Warning: Extremely Technical Content
(if this is too techy for you, stay tuned for Part 2, which is more practical than theoretical)
I mess around with small boats and sail on the upper Chesapeake Bay where the wind speeds are moderate as long as there’s not a thunderstorm. I like to experiment with different sails and rigs. I'm also a retired tech-builder (aka, old-school engineer), and this note addresses several questions I had concerning the stress on an unstayed mast on a small boat. There is a link to a "stress calculator" for determining beam deflection and stress, and it allows you to choose different materials like wood and aluminum. It seems to work pretty good, and I have been using it to look at hollow spars as well as the solid wood mast that I looked at in this note. It can also handle square and rectangular spars ("beams" in the calculator). Much of what I'm presenting here is based on Jim Michalak's excellent discussion of mast stresses in his newsletters, and hopefully you can still get them as a primary source. Thank you, Jim and Stuart Hopkins of Dabbler Sails, for educating me, and hopefully this will be useful to someone else who is asking the same questions (before they break that mast!).
Legal Disclaimer: I am not a structural or marine engineer. Use this information as a guide to help understand the factors that figure into mast stress. You should always consult a professional engineer to verify that your scantlings will provide the necessary margin of safety. And if you are really serious about this stuff, take a look at Dave Gerr's "Elements of Boat Strength" (International Marine, 2000).
1. Relate wind speed to pressure on the sail:
Pressure on sail, P, in pounds per square foot is given by
P = 0.005 x (V)**2, or 0.005 x V x V
where V is wind speed in knots
>> Rule of Thumb: You can figure on 1/2-pound of pressure per square foot of sail area at 10 knots of wind. And 2 pounds of pressure per square foot at 20 knots of wind. Would your sail and rig support you if you lie on top of it?
2. Calculate the maximum force on the mast due to wind -- pressure x area
Assume boat is level and stationary with respect to wind. Assume wind is directly into sail. And for this example, assume the area of sail is 100 SF
>> Rule of Thumb: A 100-SF sail in 10 knots of wind applies a 50 pound load to the mast. And a 20-knot gust increases the max force or load by a factor of four, to 200 pounds!
>> The capsize torque from that 20-knot gust is 200 pounds of sail load times the distance from the boat's center-of-gravity (CG) to the sail's center-of-effort (CE). For example, the capsize torque is 2,000 foot-pounds if the CG-to-CE distance is 10 feet (200 pounds x 10 feet).
To keep the boat level, you need to generate an equivalent torque in the opposite direction. For the above example, if you sit on the gunwale which is 3 feet from the CG (beam is 6 feet), you need 2,000 ft-lbs / 3 ft = 666 pounds of ballast on the windward rail to compensate for a 20-knot puff. Or, you can luff the sail. Conclusion: Never cleat the sheet in gusty winds.
3. Calculate bending of mast about the partner due to sail's wind load
>> Deflection, D in inches, of a cantilever beam fixed on one end is
D = P x (L**3) / (3 x E x I)
where
P = applied force, the sail pressure, in pound per square inch
L = distance between the sail's center-of-effort and the mast partner, in inches
E = Young's modulus of mast material in psi
I = moment of inertia of the mast in inches**4
Notice that the mast deflection is proportional to the applied load, P -- if you double the sail pressure, then the mast deflection doubles. Recall that sail pressure increases by the square of wind speed; if the wind speed doubles the sail pressure increases by a factor of four. Therefore the mast deflection (and the stress, until the mast begins to break) at 20 knots will be 4 times the deflection at 10 knots.
Also note that the deflection, D, scales with the cube of L (that is, L x L x L), where L is the height of the center-of-effort above the mast partner. If you can change just one variable, this is the one that makes the biggest difference. Double the sail CE height, and the deflection increases 8X. In the other direction, a small change in CE by lowering the sail CE a bit will result in a huge reduction in D.
If the mast is twice as stiff ("E" times "I" doubles), the mast deflection decreases by half. The moment of inertia ("I") will increase if the mast cross section increases. "Less bendy" wood will have a higher Young's modulus and increase the "E x I" (oak is more flexible than Douglas fir and has a low Young's modulus; pine is lower than oak. The Gougeon book, in Appendix B, has a list of Young's moduli for other woods).
