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magwas
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Homework Statement
I am trying to size parts of a boat, but I don't believe in my results. An attempt to solve one subproblem of this in a rulebook based way can be read here: http://www.boatdesign.net/forums/boat-design/rudder-scantling-31263.html Is it appropriate to bring that question here without modifications, as no one have answered it there?
There is a steering arrangement of a boat in the drawing below. Force and torsion on the rudder axlle (axle 1) [tex] F_{1} [/tex] and [tex]M_{1}[/tex] are known.
Calculate the forces on the wire ([tex]Fw_{2}[/tex]), the ring ([tex]F_{4}[/tex]), and the tiller axle ([tex]F_{2}[/tex]).
Axle 1 and axle 2 are made of circular steel rods, arm made of rectangular cross section plywood 4mm thick, placed horizontally.
Calculate the radius of axles, and the width of the arm.
This attempt actually stops at radius of axle 1.
[tex]$
known values: \\
F_{1} = 476 N \\
M_{1} = 15 N m \\
l_{1} = 0.1 m \\
l_{2} = 0.56 m \\
l_{3} = 0.8 m \\
l_{4} = 0.05 m \\
l_{5} = 0.2 m \\
l_{6} = 0.4 m \\
[/tex]
Homework Equations
[tex]\sum F = 0 [/tex]
[tex]\sum M = 0 [/tex]
[tex]M = F * l [/tex]
Euler-Bernoulli beam equation:
[tex]\sigma = \frac{M y}{Ix}[/tex]
Second moment of inertia in a circular beam:
[tex]Ix = \frac{1}{4} \pi r^{4} [/tex]
The Attempt at a Solution
[tex]
$ unit vector in direction for$ Fw_{2} \\
e_{Fw2}=\left(\begin{smallmatrix}\frac{l_{2}}{\sqrt{l_{1}^{2} + l_{2}^{2}}} & - \frac{l_{1}}{\sqrt{l_{1}^{2} + l_{2}^{2}}}\end{smallmatrix}\right)=\left(\begin{smallmatrix}0.984427575508482 & -0.175790638483658\end{smallmatrix}\right) \\
\lvert{e_{Fw2}}\rvert=1.0 \\
$ force in wire 2$ \\
Fw_{2}=- \frac{M_{1} e_{Fw2}}{l_{1}}=\left(\begin{smallmatrix}147.664136326272 N & - 26.3685957725486 N\end{smallmatrix}\right) \\
\lvert{Fw_{2}}\rvert=150.0 N \\
[/tex]
[tex] $ \\
unit vector in direction for wire 2 at tiller end $ \\
e_{Fw2 t}=\left(\begin{smallmatrix}- \frac{l_{1}}{\sqrt{l_{1}^{2} + l_{3}^{2}}} & \frac{l_{3}}{\sqrt{l_{1}^{2} + l_{3}^{2}}}\end{smallmatrix}\right)=\left(\begin{smallmatrix}-0.124034734589208 & 0.992277876713668\end{smallmatrix}\right) \\
\lvert{e_{Fw2 t}}\rvert=1.0 \\
$ force in wire 2 at tiller end $ \\
Fw_{2t}=e_{Fw2 t} \lvert{Fw_{2}}\rvert=\left(\begin{smallmatrix}- 18.6052101883813 N & 148.84168150705 N\end{smallmatrix}\right) \\
\lvert{Fw_{2t}}\rvert=150.0 N \\
$ force at ring $ \\
F_{4}=- Fw_{2} - Fw_{2t}=\left(\begin{smallmatrix}- 129.058926137891 N & - 122.473085734502 N\end{smallmatrix}\right) \\
\lvert{F_{4}}\rvert=177.920946336276 N \\
[/tex]
[tex] $ \\
$ equation for $ F_{3} $ using moment at $ F_{2} \\
F_{3} l_{6} - Fw_{2t} l_{1} = 0 \\
$ steering force at tiller $ \\
