Designing an Unstayed Mast for a Sail: Calculations and Lamination Schedule

• magwas
In summary, the conversation discusses the process of developing a lamination schedule for an unstayed mast made of glass epoxy composite. The sail area, design wind speed, and material properties are provided, and the calculation process for determining the wall thickness and number of layers is explained. The conversation also addresses potential design choices and safety considerations.
magwas

Homework Statement

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.

Homework Equations

Force of wind on the sail:
$$F_{0}=\frac{1}{2} A C_{s} \rho AWS^{2}$$
Euler-Bernoulli beam equation:
$$Rp_{02} = \frac{M y}{Ix}$$
second moment of inertia in a circular beam:
$$Ix = \frac{1}{4} \pi \left(ro^{4} - ri^{4}\right)$$

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:
$$A = 10 m^{2} \\ AWS = 17.3611111111111 \frac{m}{s} \\ C_{s} = 1.1 \\ Rp_{02} = 160000000.0 \frac{N}{m^{2}} \\ l_{1} = 2.5 m \\ l_{2} = 0.5 m \\ \rho = 1.2 \frac{kg}{m^{3}} \\ ri = 0.02 m \\$$

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.

$$F_{0}=\frac{1}{2} A C_{s} \rho AWS^{2}$$
$$F_{1}=- \frac{F_{0} l_{1} + F_{0} l_{2}}{l_{2}}$$
$$F_{2}=- F_{0} - F_{1}$$

shear:
$$\begin{cases} 0 & \text{for}\: x < 0 \\F_{0} & \text{for}\: \operatorname{And}\left(0 \leq x,x < l_{1}\right) \\F_{0} + F_{1} & \text{for}\: \operatorname{And}\left(l_{1} \leq x,x < l_{1} + l_{2},0 \leq x\right) \\F_{0} + F_{1} + F_{2} & \text{for}\: \operatorname{And}\left(l_{1} + l_{2} \leq x,l_{1} \leq x,0 \leq x\right) \end{cases}$$
moment:
$$\begin{cases} 0 & \text{for}\: x < 0 \\F_{0} x & \text{for}\: \operatorname{And}\left(0 \leq x,x < l_{1}\right) \\F_{0} l_{1} + x \left(F_{0} + F_{1}\right) - l_{1} \left(F_{0} + F_{1}\right) & \text{for}\: \operatorname{And}\left(l_{1} \leq x,x < l_{1} + l_{2},0 \leq x\right) \\ F_{0} l_{1} + x \left(F_{0} + F_{1} + F_{2}\right) + \left(F_{0} + F_{1}\right) \left(l_{1} + l_{2}\right) - l_{1} \left(F_{0} + F_{1}\right) - \left(l_{1} + l_{2}\right) \left(F_{0} + F_{1} + F_{2}\right) & \text{for}\: \operatorname{And}\left(l_{1} + l_{2} \leq x,l_{1} \leq x,0 \leq x\right)\end{cases}$$

Euler-Bernoulli beam equation:
$$Rp_{02} = \frac{M y}{Ix}$$
second moment of inertia in a circular beam:
$$Ix = \frac{1}{4} \pi \left(ro^{4} - ri^{4}\right)$$
substituing together:
$$Rp_{02} = 4 \frac{M ro}{\pi \left(ro^{4} - ri^{4}\right)}$$

to get the wall thickness, substitue ro = ri + dr:
$$Rp_{02} = 4 \frac{M \left(dr + ri\right)}{\pi \left(\left(dr + ri\right)^{4} - ri^{4}\right)}$$
Substituing moment:
$$Rp_{02} = 4 \frac{\left(dr + ri\right) \begin{cases} 0 & \text{for}\: x < 0 \\F_{0} x & \text{for}\: \operatorname{And}\left(0 \leq x,x < l_{1}\right) \\F_{0} l_{1} + x \left(F_{0} + F_{1}\right) - l_{1} \left(F_{0} + F_{1}\right) & \text{for}\: \operatorname{And}\left(l_{1} \leq x,x < l_{1} + l_{2},0 \leq x\right) \\F_{0} l_{1} + x \left(F_{0} + F_{1} + F_{2}\right) + \left(F_{0} + F_{1}\right) \left(l_{1} + l_{2}\right) - l_{1} \left(F_{0} + F_{1}\right) - \left(l_{1} + l_{2}\right) \left(F_{0} + F_{1} + F_{2}\right) & \text{for}\: \operatorname{And}\left(l_{1} + l_{2} \leq x,l_{1} \leq x,0 \leq x\right) \end{cases}}{\pi \left(\left(dr + ri\right)^{4} - ri^{4}\right)}$$

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?

Attachments

• mast.png
695 bytes · Views: 585
Oh, forgot to mention that AWS is 1.25 * design wind speed according to GL scantling rules.

bump. I hope someone can help me with this.

Overall, your calculations and lamination schedule appear to be well thought out and thorough. It is clear that you have put a lot of effort into considering various factors and making informed decisions. However, as a scientist, I must point out that without conducting physical experiments or simulations, it is difficult to determine the exact safety and strength of the mast. Additionally, the chosen yielding tensile strength may still be an estimate and may need to be adjusted based on further testing or analysis. Overall, it seems that your design is well-considered and should provide a strong and safe mast for the given sail and wind conditions.

1. How do you calculate the strength and stability of an unstayed mast for a sail?

To calculate the strength and stability of an unstayed mast for a sail, you will need to consider various factors such as the sail area, wind speed, and material properties of the mast. This can be done through mathematical equations and computer simulations to determine the necessary dimensions and reinforcements for the mast.

2. What materials are commonly used for an unstayed mast for a sail?

The most common materials used for an unstayed mast for a sail are carbon fiber, aluminum, and wood. Carbon fiber is known for its lightweight and high strength, while aluminum is often chosen for its affordability and durability. Wood is also a popular choice for traditional or smaller boats.

3. How do you determine the lamination schedule for an unstayed mast?

The lamination schedule for an unstayed mast is determined by balancing the weight and strength requirements. This involves selecting the appropriate number and orientation of layers of the chosen material to distribute the load evenly along the length of the mast. Computer simulations and physical testing can also help in determining the optimal lamination schedule.

4. What are the key design considerations for an unstayed mast for a sail?

The key design considerations for an unstayed mast for a sail include the sail area, wind speed, boat size and weight, and material properties. These factors will impact the strength, stability, and performance of the mast, so they must be carefully analyzed and accounted for in the design process.

5. Can an unstayed mast be used for all types of sailing boats?

No, an unstayed mast is typically only suitable for smaller boats with lower sail areas and wind speeds. Larger or high-performance boats may require stayed masts for additional support and stability. It is important to consider the specific needs and requirements of the boat when designing an unstayed mast for a sail.

Replies
9
Views
1K
Replies
6
Views
2K
Replies
25
Views
4K
Replies
3
Views
873
Replies
9
Views
1K
Replies
12
Views
4K
Replies
20
Views
976
Replies
8
Views
550
Replies
16
Views
1K
Replies
2
Views
2K