To determine the mast deflection, we first need to determine how high the sail's center-of-effort is above the mast partner. For example, assume the mast is 15-16 feet long and the partner is two feet above the bottom end of mast. For a typical 100 SF lug or gaff sail, the center-of-effort would be about 5 feet above the foot of the sail. If we place the boom gooseneck two feet above the partner, then d, the distance between the mast partner and the sail's center-of-effort, is 5 + 2 = is 7 feet or 84 inches. Two feet below the partner into the step; top the sail is about 14 feet above the step. The top of the mast is one to two feet above the gaff at full sail lift.
We need to know what kind of wood the mast is made from; this allows us to determine the Young's modulus (or modulus of elasticity) of the wood:
The moment of inertia, I (with dimension of inches to the 4th power), for a round beam is given by
I = 3.14 x (d**4) / 64
where d is the diameter (in inches) of the mast. For our example 3-inch mast,
I = 3.14 x (3 x 3 x 3 x 3) / 64
or moment of inertia (I) for the for 3 inch mast = 3.97 inch**4. From above, the deflection, D, of the mast (in inches) with an applied load (in pounds) from the wind in the sail is
D = P x (L**3) / 3 x E x I
For example, for the 3-inch well-dried Douglas fir mast that's 15 feet long with the CE at 7 feet,
D (inches) = P x (84 x 84 x 84) / (3 x 1950000 x 3.97), or
D (inches) = 0.0255 x P (pounds)
At a wind speed of 5 knots, the max pressure of a 100 SF sail is 12.5 pounds and the 3-inch fir mast will deflect 0.32 inches. At a wind speed of 10 knots, the pressure will increase by a factor of 4 to 50 pounds, and the mast deflection will increase to 1.28 inches. And at a wind speed of 20 knots, the mast deflection -- if the sail is not reefed or luffed -- will increase to 5.10 inches. Whoa! Look at that! Is my mast going to break?!?
Note 1: If you want to look at other loads or mast dimensions (e.g., diameter, or height of CE above the partner) use the link to "Beam Deflection Calculator -- Solid Round Beam", http://www.engineering.com/calculators/beams.htm. Be aware that the Young's modulus they are using for "wood" is lower than the value for Douglas fir.
Note 2: If you want to look at a solid square mast section, the moment of inertia is
I = (b**4)/12 where b = the dimension of the side in inches
Note 3: For a solid rectangular cross section, there are two moments for the narrow (x) and long (y) dimensions
Ix = b * (h**3)/12 and Iy = h * (b**3)/12
where "b" = short width and "h" = long width (height) of the rectangular beam
Note 4: For these or for hollow sections, use the Beam Deflection Calculator link above.
4. Calculate the stress in the mast at the partner -- Will it break?
The stress, S (in pounds per square inch), at the mast partner is given by
S = M x c / I
where "M" is the torque moment (in inch-pounds) due to the sail pressure on the mast; "c" is the distance from "neutral axis", or center of the mast, to the "extreme fiber", which is the outside edge of the mast. And "I" is the moment of inertia, the same that we calculated above.
The torque moment is simply the sail load, P (in pounds), times the distance (in inches) from the sail's CE to the mast partner, L; or M = P x L. So that the stress in the mast at the partner due to the sail pressure is given by
S = (P x L x c) / I
Note that the stress is proportional to the sail pressure; when sail pressure doubles, and stress doubles. Stress is also proportional to the distance above the mast partner; lowering the center of the sail will reduce the stress at the partner. The moment of inertia has the dimension of the mast "c" in it, so reducing the diameter reduces the moment of inertia faster, and has the effect of significantly increasing S.
For a wind speed of 10 knots, the pressure on mast from the 100-SF sail is 50 pounds. We assumed the center-to-partner distance of L = 84 inches. And for a 3-inch mast "c" = 1.5 inches. The moment of inertia, I, of a 3-inch solid round mast is 3.97 inch**4; and therefore the stress in the mast at the partner is
S (at 10 knots) = (50 x 84 x 1.5) / 3.97 = 1,587 pounds per square inch.
