
C
R
M
M
• Unit load to the left of C
4 m
MC = -4 + x + 4RB
0
4
)
4
(
1
;
0
=
+
−
−
−
=
Σ
B
C
C
R
x
M
M
+

28. 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. 29
Influence line of RA
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. 30
4 /EI
1
A B C
D E F
4 m 6 m
2 m
• Find fxA by conjugate beam
Real beam
0.5
1.5
0
18.67/EI
64/EI
4/EI
4/EI
Conjugate beam
EI
33
.
5
EI
67
.
10
0
EI
67
.
10 EI
M
f A
AA
64
' =
=

31. 31
x1
x2
=
−
EI
x
EI
x 1
3
1 33
.
5
12
EI
EI
EI
x 67
.
18
64
6
3
2
−
+
=
V´ x1
M´ x1
x1
EI
x
2
1
EI
33
.
5
EI
x
4
2
1
3
2 1
x
3
1
x
Conjugate beam
4/EI
4 m 8 m
64/EI
EI
33
.
5
EI
67
.
18
M´ x2
V´ x2
x2
EI
x2
EI
x
2
2
2
3
2
x
3
2 2
x
EI
67
.
18
EI
64

32. 32
C
to
B
for
EI
x
EI
x
M
f x
xA ,
33
.
5
12
'
3
1 −
=
=
EI
28
EI
64
B
to
A
for
EI
EI
x
EI
x
M
f x
xA ,
64
67
.
18
6
' 2
3
2
2 +
−
=
=
fxA
0
0
EI
16
−
EI
10
−
EI
14
−
x (m)
0
2
4
6
Point
C
F
E
D
B
A
8
12
G 10
fxA / fAA
0
-0.1562
-0.25
-0.2188
0
1
0.4375
x1
x2
1
fAA
fxA
A B C
D E F
G
4 m 6 m
2 m
EI
10
−
EI
16
−
EI
14
−
EI
28
EI
64
1
Influence line of RA
-0.219 -0.25 -0.156
0.438
1
=
AA
AA
f
f

33. 33
A B C
D E F
G
4 m 6 m
2 m
Using equilibrium conditions for the influence line of RB
RB RC
RA
x
1
RA
1 -0.219 -0.25 -0.156
1
0.438
RB
1
0.485
0.875
1.078
1
0.59
RB
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
RA
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
0
12
8
;
0
−
=
=
−
+
−
=
Σ
+

34. 34
A B C
D E F
G
4 m 6 m
2 m
RA
1 -0.219 -0.25 -0.156
1
0.438
RB RC
RA
x
1
x (m)
0
2
4
6
Point
C
F
E
D
B
A
8
12
G 10
RA
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
1
0.672
0.375
0.141
-0.0312
Using equilibrium conditions for the influence line of RC
1
8
5
.
0
0
8
)
8
(
1
4
;
0
+
−
=
=
−
−
−
=
Σ
+
x
R
R
R
x
R
M
A
C
C
A
B

35. 35
A B C
D E F
G
4 m 6 m
2 m
• Check ΣFy = 0
RB RC
RA
x
1
RA
1 -0.219 -0.25 -0.156
1
0.438
RB
1
0.49
0.875
1.08
1
0.59
RC
1
1
0.672
0.375
0.141
-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
RA
0
-0.1562
-0.25
-0.2188
0
0.4375
1
RB
0
0.485
0.875
1.078
1
0.5939
0
ΣR
1
1
1
1
1
1
1

36. 36
VG
RA
1 -0.219 -0.25 -0.156
1
0.438
1
x
VG = RA
RA
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 m
2 m
Using equilibrium conditions for the influence line of VG
VG = RA - 1
RA
1
x
A
VG
MG
• Unit load to the left of G
0
1
;
0 =
−
−
=
Σ
↑
+ G
A
y V
R
F

37. 37
VE
RC
1
1
0.672
0.375
0.141
-0.0312
1
x
0.625
0.328
0.0312
-0.141
-0.375
1
A B C
D E F
G
4 m 6 m
2 m
Using equilibrium conditions for the influence line of VE
RC
VE
ME
• Unit load to the left of E
VE = - RC
0
;
0 =
+
=
Σ
↑
+ E
C
y V
R
F
VE = 1 - RC
• Unit load to the right of E
RC
1 x
VE
ME
0
1
;
0 =
+
−
=
Σ
↑
+ C
E
y R
V
F

