INFLUENCE LINES FOR STATICALLY INDETERMINATE BEAMS !
!
!
Comparison Between Indeterminate and Determinate Influence line for Statically Indeterminate Beams Qualitative Influence Lines for Frames
1
Comparison between Indeterminate and Determinate
Indeterminate
1 A
D
B
E
Determinate
C
RA
1 A
D
B
E
C
RA
2
Indeterminate D
1
Determinate
B
E
C
B
E
C
A
1
D
A
D A
B
E
1
RA
VD
1 A
A
D
B
E
C
1 D
B
E
C
RA
VD
1 C
ME
A
D
B
E
C
ME 3
Influence Lines for Reaction Redundant R1 applied
1
4
2
j
3
1 Compatibility equation: f1 j + f11 R1 = ∆1 = 0
= R1 = − f1 j (
∆´1 = f1j
1 R1 = (
fjj
f j1 f11
1 ) f11
)
+
f11 ×R1
1 fj1
4
Redundant R2 applied
1
4
2
j
3
1 Compatibility equation. ∆ '2 + f 22 R2 = ∆ 2 = 0
=
f 2 j + f 22 R2 = ∆ 2 = 0
1
+
f2j
f22
1
R2 = − f 2 j (
fjj
R2 = (
fj2
f j2 f 22
1 ) f 22
)
×R2 5
Influence Lines for Shear
1
4
2
j
3
1
VE = ( 1
1 ) f j4 f 44
1 f44 fj4 1
6
Influence Lines for Bending Moment
4
1
1
2
j
3
1
ME = ( 1
α44
1
α 44
) f j4
1
fj4
7
Using Equilibrium Condition for Shear and Bending Moment • Influence line of Reaction 4
1
2
j
3
f 41 f11
f11 =1 f11
R1
R1 = (
f j1 f11
)
f j1
1 f 22 =1 f 22
f 42 f 22
f11 f j2 f 22
R2
R2 = (
1
f j2 f 22
)
f 33 =1 f 33
f 43 f 33
f j3 f 33
R3 1
R3 = (
f j3 f 33
) 8
• Influence line of Shear
• Unit load to the left of x
1
1 4
1
2
j
M4
1
3
R1 R1
R2
4
V4
+ ↑ ΣFy = 0; R1 − 1 − V4 = 0
R3
V4 = R1 - 1 R1 1
• Unit load to the right of M4
1
V4 = R1
4 1 1 1
V4 = R1 - 1
R1 V4
4
V4
+ ↑ ΣFy = 0; R1 − V4 = 0
V4 = R1
9
• Influence line of Bending moment • Unit load to the left of
l
x 1 4
1
R1
2
l1
R2
j
l2
1
3
l R1
4
1 M4 V4
+ Σ M4 = 0: M4 - 1 (l-x) - l R1 = 0
R3
M4 = - l + x + l R1 R1
• Unit load to right of
4
1 1
1
4
M4 = - l + x + l R1 1 1 M4
l R1
M4 V4
+ Σ M4 = 0: M4 - l R1 = 0 M4 = l R1
M4 = lR1
10
Qualitative Influence Lines for Frames
1
I 1
Influence Line of VI
Maximum positive shear
Maximum negative shear
11
1
1 I
Influence Line of MI
Maximum positive moment
Maximum negative moment
12
Influence Line for MOF D
A
MA
Ay
15 m
Dy
G
15 m
Gy
1.0 Gy 1.0 Dy 1.0 Ay
13
D
A
15 m
H
G
15 m
MA
1
1 MH
14
G A
B
C
D
E
1.0 RA 1.0 RB
MB 1 1 MG
15
G A
B
C
D
1
E
VG
1
VF
1
VH 16
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
C
2m
D
2m
B
2m
17
• Influence line of RB A
C
D
2m
2m
f AB f BB
f CB f BB
B
2m
f DB f BB
f BB =1 f BB
1
18
• Find fxB by conjugate beam A
6 kN•m
1
C
D
2m
2m
6 EI
