Flashboard Pin DesignGeneric Flashboard Calculator

This flashboard calculator is a useful tool to check the design of existing flashboards or it can be used to design new flashboards. It has three sections and an appendix. Section one is used todetermine the depth of water, overtopping the boards, at failure, for an existing flashboarddesign. Section Two will calculate the necessary flashboard pin spacing, to produce pin failure, given a pin diameter, height of boards and height of water overtopping the boards. SectionThree will calculate the necessary solid pin or pipe diameter to produce pin failure, for a knownpin centerline spacing, height of flashboards and height of water that is overtopping the boards. TheAppendix uses a known flashboard pin nomograph, from Creager and Justin, to check thespreadsheet calculations. Each section has its own inputs and its calculations are independent from the other section's calculations.

Section One:

Determine the depth of water overtopping the flashboards at their point of failure.

Please input the known height of the flashboards, the pin spacing and the pin diameterof the solid pin or the outside and inside diameters of the pipe pin.

I) Input

a) board height (H)= 24 inches <<<<<<<b) Pin spacing(L)= 24 inches <<<<<<<<<< enter data herec) Pin OD (D)= 1.125 inch <<<<<<<d) Pin ID (d)= 1 inch <<<<<<<e) Pin OR (R)= 0.5625 inchf) Pin IR (r)= 0.5 inchg) Extrem fiber stres= 75000 psi <<<at time of failure, C&J, 1950 editionh) gammaH2O= 0.0361 lbs/in^3 <<<<<<62.4 lbf/ft^3/1728 in^3/ft^3

II) Free Body Diagram:

III) Forces and Moments

a) F1= (L*H)*h*gammaH2O= 707 pounds/pin space <<< here h is from section 2b) F2= (L*H)*(H/2)*gammaH2O= 353 pounds/pin space for illustrative purposesc) M= (H/2)*(L*H)*gammaH2O + (H/3)* (L*H)*(H/2)*gammaH2O= gammaH2O*((L*H^2)/2)*(h+H/3)= gammaH2O*((L*H^2)/6)*(3*h+H)M= 11,310 ft-lbs

IV) Moment of Inertias:

For a solid round pin the radial moment of inertia about its central axis is:

Moment of Inertia (Is)= phi*(R^4)/4 0.079 in^4

For a hollow pin (a piece of pipe) the radial moment of inertia about its central axis is:

Moment of Inertia (Ip)= phi*(R^4-r^4)/4 0.030 in^4

V) Section Moduli:

For a round pin the extreme fiber stress is located at the external radius. Thus R=y

a) For a solid round pin the section modulus, S is:

Section modulus (Ss))= (phi*(R^4)/4)/R= phi*(R^3)/4 0.140 in^3

b) For a hollow round pin (pipe) the section modulus, S is:

Section modulus (Sp))= (phi*(R^4-r^4)/4)/R 0.059 in^3

V) Stresses

From the flexure formula:

sigmaFailure= (M*Y)/I

Moment at Failure= (sigmaFailure*I)/y= sigmaFailure*S

Setting A40 equal to A59 yields

sigmaFailureS=gammaH2O)*((L*(H^2)/6)*(3*h+H)

VI) Solving for h yields:

h= (((3*sigmaFailure*phi*R^3)/(2*gammaH2O*L*H^2))-H)/3

For the solid pin this reduces to:

h= (((3*sigmaFailure*phi*R^3)/(2*gammaH2O*L*H^2))-H)/3

h= 34.0 inches <<<<<<height of overtopping at failure

For the hollow pin this reduces to:

h= (((3*sigmaFailure*phi*(R^3-r^3))/(2*gammaH2O*L*H^2))-H)/3

h= 7.8 inches <<<<<<height of overtopping at failure

Determine the depth of water overtopping the flashboards at their point of failure.

William Fay P.E.Mar-09

For the solid pin this reduces to:

h= (((3*sigmaFailure*phi*R^3)/(2*gammaH2O*L*H^2))-H)/3

h= 34.0 inches <<<<<<height of overtopping at failure

For the hollow pin this reduces to:

h= (((3*sigmaFailure*phi*(R^3-r^3))/(2*gammaH2O*L*H^2))-H)/3

h= 7.8 inches <<<<<<height of overtopping at failure

Section Two:

Determine the flashboard's pin spacing (for failure with 1 ft of freeboard)

Please input the known height of the flashboards, the height of water overtopping the boards (freeboard minus 1ft)and the pin diameter of the solid pin or the outside and inside diameters of the pipe pin.

