Influence Line Diagrams
Transcript of Influence Line Diagrams
-
8/20/2019 Influence Line Diagrams
1/56
-
8/20/2019 Influence Line Diagrams
2/56
Influence Lines
ons er e r ge n g. . s
the car moves across the bridge,
change with the position of the
car and the maximum force in
each member will be at a differen
car location. The design of each
mem er mus e ase on e
maximum probable load each.
2
-
8/20/2019 Influence Line Diagrams
3/56
gure . r ge russ tructureSubjected to a Variable
Position Load
Therefore, the truss analysis
for each member would
position that causes the
reatest force or stress in
3each member.
-
8/20/2019 Influence Line Diagrams
4/56
If a structure is to be safely
designed, members must be
proportioned such that the
and live loads is less than the
available section ca acit .
Structural analysis for variable
1.Determining the positions of
response function is
maximum and
2.Computing the maximum
4
-
8/20/2019 Influence Line Diagrams
5/56
Influence Line
Definitions u u
reaction, axial force, shear force, o
.Influence Line ≡ graph of a
a function of the position of a
downward unit load movin across
the structure.
statically determinate structures
are always piecewise linear.
5
-
8/20/2019 Influence Line Diagrams
6/56
Once an influence line is
constructed:
• Determine where to lace live
load on a structure to maximiz
the drawn response function;
and
• Evaluate the maximum
magnitude of the response
function based on the loading.
6
-
8/20/2019 Influence Line Diagrams
7/56
Calculating Response
(Equilibrium Method)
1 ILD for A
0 L
1ILD for Cy
70 L
-
8/20/2019 Influence Line Diagrams
8/56
1x
MB
A
a
VB
0 < x < a
y yBF 0 V A 1= ⇒ = −∑
a yBM 0 M A a 1(a x= ⇒ = − −
MB
AyVB
a < x < La
y yBF 0 V A= ⇒ =∑
8a yB
M 0 M A a= ⇒ =
-
8/20/2019 Influence Line Diagrams
9/56
VB
1 – a/L
0a
-a/LILD for VB
–B
0
ILD for MB
9
-
8/20/2019 Influence Line Diagrams
10/56
Beam Example 1
Calculate and draw the su ort
reaction response functions.
10
-
8/20/2019 Influence Line Diagrams
11/56
Beam Example 2
Calculate and draw the response
functions for R A, M A, RC and VB.
11
-
8/20/2019 Influence Line Diagrams
12/56
BD: Link
Member
Calculate and draw the
response functions for Ax, Ay,and . NOTE: Unit loadABBV
12
.
-
8/20/2019 Influence Line Diagrams
13/56
Muller-Breslau
Principle-
influence line for a response
function is iven b the deflected
shape of the released structure
due to a unit displacement (or
rotation) at the location and in the
direction of the response.
A released structure is obtained
constraint corresponding to the
13
the original structure.
-
8/20/2019 Influence Line Diagrams
14/56
CAUTION: Principle is only valid
.
Releases:
Support reaction - remove
translational support restraint.
Internal shear - introduce an
internal glide support to allow
differential displacement
movement.
Bending moment - introduce an
internal hinge to allow differential
ro a on movemen .
14
-
8/20/2019 Influence Line Diagrams
15/56
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or displ ay.
Influence Line for Shear
15
-
8/20/2019 Influence Line Diagrams
16/56
Influence Line for Bending Moment
16
-
8/20/2019 Influence Line Diagrams
17/56
Application of Muller-
Breslau Principle
17
-
8/20/2019 Influence Line Diagrams
18/56
18
-
8/20/2019 Influence Line Diagrams
19/56
19
-
8/20/2019 Influence Line Diagrams
20/56
y = (L – x) (a/L)
1 2 =
20
-
8/20/2019 Influence Line Diagrams
21/56
Qualitative Influence
Lines
In many practical applications, it is
necessar to determine onl the
general shape of the influence
lines but not the numerical values
o e or na es. uc an
influence line diagram is known as-
gram.
numerical values of its ordinates is
known as a uantitative influ-
21ence line diagram.