18 EI
Real Beam
1
2m x
72 EI Conjugate Beam
V´x
18 EI
x2 2 EI
x EI
x3 72 18 x + − = M´x 6 EI EI EI
B
x 3
x
2x 3
72 EI
18 EI
19
x A
C
D
2m
2m
B
f xB
x3 72 18 x = M 'x = + − 6 EI EI EI
fxB
fxB / fBB
Point
x (m)
B
0
72 EI
1
D
2
37.33 EI
0.518
C
4
10.67 EI
0.148
A f BB 72 = =1 f BB 72
6
0
2m
72/EI 0
10.67/EI
37.33/EI
fBB 1
fxB
37.33 /72 = 0.518 10.67 /72 = 0.148 0
0
1 Influence line of RB 20
Influence line of RA A
C
D
2m
2m
1 kN f AA =1 f AA
f CA f AA
B
2m
f DA f AA
21
• Find fxA by conjugate beam 6 kN•m
1 kN A
C
B
D
2m
2m
Real Beam 1 kN
2m x
M 'A =
72 EI
Conjugate Beam B' y = 18 EI
6 EI
x 3
3
18 x x − = M´x EI 6 EI
2x 3
V´x x EI
18 EI
x x2 2 EI
B' y =
18 EI
22
x A
C
2m
1 kN fAA
72/EI
D
2m
61.33 /EI
B
f xA
2m
34.67 /EI
AA AA
1 kN =1
Point
x (m)
fxA
fxA / fAA
B
0
0
0
D
2
34.67 EI
0.482
C
4
61.33 EI
0.852
6
72 EI
1.0
fxA
72 /72 = 1.0 61.33 /72=0.852 f f
18 x x 3 = M 'x = − EI 6 EI
A
34.67 /72 = 0.482 Influence line of RA
23
Alternate Method: Use equilibrium conditions for the influence line of RA x A
MA
1 C
D
B + ↑ ΣFy = 0;
RA
2m
2m
2m
RB
RA + RB − 1 = 0 RB = 1 − RA
RB 0.148
1 .518 RB 1
RA = 1- RB 1.0 1 kN
0.852 0.482 RA 24
Using equilibrium conditions for the influence line of VC A
MA RA
• Unit load to the left of C
x
1 C
D
2m
2m
MC
B
2m
VC
RB
RB
+ ↑ ΣFy = 0; + RB + VC = 0
1 0.148
0.518 RB 1
VC = - RB • Unit load to the left of C 1
MC 0.852 1 -0.148
x
VC
0.482 VC
RB
+ ↑ ΣFy = 0; + VC − 1 + RB = 0
VC = 1 - RB
25
Using equilibrium conditions for the influence line of MA x
1 A
MA
C
D
B + ΣM A = 0;
RA
2m
2m
2m
RB
− M A − 1(6 − x) + 6 RB = 0 M A = −6 + x + 6 RB
1 0.148
0.518 RB 1
MA -1.112 1
-0.892 26
Using equilibrium conditions for the influence line of MC A
MA
• Unit load to the left of C
x
1 C
D
B
MC 4m
VC RA
2m
2m
2m
RB
ΣM C = 0;
+
− M C + 4 RB = 0
1
MC = 4RB
0.518
0.148
RB 1
• Unit load to the left of C x
1
C
0.592
RB
MC
1
VC 0.074 MC
4m
RB
+ ΣM C = 0; − M C − 1(4 − x) + 4 RB = 0 MC = -4 + x + 4RB
27
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
G
2@2=4 m
B
D
E
F
C
4@2 = 8 m
28
Influence line of RA A
G
2@2=4 m
B
D
E
F
C
4@2 = 8 m
f AA =1 f AA
1
f DA f AA
f EA f AA
f FA f AA
29
• Find fxA by conjugate beam A
B
1
D
E
2m
C
Real beam
0.5
1.5 4m
F
6m
4 /EI Conjugate beam 4/EI 0 10.67 EI
5.33 EI
4/EI
64/EI
0 18.67/EI
10.67 EI
f AA = M ' A =
64 EI
30
4/EI
x2
64/EI
x1 Conjugate beam
4m
