influence lines of determinate structures
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INFLUENCE LINES FOR STATICALLY DETERMINATE
STRUCTURES
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INFLUENCE LINES FOR STATICALLY DETERMINATE STRUCTURES - AN OVERVIEW
• 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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Definition of Influence Lines
INFLUENCE LINES
Variation of Reaction, Shear, Moment or Deflection
at a SPECIFIC POINT
due to a concentrated force moving on member
QUESTION
Where should a structural engineer place the live load on
the structure so that it creates the greates influence at a
specified point?
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LIVE LOADS
Common Structures
Bridges
Industrial Crane
Rails
Conveyors
Floor Girders any other structure where
loads move across its span
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ANALYSIS OF FORCES DUE TO LIVE LOADS
LIVE or MOVING LOADS
• Variable point of Application
• Temporary Loads
Shear Diagrams
P P
V1 V2
21 VV
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Influence Lines for Beams
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B
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Example
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Example
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Influence Lines for Beams
Concentrated Loads Distributed Loads
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3.3 MOVING CONCENTRATED LOAD
3.3.1 Variation of Reactions RA and RB as functions of load position
MA =0
(RB)(10) – (1)(x) = 0
RB = x/10
RA = 1-RB
= 1-x/10
x1
A B
C10 ft
3 ft
x1
A BC
RA=1-x/10 RB = x/10
x
AC
RA=1-x/10 RB = x/10
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RA occurs only at A; RB occurs only at B
Influence line
for RB
1-x/10
1
Influence
line for RA
x 10-x
x 10-x
x/10 1.0
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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)
Shear force at C is –ve, VC =-x/10
C
x 1.0
RA = 1-x/10 RB = x/10
3 ft
10 ft
A B
x/10
C
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3 < x < 10 ft (unit load to the right of C)
Shear force at C is +ve = 1-x/10
Influence line for shear at C
C
x
3 ft
A
RA = 1-x/10 RB = x/10
B
C
1
1
-ve
+ve
0.3
0.7
RA = 1-x/10
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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)
Bending moment is +ve at C
C
x
3 ft
A B
RA = 1-x/10RA = x/10
10 ft
C
x/10
x/10
x/10
(x/10)(7)(x/10)(7)
x/10
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3 < x < 10 ft (Unit load to the right of C)
Moment at C is +ve
Influence line for bending
Moment at C
C
x ft
3 ft
A
RA = 1-x/10
10 ft
C
1-x/10
1-x/10
(1-x/10)(3)
(1-x/10)(3)
1
RA = x/10
B
1-x/10(1-x/10)(3)
+ve
(1-7/10)(3)=2.1 kip-ft
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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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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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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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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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• 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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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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· 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
maximum negative shear on the left side of B
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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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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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Individual shear diagrams for each load case;
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Superimpose all of these diagrams together to determine the final shear
envelope.
Resulting superimposed shear envelope
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Individual moment diagrams for each load case;
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Superimpose all of these diagrams together to determine the final
moment envelope.
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Influence Lines for Floor Girders
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INFLUENCE LINE FOR FLOOR GIRDERS Floor systems are constructed as shown in figure below,
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INFLUENCE LINES FOR FLOOR GIRDERS (Cont’d)
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IL FOR SHEAR IN PANEL CD OF FLOOR GIRDER
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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.
