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Structure Analysis IChapter 6
Structure Analysis IChapter 6
Influence Line for Statically Determinate Structures
Influence lines
Influence lines provide a systematic procedure
of how force in a given part of structure varies
as the applied loads moves along the structure
Procedure for Analysis
• Place a unit load at various locations, x along the member, and at each location use static to determine the value of the function ( Reaction, Shear, Moment) at the specified point.
• If the IL for a vertical force reaction at a point on a beam is to be constructed, consider the reaction to be positive at the point when it acts upward on the beam
• If a shear or moment IL is to be drawn for a point, take the shear or moment at the point as +ve according to the same sign convention used for drawing shear and moment diagram.
Structural Analysis IDr. Mohammed Arafa
• All statically determinate beams will have IL that
consist of straight line segments. After some practice
one should be able to minimize computations and
locate the unit load only at points representing the end
points of each line segment
Example 1
Draw the IL for the Reaction at A
xA
xA
M
y
y
B
1011
0)1)(10()10(
0
IL Equation
Example 2
Draw the IL for the Reaction at B
Structural Analysis IDr. Mohammed Arafa
Example 3
Draw the IL for the Shear at C
Load from A to C
Load > C
Example 4
Draw the IL for the Moment at C
Load from A to C
Load > C
Example 5
Draw the IL for the Shear & Moment at B
Structural Analysis IDr. Mohammed Arafa
Example 6
Draw the IL for the Shear & Moment at C
0.5
Example 7Internal Hinge Example
Determine the shear & moment at point D
Influence Lines for Beams
For Concentrated Force For any concentrated force F acting on the beam at any
position x, the value of the function can be found by multiplying
the ordinate of the influence line at the position x by the
magnitude of F.
1 1
2 2yA F F lb
Influence Lines for Beams
For Uniform LoadConsider a portion of a beam subjected to a uniform load w0
0
0
0 0
dF w dx
dF y w dx y
ydF yw dx w ydx
:
The area under the influence line
Where
ydx
Influence Lines for Beams
For Uniform LoadIn general, the value of a function caused by a uniform distributed
load is simply the area under the influence line multiplied by the
intensity of the uniform load.
0
0 0
1 11
2 2
yA area w
L w Lw
Example 7-Continue
Structural Analysis IDr. Mohammed Arafa
Example 8
Determine the moment at C in the two cases
cMLI .
mkNM c . 5.27)25.1(10)5.2(6
Case 1
Example 8
cMLI .
mkNM c . 5.32)5.2(10)25.1(6
Case 2
Example 8
Example 9
Determine the maximum positive shear that can be developed at
point C in the beam shown due to a concentrated moving
load of 4000 lb and a uniform moving load of 2000 lb/ft
Due to Concenterated Load
0.75 4000 3000CV lb lb
Due to
0.5 10 2.5 0.75 2000
5625
C
Uniform Load
V
lb
Total MaximumShear
3000 5625 8625CV lb
Qualitative Influence LinesMüller-Breslau Principle
Example 10
Sketch the influence line for the vertical reaction at A
Sketch the influence line for the vertical reaction at A
Sketch the influence line for the vertical reaction at A
Example 11Sketch the influence line for the shear at B
Sketch the influence line for the shear at B
Sketch the influence line for the shear at B
Sketch the influence line for the moment at B
Example 11-bSketch the influence line for the moment at B
Sketch the influence line for the moment at B
Example 11-CSketch the influence line for the moment at B
- the reaction at A and C
- shear at D,
- the moment at D, E and F
Example 11-C
Example 11-C
Example 11-C
Example 11-C
Example 11-C
Example 11-C
Influence Line for Floor Girders
Structural Analysis IDr. Mohammed Arafa
Draw the influence line for the shear at panel CD
Example 12
Draw the influence line for the moment at F
Example 13
Example 14Truss Example
Determine The force in member GB
EGB
EGB
RF
RF
41.1
45cos
AGB
AGB
RF
RF
41.1
45cos
part second for the
x FGB
0 0
6 0.354
12 -0.707
18 -0.354
24 0
Example 15Truss Example
Determine The force in member GF, BF
0 40 17.3
2.3
cos30 1.15
B D GF
GF D
BF D BF D
M R F
F R
F R F R
0 20 17.3
1.15
cos30 1.15
B A GF
GF A
BF A BF A
M R F
F R
F R F R
Example 16Truss Example
Determine The force in member CG
Example 17
Structural Analysis IDr. Mohammed Arafa
IL due to Series of
Concentrated Loads
1
1(0.75) 4(0.625) 4(0.5) 5.25CV k
2
1( 0.125) 4(0.75) 4(0.625) 5.375CV k
3
1(0) 4( 1.25) 4(0.75) 2.5CV k
1
2
3
2(7.5) 4(6.5) 3(5.0) 56.0 .
2(4.5) 4(7.5) 3(6.0) 57.0 .
2(0) 4(3.0) 3(7.5) 34.5 .
C
C
C
M k ft
M k ft
M k ft
Example 18
Example 19
max
8(1.2) 3(0.4) 10.8 .BM kN m
Structural Analysis IDr. Mohammed Arafa
Determine the maximum positive moment that created at point B
Example 19
Absolute Maximum Shear and Moment
Absolute Maximum Moment
• The absolute maximum moment in a simply
supported beam occurs under one of the
concentrated forces, such that this force is
positioned on the beam so that it and the
resultant force of the system are equidistant
from the beam’s centerline.
The location of Maximum Moment is at
2
xx
Absolute Maximum Moment
Structural Analysis IDr. Mohammed Arafa
Example 20
The truck has a mass of 2 Mg and a center of gravity at G.
Determine the absolute maximum moment developed in the
simply supported bridge deck due to the truck’s weight. The
bridge has a length of 10 m.
Example 21
3The truck weight= 2 10 kg 9.8
=19.62kN
Example 21
The maximum moment occurs under the front wheel loading.
Using the right section of the bridge deck,