uticajne linije za staticki neodredjene nosace
TRANSCRIPT
1
! Comparison Between Indeterminate andDeterminate
! Influence line for Statically IndeterminateBeams
! Qualitative Influence Lines for Frames
INFLUENCE LINES FOR STATICALLY INDETERMINATE BEAMS
2
AB CED
RA AB CED
RA
Indeterminate Determinate
Comparison between Indeterminate and Determinate
11
3
AB CED
RAA
B CEDRA
AB CED
ME AB CED
ME
AB CED
VD AB CED
VD
1 1
1
Indeterminate Determinate
11
1
4
f1j
fjj
∆´1 = f1j
+
11 2 3j4
Redundant R1 applied
1
1
=
f11
fj1
×R1×R1
f11
Influence Lines for Reaction
Compatibility equation:
011111 =∆=+ Rff j
)1(11
11 ffR j−=
)(11
11 f
fR j=
5
11 2 3j4
1
=+
×R2
Redundant R2 applied
fjjf2j
1
fj2
f22
Compatibility equation.
0' 22222 =∆=+∆ Rf
022222 =∆=+ Rff j
)1(22
22 ffR j−=
)(22
22 f
fR j=
6
1 2 3j4
fj4
1
1
Influence Lines for Shear
444
)1( jE ff
V =
f44
1
1
7
444
)1( jE fMα
=
1 2 3j4
Influence Lines for Bending Moment
1 1
α44
fj4
1 1
8
R3 )(33
33 f
fR j=
R2 )(22
22 f
fR j=
R1)(
11
11 f
fR j=
1 2 3j4
1
1
� Influence line of Reaction
1
Using Equilibrium Condition for Shear and Bending Moment
111
11 =ff
11
41
ff
11
1
ff j
122
22 =ff
22
2
ff j
22
42
ff
133
33 =ff
33
3
ff j
33
43
ff
9
1 2 3j4
4V4
V4 = R1
R1V4
M41
� Unit load to the right of 4
� Influence line of Shear
V4 = R1
V4 = R1 - 1
1
R2 R3R1
1
1
1
R1
1x
V4
M41
� Unit load to the left of 4
V4 = R1 - 1
01;0 41 =−−=Σ↑+ VRFy
1R1
0;0 41 =−=Σ↑+ VRFy
10
1 2 3j4
� Influence line of Bending moment
M4 = - l + x + l R1
M4 - 1 (l-x) - l R1 = 0 + Σ M4 = 0:
R1
1x
V4
M4l1
� Unit load to the left of 4
M4 = l R1
R1V4
M4
� Unit load to right of
l
4
1
1
R2 R3R1
1
l1 l2
l
M4
4
1R1
1 1M4 = - l + x + l R1
M4 - l R1 = 0 + Σ M4 = 0:
M4 = lR1
11
Influence Line of VI
Maximum positive shear Maximum negative shear
Qualitative Influence Lines for Frames
I
1
1
12
Influence Line of MI
Maximum positive moment Maximum negative moment
I
11
13
A D G
15 m15 mAy GyDy
MA
Dy
1.0
Ay
1.0
Gy
1.0
Influence Line for MOF
14
MH
MA
1
1
A H G
15 m15 m
D
15
RA
1.0
RB
1.0
MB
1
MG
1
A E
G
B C D
16
VG1
VF1
VH1
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
BBf BBf BBf 1=BBf
� Influence line of RB
1
A BC D
2 m 2 m 2 m
ABf CBf DBfBBf
19
Real BeamA BC D
2 m 2 m 2 m
Conjugate Beam
� Find fxB by conjugate beam
11
6 kN�m
x
=−+EI
xEIEI