18.67 EI
8m
5.33 EI
2
x1 4 EI
x1 2 EI
x1
3
x1 5.33 x1 − = 12 EI EI
M´ x1 V´ x1 2
64 EI
x2
x2 2 EI
x1 3
x2 EI
2 x2 3
x2 V´ x2 3
5.33 EI
3
M´ x2 18.67 EI
2 x1 3
x 64 18.67 = 2 + − 6 EI EI EI
31
x2 A
G
4m 64 EI
fAA
1
B
D
2m
E
F
C
f xA
6m
28 EI
Point
x (m)
fxA
fxA / fAA
C
0
0
0
−
10 EI
-0.1562
16 EI
-0.25
14 EI
-0.2188
F
2
fxA
1
f AA =1 f AA
f xA
x1
5.33x x3 = M ' x1 = − , for B to C EI 12 EI 3 18.67 x2 64 x2 = M 'x 2 = − + , for A to B EI EI 6 EI
− 14 EI
− 16 EI
− 10 EI
Influence line of RA 0.438 -0.219 -0.25 -0.156
E
4
−
D
6
−
B
8
0
G
10
28 EI
A
12
64 EI
0 0.4375 1
32
Using equilibrium conditions for the influence line of RB 1 A
G
x B
RA
D
E
F
RB 4m
+ ΣM C = 0;
C
− 8 RB + x − 12 RA = 0 RB =
RC
2m
6m
1 0.438 -0.219 -0.25 -0.156
1
1
0.59
1.078 0.875
RA
0.485 RB
1
x 12 − RA 8 8
Point
x (m)
RA
RB
C
0
0
0
F
2
-0.1562 0.485
E
4
-0.25
D
6
-0.2188 1.078
B
8
G
10
A
12
0 0.4375 1
0.875
1 0.5939 0 33
Using equilibrium conditions for the influence line of RC 1 A
G
RA
x B
D
E
F
C
RB 4m
+ ΣM B = 0;
4 RA − 1( x − 8) − 8 RC = 0
RC
2m
6m
1 0.438 RA
-0.219 -0.25 -0.156
1
0.141 0.375 -0.0312
RC = 0.5 RA −
1
0.672
RC 1
x +1 8
Point
x (m)
RA
RC
C
0
0
1
F
2
-0.1562
0.6719
E
4
-0.25
0.375
D
6
-0.2188
0.1406
B
8
G
10
A
12
0 0.4375 1
0 -0.0312 0 34
• Check ΣFy = 0 1 A
G
x B
RA
D
E
F
C
RB 4m
+ ↑ ΣFy = 0; RA + RB + RC = 1
RC
2m
6m
1 0.438 RA
-0.219 -0.25 -0.156
1 1
0.59
1.08
0.875
0.49
RB
1 0.141 0.375 -0.0312
1
0.672 1
Point
RA
RB
RC
ΣR
C
0
0
1
1
F
-0.1562 0.485
0.6719
1
E
-0.25
0.375
1
D
-0.2188 1.078
0.1406
1
0
1
B G A
RC
0
0.875
1
0.4375 0.5939 -0.0312 1
0
0
1 1 35
Using equilibrium conditions for the influence line of VG 1
• Unit load to the left of G
x A
G
B
D
E
F
x
C
1 MG
A 4m
2m
RA
6m
VG
+ ↑ ΣFy = 0; RA − 1 − VG = 0
1 VG = RA - 1
0.438 -0.219 -0.25 -0.156
1
RA • Unit load to the right of G
0.438 1 -0.562
MG
A VG -0.219 -0.25 -0.156
RA VG = RA
VG
36
Using equilibrium conditions for the influence line of VE 1 A
G
B
D
• Unit load to the left of E
x F
E
C
ME VE
4m
2m
RC
+ ↑ ΣFy = 0; RC + VE = 0
6m
VE = - RC 0.141 0.375
1
0.672 1
-0.0312
• Unit load to the right of E RC
1
x
ME 0.625 0.0312
1 -0.141
VE
0.328 VE
-0.375
RC
+ ↑ ΣFy = 0; VE − 1 + RC = 0
VE = 1 - RC 37
Using equilibrium conditions for the influence line of MG 1
• Unit load to the left of G
x A
G
B
D
E
F
A 4m
2m
1
x
C
RA
6m
2m
MG VG
ΣM G = 0;
+
