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Transcript of 06_InfluenceLineBeams[1]
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!Influence Lines for Beams
! Influence Lines for Floor Girders
! Influence Lines for Trusses
!
Maximum Influence at a Point Due to a Seriesof Concentrated Loads
! Absolute Maximum Shear and Moment
INFLUENCE LINES FOR STATICALLY
DETERMINATE STRUCTURES
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Influence Line
Unit moving load
AB
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Example 6-1
Construct the influence line for
a) reaction atA andB
b) shear at point C
c) bending moment at point C
d) shear before and after supportBe) moment at point B
of the beam in the figure below.
B
A
C
4 m 4 m 4 m
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SOLUTION
Reaction atA
+ MB= 0:,0)8(1)8( =+ xAy xAy
8
11=
1
4 m
8 m 12 m
-0.5
0.5
Ay
x
A
C
4 m8 mAy By
1
x
0
4
8
12
x
1
0.5
0
-0.5
Ay
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A
C
4 m
Ay By8 m
Reaction at B
+ MA= 0:,01)8( = xBy xBy
8
1=
1x
4 m 8 m 12 m
By
x
1.5
1
0.5
0
4
8
12
x
0
0.5
1
1.5
By
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Shear at C
A
C
4 mAy By
1
x
4 m 4 m
40
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0
4-
4+
812
x VC
0
-0.5
0.5
0-0.5
xVC8
1=
xVC8
11=
4 m 8 m 12 m
VC
x
-0.5
0.5
-0.5
xVC8
1=
xVC 8
11=
A
C
4 m
Ay By
1
x
4 m 4 m
40
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Bending moment at C
A
C
4 m
Ay By
1
x
4 m 4 m
40
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0
4
8
12
x MC
0
2
0
-2xMC 2
1
4=
xMC2
1=
4 m
8 m 12 m
MC
x
2
-2
A
C
4 m
Ay By
1
x
4 m 4 m
40
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Or using equilibrium conditions:
Reaction atA
+ MB= 0:,0)8(1)8( =+ xAy xAy
8
11=
1
4 m
8 m 12 m
-0.5
0.5
Ay
x
A
C
4 m8 mAy By
1
x
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A
C
4 m
Ay By
8 m
Reaction at B
1
x
4 m 8 m 12 m
By
x
1.5
1
0.5
1
4 m
8 m 12 m
-0.5
0.5
Ay
x
01=+ yy BA
yy AB =1
Fy = 0:+
yy AB =1
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Shear at C
A
C
4 mAy By
1
x
4 m 4 m
40
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yC AV =1= yC AV
1
4 m
8 m 12 m
-0.5
0.5
Ay
x
-0.5
VC
8 m 12 mx
4 m
0.5
-0.5
B
A
C
4 m 4 m 4 m
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Bending moment at C
A
C
4 m
Ay By
1
x
4 m 4 m
40
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yC AM 4=)4(4 xAM yC =
2
1
4 m
8 m 12 m
-0.5
0.5
Ay
x
4 m
8 m 12 m
MC
x
-2
B
A
C
4 m 4 m 4 m
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Shear before support B
AC
4 m
Ay By
1
4 m 4 m
x
1
4 m
8 m 12 m
-0.5
0.5
Ay
x
-0.5
VB-= AyVB
-= Ay-1
1
Ay
8 m
VB-
MBx
Ay
8 m
VB-
MB
VB-
x
-1.0-0.5
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Shear after support B
AC
4 m
Ay By
1
4 m 4 m
x
1
4 m
8 m 12 m
-0.5
0.5
Ay
x
VB+
x
1
VB+
= 0
4 mV
B
+
MB
1
VB+
= 1
4 mV
B
+
MB
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Moment at support B
AC
4 m
Ay By
1
4 m 4 m
x
1
4 m
8 m 12 m
-0.5
0.5
Ay
x
-4
MB
x1
MB = 8Ay-(8-x) MB = 8Ay
1
Ay
8 m
VB-
MBx
Ay
8 m
VB-
MB
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P = 1
'x
Reaction
Influence Line for Beam
CA B
P = 1
Ay By
y = 1 'y LLsy