38. 38
MG
RA
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 m
2 m
Using equilibrium conditions for the influence line of MG
0.876
MG = -2 + x + 2RA
• Unit load to the left of G
RA
1
x
A
VG
MG
2 m
0
2
)
2
(
1
;
0
=
−
−
+
=
Σ
A
G
G
R
x
M
M
+
MG = 2RA
• Unit load to the right of G
RA
A
VG
MG
2 m
0
2
;
0
=
−
=
Σ
A
G
G
R
M
M
+

39. 39
RC
1
1
0.672
0.375
0.141
-0.0312
ME
1
x
1.5
0.688
1
4 m 6 m
2 m
A B C
D E F
G
Using equilibrium conditions for the influence line of ME
-0.125
0.564
RC
VE
ME
4 m
• Unit load to the left of E
ME = 4RC
+ ΣME = 0;
ME = - 4 + x+ 4RC
• Unit load to the right of E
RC
1 x
VE
ME
4 m
0
4
)
4
(
1
;
0
=
+
−
−
−
=
Σ
+
C
E
E
R
x
M
M

40. 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
2EI 3EI
2 m 2 m
20 kN

41. 41
1
fCA
fAA
fEA
fDA
1
/fAA /fAA /fAA
/fAA
fCA
fAA
fEA
fDA
A
C
B
4 m
2EI 3EI
2 m 2 m
Influence line of RA

42. 42
Conjugate Beam
A
C
B
4 m
2EI 3EI
2 m 2 m
1
Real Beam
• Find fxA by conjugate beam
1 kN
8 kN•m
x (m)
V (kN)
1 1
+
x (m)
M
(kN•m)
8
4
+
2EI
fAA = M´A = 60.44/EI
60.44/EI
12/EI
1.33EI
EI
67
.
2

43. 43
EI
67
.
2
2EI
60.44/EI
Conjugate Beam
12/EI
1.33EI
37.11/EI
17.77/EI 4.88/EI
60.44/EI
0
RA = fxA/fAA
0.614
0.294
0.081
1
0
fxA
A C B
• Quantitative influence line of RA

44. 44
A B
4 m 2 m 2 m
Using equilibrium conditions for the influence line of RB and MB
RB = 1 - RA
1
0.386
0.706
0.919
0
MB = 8RA - (8-x)(1)
0
-1.352
-1.088 -1.648
0
RA RB
MB
RA
0.614
0.294 0.081
1
0
1
x

45. 45
A B
4 m 2 m 2 m
Using equilibrium conditions for the influence line of VB
RA
0.614
0.294 0.081
1
0
-1
-0.386
-0.706
-0.919
0
VB = RA -1
RA
1
x
VB = RA - 1

46. 46
A B
4 m 2 m 2 m
C
Using equilibrium conditions for the influence line of VC and MC
RA
RA
0.614
0.294
0.081
1
0
1
x
RB
MB
MC = 4RA
MC = 4RA - (4-x)(1)
MC
VC = RA
VC = RA - 1
VC
1
-0.386 -0.706
0.324
0.456
1.176
0.294
0.081 0

47. 47
A B
4 m 2 m 2 m
10 kN
RA
0.614
0.294 0.081
1
0
10(0.081)=0.81 kN
MA (kN•m)
+
-
-13.53
4.86
V (kN)
-9.19
0.81 0.81
-
The quantitative shear and bending moment diagram and qualitative deflected curve
RB=9.19 kN
MB= 13.53 kN•m

48. 48
A B
4 m 2 m 2 m
10 kN
The quantitative shear and bending moment diagram and qualitative deflected curve
20 kN
20(.294) +1(0.081)
= 6.69 kN
V (kN)
-23.31
6.69 6.69
-
-13.31 -23.31
MA (kN•m)
+
-
-46.48
26.76
0.14
RA
0.614
0.294 0.081
1
0
RB=23.31 kN
MB=46.48 kN•m

49. 49
Example 1
Draw the influence line for
- the vertical reaction at B
A B
C D
2 m 2 m 2 m
APPENDIX
•Muller-Breslau for the influence line of reaction, shear and moment
•Influence lines for MDOF beams

50. 50
1 kN
Influence line of RA
A B
C D
2 m 2 m 2 m
1
=
AA
AA
f
f
AA
CA
f
f
AA
DA
f
f

51. 51
Conjugate Beam
x
V´x
M´x
EI
B y
18
' =
EI
x
2
2
EI
x
3
x
3
2x
A B
C D
2 m 2 m 2 m
Real Beam
1 kN
• Find fxA by conjugate beam
1 kN
6 kN•m
EI
M A
72
' =
x
EI
B y
18
' =
6 /EI
EI
18
=
−
EI
x
EI
x
6
18 3

