Subject
Year
: S0695 / STATICS
: 2008
Influence lines for Statically Determinate Structures
Session 19 - 26
6.1. Influence Lines
What is the Influence line ?
A curve of influence are generated by plotting points,
after moving the load to numerous points on the structure.
curve shown how to position live load to maximize the
value of internal forces at designated section of a
structure.
the influence line will consist of only
between critical ordinate values.
straight line segments
An influence line for a function differs from a shear, axial, or bending
moment diagram
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6.1. Influence Lines
Maximum
NOT ALL MAXIMUM FORCES OCCUR AT THE SUPPORTS
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6.1. Influence Lines
What are the Influence lines used for ?
The curve of influence line that has plotted at
that point on the structure has shown that how
much the load influences the reaction
(or shear, or other action) . This is an important
application for the design structure that resist
large live loads.
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6.1. Influence Lines
What are the Influence lines used for ?
Influence lines are used to determine where to
place loads on a structure for
maximum results, and to compute the
resulting magnitude of the reaction or other
actions (shear/moment/axial force) once the
loads are placed in their critical positions.
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6.1. Influence Lines
What are the Influence lines used for ?
• To determine the location for positioning a uniform
load of constant intensity to produce the
maximum magnitude for a particular function,
place the load along those portions of the structure for
which the ordinates to the influence line have the
same algebraic sign.
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6.1. Influence Lines
What are the Influence lines used for ?
The value for the particular function will be equal to
the magnitude of the uniform load, multiplied
by the area under the influence diagram between the
beginning and ending points of the uniform load.
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6.1. Influence Lines
How can the Influence lines be generated?
Influence lines can be generated by placing a
dimensionless unit load on a structure, and using
statics to determine what reaction, shear, or other action is
obtained at a given point on the structure.
The unit load is them moved to a new location on the structure and
the process is repeated
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6.1. Influence Lines
What are the Influence lines Process ?
1. A unit load is placed at the left end of the structure, and the
reactions are computed.
2. The structure is then cut at the point where the influence line is
desired (the center in this case), and a free body of either end
of the structure is drawn.
3. Equilibrium is used to determine the magnitude of the
resulting action, in this case the internal shear at the
center of the structure.
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6.1. Influence Lines
What are the Influence lines Process ?
4. The resulting magnitude of the action (shear) is plotted at the
place where the unit load is applied, kind of as a reminder that
were to place a unit load there, it would get that much shear
force at the center of the structure.
5. The unit load is then moved some distance to the right, and the
process is repeated. The final plot of all the resulting shears
forms the influence line for shear at the center of the structure.
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6.1. Influence Lines
P = 1 unit
x
A
B
L
x
P = 1 unit
A
B
L
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P
6.1. Influence Lines
x
ΣMA=0
P.x - VB.L = 0
1.x - VB.L = 0
P = 1 unit
A
B
L
VB = x/L
If….
x=0
x=L
VB = 0
VB = 1
ΣMB=0
- P.( L - x ) + VA.L = 0
- 1 ( L – x ) + VA.L = 0
VA = ( L - x ) / L
If….
x=0
x=L
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VA = 1
VA = 0
6.1. Influence Lines
P = 1 unit
P = 1 unit
A
B
1
0
+
0
Influence line for
VA
1
+
0
0
Influence line for
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VB
6.1. Influence Lines
This diagram
indicates that should a
concentrated load be placed over the left
reaction it will cause a reaction of
magnitude 1.0 the value of the concentrated
load.
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6.1. Influence Lines
In fact, the result of placing a load
anywhere on the beam can now be found by
simply multiplying the magnitude of the load
times the height of the influence line at the
location of the load, x.
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6.2. Qualitative Influence Lines
There are 2 methods that can be used to plot an influence Æ
1. to write an equation for the function being
determined, e.g., the equation for the shear, moment, or axial
force induced at a point due to the application of a unit
2.
load at any other location on the structure.
uses the Müller Breslau Principle, can be utilized to
draw qualitative influence lines, which are directly
proportional to the actual influence line.
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6.2. Qualitative Influence Lines
Müller Breslau Principle
The Müller Breslau Principle is another alternative
available to qualitatively develop the influence lines for
different functions.
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6.2. Qualitative Influence Lines
The Müller Breslau Principle states that Æ
the ordinate value of an influence line for any
function on any structure is proportional to the
ordinates of the deflected shape that is obtained by
removing the restraint corresponding to the function
from the structure and introducing a force that
causes a unit displacement in the positive
direction.