x 18726
3
EI6
EI18
EI18
EI72
3x
32xV´x
M´x
x
EI72
EI18
EIx
2
2
EIx
20
x (m)
0
2
4
6
Point
B
D
C
A
x
fxB / fBB
1
0.518
0.148
0
A BC D
2 m 2 m 2 m
1
fBB
fxB
0
72/EI37.33/EI10.67/EI
EIx
EIEIxMf xxB
18726
'3
−+==
EI72
EI33.37
EI67.10
0
/72 = 0.518
/72 = 0.148
1
37.33
10.670
Influence line of RB
17272
==BB
BB
ff
fxB
21
1 kN
Influence line of RA
A BC D
2 m 2 m 2 m
AA
CA
ff
AA
DA
ff
1=AA
AA
ff
22
Conjugate Beam
xV´x
M´x
EIB y
18' =
EIx
2
2
EIx
3x
32x
A BC D
2 m 2 m 2 m
Real Beam
1 kN
� Find fxA by conjugate beam
1 kN
6 kN�m
EIM A
72' =
x
EIB y
18' =
=−EIx
EIx
618 3
EI18
EI6
23
Point
B
D
C
A
fxA / fAA
0
0.482
0.852
1.0
EIx
EIxMf xxA 6
18'3
−==
x (m)
0
2
4
6
x
A BC D
2 m 2 m 2 m
1 kNfAA fxA
72/EI 61.33 /EI34.67 /EI
fxA
0
EI67.34
EI33.61
EI72
34.67
61.3372
1 kN
Influence line of RA
/72 = 1.0/72=0.852
/72 = 0.4821=
AA
AA
ff
24
AB
BA
y
RRRRF
−==−+
=Σ↑+
101
;0
A BC D
2 m 2 m 2 m
1x
RA
MA
RB
0.148.518
1RB
1RB
0.4820.852
1.01 kN
RA
Alternate Method: Use equilibrium conditions for the influence line of RA
RA = 1- RB
25
VC
1
RB
0.1480.518
1
A BC D
2 m 2 m 2 mRA
MA
RB
0.8520.482
-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
01
;0
=+−+
=Σ↑+
BC
y
RVF
RBVC
MC
VC = - RB
0;0
=++
=Σ↑+
CB
y
VRF
� Unit load to the left of C
26
BA
BA
A
RxMRxM
M
6606)6(1
;0
++−==+−−−
=Σ
A BC D
2 m 2 m 2 mRA
MA
RB
1 x
1
MA
1
RB
0.1480.518
1
-1.112 -0.892
Using equilibrium conditions for the influence line of MA
+
27
MC
C
1
RB
0.1480.518
1
A BC 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
RBVC
MC
4 m
MC = 4RB
+04
;0=+−
=Σ
BC
C
RMM
� Unit load to the left of C
4 m
MC = -4 + x + 4RB
04)4(1;0
=+−−−=Σ
BC
C
RxMM+
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 CD E F
2@2=4 m 4@2 = 8 m
G
29
Influence line of RA
1
A B CD E F
2@2=4 m 4@2 = 8 m
G
1=AA
AA
ff
AA
DA
ff
AA
EA
ff
AA
FA
ff
30
4 /EI
1
A B CD E F
4 m 6 m2 m
� Find fxA by conjugate beam
Real beam
0.51.5
0
18.67/EI
64/EI
4/EI
4/EI
Conjugate beam
EI33.5
EI67.10
0
EI67.10 EI
Mf AAA64' ==
31
x1x2
=−EI
xEI
x 13
1 33.512
EIEIEIx 67.18646
32 −+=
V´ x1
M´ x1
x1EIx
21
EI33.5
EIx
4
21
32 1x
31x
Conjugate beam
4/EI
4 m 8 m
64/EI
EI33.5
EI67.18
M´ x2
V´ x2
x2
EIx2
EIx2
22
32x
32 2x
EI67.18
EI64
32
CtoBforEI
xEI
xMf xxA ,33.512
'3
1 −==
EI28
EI64
BtoAforEIEI
xEI
xMf xxA ,6467.186
' 23
22 +−==
fxA
0
0
EI16
