Monday, 2 April 2012

Optimum Positions for HB vehicle for Maximum Longitudinal Moments and Support Reactions on Short Span Bridge Decks


Plan of Bridge Deck
The HB vehicle is placed on the bridge deck as shown above
Where   L = span of bridge or beam
                X = varying distance of axle 1 from left support A
  P = wheel load for number of units of HB vehicle under consideration
Taking moment about A
RB.L = 4P (9.6 +x) + 4P(7.8 + x) + 4P(1.8 + x) + 4Px
RB.L = 4P(9.6 + 7.8 + 1.8 +x +x +x + x)
         = 4P(19.2 + 4x)
RB = 4P  x 4(4.8 + x)
         L
 
RB = 16P(4.8 + x)
         L
RB = 16P(4.8 + x)……………………………………………………………….(1)
         L
Corresponding relation at A
 RB =16P -  16P (4.8 + x)…………………………………………………………………………………..(2)
                   L
Maximum moment should exist under axle 2 at (x + 1.8)m distance from A
1-1(4.8 + 2x)  =  1.8
    L                      L       
L – (4.8 + 2x) =  1.8
L – 4.8-2x = 1.8
L-2x = 6.6
L – 6.6 = 2x
X =  L – 6.6  =  L – 3.3
           2           2

X =    L – 3.3…………………………………………………………………………………………………………………(3)
         2
This is the distance of axle  1 from A at maximum moment.

The distance of axle 2 from A is
X + 1.8   =  L   3.3 + 1.8  =     L – 1.5
                  2                            2
(x + 1.8)  =   L – 1.5…………………………………………………………………………………………………..(4)
                    2 
 
i.e. the vehicle is placed such that the centire line of the bridge is equidistant from the axle 
 2 and the centre line of the HB vehicle.
Mmax = RA(x + 1.8) – 4P X 1.8
          = (16P  16P(4.8 +x))(x + 1.8) – 4.8 X 1.8
                             L                          
= (16P  16P(4.8 +L – 3.3))(L -3.3 + 1.8) – 7.2P
                 L            2            2         

= (16P  16P(L + 1.5))(L -1.5) – 7.2P
                 L    2            2 
= [(16P  16P. L -16P . 3)( L – 1.5)] – 7.2P
                   L  . 2   L  .   2   2
= [(8P  – 8P . 3)   (L -1.5))]  – 7.2P
                   L        2

= 8P(1  3) .  (L -1.5)  – 7.2P
               L       2 

= 8P[( L – 3) .  (L -3)] -7.2P
             L           2 
Mmax = 8P( L – 3)2 – 7.2P ………………………………………………………………(5)
                 2L          
Calculation for Max RA
From eqn 1
              
RA = 16P -    16P(4.8 + x)
                      L
RA is a maximum when X=0
Therefore;
   RA = 16P (1 – 4.8)
                          L
RA = 16P (L – 4.8)………………………………………………..(6)
              L


              




                         

3 comments:

  1. i need to know how do u form the equation 1-1/L(4.8+2x)=1.8/L ?

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  2. 1-1(4.8 + 2x) = 1.8
    L L

    where does this come from? pls help

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  3. It means maximum moment under the second axle.At a distance of x+1.8 or you could write the full moment equation and differentiate to obtain the maximum and solve for x.

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