From "Gougeon Brothers on Boat Construction" (Appendix B), the breaking strength of Douglas fir is listed at 5,700-10,200 pounds per square inch. So we can conclude that at a wind speed of 10 knots, the stress on the 3-inch mast at the partner (for a 100 SF sail) is well below the breaking strength of fir.
However, with a gust of 20 knots the sail pressure increases to 200 pounds. In this case the stress on the partner is significantly bigger, S = 6,350 pounds per square inch, which exceeds the lower bound of breaking strength of fir -- and the mast may fail. Maybe well-dried, knot-free fir, with breaking strength at the upper limit, will survive the 20 knot gust. Of course, in the real world, there are a number of ways that the actual pressure on the mast could end up being less (e.g., a lug sail that will twist off in a puff, or we luff the sail, or bear up).
At this wind speed, luffing, reefing, using a tapered mast (to "twist off" wind pressure), adding shrouds and/or backstays or another similar strategy to insure mast and rig survival will be required.
5. Bonus -- What happens if the mast is made of aluminum tubing?
For a windspeed of 10 knots, the pressure on mast from the 100-SF sail is 50 pounds. We assumed the sail has a center-to-partner distance of L = 84 inches. We will stick with the 3-inch diameter mast and choose aluminum tubing with a wall thickness of 0.125 inches. Using the stress calculator (link shown above), the stress in the hollow aluminum mast at the mast partner will be
S (at 10 knots) = 5,390 pounds per square inch
and the mast deflection (at 10 knots) is 0.80 inch -- that is, barely noticeable.
Since the ultimate strength of 6061-T6 aluminum is listed as 35,000 pounds per square inch, which is about seven times the maximum stress on the mast at 10 knots of breeze, there should be little concern about mast failure. For a gust of 20 knots, the stress will increase to 21,500 pounds per square inch, which is still in the safe region for aluminum. However, with a deflection is 3.2 inches, the rig will begin to "twist off" -- or at the very least it will look like it is going to fail -- which will hopefully be recognized by the crew.
1. Relate wind speed to pressure on the sail:
Pressure on sail, P, in pounds per square foot is given by
P = 0.005 x (V)**2, or 0.005 x V x V
where V is wind speed in knots
- At 5 knots, P = 0.125 psf
- At 10 knots, P = 0.50 psf
- At 15 knots, P = 1.12 psf
- At 20 knots, P = 2.0 psf
>> Rule of Thumb: You can figure on 1/2-pound of pressure per square foot of sail area at 10 knots of wind. And 2 pounds of pressure per square foot at 20 knots of wind. Would your sail and rig support you if you lie on top of it?
2. Calculate the maximum force on the mast due to wind -- pressure x area
Assume boat is level and stationary with respect to wind. Assume wind is directly into sail. And for this example, assume the area of sail is 100 SF
- At 5 knots, max force is (0.125 x 100) 12.5 pounds
- At 10 knots, max force is 50 pounds
- At 15 knots, max force is 112 pounds
- At 20 knots, mas force is 200 pounds
The force scales with area. If the area of the sail doubles to 200 SF, the force doubles. At 20 knots of wind, the force on the rig with a 200 SF sail will therefore be 400 pounds.
-------------->> Rule of Thumb: A 100-SF sail in 10 knots of wind applies a 50 pound load to the mast. And a 20-knot gust increases the max force or load by a factor of four, to 200 pounds!
>> The capsize torque from that 20-knot gust is 200 pounds of sail load times the distance from the boat's center-of-gravity (CG) to the sail's center-of-effort (CE). For example, the capsize torque is 2,000 foot-pounds if the CG-to-CE distance is 10 feet (200 pounds x 10 feet).
To keep the boat level, you need to generate an equivalent torque in the opposite direction. For the above example, if you sit on the gunwale which is 3 feet from the CG (beam is 6 feet), you need 2,000 ft-lbs / 3 ft = 666 pounds of ballast on the windward rail to compensate for a 20-knot puff. Or, you can luff the sail. Conclusion: Never cleat the sheet in gusty winds.