−
EI10
−
EI14
−
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
x1x2
1
fAA fxA
A B CD E FG
4 m 6 m2 m
EI10−
EI16−
EI14−
EI28
EI64
1
Influence line of RA
-0.219 -0.25 -0.156
0.438
1=AA
AA
ff
33
A B CD E FG
4 m 6 m2 m
Using equilibrium conditions for the influence line of RB
RB RCRA
x1
RA
1 -0.219 -0.25 -0.156
10.438
RB
1
0.4850.8751.07810.59
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
AB
AB
C
RxR
RxRM
812
8
0128;0
−=
=−+−=Σ+
34
A B CD E FG
4 m 6 m2 m
RA
1 -0.219 -0.25 -0.156
10.438
RB RCRA
x1
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
10.672
0.3750.141
-0.0312
Using equilibrium conditions for the influence line of RC
18
5.0
08)8(14;0
+−=
=−−−=Σ+
xRR
RxRM
AC
CA
B
35
A B CD E FG
4 m 6 m2 m
� Check ΣFy = 0
RB RCRA
x1
RA
1 -0.219 -0.25 -0.156
10.438
RB
1
0.490.8751.081
0.59
RC1
10.672
0.3750.141
-0.0312
1
;0
=++
=Σ↑+
CBA
y
RRRF
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
VG
RA
1 -0.219 -0.25 -0.156
10.438
1x
VG = RA
RA
AVG
MG
� Unit load to the right of G
0.438
-0.219 -0.25 -0.156-0.5621
A B CD E FG
4 m 6 m2 m
Using equilibrium conditions for the influence line of VG
VG = RA - 1
RA
1x
AVG
MG
� Unit load to the left of G
01;0 =−−=Σ↑+ GAy VRF
37
VE
RC1
10.672
0.3750.141
-0.0312
1x
0.6250.328
0.0312
-0.141-0.375
1
A B CD E FG
4 m 6 m2 m
Using equilibrium conditions for the influence line of VE
RCVE
ME
� Unit load to the left of E
VE = - RC
0;0 =+=Σ↑+ ECy VRF
VE = 1 - RC
� Unit load to the right of E
RC
1 x
VE
ME
01;0 =+−=Σ↑+ CEy RVF
38
MG
RA
1 -0.219 -0.25 -0.156
10.438
1x
-0.438 -0.5 -0.312
1
A B CD E FG
4 m 6 m2 m
Using equilibrium conditions for the influence line of MG
0.876
MG = -2 + x + 2RA
� Unit load to the left of G
RA
1x
AVG
MG2 m
02)2(1;0
=−−+=Σ
AG
G
RxMM+
MG = 2RA
� Unit load to the right of G
RA
AVG
MG2 m
02;0
=−=Σ
AG
G
RMM+
39
RC1
10.672
0.3750.141
-0.0312
ME
1x
1.5
0.6881
4 m 6 m2 m
A B CD E FG
Using equilibrium conditions for the influence line of ME
-0.125
0.564
RCVE
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
ME4 m
04)4(1;0
=+−−−=Σ+
CE
E
RxMM
40
Example 3
For the beam shown(a) Draw quantitative influence lines for the reaction at supports A and B, andbending 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
AC
B
4 m
2EI 3EI
2 m 2 m
20 kN
411
fCA
fAAfEA
fDA
1
