influence lines for statically determinate · pdf file1! influence lines for beams! influence...
TRANSCRIPT
1
! 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
2
Influence Line
Unit moving load
A
B
3
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
4
SOLUTION
� Reaction at A
+ ΣMB = 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
+ ΣMA = 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
6
� Shear at C
AC
4 mAy By
1x
4 m 4 m
40 <≤ x 124 ≤< x
01811 =−−− CVx
xVC 81
−=
ΣFy = 0:+
0811 =−− CVx
xVC 811−=
ΣFy = 0:+
VC
MC
VC
MC
xAy 811−=
ACx
4 m
124 ≤< x
AC
1x
4 m
40 ≤≤ x
xAy 811−=
7
04-
4+
812
x VC
0-0.50.50
-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 <≤ x 124 ≤< x
8
� Bending moment at C
AC
4 mAy By
1x
4 m 4 m
40 <≤ x 124 ≤< x
VC
MC0)4)(
811()4(1 =−−−+ xxM C
xM C 21
=
VC
MC0)4)(
811( =−− xM C
+ ΣMC = 0:
+ ΣMC = 0:xAy 8
11−=
ACx
4 m
124 ≤< x
AC
1x
4 m
40 ≤≤ x
xAy 811−=
xM C 214 −=
9
04
812
x MC
02
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 <≤ x 124 ≤< x
xM C 21
=xM C 2
14 −=
10
Or using equilibrium conditions:
� Reaction at A
+ ΣMB = 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
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
ΣFy = 0:+
yy AB −= 1
12
� Shear at C
AC
4 mAy By
1x
4 m 4 m
40 <≤ x 124 ≤< x
01 =−− Cy VA
1−= yC AV
ΣFy = 0:+
0=− Cy VA
yC AV =
ΣFy = 0:+
VC
MC
xAy 811−=
ACx
4 m
124 ≤< x
VC
MCA
C
1x
4 m
40 ≤≤ x
xAy 811−=
13
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
14
� Bending moment at C
AC
4 mAy By
1x
4 m 4 m
40 <≤ x 124 ≤< x
0)4(1)4( =+−+ Cy MxA
)4(4 xAM yC −−=
+ ΣMC = 0:
VC
MC
xAy 811−=
ACx
4 m
124 ≤< x
VC
MCA
C
1x
4 m
40 ≤≤ x
xAy 811−=
0)4( =+− Cy MA+ ΣMC = 0:
yC AM 4=
15
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
16
� 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
Ay
8 mVB
-
MBx
Ay
8 mVB
-
MB
VB-
x
-1.0-0.5
17
� 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
+
MB
1
VB+
= 1
4 mVB
+
MB
18
� 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
MB
x1
MB = 8Ay-(8-x) MB = 8Ay
1
Ay
8 mVB
-
MBx
Ay
8 mVB
-
MB
19
P = 1
'x
� Reaction
Influence Line for Beam
CA B
P = 1
Ay By
δy = 1 'yδ LLs y
B1
==δ
CA B
L
0)0()(1)1( ' =+− yyy BA δ
'yyA δ=
20
δy = 1'yδC
A B
P = 1
Ay ByLLs y
A1
==δ
CA B
L
P = 1
'x
'yyB δ=
0)1()(1)0( ' =+− yyy BA δ
21
L
AB
a
- Pinned Support
RA
RA
x
1
Lb
b
CA B
22
A B
A B
a b
L
RA
RA
x
1 1
- Fixed Support
23
� ShearCA B
P = 1
a b
L
LsB
1=
δy=1
δyL
δyR'yδ
A B
VC
VC
P = 1
Ay By
δy=1
LsA
1=
0)0()(1)()()0( ' =+−++ yyyRCyLCy BVVA δδδ
')( yyRyLCV δδδ =+
'yCV δ=
BA ssslopes =:
24
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=
25
A B
a b
L
A B
VB
VB
VB
x
1 1
- Fixed Support
26
1=+= BA θθφ
� Bending Moment
a b
L
ah
A =θ
'yδ
bh