3. Calculate bending of mast about the partner due to sail's wind load
>> Deflection, D in inches, of a cantilever beam fixed on one end is
D = P x (L**3) / (3 x E x I)
where
P = applied force, the sail pressure, in pound per square inch
L = distance between the sail's center-of-effort and the mast partner, in inches
E = Young's modulus of mast material in psi
I = moment of inertia of the mast in inches**4
Notice that the mast deflection is proportional to the applied load, P -- if you double the sail pressure, then the mast deflection doubles. Recall that sail pressure increases by the square of wind speed; if the wind speed doubles the sail pressure increases by a factor of four. Therefore the mast deflection (and the stress, until the mast begins to break) at 20 knots will be 4 times the deflection at 10 knots.
Also note that the deflection, D, scales with the cube of L (that is, L x L x L), where L is the height of the center-of-effort above the mast partner. If you can change just one variable, this is the one that makes the biggest difference. Double the sail CE height, and the deflection increases 8X. In the other direction, a small change in CE by lowering the sail CE a bit will result in a huge reduction in D.
If the mast is twice as stiff ("E" times "I" doubles), the mast deflection decreases by half. The moment of inertia ("I") will increase if the mast cross section increases. "Less bendy" wood will have a higher Young's modulus and increase the "E x I" (oak is more flexible than Douglas fir and has a low Young's modulus; pine is lower than oak. The Gougeon book, in Appendix B, has a list of Young's moduli for other woods).
To determine the mast deflection, we first need to determine how high the sail's center-of-effort is above the mast partner. For example, assume the mast is 15-16 feet long and the partner is two feet above the bottom end of mast. For a typical 100 SF lug or gaff sail, the center-of-effort would be about 5 feet above the foot of the sail. If we place the boom gooseneck two feet above the partner, then d, the distance between the mast partner and the sail's center-of-effort, is 5 + 2 = is 7 feet or 84 inches. Two feet below the partner into the step; top the sail is about 14 feet above the step. The top of the mast is one to two feet above the gaff at full sail lift.
We need to know what kind of wood the mast is made from; this allows us to determine the Young's modulus (or modulus of elasticity) of the wood:
- white pine = 0.99-1.24 x 10*6 psi (Gougeon, Appendix B)
- Sitka spruce = 1.23-1.57 x 10*6 psi (Gougeon)
- oak = 1.6 x 10*6 psi
- Douglas fir = 1.56-1.95 x 10*6 psi (Gougeon)
The moment of inertia, I (with dimension of inches to the 4th power), for a round beam is given by
I = 3.14 x (d**4) / 64
where d is the diameter (in inches) of the mast. For our example 3-inch mast,
I = 3.14 x (3 x 3 x 3 x 3) / 64
or moment of inertia (I) for the for 3 inch mast = 3.97 inch**4. From above, the deflection, D, of the mast (in inches) with an applied load (in pounds) from the wind in the sail is
D = P x (L**3) / 3 x E x I
For example, for the 3-inch well-dried Douglas fir mast that's 15 feet long with the CE at 7 feet,
D (inches) = P x (84 x 84 x 84) / (3 x 1950000 x 3.97), or
D (inches) = 0.0255 x P (pounds)
At a wind speed of 5 knots, the max pressure of a 100 SF sail is 12.5 pounds and the 3-inch fir mast will deflect 0.32 inches. At a wind speed of 10 knots, the pressure will increase by a factor of 4 to 50 pounds, and the mast deflection will increase to 1.28 inches. And at a wind speed of 20 knots, the mast deflection -- if the sail is not reefed or luffed -- will increase to 5.10 inches. Whoa! Look at that! Is my mast going to break?!?
Note 1: If you want to look at other loads or mast dimensions (e.g., diameter, or height of CE above the partner) use the link to "Beam Deflection Calculator -- Solid Round Beam", http://www.engineering.com/calculators/beams.htm. Be aware that the Young's modulus they are using for "wood" is lower than the value for Douglas fir.
