Bending Moments and Shears: Equal Span Coefficients

For two span Beam

Using the three moment equation

M_AL + 2M_B(L+L) + M_CL = (wL³/4) + (wL³/4)

Since M_A = M_B = 0

4M_BL = wL³/2

M_B = wL²/8 = 0.125wL²………………………………….......…..1

Reaction R_AB

Note: S. S. R. means simple suport reaction
          C. C. means continuity correction

S.S.R =wl/2

C.C = (|M_A| - |M_B|)/L = (-M_B)/L = wl/8                  

 Reaction R_AB = ∑R = S.S.R + C.C.

                        = wL/2-wL/8 =3wL/8

            R_AB =3wL/8 = 0.375wL…………………………………2

Reaction R_BA

S.S.R, = wL/2

C.C = (|M_B|-|M_A|)/L = (M_B)/L  

Reaction R_AB = ∑R = S.S.R + C.C. = wL/2 + wL/8 = 5wL/8

R_BA =5wL/8 = 0.625wL................................................................3

Due to symmetry, R_AB =R_CB = 3wL/8

R_BA = R_BC =5wL/8

Maximum Span Moment on Span A-B

M_x = R_AB - wx²/2...........................................................................4

At maximum moment

dM_x/d_x = R_AB - wx = 0

x = R_AB/w = 3L/8 = 0.375L

Substituting for x = 3L/8 in equ (4)

M_max = (3wL/8).(3L/8) - (w/2). (3L/8)²

              = 9wL²/64( 1-1/2) = 9wL²/128

 M_max = 9wL²/128 = 0.0703wL²...................................................5

Alternatively,

M_x = R_BA - wx²/2 - M_B

At maximum moment

dM_x/d_x = R_BA - wx = 0

x = R_BA/w = 5L/8

Substituting for x = 5L/8 in equ (4)

M_max = (5wL/8).(5L/8) - (w/2). (5L/8)² -wL²/8

= 9wL²/128

M_max = 9wL²/128 = 0.0703wL².................................................6

For three span Beam

Using the three moment equation

SPANS A-C

M_AL + 2M_B(L+L) + M_CL = wL³/4 + wL³/4

M_A = 0

4M_BL + M_CL = wL³/2………………………………………………7

SPANS B-D

M_BL + 2M_C(L+L) + M_DL = wL³/4 + wL³/4

M_D = 0

M_BL + 4M_CL = wL³/2………………………………………………8
Equating 7 and 8

4M_BL + M_CL = M_BL + 4M_CL

4M_B + M_C  = M_B + 4M_C

3M_B = 3M_C

M_B = M_C

Substituting for M_B = M_C in equation 7
4M_CL + M_CL = wL³/2

5M_CL = wL³/2

M_C = wL²/10 = 0.1wL²....……………………………………………9

Since M_B = M_C

M_B = M_C = wL²/10 = 0.1wL²

S.S.R = wL/2

C.C = (|M_A|-|M_B|)/L = (- M_B)/L = wL/10
R_AB = ∑R = S.S.R + C.C.  = wL²/2-wL/10

R_AB = 4wL/10 = 0.4wL......................................................................10
Similarly

R_BA = wL/2 + wL/10 = 6wL/10

R_BA = 6wL/10 = 0.6wL……………………………………………..11
S.S.R. = wL/2
C.C = (|M_B|-|M_C|)/L = 0

R_BC =R_CB = wL/2 = 0.5wL.................................................................12

Maximum Span Moment on Span A-B

M_x = R_AB - wx²/2...........................................................................13

At maximum moment

dM_x/d_x = R_AB - wx = 0

x = R_AB/w = 4L/10 = 0.4L

Substituting for x = 4L/10 in equ (13)

M_max = (4wL/10).(4L/10) - (w/2). (4L/10)²

              = 16wL²/100( 1-1/2) = 16wL²/200

 M_max = 16wL²/200 = 0.08wL²......................................................14

Maximum Span Moment on Span B-C

M_x = R_BC - wx²/2 - |M_B| ...............................................................15

At maximum moment

dM_x/d_x = R_BC - wx = 0

x = R_BC/w = 5L/10

Substituting for x = 5L/10 in equ (15)

M_max = (5wL/10).(5L/10) - (w/2). (5L/10)² -wL²/10

= 25wL²/200 - wL²/10

M_max = wL²/40 = 0.025wL²............................................................16

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

              

                         

Grillage Analysis:Wheel Load Transfer to Nearest Joints

Grillage Analysis

      Wheel Load Transfer to nearest joints.