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6.2. Qualitative Influence Lines
The Müller Breslau Principle Æ
Influence line for Reaction at A
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6.2. Qualitative Influence Lines
The Müller Breslau Principle Æ
Influence Line for Shear at C
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Influence Line for
Bending Moment at C
6.2. Qualitative Influence Lines
The Müller Breslau Principle Æ
Influence Line for
Shear at E
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6.2. Qualitative Influence Lines
The Müller Breslau Principle Æ
Influence Line for Bending Moment at E
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6.3. Max. Influence
The max. effect caused by Æ
a live concentrated force is determined by
multiplying the peak ordinate of the influence line
by magnitude of the force.
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6.3. Max. Influence
Shear Æ
The influence line for
a shear force at a
given location will contain a translational
discontinuity at this location. The summation
of the positive and negative shear forces at this
location is equal to unity.
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6.3. Max. Influence
Moment Æ
The influence line for a bending
moment will contain a unit rotational
discontinuity at the point where the
bending moment is being evaluated.
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6.3. Max. Influence
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6.3. Max. Influence
Except at an internal hinge
location, the slope to the shear
force influence line will be the
same on each side of the critical
section since the bending moment is
continuous at the critical section.
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6.4. Absolute Max. Shear & Moment
Shear Æ
To induce the maximum positive shear at the
quarter-point of a simply supported beam, the
influence line generated earlier is shown here:
0.75
0.25
0.25
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0.50
6.4. Absolute Max. Shear & Moment
Shear Æ
For maximum positive shear, the loads are situated such that
the highest loads are placed over the highest positive ordinates on
the influence line, as follows:
L/4
w
w
0.75
0.25
0.25
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0.50
6.4. Absolute Max. Shear & Moment
Shear Æ
For maximum negative shear, the loads are
situated such that the highest loads are placed
over the highest negative ordinates on the
influence line, as follows:
0.25
w
0.75
w
0.25
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L/4
0.50
6.4. Absolute Max. Shear & Moment
Moment Æ
To induce the maximum positive moment at the
quarter-point of a simply supported doubly-overhanging
beam, the influence line generated earlier is shown here:
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6.4. Absolute Max. Shear & Moment
Moment Æ
For maximum positive moment, the loads are situated
such that the highest loads are placed only over the
highest positive ordinates on the influence line, and not in
the negative regions, as follows:
L/4
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w
6.4. Absolute Max. Shear & Moment
Placement of mixed loads for maximum positive moment
Note that the large wheel load is placed over the
maximum positive ordinate of the influence line, while
the smaller wheel load is placed to its right. This is because
the influence line has a smaller slope, and descends more
gradually to the right of the peak, than to the left of the
peak, thus making that the higher ordinate at L/4 from the
peak.
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6.4. Absolute Max. Shear & Moment
Placement of mixed loads for maximum negative moment
To induce the maximum negative moment at the quarterpoint of a simply supported doubly-overhanging beam, the
influence line generated earlier is shown here:
w
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w
6.4. Absolute Max. Shear & Moment
Placement of mixed loads for maximum negative moment
Again, the large wheel load is placed at the
maximum negative ordinate, with the
smaller wheel load placed at L/4 away. The uniform
live loads are placed continuously throughout all
negative regions.
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Influence Lines for Beam
P = 1 unit
x
A
B
C
4m
x
P = 1 unit
A
B
C
4m
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4m
4m
4m
4m
Influence Lines for Beam
x
P = 1 unit
A
ΣMA=0
P.x - VB.8m = 0
1.x - VB.8m = 0
B
C
4m
4m
4m
VB = x/8
If….
x=0
x=8
x = 12
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VB = 0
VB = 1 t
VB = 1 ½ t
ΣMB=0
- P.( 8 - x ) + VA.8m = 0
- 1 ( 8 – x ) + VA.8m = 0
VA = ( 8 - x ) / 8
If….
x=0
x=8
x = 12
VA = 1 tm
VA = 0
VA = - ½ tm
6.1. Influence Lines
P = 1 unit
P = 1 unit
A
B
C
4m
4m
4m
1
0
+
0
1
0
Influence line for
VA
½
1½
+
Influence line for
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VB
Influence Lines for Beam
Step 1 :
Cut the structure and
determine with cross
section method.
x
P = 1 unit
A
4m
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B
C
4m
4m
Influence Lines for Beam
Step 1st
:
Cut the structure and
determine with cross
section method.