Note 2: If you want to look at a solid square mast section, the moment of inertia is
I = (b**4)/12 where b = the dimension of the side in inches
Note 3: For a solid rectangular cross section, there are two moments for the narrow (x) and long (y) dimensions
Ix = b * (h**3)/12 and Iy = h * (b**3)/12
where "b" = short width and "h" = long width (height) of the rectangular beam
Note 4: For these or for hollow sections, use the Beam Deflection Calculator link above.
4. Calculate the stress in the mast at the partner -- Will it break?
The stress, S (in pounds per square inch), at the mast partner is given by
S = M x c / I
where "M" is the torque moment (in inch-pounds) due to the sail pressure on the mast; "c" is the distance from "neutral axis", or center of the mast, to the "extreme fiber", which is the outside edge of the mast. And "I" is the moment of inertia, the same that we calculated above.
The torque moment is simply the sail load, P (in pounds), times the distance (in inches) from the sail's CE to the mast partner, L; or M = P x L. So that the stress in the mast at the partner due to the sail pressure is given by
S = (P x L x c) / I
Note that the stress is proportional to the sail pressure; when sail pressure doubles, and stress doubles. Stress is also proportional to the distance above the mast partner; lowering the center of the sail will reduce the stress at the partner. The moment of inertia has the dimension of the mast "c" in it, so reducing the diameter reduces the moment of inertia faster, and has the effect of significantly increasing S.
For a wind speed of 10 knots, the pressure on mast from the 100-SF sail is 50 pounds. We assumed the center-to-partner distance of L = 84 inches. And for a 3-inch mast "c" = 1.5 inches. The moment of inertia, I, of a 3-inch solid round mast is 3.97 inch**4; and therefore the stress in the mast at the partner is
S (at 10 knots) = (50 x 84 x 1.5) / 3.97 = 1,587 pounds per square inch.
From "Gougeon Brothers on Boat Construction" (Appendix B), the breaking strength of Douglas fir is listed at 5,700-10,200 pounds per square inch. So we can conclude that at a wind speed of 10 knots, the stress on the 3-inch mast at the partner (for a 100 SF sail) is well below the breaking strength of fir.
However, with a gust of 20 knots the sail pressure increases to 200 pounds. In this case the stress on the partner is significantly bigger, S = 6,350 pounds per square inch, which exceeds the lower bound of breaking strength of fir -- and the mast may fail. Maybe well-dried, knot-free fir, with breaking strength at the upper limit, will survive the 20 knot gust. Of course, in the real world, there are a number of ways that the actual pressure on the mast could end up being less (e.g., a lug sail that will twist off in a puff, or we luff the sail, or bear up).
At this wind speed, luffing, reefing, using a tapered mast (to "twist off" wind pressure), adding shrouds and/or backstays or another similar strategy to insure mast and rig survival will be required.
5. Bonus -- What happens if the mast is made of aluminum tubing?
For a windspeed of 10 knots, the pressure on mast from the 100-SF sail is 50 pounds. We assumed the sail has a center-to-partner distance of L = 84 inches. We will stick with the 3-inch diameter mast and choose aluminum tubing with a wall thickness of 0.125 inches. Using the stress calculator (link shown above), the stress in the hollow aluminum mast at the mast partner will be
S (at 10 knots) = 5,390 pounds per square inch
and the mast deflection (at 10 knots) is 0.80 inch -- that is, barely noticeable.
Since the ultimate strength of 6061-T6 aluminum is listed as 35,000 pounds per square inch, which is about seven times the maximum stress on the mast at 10 knots of breeze, there should be little concern about mast failure. For a gust of 20 knots, the stress will increase to 21,500 pounds per square inch, which is still in the safe region for aluminum. However, with a deflection is 3.2 inches, the rig will begin to "twist off" -- or at the very least it will look like it is going to fail -- which will hopefully be recognized by the crew.
References:
Jim Michalak, Boat Designs Newsletter, www.jimsboats.com
- "Sail Rig Spars", 9/1/1998
- "Hollow Spars", 2/1/2000
- "Ballast Calculations 3", 1/1/2006 -- wind speed to sail pressure
Meade Gougeon, "Gougeon Brothers on Boat Building," 5th Edition (McKay Press, 2005)
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