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## Homework Statement

In the steering arrangement below, axle 1 is a circular cross section rod.

It is hold by two bearings.

F1= 476 N

M1 = 15 Nm

Fw2 = 304.75 N, and we know that the moment imposed by Fw2 is equal to M1

l5 = 0.2 m

l8 = 0.1 m

The tensile yielding stress:

[tex]\sigma = 235 Mpa[/tex]

safety factor fs = 1.70

Calculate the radius of the rod.

https://www.physicsforums.com/attachment.php?attachmentid=23596&stc=1&d=1265629087

## Homework Equations

## The Attempt at a Solution

I believe I can solve it for the case when I neglect M1 and Fw2:

[tex] $ \\

$ 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 \\

$ 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]

maximum bending moment in axle 1

[tex]

M_{a1}=Fb_{1} l_{5}=47.6 N m \\

\lvert{M_{a1}}\rvert=47.6 N m \\

[/tex]

Euler-Bernoulli beam equation with safety factor:

[tex]

\frac{\sigma}{fs} = \frac{M y}{Ix} \\

[/tex]

second moment of inertia in a circular beam:

[tex]

Ix = \frac{1}{4} \pi r^{4} \\

$ substituing the two equations together, $ r = y $ and $ M = M_{a1} \\

\frac{\sigma}{fs} = 4 \frac{M_{a1}}{\pi r^{3}} \\

$ solved for r :$ \\

r = \frac{2^{\frac{2}{3}} \sqrt[3]{M_{a1}} \sqrt[3]{fs} \sqrt[3]{\frac{1}{\sigma}}}{\sqrt[3]{\pi}} \\

r=\frac{2^{\frac{2}{3}} \sqrt[3]{M_{a1}} \sqrt[3]{fs} \sqrt[3]{\frac{1}{\sigma}}}{\sqrt[3]{\pi}}=0.000125670730105656 m 470^{\frac{2}{3}} \\

\lvert{r}\rvert=0.000125670730105656 m 470^{\frac{2}{3}} \\

r= 0.00759683528577m = 7.6 mm \\

[/tex]

Here is an attempt at the "M only" case:

Note that I am using the same [tex]\sigma[/tex] here. Is it okay?

torsional shearing stress with safety factor

[tex] \frac{\sigma}{fs} = \frac{M_{1} r}{J} [/tex]

polar moment of inertia for solid cylindrical shaft

[tex] J = \frac{1}{2} \pi r^{4} [/tex]

[tex]

r=\frac{\sqrt[3]{2} \sqrt[3]{M_{1}} \sqrt[3]{fs} \sqrt[3]{\frac{1}{\sigma}}}{\sqrt[3]{\pi}}=3.46762256685552 \times 10^{-5} m \sqrt[3]{30} \times 235^{\frac{2}{3}} \\

\lvert{r}\rvert=3.46762256685552 \times 10^{-5} m \sqrt[3]{30} \times 235^{\frac{2}{3}} \\

r = 0.0041031508172 m \\

[/tex]

I have no clue how to calculate with Fw2, and how to combine the whole thing together.