Transcript of "Influence Lines of Determinate Structures "
1.
INFLUENCE LINES FOR
STATICALLY DETERMINATE
STRUCTURES
1
2.
INFLUENCE LINES FOR STATICALLY DETERMINATE
STRUCTURES - AN OVERVIEW
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•
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Introduction - What is an influence line?
Influence lines for beams
Qualitative influence lines - Muller-Breslau Principle
Influence lines for floor girders
Influence lines for trusses
Live loads for bridges
Maximum influence at a point due to a series of
concentrated loads
• Absolute maximum shear and moment
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3.
Definition of Influence Lines
QUESTION
Where should a structural engineer place the live load on
the structure so that it creates the greates influence at a
specified point?
INFLUENCE LINES
Variation of Reaction, Shear, Moment or Deflection
at a SPECIFIC POINT
due to a concentrated force moving on member
4.
LIVE LOADS
Common Structures
Bridges
Industrial Crane
Rails
Conveyors
Floor Girders
any other structure where
loads move across its span
5.
ANALYSIS OF FORCES DUE TO
LIVE LOADS
LIVE or MOVING LOADS
•
•
Variable point of Application
Temporary Loads
P
V1
P
Shear Diagrams
V1 V2
V2
10.
Influence Lines for Beams
Concentrated Loads
Distributed Loads
11.
3.3
MOVING CONCENTRATED LOAD
3.3.1 Variation of Reactions RA and RB as functions of load position
1
x
B
A
C
10 ft
3 ft
MA =0
(RB)(10) – (1)(x) = 0
RB = x/10
RA = 1-RB
= 1-x/10
1
x
A
C
RA=1-x/10
B
RB = x/10
x
A
RA=1-x/10
C
RB = x/10
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12.
RA occurs only at A; RB occurs only at B
1
Influence
line for RA
1-x/10
x
Influence line
for RB
10-x
1.0
x/10
x
10-x
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13.
3.3.2 Variation of Shear Force at C as a function of load position
0 < x < 3 ft (unit load to the left of C)
x
1.0
C
A
B
3 ft
RB = x/10
RA = 1-x/10
10 ft
C
x/10
Shear force at C is –ve, VC =-x/10
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14.
3 < x < 10 ft (unit load to the right of C)
x
B
C
A
3 ft
RB = x/10
RA = 1-x/10
C
RA = 1-x/10
Shear force at C is +ve = 1-x/10
1
0.7
-ve
Influence line for shear at C
+ve
0.3
1
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15.
3.3.3 Variation of Bending Moment at C as a function of load position
0 < x < 3.0 ft (Unit load to the left of C)
x
C
A
B
3 ft
RA = x/10
RA = 1-x/10
10 ft
x/10
(x/10)(7)
x/10
(x/10)(7)
C
x/10
x/10
Bending moment is +ve at C
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16.
3 < x < 10 ft (Unit load to the right of C)
1
x ft
B
C
A
3 ft
RA = x/10
RA = 1-x/10
10 ft
1-x/10
(1-x/10)(3)
1-x/10
(1-x/10)(3)
1-x/10
C
(1-x/10)(3)
Moment at C is +ve
(1-7/10)(3)=2.1 kip-ft
Influence line for bending
Moment at C
+ve
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17.
PROBLEM
• Beam mass=24kg/m
• Dolly can travel entire
length of beam CB
• A is pin and B is roller
• Determine Ra and Rb
and MD Max
• Ra=4.63, Rb=3.31 and
MD Max=2.46
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19.
QUALITATIVE INFLUENCED LINES - MULLER-BRESLAU’S
PRINCIPLE
• The principle gives only a procedure to determine of the influence line of a
parameter for a determinate or an indeterminate structure
• But using the basic understanding of the influence lines, the magnitudes
of the influence lines also can be computed
• In order to draw the shape of the influence lines properly, the capacity of the
beam to resist the parameter investigated (reaction, bending moment, shear
force, etc.), at that point, must be removed
• The principle states that:The influence line for a parameter (say, reaction, shear
or bending moment), at a point, is to the same scale as the deflected shape of
the beam, when the beam is acted upon by that parameter.