B
1
==
CA B
L
0)0()(1)1( ' =+ yyy BA
'yyA =
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y = 1'yC
A B
P = 1
Ay ByLLs
y
A
1==
CA B
L
P = 1
'x
'yyB =
0)1()(1)0( ' =+ yyy BA
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L
AB
a
- Pinned Support
RA
RA
x
1
L
b
b
CA B
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A B
A B
a b
L
RA
RA
x
1 1
- Fixed Support
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ShearCA B
P = 1
a b
L
LsB
1=
y=1
yL
yR'y
A B
VC
VC
P = 1
Ay By
y=1
LsA
1=
0)0()(1)()()0( ' =+++ yyyRCyLCy BVVA
')( yyRyLCV =+
'yCV =
BA ssslopes =:
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L
A B
a
VC
VCVC
x
1bL
-a
L
1
-1
Slope atA = Slope atB
- Pinned Support
b
CA B
LsSlope B
1=
LsSlope A
1=
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A B
a b
L
A B
VB
VB
VB
x
1 1
- Fixed Support
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1=+= BA
Bending Moment
a b
L
a
hA =
'y
b
hB =
1
A B
MC MC
P = 1
Ay By
h
CA B
P = 1
0)0()(1)()()0( ' =++++ yyBCACy BMMA
')( yBACM =+
'yCM =
1)( =+b
h
a
h
)(
,1)(
ba
abh
ab
bah
+
==+
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a
L
A B
Hinge
MCMC
ba
C= A + B = 1
MC
x
ab
a+b
- Pinned Support
b
CA B
L
b
A
=
L
aA =
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A B
a b
L
A B
MCMC
MB
x1
-b
- Fixed Support
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General Shear
x
VBL
-1
x
VC
-1/4
3/41
x
VD
-2/4
2/4
1
-3/4
x
VE1/4
1
A
C D E B F G H
L
L/4 L/4 L/4 L/4 L/4 L/4 L/4
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x
VBL
-1
x
VG 1
x
VF 1
x
VBR 1
A
C D E B F G H
L
L/4 L/4 L/4 L/4 L/4 L/4 L/4
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General Bending Moment
A= 3/4 B= 1/4
= sA+ sB= 1
A= 1/2 B= 1/2
= sA+ sB= 1
A= 1/4 B= 3/4
= A+ B= 1
x
MC 3L/16
x
MD 4L/16
x
ME 3L/16
A
C D E B F G H
L
L/4 L/4 L/4 L/4 L/4 L/4 L/4
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x
MB
3L/4
1
x
MG
L/4
1
x
MF
2L/4
1
A
C D E B F G H
L
L/4 L/4 L/4 L/4 L/4 L/4 L/4
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Example 6-2
Construct the influence line for
- the reaction atA, Cand E
- the shear atD
- the moment at D
- shear before and after support C- moment at point C
A B C D E
2 m 2 m 2 m 4 m
Hinge
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AC D E
2 m 2 m 2 m 4 m
B
RA
x
1
RA
SOLUTION
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AC D E
2 m 2 m 2 m 4 m
B
RC
RC
x
18/6
4/6
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A B C D
E
2 m 2 m 2 m 4 m
RE
RE
x
-2/6
2/61
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A B C D E
2 m 2 m 2 m 4 m
VD
VD
VD
x
1
2/6
-1
1
sE = sC
sE= 1/6sC= 1/6
=
-2/6
=
4/6
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Or using equilibrium conditions:
VD = 1 -RE
1
VD
x
2/6
-2/6
4/6
RE
VDMD4 m
1x
VD = -RE
RE
VDMD 4 m
RE
x
-2/6
1
2/6
A B C D E
2 m 2 m 2 m 4 m
Hinge
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A B C
D
E
2 m 2 m 2 m 4 m
4
-1.33
MD
x
(2)(4)/6 = 1.33
D = C+E= 1
C= 4/6
2
2/6 = E
MD MD
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Or using equilibrium conditions:
MD = -(4-x)+4RE
1
RE
VDMD4 m
1x
MD = 4RE
RE
VDMD 4 m
RE
x
-2/6
1
2/6
MD
x
-8/6
8/6
A B C D E
2 m 2 m 2 m 4 m
Hinge