52. 52
Point
B
D
C
A
fxA / fAA
0
0.482
0.852
1.0
x (m)
0
2
4
6
x
A B
C D
2 m 2 m 2 m
1 kN
fAA
fxA
72/EI 61.33 /EI
34.67 /EI
EI
x
EI
x
M
f x
xA
6
18
'
3
−
=
=
fxA
0
EI
67
.
34
EI
33
.
61
EI
72
34.67
61.33
72
1 kN
Influence line of RA
/72 = 1.0
/72=0.852
/72 = 0.482
1
=
AA
AA
f
f

53. 53
• Influence line of RB
1
A B
C D
2 m 2 m 2 m
1
=
BB
BB
f
f
BB
AB
f
f
BB
CB
f
f
BB
DB
f
f

54. 54
Real Beam
A B
C D
2 m 2 m 2 m
Conjugate Beam
• Find fxB by conjugate beam
1
1
6 kN•m
x
=
−
+
EI
x
EI
EI
x 18
72
6
3
EI
6
EI
18
EI
18
EI
72
3
x
3
2x
V´x
M´x
x
EI
72
EI
18
EI
x
2
2
EI
x

55. 55
x (m)
0
2
4
6
Point
B
D
C
A
x
fxB / fBB
1
0.518
0.148
0
A B
C D
2 m 2 m 2 m
1
fBB
fxB
0
72/EI
37.33/EI
10.67/EI
/72 = 0.518
/72 = 0.148
/72 = 1
1
37.33
10.67
0
fBB
Influence line of RB
fBB
= 1
72
EI
x
EI
EI
x
M
f x
xB
18
72
6
'
3
−
+
=
=
fxB
0
EI
72
EI
33
.
37
EI
67
.
10

56. 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

57. 57
A B C
D E
D
2 m 2 m 2 m 2 m
The influence line for the shear at D
1 kN
1 kN
D
VD
1 kN
1 kN
DD
ED
f
f
1
=
DD
DD
f
f

58. 58
A B C
D E
2 m 2 m 2 m 2 m
2 kN•m
1 kN
1 kN
2 k
1 kN
1 kN
2 kN•m
1 kN
2 kN
1 kN
• Using conjugate beam for find fxD
1 kN
1 kN

59. 59
A B C
D E
2 m 2 m 2 m 2 m
1 kN
1 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

60. 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
m
3
8
• Determine M´D at D
EI
3
16
EI
3
8
128/3EI
4/EI
8/EI
m
3
8
EI
3
16
EI
3
40

61. 61
=
−
=
− )
2
)(
3
8
(
)
3
2
)(
2
(
4
EI
EI
EI
EI
EI
EI
EI 3
52
)
2
)(
3
40
(
3
128
)
3
2
)(
2
( =
−
+
=
EI
EI
EI 3
76
)
2
)(
3
40
(
)
3
2
)(
2
( −
=
−
=
128/3EI
2 m 2 m 2 m 2 m
Conjugate beam
4/EI
A
B
C
D E
EI
3
8
EI
3
40
V´DL
M´DL
2/EI
2/EI
3
2
EI
3
40
EI
3
40
V´DR
M´DR
2/EI
3
2
128/3EI
2/EI
2/EI
V´E
M´E
3
2
EI
3
8

62. 62
128/3EI = M´D = fDD
V ´
x (m) θ
fxD = M ´
x (m) ∆
EI
3
76
−
EI
3
40
−
EI
3
34
−
EI
3
2
EI
3
16
−
EI
3
8
EI
3
52
EI
4
−
Influence line of VD = fxD/fDD
76
52
4
0.406 =
128
/128 = -0.594
/(128/3) = -0.094
Conjugate beam
EI
3
8
128/3EI
2 m 2 m 2 m 2 m
4/EI
A
B
C
D E
EI
3
40

63. 63
A B C
D E
2 m 2 m 2 m 2 m
The influence line for the bending moment at D
1 kN •m
1 kN •m
αDD
MD
DD
ED
f
α
DD
DD
f
α

64. 64
1 kN•m
0.5 kN
1 kN•m
0.5 kN
1 kN
0.5 kN
2 m 2 m 2 m 2 m
0.5 kN
0.5 kN
0.5 kN
1 k
A B C
D E
1 kN •m
1 kN •m
• Using conjugate beam for find fxD

65. 65
2 m 2 m 2 m 2 m
Real beam
A B C
D E
0.5 kN
1 kN
0.5 kN
1 kN •m
1 kN •m
V (kN)
x (m)
0.5
1
M
(kN •m)
x (m)
2
2/EI
Conjugate beam