/fAA /fAA /fAA
/fAA
fCA
fAAfEA
fDA
AC
B
4 m
2EI 3EI
2 m 2 m
Influence line of RA
42Conjugate Beam
AC
B
4 m
2EI 3EI
2 m 2 m1Real Beam
� Find fxA by conjugate beam
1 kN
8 kN�m
x (m)V (kN)1 1
+
x (m) M (kN�m)
84
+
2EI
fAA = M´A = 60.44/EI
60.44/EI
12/EI
1.33EIEI67.2
43
EI67.2
2EI
60.44/EI
Conjugate Beam12/EI
1.33EI
37.11/EI 17.77/EI 4.88/EI60.44/EI
0
RA = fxA/fAA
0.614 0.294 0.081
1
0
fxA
A C B
� Quantitative influence line of RA
44
A B
4 m 2 m 2 m
Using equilibrium conditions for the influence line of RB and MB
RB = 1 - RA
10.386
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
1x
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
1x
VB = RA - 1
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
1x
RB
MB
MC = 4RAMC = 4RA - (4-x)(1)
MC
VC = RAVC = RA - 1
VC1
-0.386 -0.706
0.3240.4561.176
0.294 0.081 0
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
A B
4 m 2 m 2 m
10 kNThe 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
26.760.14
RA
0.614 0.294 0.081
1
0
RB=23.31 kN
MB=46.48 kN�m
49
Example 1
Draw the influence line for - the vertical reaction at B
A BC 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
1 kN
Influence line of RA
A BC D
2 m 2 m 2 m
1=AA
AA
ff AA
CA
ff
AA
DA
ff
51
Conjugate Beam
xV´x
M´x
EIB y
18' =
EIx
2
2
EIx
3x
32x
A BC D
2 m 2 m 2 m
Real Beam
1 kN
� Find fxA by conjugate beam
1 kN
6 kN�m
EIM A
72' =
x
EIB y
18' =
6 /EIEI18
=−EIx
EIx
618 3
52
Point
B
D
C
A
fxA / fAA
0
0.482
0.852
1.0
x (m)
0
2
4
6
x
A BC D
2 m 2 m 2 m
1 kNfAA fxA
72/EI 61.33 /EI34.67 /EI
EIx
EIxMf xxA 6
18'3
−==
fxA
0
EI67.34
EI33.61
EI72
34.67
61.3372
1 kN
Influence line of RA
/72 = 1.0/72=0.852
/72 = 0.4821=
AA
AA
ff
53
� Influence line of RB
1
A BC D
2 m 2 m 2 m
1=BB
BB
ff
BB
AB
ff
BB
CB
ff
BB
DB
ff
54
Real BeamA BC D
2 m 2 m 2 m
Conjugate Beam
� Find fxB by conjugate beam
11
6 kN�m
x
=−+EI
xEIEI
x 18726
3
EI6
EI18
EI18
EI72
3x
32xV´x
M´x
x
EI72
EI18
EIx
2
2
EIx
55
x (m)
0
2
4
6
Point
B
D
C
A
x
fxB / fBB
1
0.518
0.148
0
A BC D
2 m 2 m 2 m
1
fBB
fxB
0
72/EI37.33/EI10.67/EI
/72 = 0.518
/72 = 0.148/72 = 1
1
37.33
10.670
fBB
Influence line of RB
fBB= 1
72
EIx
EIEIxMf xxB
18726
'3
−+==
fxB
0
EI72
EI33.37
EI67.10
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 beamEI is constant.Plot numerical values every 2 m.
