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# L8 - Influence Line Diagrams for statically determinate structures by SaiDeepika2

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```									                                         Influence Lines
Consider the bridge in Fig. 1. As
Influence Lines for                    the car moves across the bridge,
Determinate Structures                   the forces in the truss members
change with the position of the
Introduction                             car and the maximum force in
each member will be at a different
Previous developments have               car location. The design of each
been limited to structures               member must be based on the
member will experience.
Structures are also subjected to
vary on the structure. This
chapter focuses on such loads for
statically determinate structures.
1                                      2

If a structure is to be safely
designed, members must be
proportioned such that the
and live loads is less than the
available section capacity.
Figure 1. Bridge Truss Structure
Subjected to a Variable          Structural analysis for variable
1.Determining the positions of
Therefore, the truss analysis               the loads at which the
for each member would                       response f    ti i
function is
involve determining the load                maximum; and
position that causes the                  2.Computing the maximum
greatest force or stress in                 value of the response function.
each member.                  3                                             4

1
Influence Line                   Once an influence line is
Definitions                    constructed:
• Determine where to place live
Response Function ≡ support
load on a structure to maximize
reaction, axial force, shear force, or
the drawn response function;
bending moment.
and
Influence Line ≡ graph of a
• Evaluate the maximum
response function of a structure as
magnitude of the response
a function of the position of a
structure
the structure.
NOTE: Influence lines for
statically determinate structures
are always piecewise linear.
5                                       6

1
Calculating Response                          x
MB
Functions                                    a
0<x<a
(Equilibrium Method)                              VB
Ay

∑ Fy = 0 ⇒ V B = A y − 1
∑ Ma = 0 ⇒ M B = A y a −1(a − x)

1          ILD for Ay                                      MB
a
VB        a<x<L
Ay
0                   L
ILD for Cy          1
∑ Fy = 0 ⇒ V B = A y
0                   L
7
∑ Ma = 0 ⇒ M B = A y a        8

2
1 – a/L
VB                                       Beam Example 1
0
a         L

ILD for VB
/L
-a/L

MB       a (1 – a/L)

0       a             L            Calculate and draw the support
ILD for MB                     reaction response functions.

9                                     10

Beam Example 2
Frame Example

Member

Calculate and draw the response
functions for RA, MA, RC and VB.

Calculate and draw the
response functions for Ax, Ay,
AB
and VB . NOTE: Unit load
traverses span AC.
11                                        12

3
CAUTION: Principle is only valid
Muller-Breslau                                                                                         for force response functions.
Principle
Releases:
Muller-Breslau Principle ≡ The
Support reaction - remove
influence line for a response
restraint.
translational support restraint
function is given by the deflected
shape of the released structure                                                                            Internal shear - introduce an
due to a unit displacement (or                                                                             internal glide support to allow
rotation) at the location and in the                                                                       differential displacement
direction of the response                                                                                  movement.
function.
Bending moment - introduce an
A released structure is obtained                                                                           internal hinge to allow differential
by removing the displacement                                                                               rotation movement.
constraint corresponding to the
response function of interest from
the original structure.         13                                                                                                            14

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

Influence Line for Shear                                                                              Influence Line for Bending Moment

15                                      16

4
Application of Muller-
Breslau Principle

17                       18

y = (L – x) (a/L)

θ1 + θ2 = 1

19                       20

5
Qualitative Influence                  NOTE: An advantage of
Lines                          constructing influence lines using
the Muller-Breslau Principle is
that the response function of
In many practical applications, it is   interest can be determined
necessary to determine only the         directly. It does not require
general shape of the influence          determining the influence lines
lines but not the numerical values      for other functions, as was the
of the ordinates. Such an               case with the equilibrium
influence line diagram is known as      method.
a qualitative influence line dia-
gram.
gram
An influence line diagram with
numerical values of its ordinates is
known as a quantitative influ-
ence line diagram.               21                                    22

Influence Lines for
Trusses
In a gable-truss frame building,
to the top chord joints through roof
purlins as shown in Fig. T.1.
Similarly, highway and railway
bridge truss-structures transmit
floor or deck loads via stringers to
floor beams to the truss joints as
shown schematically in Fig. T.2.
y       g

Fig. T.2. Bridge Truss

Fig. T.1. Gable Roof Truss     23                                   24

6
These load paths to the truss joints   Due to the load transfer
provide a reasonable assurance         process in truss systems, no
that the primary resistance in the     discontinuity will exist in the
truss members is in the form of        member force influence line
axial force. Consequently,             diagrams. Furthermore, since
influence lines for axial member       we are restricting our attention to
forces are developed by placing a      statically determinate struc-
unit load on the truss and making      tures, the influence line
judicious use of free body             diagrams will be piecewise
diagrams and the equations of          linear.
statics.

