The Golden Spiral and The Fibonacci Numbers

The Golden Spiral Approximately derived from the Fibonacci Numbers :

1² + 1² + 2² + 3² + 5² + 8² +13² + F(n)²

Shells: Example of the Golden Spiral in Nature

Shells: Example of the Golden Spiral in Nature

The Cosmos: Example of the Golden Spiral in Nature

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

R_B.L = 4P (9.6 +x) + 4P(7.8 + x) + 4P(1.8 + x) + 4Px

R_B.L = 4P(9.6 + 7.8 + 1.8 +x +x +x + x)
         = 4P(19.2 + 4x)

R_B = 4P  x 4(4.8 + x)

         L

 

R_B = 16P(4.8 + x)

         L

R_B = 16P(4.8 + x)……………………………………………………………….(1)

         L

Corresponding relation at A

 R_B =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.

M_max = R_A(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 

M_max = 8P( L – 3)² – 7.2P ………………………………………………………………(5)

                 2L          

Calculation for Max R_A

From eqn 1

              

R_A = 16P -    16P(4.8 + x)

                      L

R_A is a maximum when X=0

Therefore;

   R_A = 16P (1 – 4.8)

                          L

R_A = 16P (L – 4.8)………………………………………………..(6)

              L