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9785f4b2-92de-4cd9-aa22-0231e5e69df0 61 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 50 knot * : 25.7 m s / *

, density of air @ -20C

, wind velocity

These are used to calculate the dynamic pressure of air:

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P.d 12 / ρ * V.w 2^* :

P.d 463 Pa *

Then the dynamic pressure and sail area is used to calculate the total wind force on the sail. This assumes the sails are set perpendicular to the wind, such as when running:

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A.s 39.5 m 2^* :

, one mast

F.w P.d A.s * :

F.w kN 18.3

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 5.5

F.c F.w F.d - : kN 12.8

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 57.3

R.b R.a F.c + F.d + : kN 75.6

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

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x V.ab R.a - : kN 57.3 -

, x: 0 -> Lab

1 V.ab 57289.4591 N * -

x V.bc R.a - R.b + : kN 18.3

, x: Lab -> Lab + Lbc

x V.cd R.a - R.b + F.c - : kN 5.5

, x: Lab + Lbc -> Ltot

V

x V x V.ab 1000 / x L.ab ≤ (x V.bc 1000 / x L.ab L.bc + ≤ x V.cd 1000 / cases :

V.Cab Lab V.ab : 57289.4591 N * -

V.Cbc L.ab V.bc : 18294.113 N *

V.Ccd L.ab Lbc + V.cd : 5534.3535 N *

x m * V

Graph 1: Shear on Mast

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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 + ≡

, V is a constant

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 M.ab + L.ab L.bc + M.bc + ≡

, x: Lab + Lbc -> Ltot

M

x M.ab V.Cab x * :

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

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

x m x 0 m * ≥ (x L.ab < (& (x M.ab * x L.ab ≥ (x L.ab L.bc + < (& (x M.bc * + x L.ab L.bc + ≥ (x L.tot ≤ (& (x M.cd * + 1000 / :

x m * m

Graph 2: Moment on Mast

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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

Lower Section of Mast

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Upper Section of Mast

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Diagram 3: Cross Section of Mast

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In free standing masts there is a great deal more shear and tension in the materials due to no halyards to offset the forces. Since our plan is to use E-glass instead of carbon fiber in the construction of the mast we have to compensate for the short comings of E-glass when compared to carbon. The easiest way to do this is to use a thin mast for the upper section and thicken it up for the lower section. Diagram 3 shows a planned cross section of the mast above and below a point Ldiv. The "core" will be the full length, acting as the mast in the upper part, and the in the lower section being the base to built up the mast to handle the extreme shear and tension. In building up the lower mast we can keep a minimum weight mast and also facilitate a better shaft and bushing arrangement by giving the bushing more surface area.

Straping is chosen as the main material to handle the stress caused the the moments. Most of straping's fibers are in the longitudinal direction, and it can be easily pretensioned, giving it up to 40% more stength than in the relaxed state.

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LS materials.ods wood B16 C19 importData.ODS :

HDRu materials.ods wood D13 K15 importData.ODS :

BODY materials.ods wood D16 K19 importData.ODS :

m M.cd M.bc M.ab V.Ccd V.Cbc V.Cab V V.cd V.bc V.ab R.b R.a F.c F.d L.tot L.cd L.cd2 L.cd1 L.bc L.ab F.w A.s P.d V.w ρ lastError m M.cd M.bc M.ab V.Ccd V.Cbc V.Cab V V.cd V.bc V.ab R.b R.a F.c F.d L.tot L.cd L.cd2 L.cd1 L.bc L.ab F.w A.s P.d V.w ρ lastError m M.cd M.bc M.ab V.Ccd V.Cbc V.Cab V.cd V.bc V.ab R.b R.a F.c F.d L.tot L.cd L.cd2 L.cd1 L.bc L.ab F.w A.s P.d V.w ρ

BODY
S:BODY
#
#
#
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F

LS Douglas Fir Douglas Fir Western Cedar Western Cedar 14 mat transpose :

HDRu Grade Bending Tens|| Shear|| Comp|- Comp|| Elstic Elstic Density MPa MPa MPa MPa MPa MPa MPa kg/m^3 Fb Ft Fv Fol Fo E Emin 12% 39 mat :

BODY SS 10.3 6.9 1.2 4.3 11.7 131004757545 No1 8.3 5.5 1.2 4.3 10.7 124114551545 SS 6.9 4.1 1.1 2.9 6.9 75842758368 No1 5.0 2.9 1.1 2.9 5.7 6895255136849 mat :

BODY LS m M.cd M.bc M.ab V.Ccd V.Cbc V.Cab V V.cd V.bc V.ab R.b R.a F.c F.d L.tot L.cd L.cd2 L.cd1 L.bc L.ab F.w A.s P.d V.w ρ lastError BODY HDRu LS m M.cd M.bc M.ab V.Ccd V.Cbc V.Cab V V.cd V.bc V.ab R.b R.a F.c F.d L.tot L.cd L.cd2 L.cd1 L.bc L.ab F.w A.s P.d V.w ρ lastError m M.cd M.bc M.ab V.Ccd V.Cbc V.Cab V.cd V.bc V.ab R.b R.a F.c F.d L.tot L.cd L.cd2 L.cd1 L.bc L.ab F.w A.s P.d V.w ρ

BODY
SS	10.3	6.9	1.2	4.3	11.7	13100	4757	545
No1	8.3	5.5	1.2	4.3	10.7	12411	4551	545
SS	6.9	4.1	1.1	2.9	6.9	7584	2758	368
No1	5	2.9	1.1	2.9	5.7	6895	2551	368
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Grade	Bending	Tens\|\|	Shear\|\|	Comp\|-	Comp\|\|	Elstic	Elstic	Density
	MPa	MPa	MPa	MPa	MPa	MPa	MPa	kg/m^3
	Fb	Ft	Fv	Fol	Fo	E	Emin	12%
Douglas Fir
Douglas Fir
Western Cedar
Western Cedar
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