25                                     26

Example Truss Structure              Use of Influence Lines

Point Response Due to a
Single Moving
Each ordinate of an influence
line gives the value of the
response function due to a
Calculate and draw the response
unit magnitude placed on the
functions for Ax, Ay, FCI and FCD.
structure at the location of that
ordinate. Thus,

27                                     28

7
1. The value of a response
P                        function due to any single
A    B                C   D           concentrated load can be
obtained by multiplying the
x                                magnitude of the load by the
ordinate of the response
yB                               function influence line at the
D

A    B            C
2. Maximum positive value of
the response function is
ILD for MB
-yD         obtained by multiplying the
positive ordinate. Similarly, the
+
(M B ) max ⇒ place P at B                maximum negative value is
obtained by multiplying the
−
(M B ) max ⇒ place P at D                point load by the maximum
29
negative ordinate.            30

Point Response Due to a                     dR = dP y = w dx y
Uniformly Distributed Live             where y is the influence line
Load                                   ordinate at x, which is the point of
application of dP.
Influence lines can also be
employed to determine the              To determine the total response
values of response functions of        function value at a point for a
structures due to distributed          distributed load between x = a to x
loads. This follows directly from      = b, simply integrate:
point forces by treating the
uniform load over a differential                 b                b
segment as a differential point
force, i.e., dP = w dx. Thus, a
R=   ∫ w ydx = w ∫ ydx
response function R at a point                   a                a
can be expressed as
31                                    32

8
in which the last integral expres-
sion represents the area under the    2. To determine the maximum
segment of the influence line,           positive (or negative) value of a
which corresponds to the loaded          response function due to a
portion of the beam.                     uniformly d st buted live load,
u o y distributed e oad,
the load must be placed over
SUMMARY                          those portions of the structure
where the ordinates of the
1. The value of a response               response function influence line
function due to a uniformly           are positive (or negative).
portion of the structure can be
obtained by multiplying the load   Points 1 and 2 are schematically
intensity by the net area under    demonstrated on the next slide for
the corresponding portion of the   moment MB considered in the point
response function influence 33     load case.
34
line.

35                                     36

9
Where should a CLL
on the typical ILD’s shown below               Beam Shear ILD
to maximize the response
functions?

Typical Interior
Bending Moment ILD

Typical End Shear
(Reaction) ILD
Possible Truss Member ILD
37                                       38

To calculate the response
Highway and                     function for a given position of
simply multiply the value of each
Live loads due to vehicular traffic    series load Pi by the magnitude
on highway and railway bridges           f h i fl        line diagram
of the influence li di
are represented by a series of         ordinate  yi at the position of Pi ,
specified spacing between the                R = ∑ Pi yi
loads. In this section, we discuss                i
the use of influence lines to          The ordinate magnitude yi can be
determine: (1) the value of the
( )                       calculated from the slope of the
response function for a given          influence line diagram (m) via
position of a series of concentrated
loads and (2) the maximum value           yi = m x i
of the response function due to a
series of moving concentrated 39                                         40

10
where x i is the distance to point i             For example, consider the ILD
xi
measured from the zero y-axis                    shown on the next slide
intercept, as shown in the                       subjected to the given wheel
m
L dP ii 1
1
yb
ya
VB1 = 8( 1 20) + 10( 1 16) + 15( 1 13) +5( 1 8)
30          30         30         30
x
=   (  1 )(8(20) + 10(16) + 15(13) + 5(8))
a                                           30
b                             = m∑Pi xi = 18 5k
18.5k
i
ya y b
=    ⇒ similar triangles
a    b
y          y
∴ ya = b a ; m = b                        41                                                42
b          b

2/3

10 ft.

ft.
20 ft

-1/3                                             Position 1

ILD for Internal Shear SB

Position 2
43                                                44

11
influence line ordinate before the
V = 1 (−8(6) + 10(20) + 15(17) + 5(12))
B2                                       lighter loads in the series. In such
30
= 15.6k                               a case, it may not be necessary to
Thus, load position 1 results in the
,      p                              Instead the analysis can be
maximum shear at point B.                 ended when the value of the
response function begins to
NOTE: If the arrangement of               decrease; i.e., when the value of
loads is such that all or most of the     the response function is less than
of the ends of the series, then the       p
process is known as the
analysis can be expedited by              “Increase-Decrease Method”.
selecting a direction of movement
for the series so that the heavier
45                                        46

CAUTION: This criterion is not            Zero Ordinate Location
valid for any general series of
Linear Influence Line
and shape of the influence line,
1
h     l    f h
the value of the response                                 m+
function, after declining for some
x-
increasing again for subsequent
m-
1
higher maximum.                                      x+                      b-

L

47                                        48

12
b −b          Example Truss Problem:
x + = −b+       ; m+ = − +          Application of Loads to
m+         L
Maximize Response
b −b
x − = −b−      ; m− = + −
m−         L

NOTE: Both of these solutions
are obtained from
y = mx + b
with y = 0.

ML

49
CM     50

Place                                Force and Moment
UDL = 1.0 k/ft;                     Envelopes
ULL = 4.0 k/ft;                A plot of the maximum
CLL = 20 kips                  response function as a
to maximize the tension and         function of the location of the
compression axial forces in         response function is referred
members CM and ML.                  to as the envelope of the
maximum values of a
Calculate the magnitudes of the     response function for the
tension and compression forces.     particular load case being
id d
considered.

51                                     52

13
For a single concentrated           For a uniformly distributed
force for a simply sup-             load on for a simply sup-
ported beam:                        ported beam:

(V)+ =
w
max
⎛ a⎞
(V) + = P ⎜1− ⎟                      ( )max        ( L − a )2
2L
⎝       L⎠
w a2
(V)− = − P
a                          (V)− = −
max
max                                              2L
L
⎛ a⎞                               w a
M max = P a ⎜1− ⎟                     M ma =
max        (L −a )
⎝ L⎠                                2

Plot is obtained by treating “a”    Plot is obtained by treating “a”
as a variable.                      as a variable.
53                                  54

55                                  56

14

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