F_{3}=\frac{Fw_{2t} l_{1}}{l_{6}}=\left(\begin{smallmatrix}- 4.65130254709532 N & 37.2104203767625 N\end{smallmatrix}\right) \\
\lvert{F_{3}}\rvert=37.5 N \\
$ equation for $ F_{2} $ using moment at joint of tiller and wire 2 \\
F_{3} \left(l_{1} + l_{6}\right) - F_{2} l_{1} = 0 \\
$ force on axle 2 $ \\
F_{2}=\frac{F_{3} l_{1} + F_{3} l_{6}}{l_{1}}=\left(\begin{smallmatrix}- 23.2565127354766 N & 186.052101883813 N\end{smallmatrix}\right) \\
\lvert{F_{2}}\rvert=187.5 N \\
F_{2}=\frac{F_{3} l_{1} + F_{3} l_{6}}{l_{1}}=\left(\begin{smallmatrix}- 23.2565127354766 N & 186.052101883813 N\end{smallmatrix}\right) \\
\lvert{F_{2}}\rvert=187.5 N \\
$ Fb_{1} = Fb_{2} = F_{1}/2 \\
Fb_{1}=\frac{1}{2} F_{1}=238 N \\
\lvert{Fb_{1}}\rvert=238 N \\
Fb_{2}=\frac{1}{2} F_{1}=238 N \\
\lvert{Fb_{2}}\rvert=238 N \\
[/tex]
[tex] $ \\
shear in axle 1 $ \\
shear = \begin{cases} 0 & \text{for}\: x < 0 \\Fb_{1} & \text{for}\: \operatorname{And}\left(0 \leq x,x < l_{5}\right) \\Fb_{1} - F_{1} & \text{for}\: \operatorname{And}\left(l_{5} \leq x,x < 2 l_{5},0 \leq x\right) \\\operatorname{And}\left(2 l_{5} \leq x,l_{5} \leq x,0 \leq x\right) & \text{for}\: Fb_{1} + Fb_{2} - F_{1} \end{cases} \\
moment = \begin{cases} 0 & \text{for}\: x < 0 \\Fb_{1} x & \text{for}\: \operatorname{And}\left(0 \leq x,x < 0.2 m\right) \\x \left(Fb_{1} - F_{1}\right) + 0.2 F_{1} m & \text{for}\: \operatorname{And}\left(0.2 m \leq x,x < 0.4 m\right) \\0.4 m \leq x & \text{for}\: x \left(Fb_{1} + Fb_{2} - F_{1}\right) + 0.2 F_{1} m - 0.4 Fb_{2} m \end{cases} \\
[/tex]
[tex] $ \\
maximum bending moment in axle 1 $ \\
M_{a1}=Fb_{1} l_{5}=47.6 N m \\
\lvert{M_{a1}}\rvert=47.6 N m \\
[/tex]
[tex] \\
$ Euler-Bernoulli beam equation: $ \
\sigma = \frac{M y}{Ix} \\
$ second moment of inertia in a circular beam: $ \
Ix = \frac{1}{4} \pi r^{4} \\
$ substituing the two equations together, $ r = y $ and $ M = M_{a1} \\
\sigma = 4 \frac{M_{a1}}{\pi r^{3}} \\
$ solved for r :$ \\
r = \frac{2^{\frac{2}{3}} \sqrt[3]{M_{a1}} \sqrt[3]{\frac{1}{\sigma}}}{\sqrt[3]{\pi}} \\
[/tex]
Now I have a major problem here. I don't know which value to substitute for [tex]\sigma[/tex].
Looking up physical properties of materials, the ones expressed in [tex]\frac{N}{m^{2}}[/tex] are:
Ultimate tensile strength
Yielding tensile strength
Modulus of Elasticity (I think I understand that one)
Compressive yield strength
Bulk modulus
Fatique strength
Shear modulus
Shear strength
Flexural modulus
I bet at yielding tensile strength, but not sure.
I calculate with 200Mpa
[tex]
r=\frac{2^{\frac{2}{3}} \sqrt[3]{M_{a1}} \sqrt[3]{\frac{1}{\sigma}}}{\sqrt[3]{\pi}}=0.247448994460805 m 2^{\frac{2}{3}} \sqrt[3]{5} \\
\lvert{r}\rvert=0.247448994460805 m 2^{\frac{2}{3}} \sqrt[3]{5} \\
r= 0.671679909773 \\
[/tex]
I found this value rather off the scale, so there should be something wrong with my calculations.
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