– The capacity of the beam to resist that parameter, at that point, must be
removed.
– Then allow the beam to deflect under that parameter
– Positive directions of the forces are the same as before
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20.
PROBLEM
• Find MD Max
when
• Conc moving
load=4000 lbs
• UD Moving
load=300 lbs/ft
• Beam wt=200
lbs/ft
• MD Max=22.5 k-ft
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27.
SHEAR AND MOMENT ENVELOPES DUE TO
UNIFORM DEAD AND LIVE LOADS
•
•
The shear and moment envelopes are graphs which show the variation in
the minimum and maximum values for the function along the structure due
to the application of all possible loading conditions. The diagrams are
obtained by superimposing the individual diagrams for the function based
on each loading condition. The resulting diagram that shows the upper and
lower bounds for the function along the structure due to the loading
conditions is called the envelope.
The loading conditions, also referred to as load cases, are determined by
examining the influence lines and interpreting where loads must be placed
to result in the maximum values. To calculate the maximum positive and
negative values of a function, the dead load must be applied over the entire
beam, while the live load is placed over either the respective positive or
negative portions of the influence line. The value for the function will be
equal to the magnitude of the uniform load, multiplied by the area under the
influence line diagram between the beginning and ending points of the
uniform load.
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28.
• For example, to develop the shear and moment
envelopes for the beam shown, first sketch the
influence lines for the shear and moment at various
locations. The influence lines for Va-R, Vb-L, Vb-R, Mb,
Vs, and Ms are shown
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29.
These influence lines are used to determine where to place the
uniform live load to yield the maximum positive and negative
values for the different functions. For example;
Support removed, unit load applied, and resulting influence
line for support reaction at A
· The maximum value for the positive reaction at A, assuming no
partial loading, will occur when the uniform load is applied on the
beam from A to B (load case 1)
Load case 1
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30.
· The maximum negative value for the reaction at A will occur if a
uniform load is placed on the beam from B to C (load case 2)
· Load case 1 is also used for:
maximum positive value of the shear at the right of
support A
maximum positive moment Ms
· Load case 2 is also used for:
maximum positive value of the shear at the right of
support B
maximum negative moments at support B and Ms
· Load case 3 is required for:
maximum positive reaction at B
60
maximum negative shear on the left side of B
31.
Load case 3
· Load case 4 is required for the maximum positive shear force at
section s
Load case 4
· Load case 5 is required for the maximum negative shear force at
section s
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32.
Load case 5
To develop the shear and moment envelopes,
construct the shear and moment diagrams for each
load case. The envelope is the area that is enclosed
by superimposing all of these diagrams. The
maximum positive and negative values can then be
determined by looking at the maximum and
minimum values of the envelope at each point.
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33.
Individual shear diagrams for each load case;
63
34.
Superimpose all of these diagrams together to determine the final shear
envelope.
Resulting superimposed shear envelope
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35.
Individual moment diagrams for each load case;
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36.
Superimpose all of these diagrams together to determine the final
moment envelope.
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42.
INFLUENCE LINES FOR FLOOR GIRDERS
Force Equilibrium Method:
Draw the Influence Lines for: (a) Shear in panel CD of
the girder; and (b) the moment at E.
x
A´
B´
C´
D´
E´
F´
A
B
C
D
E
5 spaces @ 10´ each = 50 ft
F
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43.
Place load over region A´B´ (0 < x < 10 ft)
Find the shear over panel CD
VCD= - x/50
At x=0, VCD = 0
At x=10, VCD = -0.2
C
F
D
Shear is -ve
Find moment at E = +(x/50)(10)=+x/5
At x=0, ME=0
At x=10, ME=+2.0
E
+ve moment
RF=x/50
F
RF=x/50
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44.
Continuation of the Problem
x
-ve
0.2
I. L. for VCD
+ve
2.0
I. L. for ME
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45.