M G + 1(2 − x) − 2 RA = 0
1
MG = -2 + x + 2RA
0.438 -0.219 -0.25 -0.156
1
RA
• Unit load to the right of G A
0.876
RA
1 MG -0.438 -0.5
-0.312
+
2m
MG VG
ΣM G = 0; M G − 2 RA = 0
MG = 2RA
38
Using equilibrium conditions for the influence line of ME 1 A
G
B
D
• Unit load to the left of E
x F
E
4m
C
ME VE
4m
2m
0.141
6m
0.375
0.564
1
ME = 4RC
1
1 1.5
-0.125
+ ΣME = 0;
0.672
-0.0312
RC
RC
• Unit load to the right of E x 1 ME VE
0.688
4m
RC
+ ΣM E = 0;
ME
− M E − 1(4 − x) + 4 RC = 0
ME = - 4 + x+ 4RC 39
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 20 kN
10 kN
2EI
3EI B
A 4m
C
2m
2m
40
Influence line of RA 3EI
2EI
B
A C
4m
fAA
fDA
2m
2m
fCA
fEA
fCA /fAA
fEA /f
1
fAA /fAA
1
fDA /f
AA
AA
41
• Find fxA by conjugate beam 8 kN•m
3EI
2EI
B
A 1
C
4m
1 V (kN)
1 kN
2m
2m
Real Beam
1 +
x (m)
8
4 +
M (kN•m)
2EI
1.33EI
x (m)
2.67 EI
60.44/EI 12/EI
fAA = M´A = 60.44/EI
Conjugate Beam 42
• Quantitative influence line of RA 2EI
2.67 EI
1.33EI
60.44/EI A
12/EI
60.44/EI
1
B
C
37.11/EI
0.614
17.77/EI
0.294
Conjugate Beam
4.88/EI
0
0.081
0
fxA
RA = fxA/fAA
43
Using equilibrium conditions for the influence line of RB and MB x
1
B
A RA 1
0
4m
0.614
0.386
2m
0.294
0.706
2m
0.081
MB
RB
0
0.919
RA
1 RB = 1 - RA
0
0 -1.088
-1.648
-1.352
MB = 8RA - (8-x)(1) 44
Using equilibrium conditions for the influence line of VB x
1
B
A RA 1
4m
0.614
0.294
VB = RA - 1
2m
2m
0.081
0
RA
0
VB = RA -1 -0.386
-0.706
-0.919
-1
45
Using equilibrium conditions for the influence line of VC and MC x
1
C
B
A RA 1
4m
2m
0.614
0.294
2m
MB
RB
0.081
0
0.081
0
RA
VC = RA - 1 VC = RA 1 -0.386 MC = 4RA - (4-x)(1) 0.456
0.294
VC
-0.706 MC = 4RA 1.176 0.324
MC
46
The quantitative shear and bending moment diagram and qualitative deflected curve 10 kN MB= 13.53 kN•m B A 4m
10(0.081)=0.81 kN
1
V (kN)
0.81
0.614
2m
0.294
2m
RB=9.19 kN
0.081
0
RA
0.81 -9.19
MA (kN•m)
+
4.86 -13.53
47
The quantitative shear and bending moment diagram and qualitative deflected curve 10 kN 20 kN MB=46.48 kN•m B A 4m
20(.294) +1(0.081) = 6.69 kN 1
V (kN)
6.69
0.614
2m
0.294
2m
RB=23.31 kN
0.081
0
RA
6.69 -13.31
MA (kN•m)
26.76 +
-23.31
-23.31
0.14 -46.48
48
APPENDIX •Muller-Breslau for the influence line of reaction, shear and moment •Influence lines for MDOF beams Example 1 Draw the influence line for - the vertical reaction at B
A
C
2m
D
2m
B
2m
49
Influence line of RA A
C