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A
B
C
D
E
2 m 2 m 2 m 4 m
VCL
VCL
VCLx
-1-1
O i ilib i diti
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A B C D E
2 m 2 m 2 m 4 m
Or using equilibrium conditions:
1
RA
x
1
VCL= RA - 1
VCL
x
-1-1
VCL= RA
RA
1
VCL
MB
RA VCL
MB
V
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A
B
C
D
E
2 m 2 m 2 m 4 m
VCR
x
VCR
VCR
1
0.3330.667
Or using equilibrium conditions:
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A B C D E
2 m 2 m 2 m 4 m
Or using equilibrium conditions:
1
VCR
x
0.333
10.667
RE
x
-2/6 = -0.333
1
2/6=0.33
VCR= -RE
RE
VCR
MC
VCR= 1 -RE
RE
VCR
MC
1
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A B C D E
2 m 2 m 2 m 4 m
MC MC
MC
x1
-2
Or using equilibrium conditions:
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A B C D E
2 m 2 m 2 m 4 m
MC
x1
-2
Or using equilibrium conditions:
1
RE
x
-2/6 = -0.333
1
2/6=0.33
MC= 6RE
RE
VCR
MC
6 m
'6 xRM AC =
6 m
'x
RE
VCR
MC
1
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Example 6-3
Construct the influence line for
- the reaction atA and C
- shear atD, EandF
- the moment at D, EandF
HingeA B CD E F
2 m 2 m 2 m 2 m2 m 2 m
SOLUTION
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SOLUTION
RA
RA
x
1 1
-1
0.5
-0.5
2 m 2 m 2 m 2 m
A
B CD E
2 m 2 m
F
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2 m 2 m 2 m 2 m
A BCD E
2 m 2 m
F
RC x
RC
10.5
1.52
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2 m 2 m 2 m 2 m
A B CD E
2 m 2 m
F
VD
VD
VD
x
-1
0.5
-0.5
1 1=
=
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A B CD E
2 m 2 m 2 m 2 m2 m 2 m
F
VE
VE
VE
x1
-0.5-1
0.5
-0.5
=
=
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2 m 2 m 2 m 2 m
A B CD E
2 m 2 m
F
VF
VF
VF
x
1 =
=
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2 m 2 m 2 m 2 m
A B CD E
2 m 2 m
F
MD MD
MD
x
-2
D = 1
-1
1
2
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2 m 2 m 2 m 2 m
A B CD E
2 m 2 m
F
E= 1
ME ME
ME
x
C= 0.5B = 0.5
(2)(2)/4 = 1
-2
-1
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2 m 2 m 2 m 2 m
A B CD E
2 m 2 m
FME ME
MF
xF= 1
-2
Example 6 4
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Example 6-4
Determine the maximum reaction at supportB, the maximum shear at point Cand
the maximum positive moment that can be developed
at point Con the beam shown due to
- a single concentrate live load of 8000 N
- a uniform live load of 3000 N/m
- a beam weight (dead load) of 1000 N/m
4 m 4 m 4 m
AB
C
SOLUTION 8000 N3000 N/m
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0.5(12)(1.5) = 9
RB
x
1
1.5
0.5
= 48000 N = 48 kN
(RB)max + (8000)(1.5)= (1000)(9) + (3000)(9)
1000 N/m
4 m 4 m 4 m
ABC
3000 N/m 3000 N/m8000 N
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0.5(4)(-0.5) = -10.5(4)(0.5) = 10.5(4)(-0.5) = -1
= (1000)(-2+1)
4 m 4 m 4 m
ABC
VC
x
0.5
-0.5
1000 N/m
(VC)max + (8000)(-0.5)
= -11000 N = 11 kN
+ (3000)(-2)
-0.5
8000 N3000 N/m3000 N/m
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1000 N/m
(1/2)(4)(2) = 4
+(1/2)(8)(2) = 8
4 m 4 m 4 m
ABC
(MC)max positive = (8000)(2)
= 44000 Nm = 44 kNm
+ (8-4)(1000)+ (3000)(8)
MC
x
2
-2
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