Place load over region B´C´ (10 ft < x < 20ft)
VCD = -x/50 kip
At x = 10 ft
VCD = -0.2
At x = 20 ft
VCD = -0.4
ME = +(x/50)(10)
= +x/5 kip.ft
At x = 10 ft, ME = +2.0 kip.ft
At x = 20 ft, ME = +4.0 kip.ft
C
D
Shear is -ve
E
D
Moment is +ve
F
RF = x/50
F
RF = x/50
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46.
x
B´
C´
-ve
0.2
0.4
I. L. for VCD
+ve 4.0
2.0
I. L. for ME
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47.
Place load over region C´D´ (20 ft < x < 30 ft)
When the load is at C’ (x = 20 ft)
C
D
Shear is -ve
RF=20/50
=0.4
VCD = -0.4 kip
When the load is at D´ (x = 30 ft)
A
B
C
RA= (50 - x)/50
D
Shear is +ve
VCD= + 20/50
= + 0.4 kip
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48.
ME = + (x/50)(10) = + x/5
x
Load P
E
+ve moment
A´
B´
C´
A
B
D´
C
0.2
RF= x/50
+ve
-ve 0.4
D
I. L. for VCD
+ve
2.0
4.0
6.0
I. L. for ME
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49.
Place load over region D´E´ (30 ft < x < 40 ft)
VCD= + (1-x/50) kip
A
B
RA= (1-x/50)
C
D
E
Shear is +ve
ME= +(x/50)(10)
= + x/5 kip.ft
E
Moment is +ve
RF= x/50
At x = 30 ft, ME = +6.0
At x = 40 ft, ME = +8.0
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50.
Problem continued
x
A´
B´
C´
D´
0.4
E´
+ve
0.2
I. L. for VCD
2.0
4.0
+ve
8.0
6.0
I. L. for ME
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51.
Place load over region E´F´ (40 ft < x < 50 ft)
VCD = + 1-x/50 At x = 40 ft, VCD= + 0.2
At x = 50 ft, VCD = 0.0
x
1.0
A
B
C
D
E
RA= 1-x/50
Shear is +ve
ME= + (1-x/50)(40) = (50-x)*40/50 = +(4/5)(50-x)
x
A
B
RA=1-x/50
C
D
E
F
At x = 40 ft, ME= + 8.0 kip.ft
At x = 50 ft, ME = 0.0
Moment is +ve
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52.
x
A´
B´
1.0
C´
D´
E´
0.4
0.2
-ve
+ve
F´
0.2
0.4
I. L. for VCD
2.0
4.0
6.0
+ve
8.0
I. L. for ME
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61.
INFLUENCE LINES FOR TRUSSES
Draw the influence lines for: (a) Force in Member GF; and
(b) Force in member FC of the truss shown below in Figure below
G
F
E
20 ft
10(3)1/3
600
A
B
20 ft
D
C
20 ft
20 ft
100
62.
Place unit load over AB
(i) To compute GF, cut section (1) - (1)
(1)
G
1-x/20
1
F
E
x/20
x
600
A
B
RA= 1- x/60
At x = 0,
FGF = 0
At x = 20 ft
FGF = - 0.77
C
(1)
D
RD=x/60
Taking moment about B to its right,
(RD)(40) - (FGF)(103) = 0
FGF = (x/60)(40)(1/ 103) = x/(15 3) (-ve)
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63.
(ii) To compute FFC, cut section (2) - (2)
G
x
(2)
F
E
1
x/20
1-x/20
reactions at nodes
A
300
600
B
RA =1-x/60
D
C
(2)
RD=x/60
Resolving vertically over the right hand section
FFC cos300 - RD = 0
FFC = RD/cos30 = (x/60)(2/3) = x/(30 3) (-ve)
102
64.
At x = 0, FFC = 0.0
At x = 20 ft, FFC = -0.385
20 ft
I. L. for FGF
-ve
0.77
-ve
0.385
I. L. for FFC
103
65.