x
P = 1 unit
A
B
C
4m
4m
4m
P = 1 at Right
side Æ Determine at Left
side
When
MC
Vc = - VB
B
C
4m
4m
VC
VB
Mc = - 4.VB
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Influence Lines for Beam
Step 2nd :
Cut the structure and
determine with cross
section method.
x
P = 1 unit
A
B
C
4m
4m
4m
When P = 1 at Left
side Æ Determine at
Right side
VC
Vc = V A
A
C
4m
Mc = 4.VA
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MC
VA
Step 3rd :
Draw the
influence
lines
x
P = 1 unit
A
P = 1 at
Right side Æ
B
C
4m
4m
4m
1
When
0
0
+
Influence line for
-
½
1½
1
Vc = - VB
+
0
Vc = VA
When P = 1 at
Left side Æ
Mc = - 4.VB
Mc = 4.VA
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½
0
0
+
-
-
-
½
-2
-2
Influence line for
VB
0
-
+
VA
-
½
Influence Lines
Shear force C
2
Influence Lines
Bending Moment C
Influence Lines for Trusses
E
4
3m
7
A
G
F
8
5
9
1
10
12
3
D
3x3m
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11
2
C
H
6
13
B
Influence Lines for Trusses
Step 1st :
Idealized Structure
4
5
8
7
A
9
1
10
2
3x3m
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6
11
12
13
3m
3
B
Influence Lines for Trusses
Step 2nd :
4
Put a unit load is placed
at the left end of the
structure, and
5
8
7
9
x
compute the reaction
11
12
2
3
3x3m
ΣMA=0
P.x - VB.L = 0
1.x - VB.L = 0
VB = x/L
If….
x=0
x=L
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3m
13
P = 1 unit
1
A
10
6
VB = 0
VB = 1
B
ΣMB=0
- P.( L - x ) + VA.L = 0
- 1 ( L – x ) + VA.L = 0
VA = ( L - x ) / L
If….
x=0
x=L
VA = 1
VA = 0
4
Step 3rd
:
7
P = 1 unit 8
Draw the
influence line of
reactions
A
1
5
9
10
2
6
11
P = 1 unit
12
3m
13
3
3x3m
B
1
+
Influence line for
VA
1 Influence line for
+
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VB
Influence Lines for Trusses
Step 4th :
4
Determine the influence
line of member forces
at section I - I
5
8
7
9
x
11
12
2
3
3x3m
ΣMA=0
P.x - VB.L = 0
1.x - VB.L = 0
VB = x/L
If….
x=0
x=L
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3m
13
P = 1 unit
1
A
10
6
VB = 0
VB = 1
B
ΣMB=0
- P.( L - x ) + VA.L = 0
- 1 ( L – x ) + VA.L = 0
VA = ( L - x ) / L
If….
x=0
x=L
VA = 1
VA = 0
Influence Lines for Trusses
Step 4th :
Determine the influence
line of member forces
at section I - I
I
4
5
8
7
9
x
A
6
10
11
3m
13
P = 1 unit
1
2
3
I
3x3m
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12
B
Influence Lines for Trusses
Step 4.a :
Cut the structure and determine with cross section method.
When
P = 1 at Right side Æ Determine at Left side
I
4
5
8
7
9
x
A
5
10
10
6
11
12
13
P = 1 unit
1
2
2
I
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I
I
3
B
Influence Lines for Trusses
I
Step 4.a :
5
Cut the structure and
determine with cross
section method.
6
F
10
P = 1 at Right
side Æ Determine at Left
side
11
12
13
When
2
3
I
ΣMF=0
S2.3m - VB.6m = 0
S2 = 2 VB
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B
Influence Lines for Trusses
I
Step 4.a :
5
Cut the structure and
determine with cross
section method.
10
P = 1 at Right
side Æ Determine at Left
side
6
11
12
13
When
2
I
3
D
ΣMD=0
-S5.3m - VB.3m = 0
S5 = - VB
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B
Influence Lines for Trusses
I
Step 4.a :
5
Cut the structure and
determine with cross
section method.
10
P = 1 at Right
side Æ Determine at Left
side
6
11
12
13
When
2
ΣV=0
S10.(sin 45o) + VB = 0
½ √2 .S10 + VB = 0
I
3
D
B
S10 = - √2 VB
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Influence Lines for Trusses
Step 4.b :
Cut the structure and determine with cross section method.
When
P = 1 at Left side Æ Determine at Right side
I
4
5
8
7
I
9
5
10
10
6
11
12
13
x
P = 1 unit
A
1
2
2
I
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I
3
B
Influence Lines for Trusses
Step 4.b :
Cut the structure
and determine with
cross section
method.