-
8/20/2019 Influence Line Diagrams
22/56
NOTE: An advanta e of
constructing influence lines usin
the Muller-Breslau Principle is
that the response function of
interest can be determined
.determining the influence lines
for other functions as was the
case with the equilibrium
method.
22
-
8/20/2019 Influence Line Diagrams
23/56
Influence Lines for
Trusses
In a gable-truss frame building,
to the top chord joints through roo
purlins as shown in Fig. T.1.
Similarly, highway and railway
brid e truss-structures transmit
floor or deck loads via stringers to
floor beams to the truss joints asshown schematically in Fig. T.2.
Fig. T.1. Gable Roof Truss
23
-
8/20/2019 Influence Line Diagrams
24/56
Fig. T.2. Bridge Truss
24
-
8/20/2019 Influence Line Diagrams
25/56
These load paths to the truss joints
provide a reasonable assurance
that the primary resistance in the
axial force. Consequently,
influence lines for axial member
forces are developed by placing a
unit load on the truss and making
judicious use of free body
diagrams and the equations of.
25
-
8/20/2019 Influence Line Diagrams
26/56
Due to the load transfer
process in truss systems, no
discontinuity will exist in the
diagrams. Furthermore, since
we are restrictin our attention to
statically determinate struc-
tures, the influence line
diagrams will be piecewise
linear .
26
-
8/20/2019 Influence Line Diagrams
27/56
Example Truss Structure
u w
functions for Ax, Ay, FCI and FCD
27
-
8/20/2019 Influence Line Diagrams
28/56
Use of Influence Lines
Single Moving
Concentrated Load
Each ordinate of an influence
response function due to a
sin le concentrated load of
unit magnitude placed on the
structure at the location of tha
or na e. us,
28
-
8/20/2019 Influence Line Diagrams
29/56
A B C D
x
D
B
A B C
- B
place P at BB max(M ) ⇒−
29B max ⇒
-
8/20/2019 Influence Line Diagrams
30/56
1. The value of a response
concentrated load can be
obtained by multiplying the
magnitude of the load by the
ordinate of the response
position of the load.
. ax mum pos ve va ue o
the response function is
point load by the maximum
positive ordinate. Similarly, th
maximum negative value isobtained by multiplying the
30
po n oa y e max mum
negative ordinate.
-
8/20/2019 Influence Line Diagrams
31/56
Point Res onse Due to a
Uniformly Distributed Live
Load
Influence lines can also be
emp oye o e erm ne evalues of response functions of
loads. This follows directly from
point forces by treating the
uniform load over a differential
segment as a differential point
orce, .e., = x. us, aresponse function R at a pointw
31
-
8/20/2019 Influence Line Diagrams
32/56
-
8/20/2019 Influence Line Diagrams
33/56
in which the last inte ral ex res-
sion represents the area under the
segment of the influence line,
which corresponds to the loaded
portion of the beam.
SUMMARY
1. The value of a response
function due to a uniformly
portion of the structure can be
intensity by the net area under
the corresponding portion of th
33response function influence
line.
-
8/20/2019 Influence Line Diagrams
34/56
2. To determine the maximum
positive (or negative) value of a
response unc on ue o a
uniformly distributed live load,
those portions of the structure
where the ordinates of the
response function influence line
are positive (or negative).
Points 1 and 2 are schematically
demonstrated on the next slide formoment MB considered in the poin
34
.
-
8/20/2019 Influence Line Diagrams
35/56
35
-
8/20/2019 Influence Line Diagrams
36/56
36
-
8/20/2019 Influence Line Diagrams
37/56
Where should a CLL
(Concentrated Live Load), a ULL
(Uniform Live Load) and UDL
on the typical ILD’s shown below
to maximize the res onse
functions?
Typical End Shear(Reaction) ILD
37
-
8/20/2019 Influence Line Diagrams
38/56
Beam Shear ILD
Typical Interior
Bending Moment ILD
Possible Truss Member ILD
38
-
8/20/2019 Influence Line Diagrams
39/56
Live Loads for
Railroad BridgesLive loads due to vehicular traffic
on highway and railway bridges
are represen e y a ser es omoving concentrated loads with
loads. In this section, we discuss
the use of influence lines to
determine: (1) the value of the
response function for a given
loads and (2) the maximum value
39
series of moving concentrated
loads.