Place unit load over BC (20 ft < x <40 ft)
[Section (1) - (1) is valid for 20 < x < 40 ft]
(i) To compute FGF use section (1) -(1)
G
(1)
F
E
(40-x)/20
x
(x-20)/20
1
reactions at nodes
A
20 ft
RA=1-x/60
B
C
(1)
(x-20)
(40-x)
D
RD=x/60
Taking moment about B, to its left,
(RA)(20) - (FGF)(103) = 0
FGF = (20RA)/(103) = (1-x/60)(2 /3)
At x = 20 ft, FFG = 0.77 (-ve)
At x = 40 ft, FFG = 0.385 (-ve)
104
66.
To compute FFC, use section (2) - (2)
Section (2) - (2) is valid for 20 < x < 40 ft
G
(2)
F
(40-x)/20
(x-20)/20
1
x
E
FFC
300
600
A
B
RA =1-x/60
C
(2)
D
RD=x/60
Resolving force vertically, over the right hand section,
FFC cos30 - (x/60) +(x-20)/20 = 0
FFC cos30 = x/60 - x/20 +1= (1-2x)/60 (-ve)
FFC = ((60 - 2x)/60)(2/3) -ve
105
67.
At x = 20 ft, FFC = (20/60)(2/ 3) = 0.385 (-ve)
At x = 40 ft, FFC = ((60-80)/60)(2/ 3) = 0.385 (+ve)
-ve
0.385
0.77
I. L. for FGF
0.385
-ve
I. L. for FFC
106
68.
Place unit load over CD (40 ft < x <60 ft)
(i) To compute FGF, use section (1) - (1)
G
(1)
F
E
(x-40) 1
x
(60-x)/20
A
20 ft
RA=1-x/60
B
(1)
C
reactions at nodes
(60-x)
(x-40)/20
D
RD=x/60
Take moment about B, to its left,
(FFG)(103) - (RA)(20) = 0
FFG = (1-x/60)(20/103) = (1-x/60)(2/3) -ve
At x = 40 ft, FFG = 0.385 kip (-ve)
At x = 60 ft, FFG = 0.0
107
69.
To compute FFG, use section (2) - (2)
reactions at nodes
G
F
E
(60-x)/20
(x-40)/20
FFC
x
1
300
600
A
B
RA =1-x/60
C
x-40
60-x
(2)
D
RD=x/60
Resolving forces vertically, to the left of C,
(RA) - FFC cos 30 = 0
FFC = RA/cos 30 = (1-x/10) (2/3) +ve
108
70.
At x = 40 ft, FFC = 0.385 (+ve)
At x = 60 ft, FFC = 0.0
-ve
0.385
0.770
I. L. for FGF
+ve
-ve
0.385
I. L. for FFC
109
76.
Live Loads for Highway BRIDGES
•Caused by Traffic
•Heaviest: Series of Trucks
Specifications: AASHTO
American Association of State and
Highway Transportation Officials
Load Desigantions
15 Tons (10-20)
H 15-44
2 Axle Truck
20 Tons
HS 20-44
Year of
Specifications
2 Axle Truck &
1 Axle Semitrailer
Year of
Specifications
Simplification: Uniform Load plus a Concnetrated force placed at critical
locations
77.
Live Loads for Railroad BRIDGES
Specifications: AREA
American Railroad Engineers
Associations
Load Desigantions
E-72
Devised by
Theodore Cooper
1894
Loading on Driving
Axle
M-72
Devised by D.B.
Steinman
Loading on Driving
Axle
78.
Maximum Influence due to a Series of
Concentrated Loads - Shear
79.
Maximum Influence due to a Series of
Concentrated Loads - Shear
80.
Maximum Influence due to a Series of
Concentrated Loads - Shear
81.
Maximum Influence due to a Series of
Concentrated Loads - Shear
V12 5.375 5.25 0.125
V23 2.5 5.375 2.875
82.
Maximum Influence due to a Series of
Concentrated Loads - Shear
83.
Maximum Influence due to a Series of
Concentrated Loads - Moment
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