V

9785f4b2-92de-4cd9-aa22-0231e5e69df0 33 4 5 decimal

&[DATE] &[TIME] - &[FILENAME]

MAST DESIGN

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1: Mast Sheer and Moment

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Diagram 1: Mast Forces

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Diagram 1 shows the forces on the mast due to the wind force on the sails and the resulting forces in the trunk. To start we need to determine the force of the wind. Our design conditions are worst case, giving the following:

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ρ 1.4 kg m 3^/ * :

V.w 31 m s / * :

, density of air @ -20C

, wind velocity @ 60 knots

These are used to calculate the dynamic pressure of air:

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P.d 12 / ρ * V.w 2^* :

P.d 672.7 Pa *

Then the dynamic pressure and sail area is used to calculate the total wind force on the sail:

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A.s 39.5 m 2^* :

, one mast

F.w P.d A.s * :

F.w kN 26.6

The dimensions of the mast as well as the centroid of the sail area were taken from the CAD program and are as follows:

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L.ab 1.52 m * :

L.bc 1.16 m * :

L.cd1 3.6 m * :

L.cd2 8.3 m * :

L.cd L.cd1 L.cd2 + : 11.9 m *

L.tot L.ab L.bc + L.cd + : 14.6 m *

Diagram 2: Mast Dimensions

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The moments about the centroid on the sail sum to zero, so:

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F.c F.w F.d - ≡

eq. 1

The sum of the forces Fc and Fd equal the total force on the sail Fw, so:

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F.d L.cd2 * F.c L.cd1 * ≡

eq. 2

Substituting eq 1 for Fg in eq 2. we get:

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F.d F.w F.d - (L.cd1 L.cd2 / * ≡

F.d F.w L.cd1 L.cd1 L.cd2 + / * :

rearranges to

F.d kN 8

F.c F.w F.d - : kN 18.5

The moments about the deck sum to zero, so:

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R.a L.ab * F.c L.bc * F.d L.bc L.cd + (* + ≡

R.a F.c L.bc * F.d L.bc L.cd + (* + (L.ab / : kN 83.2

And all these forces should sum to zero, so:

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R.b R.a F.c + F.d + : kN 109.8

The shear V is constant between point loads, giving the following:

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x V.ab R.a - : kN 83.2 -

, x: 0 -> Lab

x V.bc R.a - R.b + : kN 26.6

, x: Lab -> Lab + Lbc

x V.cd R.a - R.b + F.c - : kN 8

, x: Lab + Lbc -> Ltot

V

x V x 1.52 ≤ x V.ab 1000 / x 1.52 1.16 + ≤ x V.bc 1000 / x V.cd 1000 / if if :

0 V N 83.2 -

L.ab 2 / V #

x c 0 x ≤ (x 14.6 ≤ (* :

L.ab L.bc 2 / + V #

x p x V x c cases :

L.ab L.bc + L.cd 2 / + V #

14.6 p 8 N *

L.tot V #

x V

Graph 1: Shear V(x) VS Distance

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The relationship between shear and moment here is:

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x M V x x.1 x int ≡

M x diff x V ≡

x M V x x.1 - (* x.1 M + ≡

Starting from the end of the beam this gives the following:

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x M.ab x V.ab x * :

, x: 0 -> Lab

x M.bc X V.bc x L.ab - (* L.ab M.ab + :

, x: Lab -> Lab + Lbc

x M.cd x V.cd x L.ab - L.bc - (* L.ab L.bc + M.bc + :

, x: Lab + Lbc -> Ltot

M

x M x 1.52 ≤ x M.ab 1000 / x * x 1.52 1.16 + ≤ x M.bc 1000 / x M.cd 1000 / if if :

L.ab 1.52 m *

L.bc 1.16 m *

x M x M.ab 1000 / x * x 1.52 ≤ x M.bc 1000 / x 1.52 1.16 + ≤ x M.cd 1000 / cases :

x M x p 2 1 sys

xx 1 - :

xx M 11 mat 83.2112 kg m * s 2^/ * - 11 mat

x p

Don't understand the two different p(x) plots ?

Graph 2: Moment M(x) VS Distance

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To use the calculated shear and moment values to determine the stresses and deflection of the mast we must calculate the moment of inertia of the cross section of the mast. Here are the formulas for circular sections:

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I.cir 0.78 r 4^* ≡

, solid circle

I.thick 0.78 r.o 4^r.i 4^- (* ≡

, thick walled pipe

I.thin 3.14 r 3^* t * ≡

, thin walled pipe of thickness t