A
When P = 1 at Left
side Æ Determine
at Right side
ΣMF=0
F
4
8
7
9
1
I
5
10
2
I
-S2.3m + VA.3m = 0
S2 = VA
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Influence Lines for Trusses
Step 4.b :
Cut the structure
and determine with
cross section
method.
A
When P = 1 at Left
side Æ Determine
at Right side
ΣMD=0
4
5
8
7
I
9
1
10
2
S5.3m + VA.6m = 0
S5 = - 2 VA
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I D
Influence Lines for Trusses
Step 4.b :
Cut the structure
and determine with
cross section
method.
When P = 1 at Left
side Æ Determine
at Right side
F
4
8
7
A
9
1
I
5
10
2
I
ΣV=0
-S10.(sin 45o) + VA = 0
½ √2 .- S10 + VA = 0
S10 = √2 VA
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I
F
4
Step 5th :
Draw the
influence
lines
5
8
7
10
9
1
A
6
11
2
¾
D
B
I
½
+
0
0
¼
0
0
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3m
13
3
3x3m
1
12
½
Influence Line of VA
1
+
0
0
Influence Line of VB
I
F
4
Step 5th :
Draw the
influence
A
lines
P = 1 at
Right side Æ
5
8
7
6
10
9
1
11
2
When
2/3
D
S2 = 2 VB
When P = 1 at
Left side Æ
0
S2 = VA
1/3
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0
2/3
1/3
+
Influence Line of VA
1
+
0
2.1/3 = 2/3
0
B
I
+
0
3m
13
3
3x3m
1
12
Influence Line of VB
1/3 = 1/3
0
Influence Line of S2
I
F
4
Step 5th :
Draw the
influence
A
lines
P = 1 at
Right side Æ
5
8
7
10
9
1
2/3
1
12
3m
13
3
D
3x3m
B
I
1/3
+
0
0
1/3
When P = 1 at
Left side Æ
11
2
When
S5 = - VB
6
2/3
Influence Line of VA
1
+
0
0
Influence Line of VB
S5 = - 2 VA
0
-1/3 = -1/3
0
-2.1/3 = -2/3
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Influence Line of S5
I
F
4
Step 5th :
Draw the
influence
A
lines
P = 1 at
Right side Æ
5
8
7
10
9
1
2/3
1
S10 =
3m
13
D
B
I
1/3
+
0
0
2/3
Influence Line of VA
1
+
0
0
Influence Line of VB
1/3 . √2 = 1/3 √2
√2 VA
0
+
-1/3 .√2 = -1/3 √2
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12
3
3x3m
1/3
When P = 1 at
Left side Æ
11
2
When
S10 = - √2 VB
6
0
Influence Line of S10
I
F
4
Step 5th :
Draw the
influence
lines
5
8
7
10
9
x
11
12
2
3
D
3x3m
¾
1
B
I
½
+
0
0
½
¼
0
3m
13
P = 1 unit
1
A
6
2.1/3 = 2/3
1
+
0
1/3 = 1/3
+
-
1/3 . √2 = 1/3 √2
-2.1/3 = -2/3
+
-1/3 = -1/3
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-1/3 .√2 = -1/3 √2
Influence Line of VA
Influence Line of VB
Influence Line of S2
Influence Line of S5
Influence Line of S10
Influence Lines of S2
E
F
4
3m
7
A
P = 1 unit8
5
9
1
2/3
Bina Nusantara
10
6
11
+
3x3m
12
H
P =13
1 unit
3
2
C
0
G
B
D
1/3
0
Influence Lines of S5
E
F
4
3m
7
A
P = 1 unit8
G
5
9
1
10
6
11
0
-1/3
B
D
3x3m
0
-2/3
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P =13
1 unit
3
2
C
12
H
Influence Lines of S10
E
F
4
3m
7
A
G
5
P = 1 unit8
9
1
10
6
11
12
3x3m
P =13
1 unit
3
2
C
H
B
D
1/3
0
-1/3
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+
0
Influence Lines for Girder
x
P = 1 unit
A
B
4m
x
P = 1 unit
A
B
4m
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4m
4m
4m
4m
Influence Lines for Girder
P = 1 unit
A
1
2
3
4m
The Unit
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4
5
6
4m
7
8
9 B10 11 12 13
4m
load applied to the stringer
Influence Lines for Girder
x
P = 1 unit
A
B
4m
4m
4m
Solution :
Cut the structure and determine with
cross section method for each
section according the stringer.
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