First Step

Cutting section along e-f

R_e x l₁ = P x b

R_e = P x b/l₁

Similarly;      

R_f = P x a/l₁

Second Step

Considering beam 1-4

R₁ = R_e x d/l₂

Thus:

R₁ = P x b x d/ (l₁ x l₂) = (b x d/l₁ x l₂) x P

Similarly:

R₄ = R_e x c/l₂

R₄ = P x b x c/ (l₁ x l₂) = (b x c/l₁ x l₂) x P

 

Considering beam 2-3

R₂ = R_f x d / l₂

R₂ = P x a x d / (l₁ x l₂) = (a x d /l₁ x l₂) x P

Also:

R₃ = R_f x c / l₂

R₃ = P x a x c / (l₁ x l₂) = (a x c / l₁ x l₂) x p

R x l₁ x l₂ = P x b x d

R₁ = (b x d / l₁ x l₂) x P

                                       Similar argument can be applied to the other nodes.

Vortex Induced Vibration-Oil Platform Riser (English Units)

CROSS FLOW VORTEX INDUCED VIBRATION ANALYSIS
Global Variables - Input
Prepared by:				Checked by:
Pipe Outside Diameter, Do =						18	in
Pipe Wall Thickness,t =						0.562	in
Concrete weight coating thickness, tc =						1.5	in
Young Modulus -Steel, E =						2.90E+07	psi
Sea current velocity for a 100 yr Return period storm, U =						3	ft/s
Specified minimum yield strength of pipe, Sy =						52000	psi
Strouhal number, St =						0.219
Kinematic viscosity of sea water at 70oC, vk						0.0000113	ft2/s
End fixity constant, C =(kπ)/2						2.55
Added mass coefficient, Cm						1
Concrete weight coating density, ρoc =						140	lbs/ft3
Density of Steel, ρos =						490	lbs/ft3
Density of Seawater, ρow =						64	lbs/ft3
Logarithm Decrement, δ						0.05
ψT =	Normal Safety Class coefficient					2
ψD =	Period transformation Factor					1
ψR =	Natural frequency Reduction Factor					1
ψU =	Extreme Current Variability factor					1
Pipe Inside Diameter, Di = Do - 2t						16.876	in
Total pipe outside diameter with concrete coating, Dtot = Do+2tc						21	in
Moment of inertia of cross section, I						1171.637829	in4
Damping ratio, ζ = δ2/(δ2+4π2)						0.007956464
Weight of displaced water, ww =(π/4)*Dtot2*ρow						153.958	lbs/ft
Weight of contents, wocn =(π/4)*Di2*ρocn						99.4266266	lbs/ft
Dry weight of pipe with concrete coating, ws
=(π/4)*((Do2-Di2)*ρos +(Dtot2-Do2))*ρoc						194.1292651	lbs/ft
Specific weight of pipe, (ρs/ρw) = (ws/ww)						1.260923532
Dynamic mass, wd =(ww + wocn + ws) / g						13.89794695	slugs/ft
Stability Parameter, Ks =π2 ζ ((ρs/ρw) -Cm)						0.020494895
Free Span Reduced Velocity, VR,ONSET CROSS FLOW
=((π3)*((ρs/ρw + Cm)/(1.5+St2+(ρs/ρw + Cm)-1.5Ks2)))^0.5						3.797567143
Limiting criteria for the onset of Cross flow Vibration,
fn ≥ (U/(VR,ONST*Dtot))*ΨT *ΨD*ΨR*ΨU						0.902833656	Hz
Maximum Allowable Free Span length, L							ft
=(((C*((EI/M)^0.5)))/fn)^0.5)						107.8787189	ft