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A C D E F B
B´ A´ D´ C´ E´ F´
x
5 spaces @ 10´ each = 50 ft
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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
Find moment at E = +(x/50)(10)=+x/5
At x=0, ME=0
At x=10, ME=+2.0
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D C
Shear is -ve RF=x/50
F
F
RF=x/50
E
+ve moment
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Continuation of the Problem
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-ve 0.2
2.0 +ve
x
I. L. for VCD
I. L. for ME
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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
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D F C
Shear is -ve RF = x/50
D F
RF = x/50
E
Moment is +ve
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0.4 0.2 -ve
x
B´ C´
I. L. for VCD
+ve 4.0
2.0
I. L. for ME
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Place load over region C´D´ (20 ft < x < 30 ft)
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When the load is at C’ (x = 20 ft)
C D
RF=20/50
=0.4
Shear is -ve
VCD = -0.4 kip
When the load is at D´ (x = 30 ft)
A
RA= (50 - x)/50
B C D Shear is +ve
VCD= + 20/50
= + 0.4 kip
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ME = + (x/50)(10) = + x/5
E
RF= x/50 +ve moment
-ve
A B C
D
A´ B´ C´ D´
x
+ve
0.4 0.2
I. L. for VCD
+ve
2.0 4.0 6.0
I. L. for ME
Load P
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Place load over region D´E´ (30 ft < x < 40 ft)
A B C D
E
RA= (1-x/50) Shear is +ve
VCD= + (1-x/50) kip
RF= x/50 Moment is +ve
E
ME= +(x/50)(10)
= + x/5 kip.ft
At x = 30 ft, ME = +6.0
At x = 40 ft, ME = +8.0
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A´ B´ C´ D´ E´
x
0.4 0.2 +ve
+ve 8.0
6.0 4.0 2.0
I. L. for VCD
I. L. for ME
Problem continued
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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)
B C D E F A
x
RA=1-x/50 At x = 40 ft, ME= + 8.0 kip.ft
At x = 50 ft, ME = 0.0
Moment is +ve
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A´ B´ C´ D´ E´ F´
x 1.0
0.2 0.4
0.4 0.2
2.0 4.0
6.0 8.0
I. L. for VCD
I. L. for ME
-ve
+ve
+ve
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Draw IL for moment at point F for the floor girder
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Influence Lines for Trusses
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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
20 ft 20 ft 20 ft
F
B C D
G
A
E
600
20 ft
10(3)1/3
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Place unit load over AB
(i) To compute GF, cut section (1) - (1)
Taking moment about B to its right,
(RD)(40) - (FGF)(103) = 0
FGF = (x/60)(40)(1/ 103) = x/(15 3) (-ve)
At x = 0,
FGF = 0
At x = 20 ft
FGF = - 0.77
(1)
(1)
A B C D
G F E
x
1-x/20 x/20 1
600
RA= 1- x/60 RD=x/60
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(ii) To compute FFC, cut section (2) - (2)
Resolving vertically over the right hand section
FFC cos300 - RD = 0
FFC = RD/cos30 = (x/60)(2/3) = x/(30 3) (-ve)
reactions at nodes
x 1
1-x/20
x/20
(2)
(2)
300
600
A B C D
G F E
RA =1-x/60 RD=x/60
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At x = 0, FFC = 0.0
At x = 20 ft, FFC = -0.385
I. L. for FGF
I. L. for FFC
0.77
20 ft
-ve
0.385
-ve
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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)
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)
(1)
(1)
A B C D
G F E
x
(40-x)/20
(x-20)/20 1
reactions at nodes
20 ft
RA=1-x/60 RD=x/60 (x-20) (40-x)
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To compute FFC, use section (2) - (2)
Section (2) - (2) is valid for 20 < x < 40 ft
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
x 1
(2)
300
600
A B C D
G F E
RA =1-x/60 RD=x/60
(40-x)/20 (x-20)/20
(2)
FFC
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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
0.77 0.385
-ve
0.385
I. L. for FGF
I. L. for FFC
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Place unit load over CD (40 ft < x <60 ft)
(i) To compute FGF, use section (1) - (1)
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
(1)
(1)
A B C D
G F E
x
(60-x)/20 (x-40)/20
1
reactions at nodes
20 ft
RA=1-x/60 RD=x/60
(x-40) (60-x)
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To compute FFG, use section (2) - (2)
Resolving forces vertically, to the left of C,
(RA) - FFC cos 30 = 0
FFC = RA/cos 30 = (1-x/10) (2/3) +ve
x 1
(2)
300
600
A B C D
G F E
RA =1-x/60
(60-x)/20 (x-40)/20
FFC
RD=x/60
x-40 60-x
reactions at nodes
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At x = 40 ft, FFC = 0.385 (+ve)
At x = 60 ft, FFC = 0.0
-ve
0.770 0.385
-ve
+ve
I. L. for FGF
I. L. for FFC
0.385
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Influence lines for wheel loads
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Live Loads for Highway BRIDGES
•Caused by Traffic
•Heaviest: Series of Trucks
Specifications: AASHTO American Association of State and
Highway Transportation Officials
Load Desigantions
H 15-44 HS 20-44
2 Axle Truck
15 Tons (10-20)
Year of
Specifications
20 Tons
2 Axle Truck &
1 Axle Semitrailer
Year of
Specifications
Simplification: Uniform Load plus a Concnetrated force placed at critical
locations
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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
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Maximum Influence due to a Series of Concentrated Loads - Shear
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Maximum Influence due to a Series of Concentrated Loads - Shear
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Maximum Influence due to a Series of Concentrated Loads - Shear
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Maximum Influence due to a Series of Concentrated Loads - Shear
125.025.5375.521 V
875.2375.55.232 V
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Maximum Influence due to a Series of Concentrated Loads - Shear
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Maximum Influence due to a Series of Concentrated Loads - Moment
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