Influence Line Diagrams

8/20/2019 Influence Line Diagrams

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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.

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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.


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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

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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.

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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.

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Calculating Response

(Equilibrium Method)

1 ILD for A

0 L

1ILD for Cy

70 L


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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= ⇒ =


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VB

1 – a/L

0a

-a/LILD for VB

–B

0

ILD for MB

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Beam Example 1

Calculate and draw the su ort

reaction response functions.

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Beam Example 2

Calculate and draw the response

functions for R A, M A, RC and VB.

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BD: Link

Member

Calculate and draw the

response functions for Ax, Ay,and . NOTE: Unit loadABBV

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.


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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

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the original structure.


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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 .

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Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or displ ay.

Influence Line for Shear

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Influence Line for Bending Moment

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Application of Muller-

Breslau Principle

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y = (L – x) (a/L)

1 2 =

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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.


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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.

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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

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Fig. T.2. Bridge Truss

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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.

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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 .

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Example Truss Structure

u w

functions for Ax, Ay, FCI and FCD

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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,

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A B C D

x

D

B

A B C

- B

place P at BB max(M ) ⇒−

29B max ⇒


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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

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po n oa y e max mum

negative ordinate.


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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

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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.


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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

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.


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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

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Beam Shear ILD

Typical Interior

Bending Moment ILD

Possible Truss Member ILD

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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

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series of moving concentrated

loads.


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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

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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

∴ = =


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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 = =∑

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2/3

10 ft.

20 ft.

-1/3

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Position 1

Position 2

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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

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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” .

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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 .

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Zero Ordinate Location

Linear Influence Line

b+

1

x-

1

m-

-

L

+

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b b b − −−x ; mm

L

+ ++

= =

b b bx ; m

m+ −−

− −−

−−= =

: u

are obtained fromy = mx + b

with y = 0.

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Example Truss Problem:

Application of Loads to

Maximize Response

ML

50CM


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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

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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.

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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”

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as a var a e.


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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”

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as a var a e.


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Influence Line Diagrams

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