-
8/20/2019 Influence Line Diagrams
40/56
To calculate the response
unc on or a g ven pos on o
the concentrated load series,
series load by the magnitude
of the influence line diagramiP
ordinate at the position of ,
i.e.iy iP
= i i
i
y
calculated from the slope of the
influence line diagram (m) via
= miy ix
40
-
8/20/2019 Influence Line Diagrams
41/56
where is the distance to point ix
measure rom e zero y-ax s
intercept, as shown in the.
m
yb
ya
1
x
b
a similar trianglesa b= ⇒
41 b b
ay y
y a ; m
b b
∴ = =
-
8/20/2019 Influence Line Diagrams
42/56
For example, consider the ILDix
s own on e nex s e
subjected to the given wheel
Load Position 1:
1 1 1 130 30 30 30
130( )(8(20) 10(16) 15(13) 5(8))= + + +
i ii
m P x 18.5k = =∑
42
-
8/20/2019 Influence Line Diagrams
43/56
2/3
10 ft.
20 ft.
-1/3
43
-
8/20/2019 Influence Line Diagrams
44/56
Position 1
Position 2
44
-
8/20/2019 Influence Line Diagrams
45/56
Load Position 2:
B230
V ( 8(6) 10(20) 15(17) 5(12= − + + +
=
Thus, load position 1 results in the
.
NOTE: If the arrangement of
loads is such that all or most of the
heavier loads are located near oneo e en s o e ser es, en e
analysis can be expedited by
for the series so that the heavier
loads will reach the maximum
45
-
8/20/2019 Influence Line Diagrams
46/56
lighter loads in the series. In sucha case, it may not be necessary to
examine all the loading positions.
Instead, the analysis can be
w v uresponse function begins to
. .,
the response function is less than
the preceding load position. This
process is known as the
“ Increase-Decrease Method” .
46
-
8/20/2019 Influence Line Diagrams
47/56
CAUTION: This criterion is not
va or any genera ser es o
loads. In general, depending on, ,
and shape of the influence line,
the value of the response
function, after declining for some
loading positions, may start
ncreas ng aga n or su sequen
loading positions and may attain .
47
-
8/20/2019 Influence Line Diagrams
48/56
Zero Ordinate Location
Linear Influence Line
b+
1
x-
1
m-
-
L
+
48
-
8/20/2019 Influence Line Diagrams
49/56
b b b − −−x ; mm
L
+ ++
= =
b b bx ; m
m+ −−
− −−
−−= =
: u
are obtained fromy = mx + b
with y = 0.
49
-
8/20/2019 Influence Line Diagrams
50/56
Example Truss Problem:
Application of Loads to
Maximize Response
ML
50CM
-
8/20/2019 Influence Line Diagrams
51/56
Place
UDL = 1.0 k/ft;ULL = 4.0 k/ft
CLL = 20 kips
compression axial forces in
members CM and ML.
Calculate the magnitudes of the
51
-
8/20/2019 Influence Line Diagrams
52/56
Force and Moment
Envelopes
response function as a
function of the location of the
response function is referred
to as the envelope of the
maximum values of a
response function for the
considered.
52
-
8/20/2019 Influence Line Diagrams
53/56
For a single concentrated
force for a simply sup-
ported beam:
maxa
(V) P 1+ ⎛ ⎞= −⎜ ⎟
a−max
L−
maxM P a 1L
= −⎜ ⎟⎝ ⎠
Plot is obtained by treating “a”
53
as a var a e.
-
8/20/2019 Influence Line Diagrams
54/56
For a uniformly distributed
load on for a simply sup-
ported beam:
( )2
maxw
(V) L a2L
+ = −
2w aV − = −
2L
( )maxM L a2
= −
Plot is obtained by treating “a”
54
as a var a e.
-
8/20/2019 Influence Line Diagrams
55/56
55
-
8/20/2019 Influence Line Diagrams
56/56