66. 66
2/EI
4/EI
m
3
8
EI
3
8
2/EI
0
4/EI
m
3
8
2 m 2 m 2 m 2 m
2/EI
Conjugate beam
EI
3
4
EI
3
8
EI
3
32
EI
3
4

67. 67
EI
EI
EI 3
26
)
2
)(
4
(
)
3
2
)(
1
( =
+
=
=
−
=
− )
2
)(
3
4
(
)
3
2
)(
1
(
2
EI
EI
EI
2 m 2 m 2 m 2 m
2/EI
Conjugate beam
EI
3
32
EI
3
4
EI
4
M´D
V´D
1/EI
1/EI
3
2
EI
4
1/EI
1/EI
V´E
M´E
3
2
EI
3
4

68. 68
/(32/3) = -0.188
-2
Influence line of MD
813
.
0
32
26
=
xD
xD
f
α
αDD = 32/3EI
2 m 2 m 2 m 2 m
2/EI
Conjugate beam
4
3EI
4
EI
32
3EI
x (m)
V ´ θ
EI
4
EI
3
8
−
EI
3
1
EI
3
4
EI
3
17
−
EI
5
fxD = M ´
x (m) ∆
-2/EI
θD = 0.469 + 0.531 = 1 rad
θDL = 5/(32/3) = 0.469 rad.
θDR = -17/32 = -0.531 rad.
26/3EI

69. 69
Example 3
Draw the influence line for the reactions at supports for the beam shown in the
figure below. EI is constant.
A D
B C
5 m 5 m 5 m 5 m
5 m 5 m
G
E F

70. 70
Influence line for RD
A D
B C G
E F
5 m 5 m 5 m 5 m
5 m 5 m
1
fBD
fCD fDD fED fFD
fXD
1
fXD/fDD = Influence
line for RD
DD
BD
f
f 1
=
DD
DD
f
f
DD
CD
f
f
DD
ED
f
f
DD
FD
f
f

71. 71
Conjugate beam
15 + (2/3)(15)
EI
5
.
112
EI
15
• Use the consistency deformation method
1
+
x RG
- Use conjugate beam for find ∆´G and fGG
∆´G + fGGRG = 0 ------(1)
1
Real beam
15 m 15 m
A G
1
15
1
Real beam
30 m
A G
1
30
1
A G
3@5 =15 m 3@5 =15 m
fGG
∆´G
1
=
112.5/EI
EI
M C
C
5
.
2812
'
' =
=
∆
Conjugate beam
20 m
EI
15 EI
450
EI
M
f G
GG
9000
'
'
' =
=
EI
450

72. 72
x RG = -0.3125 kN
0
9000
5
.
2812
=
+ G
R
EI
EI
↓
−
= ,
3125
.
0 kN
RG
Substitute ∆´G and fGG in (1) :
1
A G
5.625
0.6875 0.3125
1
1
15
=
1
+
1
30

73. 73
8.182 m
6.818 m
EI
16
.
35
EI
98
.
15
EI
01
.
23
)
3
(
2
3125
.
0 2
2
2 x
x
−
2
2 '
13
.
28
x
M
x
EI
=
+
)
2
(
6875
.
0
625
.
5 1
2
1
1 x
EI
x
x −
=
)
3
2
(
2
6875
. 1
2
1 x
EI
x
+
1
fBD
fCD fDD fED fFD
A G
3@5 =15 m 3@5 =15 m
Real beam
5.625
0.6875 0.3125
• Use the conjugate beam for find fXD
28.13
EI
x1
x2
x1 = 5 m -----> fBD = M´1= 56/EI
x1 = 10 m -----> fCD = M´1= 166.7/EI
x1 = 15 m -----> fDD = M´1= 246.1/EI
x2 = 5 m -----> fFD = M´2= 134.1/EI
x2 = 10 m -----> fED = M´2= 229.1/EI
x2 = 15 m -----> fDD = M´2= 246.1/EI
A G Conjugate beam
EI
625
.
5
EI
688
.
4
−
A
x1
(5.625-0.6875x1)/EI
V´1
M´1
EI
625
.
5 EI
x
x
2
1
1 6875
.
0
625
.
5 −
EI
x
2
6875
.
0
2
1
G
x2
0.3125x2
V´2
M´2
EI
13
.
28
EI
x
2
3125
.
0
2
2