D
2m
2m
1 kN f AA =1 f AA
B
f CA f AA
2m
f DA f AA
50
• Find fxA by conjugate beam 6 kN•m
1 kN A
C
B
D
2m
2m
Real Beam 1 kN
2m x
M 'A =
72 EI
Conjugate Beam B' y =
6 /EI
18 EI
x 3
3
18 x x − = M´x EI 6 EI
2x 3
V´x x EI
18 EI
x x2 2 EI
B' y =
18 EI
51
x A
C
2m
1 kN fAA
72/EI
D
2m
61.33 /EI
B
f xA
2m
34.67 /EI
Point
x (m)
fxA
fxA / fAA
B
0
0
0
D
2
34.67 EI
0.482
4
61.33 EI
0.852
6
72 EI
1.0
fxA C 72/72 = 1.0 61.33 /72=0.852
1 kN f AA =1 f AA
18 x x 3 = M 'x = − EI 6 EI
A
34.67 /72 = 0.482 Influence line of RA
52
• Influence line of RB A
C
D
2m
2m
f AB f BB
f CB f BB
B
2m
f DB f BB
f BB =1 f BB
1
53
• Find fxB by conjugate beam A
6 kN•m
1
C
D
2m
2m
6 EI
B 1
2m
18 EI
Real Beam
x 72 EI 18 EI
x2 2 EI
x EI
Conjugate Beam
x 72 EI
x3 72 18 x + − = M´x 6 EI EI EI
V´x
x 3
2x 3
18 EI 54
x A
C
D
2m
2m
B
f xB
2m
x3 72 18 x = M 'x = + − 6 EI EI EI
Point
x (m)
fxB
fxB / fBB
B
0
72 EI
1
D
2
37.33 EI
0.518
C
4
10.67 EI
0.148
A
6
0
72/EI 0
10.67/EI
37.33/EI
fBB 1
37.33 /72 = 0.518 10.67 /72 = 0.148 0
fxB
72 /72 = 1 fBB =1 fBB
0
1 Influence line of RB 55
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
D 2m
B 2m
E 2m
C 2m
56
The influence line for the shear at D 1 kN A
D
B
E
C
D 1 kN 2m
2m f DD =1 f DD
2m
2m
1 kN D 1 kN
f ED f DD
VD
57
• Using conjugate beam for find fxD 1 kN A
D
2m
1 kN
B
E
2m
2m
C
2m
2 kN•m 1 kN
1 kN 1 kN
2 kN•m 2k
1 kN
2 kN
1 kN
1 kN
58
1 kN A
D
B
1 kN 2m
E
C
2kN 1 kN
V( kN)
2m
2m
Real beam 1 kN
2m
1 x (m) -1 4
M (kN •m)
x (m) 4/EI Conjugate beam M´D
59
• Determine M´D at D 4/EI A
M´D D
B 2m
2m
C
E 2m
Conjugate beam
2m
8/EI 8 m 3
4/EI 0
8/EI 8 m 3
128/3EI 40 3EI
16 3EI
8 3EI
4/EI 16 3EI 60
4/EI A D 128/3EI
40 3EI
B 2m
2m
C
E 2m
2m
Conjugate beam
8 3EI
2/EI 2/EI M´DL 40 3EI
2 3
=(
2 2 40 76 )( ) − ( )(2) = − EI 3 3EI 3EI
V´DL
2/EI 128/3EI 40 3EI
2 3
M´DR = (
2 2 128 40 52 )( ) + −( )(2) = EI 3 3EI 3EI 3EI
V´DR
2/EI −
4 2 2 8 = ( )( ) − ( )(2) = M´E EI EI 3 3EI
2/EI
2 V´E 3
8 3EI 61
4/EI A 40 3EI
D 128/3EI
B 2m
2m
E 2m
V´
−
40 3EI
34 3EI 52 3EI
−
128/3EI = M´D = fDD
16 3EI
fxD = M ´ −
0.406 =
2m
2 3EI −
52 128
C
76 3EI
−
Conjugate beam
8 3EI 8 3EI x (m)
θ
x (m)
∆
4 EI
Influence line of VD = fxD/fDD 76 /128 = -0.594
4 /(128/3) = -0.094
62
The influence line for the bending moment at D αDD A
D
1 kN •m
B
E
C
1 kN •m 2m
2m
2m
2m
f DD
α DD
f ED
MD
α DD
63
• Using conjugate beam for find fxD
