influence lines beams
DESCRIPTION
beam influence linesTRANSCRIPT
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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 Series
of Concentrated Loads! Absolute Maximum Shear and Moment
INFLUENCE LINES FOR STATICALLY DETERMINATE STRUCTURES
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Influence Line
Unit moving load
A
B
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Example 6-1
Construct the influence line fora) reaction at A and Bb) shear at point Cc) bending moment at point Cd) shear before and after support Be) moment at point B
of the beam in the figure below.
B
AC
4 m 4 m 4 m
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SOLUTION
Reaction at A
+ MB = 0: ,0)8(1)8( =+ xAy xAy 811=
1
4 m
8 m 12 m
-0.5
0.5
Ay
x
AC
4 m8 mAy By
1x
04812
x
10.50
-0.5
Ay
-
5
AC
4 mAy By
8 m
Reaction at B
+ MA = 0: ,01)8( = xBy xBy 81
=
1x
4 m 8 m 12 m
By
x
1.51
0.5
04812
x
00.51
1.5
By
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Shear at C
AC
4 mAy By
1x
4 m 4 m
40
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7
04-
4+
812
x VC0
-0.50.50
-0.5
xVC 81
=
xVC 811=
4 m 8 m 12 m
VC
x
-0.5
0.5
-0.5xVC 8
1=
xVC 811=
AC
4 mAy By
1x
4 m 4 m
40
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Bending moment at C
AC
4 mAy By
1x
4 m 4 m
40
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04
812
x MC02
0-2
xM C 214 =
xM C 21
=
4 m
8 m 12 m
MC
x
2
-2
AC
4 mAy By
1x
4 m 4 m
40
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Or using equilibrium conditions:
Reaction at A
+ MB = 0: ,0)8(1)8( =+ xAy xAy 811=
1
4 m
8 m 12 m
-0.5
0.5
Ay
x
AC
4 m8 mAy By
1x
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11
AC
4 mAy By
8 m
Reaction at B1
x
4 m 8 m 12 m
By
x
1.51
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
AC
4 mAy By
1x
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
AC
4 m 4 m 4 m
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Bending moment at C
AC
4 mAy By
1x
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
AC
4 m 4 m 4 m
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Shear before support B
AC
4 mAy By
1
4 m 4 m
x
1
4 m
8 m 12 m
-0.5
0.5Ay
x
-0.5
VB- = AyVB- = Ay-1
1
Ay8 m
VB-
MBx
Ay8 m
VB-
MB
VB-x
-1.0-0.5
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Shear after support B
AC
4 mAy By
1
4 m 4 m
x
1
4 m
8 m 12 m
-0.5
0.5Ay
x
VB+x
1
VB+ = 0
4 mVB+
MB1
VB+ = 1
4 mVB+
MB
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Moment at support B
AC
4 mAy By
1
4 m 4 m
x
1
4 m
8 m 12 m
-0.5
0.5Ay
x
-4
MBx
1
MB = 8Ay-(8-x) MB = 8Ay
1
Ay8 m
VB-
MBx
Ay8 m
VB-
MB
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P = 1
'x
Reaction
Influence Line for Beam
CA B
P = 1
Ay By
y = 1 'y LLs yB
1==
CA B
L
0)0()(1)1( ' =+ yyy BA
'yyA =
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y = 1'yCA B
P = 1
Ay ByLLs yA
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
Lb
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
x1
bL
-a
L
1
-1Slope at A = Slope at B
- 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
ah
A =
'y
bh
B =
1
A BMC MC
P = 1
Ay By
h
CA BP = 1
0)0()(1)()()0( ' =++++ yyBCACy BMMA
')( yBACM =+
'yCM =
1)( =+bh
ah
)(,1)(
baabh
abbah
+==
+
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a
L
A B
Hinge
MC MC
baC = A + B = 1
MC
x
aba+b
- Pinned Support
b
CA B
Lb
A =La
A =
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A B
a b
L
A B
MC MC
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
AC 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 1x
VBR 1
AC 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
AC 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/41
x
MF
2L/41
AC 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 at A, C and E- the shear at D- 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 DE
2 m 2 m 2 m 4 m
RE
REx
-2/6
2/61
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A B C D E
2 m 2 m 2 m 4 m
VD
VD
VDx
12/6
-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
VDx
2/6
-2/6
4/6
RE
VDMD
4 m
1 x
VD = -RE
RE
VDMD
4 m
REx
-2/6
12/6
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
VDMD
4 m
1 x
MD = 4RE
RE
VDMD
4 m
REx
-2/6
12/6
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
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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
VCLx
-1-1
VCL = RA
RA
1
VCL
MB
RA VCL
MB
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A
B
C
D
E
2 m 2 m 2 m 4 m
VCRx
VCR
VCR
10.333 0.667
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A B C D E
2 m 2 m 2 m 4 m
Or using equilibrium conditions:1
VCRx
0.3331
0.667
REx
-2/6 = -0.333
12/6=0.33
VCR = -RE RE
VCR
MC
VCR = 1 -RE RE
VCR
MC1
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A B C D E
2 m 2 m 2 m 4 m
MC MC
MCx
1
-2
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A B C D E
2 m 2 m 2 m 4 m
MCx
1
-2
Or using equilibrium conditions:
1
REx
-2/6 = -0.333
12/6=0.33
MC = 6RE RE
VCR
MC6 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 at A and C- shear at D, E and F- the moment at D, E and F
HingeA B CD E F
2 m 2 m 2 m 2 m2 m 2 m
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SOLUTION
RA
RA
x
1 1
-1
0.5
-0.5
2 m 2 m 2 m 2 m
AB CD E
2 m 2 m
F
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2 m 2 m 2 m 2 m
A B CD 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
xC = 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
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Example 6-4
Determine the maximum reaction at support B, the maximum shear at point C andthe maximum positive moment that can be developedat point C on 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
A BC
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0.5(12)(1.5) = 9
SOLUTION
RB
x
11.5
0.5
= 48000 N = 48 kN
(RB)max + (8000)(1.5)= (1000)(9) + (3000)(9)
8000 N
1000 N/m
3000 N/m
4 m 4 m 4 m
ABC
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0.5(4)(-0.5) = -10.5(4)(0.5) = 1
0.5(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
3000 N/m 3000 N/m 8000 N
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8000 N 3000 N/m
1000 N/m
(1/2)(4)(2) = 4
+(1/2)(8)(2) = 8
4 m 4 m 4 m
ABC
3000 N/m
(MC)max positive = (8000)(2)
= 44000 Nm = 44 kNm
+ (8-4)(1000)+ (3000)(8)
MC
x
2
-2