A B CD E
2 m 2 m 2 m 2 m
57
A B CD ED
2 m 2 m 2 m 2 m
The influence line for the shear at D
1 kN
1 kN
DVD
1 kN
1 kNDD
ED
ff
1=DD
DD
ff
58
A B CD E
2 m 2 m 2 m 2 m
2 kN�m
1 kN
1 kN2 k
1 kN
1 kN2 kN�m
1 kN2 kN1 kN
� Using conjugate beam for find fxD
1 kN
1 kN
59
A B CD 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 kN2kN
60
2 m 2 m 2 m 2 m
M´D
Conjugate beam
4/EI
AB
CD E
4/EI
0
8/EI
m38
� Determine M´D at D
EI316
EI38
128/3EI
4/EI
8/EI
m38
EI316
EI340
61
=−=− )2)(3
8()32)(2(4
EIEIEI
EIEIEIEI 352)2)(
340(
3128)
32)(2( =−+=
EIEIEI 376)2)(
340()
32)(2( −=−=
128/3EI
2 m 2 m 2 m 2 m
Conjugate beam
4/EI
AB
CD E
EI38
EI340
V´DL
M´DL
2/EI2/EI
32
EI340
EI340 V´DR
M´DR
2/EI
32
128/3EI
2/EI2/EI
V´E
M´E
32
EI38
62
128/3EI = M´D = fDD
V ´x (m) θ
fxD = M ´ x (m) ∆
EI376
−
EI340
−EI334
−
EI32
EI316
−
EI38
EI352
EI4
−
Influence line of VD = fxD/fDD
76
52
4
0.406 =128
/128 = -0.594 /(128/3) = -0.094
Conjugate beam
EI38128/3EI
2 m 2 m 2 m 2 m
4/EI
AB
CD E
EI340
63
A B CD E
2 m 2 m 2 m 2 m
The influence line for the bending moment at D
1 kN �m1 kN �m
αDD
MD
DD
EDfα
DD
DDfα
64
1 kN�m
0.5 kN1 kN�m
0.5 kN1 kN0.5 kN
2 m 2 m 2 m 2 m
0.5 kN 0.5 kN
0.5 kN1 k
A B CD E
1 kN �m1 kN �m
� Using conjugate beam for find fxD
65
2 m 2 m 2 m 2 m
Real beamA B CD 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
66
2/EI
4/EI
m38
EI38
2/EI
0
4/EI
m38
2 m 2 m 2 m 2 m
2/EI
Conjugate beam
EI34
EI38
EI332
EI34
67
EIEIEI 326)2)(4()
32)(1( =+=
=−=− )2)(3
4()32)(1(2
EIEIEI
2 m 2 m 2 m 2 m
2/EI
Conjugate beam
EI332
EI34
EI4
M´D
V´D
1/EI1/EI
32
EI4
1/EI1/EI
V´E
M´E
32
EI34
68/(32/3) = -0.188-2Influence line of MD
813.03226
=
xD
xDfα
αDD = 32/3EI
2 m 2 m 2 m 2 m
2/EI
Conjugate beam
43EI
4EI
323EI
x (m)V ´ θEI4
EI38
−
EI31
EI34
EI317
−
EI5
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
Example 3
Draw the influence line for the reactions at supports for the beam shown in thefigure below. EI is constant.
A DB C
5 m 5 m 5 m 5 m5 m 5 m
GE F
70
Influence line for RD
A DB C GE F
5 m 5 m 5 m 5 m5 m 5 m
1
fBDfCD fDD fED fFD
fXD
1
fXD/fDD = Influence line for RD
DD
BD
ff 1=
DD
DD
ff
DD
CD
ff
DD
ED
ff
DD
FD
ff
71Conjugate beam
15 + (2/3)(15)
EI5.112
EI15
� 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 mA G
1
30
1
A G3@5 =15 m 3@5 =15 m
fGG
∆´G
1
=
112.5/EI
EIM CC
5.2812'' ==∆
Conjugate beam
20 m
EI15 EI
450
EIMf GGG
9000''' ==
EI450
72
x RG = -0.3125 kN
090005.2812=+ GR
EIEI
↓−= ,3125.0 kNRG
Substitute ∆´G and fGG in (1) :
1
A G5.625
0.6875 0.3125
11
15
=
1
+
1
30
73
8.182 m
6.818 m
EI16.35