A
D
1 kN •m 2m
B
E
C
1 kN •m 2m
2m
2m
1 kN•m 0.5 kN
0.5 kN 0.5 kN
1 kN•m 1k
0.5 kN
1 kN
0.5 kN
0.5 kN
64
1 kN •m 1 kN •m A D B 0.5 kN 2m V (kN)
2m
1 kN 2m
E
C
Real beam 0.5 kN
2m
0.5 x (m) 1 2
M (kN •m)
x (m) 2/EI Conjugate beam 65
2/EI Conjugate beam
2m
2m
2m
2m
4/EI 8 m 3
2/EI 0
4/EI 8 m 3
4 3EI
8 3EI
4 3EI
2/EI
32 3EI
8 3EI 66
2/EI Conjugate beam 4 EI
4 3EI
32 3EI
2m
2m 1/EI
2m
1/EI M´D = (
4 EI
2m
1 2 4 26 )( ) + ( )(2) = EI 3 EI 3EI
2 V´D 3
1/EI −
2 1 2 4 = ( )( ) − ( )(2) = M´E EI EI 3 3EI
V´E
1/EI
2 3
4 3 EI 67
2/EI Conjugate beam 4 EI 2m
4 3EI
32 3EI 2m
2m
2m
5 1 EI αDD = 32/3EI 3EI
4 EI V´ 17 − 3EI
−
4 3EI x (m)
8 3EI
θ
26/3EI fxD = M ´
x (m) 26 = 0.813 32
f xD
α xD
∆
-2/EI θDL = 5/(32/3) = 0.469 rad. θDR = -17/32 = -0.531 rad.
θD = 0.469 + 0.531 = 1 rad
Influence line of MD 68 -2 /(32/3) = -0.188
Example 3 Draw the influence line for the reactions at supports for the beam shown in the figure below. EI is constant.
A 5m
B
C 5m
D 5m
E 5m
F 5m
G 5m
69
Influence line for RD A 5m
B
C 5m
fBD
D 5m
fCD
E
F
5m
5m
fDD
fED
G 5m
fFD
fXD
1
f BD f DD
f CD f DD
f DD = 1 f ED f DD f DD
f FD f DD
fXD/fDD = Influence line for RD
1
70
• Use the consistency deformation method A
G 3@5 =15 m
3@5 =15 m
=
1 ∆´G
+
1
∆´G + fGGRG = 0
15
15 EI
------(1)
x RG
1
- Use conjugate beam for find ∆´G and fGG 30 Real beam Real beam G A
A 1
fGG
15 m
112.5 EI
1
30 m
15 m
2812.5 ∆'C = M 'C = EI
Conjugate beam 15 + (2/3)(15)
G
15 EI
1
450 EI
1 f 'GG = M ' 'G =
Conjugate beam
112.5/EI
20 m
9000 EI
450 71 EI
Substitute ∆´G and fGG in (1) :
2812.5 9000 + RG = 0 EI EI RG = −0.3125 kN , ↓
5.625 A
G 1
0.3125
=
0.6875
15 1
+
1 30
x RG = -0.3125 kN
1
1 72
• Use the conjugate beam for find fXD fCD 5.625 fBD A 3@5 =15 m
fDD
fED
fFD
G
Real beam
3@5 =15 m
23.01 1 0.3125 EI 5.625 x2 6.818 m EI G Conjugate beam A 8.182 m 28.13 − 4.688 35.16 2 x1 0.6875 x1 EI 15 . 98 EI EI 2 EI 5.625 x1 − 0.6875 x12 x2 EI EI V´2 5.625 (5.625-0.6875x1)/EI M´2 G EI 2 2 A x − x x 5 . 625 0 . 6875 x x2 0 . 3125 1 1 1 2 M´ 0.3125x2 = ( ) − ( ) 1 28.13 x1 EI 2 2 3 2 2 V´1 EI 0 . 3125 x 2 .6875 x1 2 x1 28 . 13 + ( ) + x2 = M ' x 2 2 EI 2 EI 3 EI
0.6875
x1 = 5 m -----> fBD = M´1= 56/EI
x2 = 5 m -----> fFD = M´2= 134.1/EI
x1 = 10 m -----> fCD = M´1= 166.7/EI
x2 = 10 m -----> fED = M´2= 229.1/EI
x1 = 15 m -----> fDD = M´1= 246.1/EI
x2 = 15 m -----> fDD = M´2= 246.1/EI
