(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

There is an unstayed mast which is 3m long, supported at the bottom and at 0.5m from the bottom.

Sail area is 10 m^2, design wind speed is 50 km/h.

It is made from glass epoxy composite.

The base fiber is 300 g/m^2 glass textile. E-glass, 2/2 twill.

The inner diameter is 40mm.

Develop the lamination schedule for the mast.

2. Relevant equations

Force of wind on the sail:

[tex]F_{0}=\frac{1}{2} A C_{s} \rho AWS^{2}[/tex]

Euler-Bernoulli beam equation:

[tex]Rp_{02} = \frac{M y}{Ix}[/tex]

second moment of inertia in a circular beam:

[tex]Ix = \frac{1}{4} \pi \left(ro^{4} - ri^{4}\right)[/tex]

3. The attempt at a solution

My first problem is to come up with a sensible yielding tensile strength for the material.

I have the following data:

1.) Flexural "Maximum resistance" of a composit made of E Glass, 2/2 Twill, 300 g/m² is 662 MPa

2..) ABS standard gives 124 MPa for "Tensile Strength" for "basic laminate", which is defined as:

"""

The basic laminate consists of an unsaturated general-purpose polyester resin and alternate plies of

E-glass, fiberglass mat and fiberglass-woven roving fabricated by the contact or hand lay-up process.

The minimum glass content of this laminate is 35% by weight.

"""

3. minimum ultimate tensile strength in fiber direction of a laminate made by 300 g/m^2 plain weave is 70.4 ksi = 485 Mpa

Somewhat arbitrarily I have choosen 160 MPa as yielding tensile strength. (I call it Rp02, because this number is actually yield strength of T-66 Al).

Line of thought:

- as the composite is stiff, yield strength must be close to ultimate strength

- both composites close to the choosen one have far more than 400Mpa as ultimate tensile strength. Choosing a value less than half of that should account for both difference in yielding and ultimate strength and safety factor. This is also not much more than "basic laminate" which is obviously weaker than our material due to the reinforcement material used.

Is the choosen value a safe one?

known values:

[tex]$

A = 10 m^{2} \\

AWS = 17.3611111111111 \frac{m}{s} \\

C_{s} = 1.1 \\

Rp_{02} = 160000000.0 \frac{N}{m^{2}} \\

l_{1} = 2.5 m \\

l_{2} = 0.5 m \\

\rho = 1.2 \frac{kg}{m^{3}} \\

ri = 0.02 m \\

[/tex]

Though F0 acts in the center of the sail, I have choosen to account for it at the top of the mast for additional safety margin and ease of calculation.

[tex]F_{0}=\frac{1}{2} A C_{s} \rho AWS^{2}[/tex]

[tex]F_{1}=- \frac{F_{0} l_{1} + F_{0} l_{2}}{l_{2}}[/tex]

[tex]F_{2}=- F_{0} - F_{1}[/tex]

shear:

[tex]$\begin{cases} 0 & \text{for}\: x < 0 \\F_{0} & \text{for}\: \operatorname{And}\left(0 \leq x,x < l_{1}\right) \\F_{0} + F_{1} & \text{for}\: \operatorname{And}\left(l_{1} \leq x,x < l_{1} + l_{2},0 \leq x\right)

\\F_{0} + F_{1} + F_{2} & \text{for}\: \operatorname{And}\left(l_{1} + l_{2} \leq x,l_{1} \leq x,0 \leq x\right) \end{cases}$[/tex]

moment:

[tex]\begin{cases} 0 & \text{for}\: x < 0 \\F_{0} x & \text{for}\: \operatorname{And}\left(0 \leq x,x < l_{1}\right) \\F_{0} l_{1} + x \left(F_{0} + F_{1}\right) - l_{1} \left(F_{0} + F_{1}\right) & \text{for}\: \operatorname{And}\left(l_{1} \leq x,x < l_{1} + l_{2},0 \leq x\right)

\\ F_{0} l_{1} + x \left(F_{0} + F_{1} + F_{2}\right) + \left(F_{0}

+ F_{1}\right) \left(l_{1} + l_{2}\right) - l_{1} \left(F_{0} + F_{1}\right) - \left(l_{1} + l_{2}\right) \left(F_{0} + F_{1} + F_{2}\right) & \text{for}\: \operatorname{And}\left(l_{1} + l_{2} \leq x,l_{1} \leq x,0 \leq x\right)\end{cases}[/tex]

Euler-Bernoulli beam equation:

[tex]Rp_{02} = \frac{M y}{Ix}[/tex]

second moment of inertia in a circular beam:

[tex]Ix = \frac{1}{4} \pi \left(ro^{4} - ri^{4}\right)[/tex]

substituing together:

[tex] Rp_{02} = 4 \frac{M ro}{\pi \left(ro^{4} - ri^{4}\right)} [/tex]

to get the wall thickness, substitue ro = ri + dr:

[tex]Rp_{02} = 4 \frac{M \left(dr + ri\right)}{\pi \left(\left(dr + ri\right)^{4} - ri^{4}\right)}$[/tex]

Substituing moment:

[tex]Rp_{02} = 4 \frac{\left(dr + ri\right) \begin{cases} 0 & \text{for}\: x < 0 \\F_{0} x & \text{for}\: \operatorname{And}\left(0 \leq x,x < l_{1}\right) \\F_{0} l_{1} + x \left(F_{0} + F_{1}\right) - l_{1} \left(F_{0} + F_{1}\right) & \text{for}\: \operatorname{And}\left(l_{1} \leq x,x < l_{1} + l_{2},0 \leq x\right)

\\F_{0} l_{1} + x \left(F_{0} + F_{1} + F_{2}\right) + \left(F_{0} + F_{1}\right) \left(l_{1} + l_{2}\right) - l_{1} \left(F_{0} + F_{1}\right) - \left(l_{1} + l_{2}\right)

\left(F_{0} + F_{1} + F_{2}\right) & \text{for}\: \operatorname{And}\left(l_{1} + l_{2} \leq x,l_{1} \leq x,0 \leq x\right) \end{cases}}{\pi \left(\left(dr + ri\right)^{4} - ri^{4}\right)}[/tex]

Solving it for dr and choosing the positive real solution for each 0.1m gives:

0.0 None

0.1 0.000965014828703786

0.2 0.00188223960651223

0.3 0.00275403303707307

0.4 0.00358340696727762

0.5 0.00437357386074760

0.6 0.00512769600546226

0.7 0.00584875779631873

0.8 0.00653950969566895

0.9 0.00720245225745922

1.0 0.00783984146648781

1.1 0.00845370456306660

1.2 0.00904586024950311

1.3 0.00961793994178479

1.4 0.0101714083263240

1.5 0.0107075823892620

1.6 0.0112276485919169

1.7 0.0117326781397463

1.8 0.0122236404327593

1.9 0.0127014148504254

2.0 0.0131668010473501

2.1 0.0136205279367525

2.2 0.0140632615282017

2.3 0.0144956117704763

2.4 0.0149181385332672

2.5 0.0153313568446162

2.6 0.0131668010473501

2.7 0.0107075823892620

2.8 0.00783984146648781

2.9 0.00437357386074760

Layer thickness is 0.25mm for each 100 g/m^2 according to ABS standard, so dividing the above numbers with 0.00075 and rounding up gives the number of layers:

0.1 2

0.2 3

0.3 4

0.4 5

0.5 6

0.6 7

0.7 8

0.8 9

0.9 10

1.0 11

1.1 12

1.2 13

1.3 13

1.4 14

1.5 15

1.6 15

1.7 16

1.8 17

1.9 17

2.0 18

2.1 19

2.2 19

2.3 20

2.4 20

2.5 21

2.6 18

2.7 15

2.8 11

2.9 6

I add 45 degree layers of 100gr cloth on top of all 300g 0 degree layers for buckling resistance and easier layout.

I use 21 layers from 2.4 - 3.0 to make the bottom uniform.

These modifications should add to the strength considerably.

So the laminating schedule is the following (L denotes a combination of 300gr 0degree and 100g 45 degree layers):

2*L 0.0-3.0

1*L 0.2-3.0

1*L 0.3-3.0

1*L 0.4-3.0

1*L 0.5-3.0

1*L 0.6-3.0

1*L 0.6-3.0

1*L 0.7-3.0

1*L 0.8-3.0

1*L 0.9-3.0

1*L 1.0-3.0

1*L 1.1-3.0

1*L 1.2-3.0

1*L 1.4-3.0

1*L 1.5-3.0

1*L 1.7-3.0

1*L 1.8-3.0

1*L 2.0-3.0

1*L 2.1-3.0

1*L 2.3-3.0

1*L 2.4-3.0

Questions:

- Is the choice of tensile strength sensible?

- Are all the design choices (putting sail force at the top, adding layers to the schedule) safe?

- Are there any flaws in the calculations or logic?

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# Homework Help: Sizing a mast

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