B =θ
1
A BMC
MC
P = 1
Ay By
h
CA BP = 1
0)0()(1)()()0( ' =++++ yyBCACy BMMA δθθ
')( yBACM δθθ =+
'yCM δ=
1)( =+bh
ah
)(,1)(
baabh
abbah
+==
+
27
a
L
A B
Hinge
MCMC
baφC = θA + θB = 1
MC
x
aba+b
- Pinned Support
b
CA B
Lb
A =θLa
A =θ
28
A B
a b
L
A B
MCMC
MB
x1
-b
- Fixed Support
29
� 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
30
x
VBL
-1
x
VG 1
x
VF 1x
VBR1
AC D E B F G H
L
L/4 L/4 L/4 L/4 L/4 L/4 L/4
31
� General Bending Moment
θA = 3/4 θB = 1/4
φ = sA + sB = 1
θA = 1/2 θB = 1/2
φ = sA + sB = 1
θA = 1/4 θB = 3/4
φ = θA + θB = 1
x
MC 3L/16
x
MD4L/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
32
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
33
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
34
AC D E
2 m 2 m 2 m 4 m
B
RA
x
1
RA
SOLUTION
35
AC D E
2 m 2 m 2 m 4 m
B
RC
RC
x
18/6
4/6
36
A B C DE
2 m 2 m 2 m 4 m
RE
RE
x
-2/6
2/61
37
A B C D E
2 m 2 m 2 m 4 m
VD
VD
VD
x1
2/6
-1
1
� sE = sC
sE = 1/6sC = 1/6
=
-2/6
=
4/6
38
Or using equilibrium conditions:
VD = 1 -RE
1
VD
x2/6
-2/6
4/6
RE
VDMD
4 m
1 x
VD = -RE
RE
VDMD
4 m
RE
x
-2/6
12/6
A B C D E
2 m 2 m 2 m 4 m
Hinge
39
A B C
D
E
2 m 2 m 2 m 4 m
4
-1.33
MD
x
(2)(4)/6 = 1.33
φD = θC+θE = 1
θC = 4/6
2
2/6 = θE
MD MD
40
Or using equilibrium conditions:
MD = -(4-x)+4RE
1
RE
VDMD
4 m
1 x
MD = 4RE
RE
VDMD
4 m
RE
x
-2/6
12/6
MD
x
-8/6
8/6
A B C D E
2 m 2 m 2 m 4 m
Hinge
41
A
B
C
D
E
2 m 2 m 2 m 4 m
VCL
VCL
VCL
x
-1-1
42
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
43
A
B
C
D
E
2 m 2 m 2 m 4 m
VCR
x
VCR
VCR
10.333 0.667
44
A B C D E
2 m 2 m 2 m 4 m
Or using equilibrium conditions:1
VCR
x
0.3331
0.667
RE
x
-2/6 = -0.333
12/6=0.33
VCR = -RE
RE
VCR
MC
VCR = 1 -RE
RE
VCR
MC
1
45
A B C D E
2 m 2 m 2 m 4 m
MC MC
MC
x1
-2
46
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
12/6=0.33
MC = 6RE
RE
VCR
MC
6 m
'6 xRM AC −=
6 m
'x
RE
VCR
MC
1
47
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
48
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
49
2 m 2 m 2 m 2 m
A B CD E
2 m 2 m
F
RC x
RC
10.5
1.52
50
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=
=
51
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
=
=
52
2 m 2 m 2 m 2 m
A B CD E
2 m 2 m
F
VF
VF
VF
x
1 =
=
53
2 m 2 m 2 m 2 m
A B CD E
2 m 2 m
F
MDMD
MD
x
-2
θD = 1-1
1
2
54
2 m 2 m 2 m 2 m
A B CD E
2 m 2 m
F
φE = 1
ME ME
ME
xθC = 0.5θB = 0.5
(2)(2)/4 = 1
-2-1
55
2 m 2 m 2 m 2 m
A B CD E
2 m 2 m
FME ME
MF
xθF = 1
-2
56
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
57
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
58
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
59
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 N�m = 44 kN�m
+ (8-4)(1000)+ (3000)(8)
MC
x
2
-2