73
• Influence Line for RD 56 EI
166.7 EI
246.1 229.2 134.1 EI EI EI
fXD
1 246.1 166.7 229.2 134.1 56 246.1 246.1 246.1 246.1 246.1
fXD/fDD
1
0.228 0.677
1.0 0.931 0.545
Influence Line for RD
74
Influence line for RG A 5m
B
C 5m
D 5m
E 5m
F 5m
fEG fBG
5m
fFG
f CG f GG
fGG fXG
fCG
1 f EG f GG
f BG f GG
G
f FG f GG
f GG =1 f GG
fXG/fGG 1
75
• Use consistency deformations fXG 1
3@5 =15 m
+
∆´D
=
3@5 =15 m
1
fDD
X RD
1
30
A
∆´D + fDDRD = 0
- Use conjugate beam for find ∆´D and fDD 15 Real beam Real beam G A 30 m
1 30 EI
------(2)
1
1
450 EI
9000 EI
Conjugate beam 20 m
15 EI
15 m
1
G 15 m
112.5 EI Conjugate beam
450/EI
15 + (2/3)(15)
76
112.5 EI
1.5 EI
M´
15 m
V´
∆'D = M ' =
9000 EI
15 EI
112.5 EI M´´
450/EI
V´´
112.5 15 9000 450 2812.5 ( )+ − (15) = EI 3 EI EI EI
f DD = M ' ' =
112.5 2 1125 ( × 15) = EI 3 EI
2812.5 1125 + RD = 0, RD = −2.5kN = 2.5kN , ↓ Substitute ∆´D and fDD in (2) : EI EI 7.5
1.5
1
=
30
2.5
15
1
+
1
x RD = -2.5 kN
1
1
77
• Use the conjugate beam for find fXG 7.5 fBG
fCG
1.5 3@5 =15 m
Real beam
2.5
15 EI
112.5 EI
fGG = M´G = 1968.56/EI G
A − 7.5 EI 18.75 EI
10 m 15 + (10/3) = 18.33 m
Conjugate beam
168.75/EI
25 + (2/3)(5) = 28.33 m
18.75 2 62.5 ( × 5) = − EI 3 EI 18.75 EI 6.67 m
M´1 = f BG = −
A − 7.5 EI 18.75 EI
1
3@5 =15 m
75 EI
5m
fGG
fFG
fEG
V´1
A − 7.5 EI
M´2 = f CG = −
− 18.75 EI
V´2
18.75 125.06 (6.67) = − EI EI 78
x2 2 x
2
x3 x3 1968.56 168.75 ( )+ x3 =M´ − 2 3 EI EI
V´2
fGG = M´G = 1968.56/EI G
x
168.75/EI
x = 5 m -----> fFG = M´= 1145.64/EI x = 10 m -----> fEG = M´ = 447.73/EI 1968.56 1145 . 64 447.73 EI EI EI fXG − 62.5 − 125 1 EI EI 1145.64 1968.56 447.73 1968.56 1968.56 1968.56 fXG/fGG − 125 − 62.5 1 1968.56 1968.56 1.0 0.582 0.227 Influence line for RG 79 -0.032 -0.064
Using equilibrium condition for the influence line for Ay 1 x A
MA
B
5m
C 5m
D 5m
Ay
E
5m RD
F 5m
G 5m RG
+ ↑ ΣFy = 0 : RA = 1 − RD − RC
1
1 Unit load 0.228 0.678
1.0 0.929 0.542 Influence Line for RD 0.227 0.582
-0.032 -0.064 1.0
0.804 0.386 -0.156 -0.124
1.0 Influence line for RG
Influence line for Ay 80
Using equilibrium condition for the influence line for MA 1 x A
MA
B
5m
C 5m
D 5m
Ay
E
5m RD
F 5m
G 5m RG
+ ΣM A = 0 : 1x − 15RD − 30 RC
5
10
15
20
25
30 1x
0.228 0.678
1.0 0.929 0.542 0.227 0.582
-0.032 -0.064 2.54 1.75
RD
x 15
RG
x 30
1.0
Influence line for MA -0.745 -0.59
81