74. 74
• Influence Line for RD
Influence Line for RD
0.2280.677 1.0 0.931 0.545
1
fXD
EI
56
EI
7
.
166
EI
1
.
246
EI
2
.
229
EI
1
.
134
1
fXD/fDD
1
.
246
56
1
.
246
7
.
166
1
.
246
1
.
246
1
.
246
2
.
229
1
.
246
1
.
134

75. 75
Influence line for RG
A D
B C G
E F
5 m 5 m 5 m 5 m
5 m 5 m
fXG
1
fBG fCG
fGG
fEG
fFG
fXG/fGG
1
GG
BG
f
f
GG
CG
f
f
GG
EG
f
f
GG
FG
f
f 1
=
GG
GG
f
f

76. 76
Conjugate beam
20 m
EI
450
EI
30
• Use consistency deformations
1
=
∆´D + fDDRD = 0 ------(2)
- Use conjugate beam for find ∆´D and fDD
1
+
1
Real beam
30 m
A G
1
30
1
Real beam
15 m 15 m
A G
1
15
X RD
fXG
1
3@5 =15 m 3@5 =15 m
∆´D
fDD
Conjugate beam
15 + (2/3)(15)
EI
15
EI
5
.
112
450/EI
EI
9000

77. 77
EI
EI
EI
EI
M
D
5
.
2812
)
15
(
450
9000
)
3
15
(
5
.
112
'
' =
−
+
=
=
∆
EI
EI
M
fDD
1125
)
15
3
2
(
5
.
112
'
' =
×
=
=
↓
=
−
=
=
+ ,
5
.
2
5
.
2
,
0
1125
5
.
2812
kN
kN
R
R
EI
EI
D
D
Substitute ∆´D and fDD in (2) :
x RD = -2.5 kN
=
1
1
1
30
+
1
1
15
1.5
7.5
2.5
15 m
V´
M´
EI
5
.
112
EI
5
.
1
450/EI
EI
9000
M´´
V´´
EI
15 EI
5
.
112

78. 78
EI
EI
fBG
5
.
62
)
5
3
2
(
75
.
18
−
=
×
−
=
=
• Use the conjugate beam for find fXG
Real beam
1
fBG fCG
fGG
fEG
fFG
3@5 =15 m 3@5 =15 m
1.5
7.5
2.5
fGG = M´G = 1968.56/EI
168.75/EI
EI
EI
fCG
06
.
125
)
67
.
6
(
75
.
18
−
=
−
=
=
A G Conjugate beam
EI
5
.
7
−
EI
15
10 m
15 + (10/3) = 18.33 m
25 + (2/3)(5) = 28.33 m
EI
5
.
112
EI
75
EI
75
.
18
A
5 m
V´1
M´1
EI
5
.
7
−
EI
75
.
18
A
6.67 m
V´2
M´2
EI
75
.
18
EI
5
.
7
−
EI
75
.
18
−

79. 79
=
−
+ 3
3
2
3 75
.
168
56
.
1968
)
3
(
2
x
EI
EI
x
x
x = 5 m -----> fFG = M´= 1145.64/EI
x = 10 m -----> fEG = M´ = 447.73/EI
Influence line for RG
-0.064
-0.032
0.227 0.582
1.0
M´ G
x
fGG = M´G = 1968.56/EI
168.75/EI
x
V´2
2
2
x
1
fXG
EI
5
.
62
−
EI
125
−
EI
73
.
447
EI
64
.
1145
EI
56
.
1968
56
.
1968
5
.
62
−
56
.
1968
125
−
56
.
1968
73
.
447
56
.
1968
64
.
1145
56
.
1968
56
.
1968
1
fXG/fGG

80. 80
Using equilibrium condition for the influence line for Ay
A D
B C G
E F
5 m 5 m 5 m 5 m
5 m 5 m
1
x
MA
Ay RD RG
Unit load
1 1
Influence Line for RD
0.2280.678 1.0 0.929 0.542
Influence line for RG
-0.064
-0.032
0.227 0.582
1.0
0.386
Influence line for Ay
0.804
-0.156 -0.124
1.0
C
D
A
y R
R
R
F −
−
=
=
Σ
↑
+ 1
:
0

81. 81
Using equilibrium condition for the influence line for MA
A D
B C G
E F
5 m 5 m 5 m 5 m
5 m 5 m
1
x
MA
Ay RD RG
x 15
x 30
RD
0.2280.678 1.0 0.929 0.542
1x
5 10 15 20 25 30
RG
-0.064
-0.032
0.227 0.582
1.0
Influence line for MA
2.54
1.75
-0.745 -0.59
C
D
A R
R
x
M 30
15
1
:
0 −
−
=
Σ
+