EI98.15
EI01.23
)3
(2
3125.0 22
2 xx−
22 '13.28xMx
EI=+
)2
(6875.0625.5 12
11 xEI
xx −=
)3
2(2
6875. 12
1 xEI
x+
1
fBDfCD fDD fED fFD
A G3@5 =15 m 3@5 =15 m
Real beam5.625
0.6875 0.3125
� Use the conjugate beam for find fXD
28.13EIx1
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 beamEI625.5
EI688.4−
Ax1
(5.625-0.6875x1)/EI
V´1
M´1
EI625.5 EI
xx 211 6875.0625.5 −EI
x2
6875.0 21
G
x2
0.3125x2
V´2M´2
EI13.28
EIx
23125.0 2
2
74
� Influence Line for RD
Influence Line for RD
0.228 0.677 1.0 0.931 0.545
1
fXD
EI56
EI7.166
EI1.246
EI2.229
EI1.134
1
fXD/fDD
1.24656
1.2467.166
1.2461.246
1.2462.229
1.2461.134
75
Influence line for RG
A DB C GE F
5 m 5 m 5 m 5 m5 m 5 m
fXG
1fBG fCG
fGGfEGfFG
fXG/fGG
1GG
BG
ff
GG
CG
ff
GG
EG
ff
GG
FG
ff 1=
GG
GG
ff
76Conjugate beam
20 m
EI450
EI30
� Use consistency deformations
1
=
∆´D + fDDRD = 0 ------(2)
- Use conjugate beam for find ∆´D and fDD
1
+1
Real beam
30 mA G
1
30
1
Real beam
15 m 15 m
A G
1
15
X RD
fXG
13@5 =15 m 3@5 =15 m
∆´D
fDD
Conjugate beam15 + (2/3)(15)
EI15
EI5.112
450/EI
EI9000
77
EIEIEIEIMD
5.2812)15(4509000)3
15(5.112'' =−+==∆EIEI
MfDD1125)15
32(5.112'' =×==
↓=−==+ ,5.25.2,011255.2812 kNkNRREIEI DDSubstitute ∆´D and fDD in (2) :
x RD = -2.5 kN
=
1
11
30+
11
15
1.5
7.5
2.5
15 mV´
M´EI
5.112
EI5.1
450/EI
EI9000
M´´
V´´
EI15 EI
5.112
78
EIEIfBG
5.62)532(75.18
−=×−==
� Use the conjugate beam for find fXG
Real beam
1fBG fCG
fGGfEGfFG
3@5 =15 m 3@5 =15 m1.5
7.5
2.5
fGG = M´G = 1968.56/EI
168.75/EI
EIEIfCG
06.125)67.6(75.18−=−==
A G Conjugate beam
EI5.7−
EI15
10 m
15 + (10/3) = 18.33 m
25 + (2/3)(5) = 28.33 m
EI5.112
EI75
EI75.18
A5 m
V´1
M´1
EI5.7−
EI75.18
A6.67 m
V´2
M´2
EI75.18
EI5.7−
EI75.18−
79
=−+ 33
23 75.16856.1968)
3(
2x
EIEIxx
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´ Gx
fGG = M´G = 1968.56/EI
168.75/EI
x
V´2
2
2x
1
fXG
EI5.62−
EI125−
EI73.447
EI64.1145
EI56.1968
56.19685.62−
56.1968125−
56.196873.447
56.196864.1145
56.196856.1968
1
fXG/fGG
80
Using equilibrium condition for the influence line for Ay
A DB C GE F
5 m 5 m 5 m 5 m5 m 5 m
1x
MA
Ay RD RG
Unit load
1 1
Influence Line for RD
0.228 0.678 1.0 0.929 0.542
Influence line for RG-0.064-0.032
0.227 0.582 1.0
0.386Influence line for Ay
0.804
-0.156 -0.124
1.0
CDAy RRRF −−==Σ↑+ 1:0
81
Using equilibrium condition for the influence line for MA
A DB C GE F
5 m 5 m 5 m 5 m5 m 5 m
1x
MA
Ay RD RG
x 15
x 30
RD
0.228 0.678 1.0 0.929 0.542
1x5 10 15 20 25 30
RG-0.064-0.032
0.227 0.582 1.0
Influence line for MA
2.541.75
-0.745 -0.59
CDA RRxM 30151:0 −−=Σ+