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L8 - Influence Line Diagrams for statically determinate structures by SaiDeepika2

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									                                         Influence Lines
   Live Load Forces:
                                         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
subjected to fixed loads.                maximum probable load each
                                         member will experience.
Structures are also subjected to
live loads whose position may
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
                                         maximum force produced by dead
                                         and live loads is less than the
                                         available section capacity.
  Figure 1. Bridge Truss Structure
        Subjected to a Variable          Structural analysis for variable
        Position Load                    loads consists of two steps:
                                          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

         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.
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
                                             function based on the loading.
downward unit load moving across
the structure.
NOTE: Influence lines for
statically determinate structures
are always piecewise linear.
                                     5                                       6

 Calculating Response                          x
      Functions                                    a
         (Equilibrium Method)                              VB

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

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

            1 – a/L
 VB                                       Beam Example 1
                  a         L

                        ILD for VB

 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

                                                         BD: Link

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

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

                                                                                                           CAUTION: Principle is only valid
    Muller-Breslau                                                                                         for force response functions.
Muller-Breslau Principle ≡ The
                                                                                                           Support reaction - remove
influence line for a response
                                                                                                           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.
                                                                                                           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

Application of Muller-
  Breslau Principle

                         17                       18

                              y = (L – x) (a/L)

                              θ1 + θ2 = 1

                         19                       20

 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-
An influence line diagram with
numerical values of its ordinates is
known as a quantitative influ-
ence line diagram.               21                                    22

    Influence Lines for
In a gable-truss frame building,
roof loads are usually transmitted
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

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.

                                 25                                     26

  Example Truss Structure              Use of Influence Lines

                                       Point Response Due to a
                                       Single Moving
                                       Concentrated Load
                                       Each ordinate of an influence
                                       line gives the value of the
                                       response function due to a
                                       single concentrated load of
  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

                                        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
                                          position of the load.

    A    B            C
                                          2. Maximum positive value of
                                          the response function is
             ILD for MB
                              -yD         obtained by multiplying the
                                          point load by the maximum
                                          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
                                          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

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).
   distributed load applied over a
   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.

                                35                                     36

 Where should a CLL
 (Concentrated Live Load), a ULL
 (Uniform Live Load) and UDL
 (Uniform Dead Load) be placed                   Typical Interior
 on the typical ILD’s shown below               Beam Shear ILD
 to maximize the response

                                               Typical Interior
                                             Bending Moment ILD

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

      Live Loads for
                                       To calculate the response
       Highway and                     function for a given position of
     Railroad Bridges                  the concentrated load series,
                                       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 ,
moving concentrated loads with         i.e.
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


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


    10 ft.

                      20 ft

               -1/3                                             Position 1

   ILD for Internal Shear SB

                                                                Position 2
             Wheel Loads
                                            43                                                44

Load Position 2:
                                          influence line ordinate before the
V = 1 (−8(6) + 10(20) + 15(17) + 5(12))
 B2                                       lighter loads in the series. In such
    = 15.6k                               a case, it may not be necessary to
                                          examine all the loading positions.
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
heavier loads are located near one        the preceding load position. This
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
loads will reach the maximum
                                   45                                        46

CAUTION: This criterion is not            Zero Ordinate Location
valid for any general series of
loads. In general, depending on
                                           Linear Influence Line
the load magnitudes, spacing,             b+
and shape of the influence line,
 h     l    f h
the value of the response                                 m+
function, after declining for some
loading positions, may start
increasing again for subsequent
loading positions and may attain a
higher maximum.                                      x+                      b-


                                   47                                        48

                       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.


                                                                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

                                51                                     52

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

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


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