VII) Input

a) board height (H)= 24 inches <<<<<<<b) Overtopping(h)= 24 inches <<<<<<<<<< enter data herec) Pin OD (D)= 1.125 inch <<<<<<<d) Pin ID (d)= 1 inch <<<<<<<e) Pin OR (R)= 0.5625 inchf) Pin IR (r)= 0.5 inchg) Extrem fiber stres= 75000 psi <<<at time of failure, C&J, 1950 editionh) gammaH2O= 0.0361 lbs/in^3 <<<<<<62.4 lbf/ft^3/1728 in^3/ft^3

VIII) For a solid pin

L= (3*sigmaFailure*phi*R^3)/(2*gammaH2O*H^2*(3*h+H))

L= 31 8/16 inches2.62 feet

IX) For a pipe pin

L= (3*sigmaFailure*phi*(R^4-r^4)/R)/(2*gammaH2O*H^2*(3*h+H))

L= 11 13/16 inches0.99 feet

Determine the flashboard's pin spacing (for failure with 1 ft of freeboard)

Please input the known height of the flashboards, the height of water overtopping the boards (freeboard minus 1ft)

Section Three:

Determine the flashboard's pin diameter at their point of failure.

Please input the known height of the flashboards, the desired height of water overtopping the boards at incipient failure and the pin spacing.

XII) Input

a) board height (H)= 4 inches <<<<<<<b) Overtopping(h)= 39.6 inches <<<<<<<<<< enter data herec) Pin spacing (L)= 48 inchesd) Extrem fiber stres= 75000 psi <<<at time of failure, C&J, 1950 editione) gammaH2O= 0.0361 lbs/in^3 <<<<<<62.4 lbf/ft^3/1728 in^3/ft^3

For this case, the equation reduces to a cubic. It is easiest solved by setting the equation equalto a residual, R. The equation can be iteratively resolved until the residual is near zero.

XIII) For a solid pin:

Residual=D^3-((16*gammaH2O*L*H^2*(3*h+H))/(12*pi*sigmaFailure))

Guess a D in inches= 0.4375

Residual= 0.006637

XIV) For a hollow, round pin (pipe):

First solve the problem for a solid pin. Then find the section modulus for the solid pin. Look in the attached pipe properties table for a pipe with the equivalent section modulus.

Residual=D^3-((16*gammaH2O*L*H^2*(3*h+H))/(12*pi*sigmaFailure))

Guess a D in inches= 0.5 inches <<<<<<insert guessed diameter until residual is near zero

Residual= 0.047896

For a solid, round, pin the section modulus, S is:

Section modulus (Ss))= (phi*(R^4)/4)/R= phi*(R^3)/4 0.008 in^3 <<<<look this number up in table

Determine the flashboard's pin diameter at their point of failure.

For this case, the equation reduces to a cubic. It is easiest solved by setting the equation equal

<<<<<<insert guessed diameter until residual is near zero

in^3 <<<<look this number up in table

APPENDIX ONE

Flashboard Pin Calculator Check

This nomograph is from the 1950 version of Creager and Justin. It predicts that for 3 feet tallflashboards, with 2 feet of overtopping, a 50,000 psi extreme fiber stress and 5 foot distancebetween pin spacing, a 2 1/2 inch standard pipe would work as a pin and a 2 1/8 solid pin canalso be used the section modulus for the solid pin is 1.1

Let us put these values in Section XIV of the pin calculator and predict the section modulus and solidpin size to see if we can predict the values obtained from the nomograph.

XII) Input

a) board height (H)= 36 inches <<<<<<<b) Overtopping(h)= 24 inches <<<<<<<<<< enter data herec) Pin spacing (L)= 60 inchesd) Extrem fiber stres= 50,000 psi <<<at time of failure, C&J, 1950 editione) gammaH2O= 0.0361 lbs/in^3 <<<<<<62.4 lbf/ft^3/1728 in^3/ft^3

For this case, the equation reduces to a cubic. It is easiest solved by setting the equation equalto a residual, R. The equation can be iteratively resolved until the residual is near zero.

XIV) For a hollow, round pin (pipe):

First solve the problem for a solid pin. Then find the section modulus for the solid pin. Look in the attached pipe properties table for a pipe with the equivalent section modulus.

Residual=D^3-((16*gammaH2O*L*H^2*(3*h+H))/(12*pi*sigmaFailure))

Guess a D in inches= 2.175 inches <<<<<<insert guessed diameter until B154 is near zero

Residual= -0.009683

For a solid, round, pin the section modulus, S is:

Section modulus (Ss))= (phi*(R^4)/4)/R= phi*(R^3)/4 1.010 in^3 <<<<look this number up in table

Now, if we enter Table One with a section modulus S= 1.010 and pick the nearest schedule 40 pipe it is:

2 1/2 inch

For this case, the equation reduces to a cubic. It is easiest solved by setting the equation equal

in^3 <<<<look this number up in table

Now, if we enter Table One with a section modulus S= 1.010 and pick the nearest schedule 40 pipe it is: