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Stresses in a rod

  1. Feb 8, 2010 #1
    1. The problem statement, all variables and given/known data

    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



    2. Relevant equations



    3. 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]
    attachment.png

    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.
     

    Attached Files:

  2. jcsd
  3. Feb 8, 2010 #2

    nvn

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    magwas: For the given value of F1, your answer for the Ma1 bending moment case is correct; r = 7.60 mm.

    For shear stress, use tau, not sigma. For shear yield strength, you could perhaps use Ssy = 0.577*Sty = 0.577(235 MPa) = 135.6 MPa. Therefore, change sigma in your "M1 only" case to Ssy. Hence, your "M1 only" case answer would become r = 4.929 mm, instead of 4.103 mm.
     
  4. Feb 8, 2010 #3
    Thank you again for helping me out with this.
    I have tried to look up shear yield strength at http://www.matweb.com/search/DataSheet.aspx?MatGUID=d1bdbccde4da4da4a9dbb8918d783b29 but I cannot identify it there.
    Do we just use 0.577*Sty as an empirical value? If so, what is the limit of applicability?

    Now I am trying to come out with something for bending moment when Fw2 is also playing. I am wrestling with the math right now, however my line of thought is the following:
    1) I broke up Fw2 to two components: one being a moment, which is equal but opposite to M1, the other being a force which is equal and same direction as Fw2, but attacks at the end of the rod. It is correct to do so?
    2) write the moment equation with forces as vectors, and lengths as signed distance from turning point.
    3) write shear using the forces found above, still as vectors. I can draw up absolute value and angle of shear.
    4) moment is the integral of shear along x (this is where I am struggling with my math package right now, but I will eventually win). It is also a vector function, but absolute value and angle can be computed.
    5) now I am looking for the point where absolute value of moment is highest. I will find it by differentiating abs(moment), and solving it.
    I will show my computation when I have done it, my questionfor now is whether this line of thought is justifiable.

    I still don't have a clue about how should I combine shear stress and tensile stress. Add them up somehow?
    Or are they so independent that I can calculate rod radius independently for them and then use the higher?
     
  5. Feb 9, 2010 #4
    Here is my attempt at calculating tensile stress in the way I have outlined above, and the presumably correct calculation for shear stress.
    How should I combine the two? I was thinking for a moment that tensile stress is normal to shear stress, so maybe they are really independent, but I don't believe in that. I just can't imagine a rod bent nearly to its limits, and withstanding the same twisting force as without bend.

    [tex]
    F1 = 476 N[/tex]

    [tex]Fw2= 300.0 N - 53.5714285714286 \mathbf{\imath} N[/tex]

    moments around Fb1: [tex] Fw_{2} l_{8} - F_{1} l_{5} - 2 Fb_{2} l_{5} = 0 [/tex]

    moments around Fb2: [tex] F_{1} l_{5} + Fw_{2} \left(l_{8} + 2 l_{5}\right) + 2 Fb_{1} l_{5} = 0 [/tex]

    [tex]Fb_{1}=- \frac{F_{1} l_{5} + Fw_{2} l_{8} + 2 Fw_{2} l_{5}}{2 l_{5}}=- 613.0 N + 66.9642857142857 \mathbf{\imath} N[/tex]
    [tex]Fb_{2}=\frac{Fw_{2} l_{8} - F_{1} l_{5}}{2 l_{5}}=- 163.0 N - 13.3928571428571 \mathbf{\imath} N[/tex]
    shear in axle 1
    [tex]shear = \begin{cases} 0 & \text{for}\: x < 0 \\Fw_{2} & \text{for}\: \operatorname{And}\left(0 \leq x,x < l_{8}\right) \\Fb_{1} + Fw_{2} & \text{for}\: \operatorname{And}\left(l_{8} \leq x,x < l_{5} + l_{8},0 \leq x\right) \\F_{1} + Fb_{1} + Fw_{2} & \text{for}\: \operatorname{And}\left(l_{5} + l_{8} \leq x,x < l_{8} + 2 l_{5},l_{8} \leq x,0 \leq x\right) \\\operatorname{And}\left(l_{8} + 2 l_{5} \leq x,l_{5} + l_{8} \leq x,l_{8} \leq x,0 \leq x\right) & \text{for}\: F_{1} + Fb_{1} + Fb_{2} + Fw_{2} \end{cases}[/tex]
    [tex]moment = \begin{cases} 0 & \text{for}\: x < 0 \\Fw_{2} x & \text{for}\: \operatorname{And}\left(0 \leq x,x < l_{8}\right) \\Fw_{2} l_{8} + x \left(Fb_{1} + Fw_{2}\right) - l_{8} \left(Fb_{1} + Fw_{2}\right) & \text{for}\: \operatorname{And}\left(l_{8} \leq x,x < l_{5} + l_{8},0 \leq x\right) \\Fw_{2} l_{8} + x \left(F_{1} + Fb_{1} + Fw_{2}\right) + \left(Fb_{1} + Fw_{2}\right) \left(l_{5} + l_{8}\right) - l_{8} \left(Fb_{1} + Fw_{2}\right) - \left(l_{5} + l_{8}\right) \left(F_{1} + Fb_{1} + Fw_{2}\right) & \text{for}\: \operatorname{And}\left(l_{5} + l_{8} \leq x,x < l_{8} + 2 l_{5},l_{8} \leq x,0 \leq x\right) \\\operatorname{And}\left(l_{8} + 2 l_{5} \leq x,l_{5} + l_{8} \leq x,l_{8} \leq x,0 \leq x\right) & \text{for}\: Fw_{2} l_{8} + x \left(F_{1} + Fb_{1} + Fb_{2} + Fw_{2}\right) + \left(Fb_{1} + Fw_{2}\right) \left(l_{5} + l_{8}\right) + \left(l_{8} + 2 l_{5}\right) \left(F_{1} + Fb_{1} + Fw_{2}\right) - l_{8} \left(Fb_{1} + Fw_{2}\right) - \left(l_{5} + l_{8}\right) \left(F_{1} + Fb_{1} + Fw_{2}\right) - \left(l_{8} + 2 l_{5}\right) \left(F_{1} + Fb_{1} + Fb_{2} + Fw_{2}\right) \end{cases}[/tex]
    attachment.php?attachmentid=23617&stc=1&d=1265723233.png
    blue is magnitude of shear, green is angle of shear in degrees, red is magnitude of moment, cyan or whatnot is angle of moment

    we can see that maximum of moment is at 0.3 = l_{5} + l_{8}
    maximum bending moment in axle 1
    [tex]M_{a1}=\lvert{Fb_{1} l_{5} + Fw_{2} l_{5} + Fw_{2} l_{8}}\rvert=32.7098569990448 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}[/tex]
    substituing the two equations together, [tex] r = y | and | M = M_{a1}[/tex]
    [tex] \frac{\sigma}{fs} = 4 \frac{M_{a1}}{\pi r^{3}} [/tex]
    solved for r :
    [tex]r = \frac{2^{\frac{2}{3}} \sqrt[3]{M_{a1}} \sqrt[3]{fs} \sqrt[3]{\frac{1}{\sigma}}}{\sqrt[3]{\pi}}[/tex]
    [tex]r=\frac{2^{\frac{2}{3}} \sqrt[3]{M_{a1}} \sqrt[3]{fs} \sqrt[3]{\frac{1}{\sigma}}}{\sqrt[3]{\pi}}=0.000110898252429377 m 470^{\frac{2}{3}}[/tex]
    [tex] r= 6.70383434932 mm[/tex]
    [tex]\tau=0.577 \sigma=135595000.0 \frac{N}{m^{2}}[/tex]
    torsional shearing stress with safety factor [tex]\frac{\tau}{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}{\tau}}}{\sqrt[3]{\pi}}=0.00158617752186715 m \sqrt[3]{30}[/tex]
    [tex]r = 4.92862235616 mm[/tex]
     

    Attached Files:

  6. Feb 9, 2010 #5

    nvn

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    It appears you have one or more mistakes. E.g. (for example), in your "moments about Fb1" equation in post 4, because you are using counterclockwise positive (which is good), the F1*L5 term should be added, not subtracted. Likewise, in your "moments about Fb2" equation, the F1*L5 term should be subtracted, not added. See if you can go ahead and implement these corrections. These are the only two equations I looked at, for now, because I have very limited time, at the moment. I will try to look at this again over the next several days, when I have time.
     
  7. Feb 9, 2010 #6
    I am using forces as vectors in the plane normal to the axle.
    Hence the direction of the force is encoded in it, and the sign of the moment is determined whether it is below or above the turning point (the arm of force is negative or positive). I not entirely sure that what I am doing is valid, but if it is, then I believe this is the way to do it. Did I miss something?
    Notice that both Fb1 and Fb2 have a large negative real part. It means they are more or less opposite to F1.
     
  8. Feb 9, 2010 #7
    And thank you for your continuing support again.
     
  9. Feb 9, 2010 #8

    nvn

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    No, in your "moments about Fb1" equation in post 4, the F1*L5 term should be added, not subtracted. Likewise, in your "moments about Fb2" equation, the F1*L5 term should be subtracted, not added. Probably your mistake is that you did not make a proper free-body diagram of axle 1, including Fw2; you did not show one herein. When you do it correctly, using your given values of F1 and Fw2, the answer for your Ma1 bending moment case should be r1 = 8.326 mm. Your "M1 only" torsional shear stress case is already correct.

    We just use 0.577*Sty as an empirical value for shear yield strength. It is applicable only to ductile materials. Mild steel is ductile. Your line of thought outlined in post 3 is correct.

    You do not need to combine shear stress and tensile stress for axle 1. In this particular given problem, torsional shear stress has essentially no effect upon the outcome, and can be ignored, because it would increase the axle 1 radius by only 0.8 %. Therefore, only compute the Ma1 bending moment case, then multiply the resulting axle 1 radius value by 1.008.
     
  10. Feb 12, 2010 #9
    The drawing contains the profile view of the rod, and a perspective view of the same.
    I have noted some orthogonals to make things clearer.
    Discovered a mistake. I think the bending component of Fw2 is the one orthogonal to the rod, that is
    paralell to the arm, that is
    [tex](Fw2 \cdot e_{arm} ) e_{arm}[/tex] where e_arm is a unit vector in the direction of the arm.

    I don't see however why the sign of the moment by F1 should be negated.
    I have depicted F1 on the other side in the perspective view to underline that I think that only the point of
    attack is significant with F1. As it is a vector in the plane orthogonal to the rod, its direction is encoded in
    the vector (orthogonal to e_arm). This is why I think the "push vs pull" issue does not arise the same way it would with a 2 dimensional case with a scalar force.
    What the problem with this line of thought?

    attachment.php?attachmentid=23679&stc=1&d=1265990779.png
     

    Attached Files:

    Last edited: Feb 12, 2010
  11. Feb 12, 2010 #10
    instead of F2, read F1 and l_F2 -> l_F1 in the picture above
     
  12. Feb 13, 2010 #11

    nvn

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    Both components of Fw2, parallel and perpendicular to the rudder lever arm, cause bending in axle 1. However, the axle 1 bending moment about the steering system global x axis currently increases the axle 1 moment by only 0.03 % (relative to the moment about the global y axis).

    F1 should be drawn pointing toward the rear of the boat. If you draw F1 in the opposite direction, then you must input a negative value, F1 = -476 N.

    Using F1 = 476 N, alpha = 60 deg, and Fw2 = 255.02 N, when you work the problem correctly, the answer for the Ma1 bending moment case should be an axle 1 radius of r1 = 8.220 mm. You can multiply this answer by 1.008 to account for the M1 torsional shear stress case, thus obtaining r1 = 8.286 mm.
     
  13. Feb 15, 2010 #12
    Thank you yet again.
    I have the same result, if:
    I use cross products correctly: l x F instead of F x l. This way Fw2 will point to the correct direction for a force which is against positive (counterclockwise) M1
    I transfer Fw2 to the end of axle1 for purposes of calculations of bend. I have made the same mistake here as in the other example. I will try to remember that
    [tex]\sum F = 0[/tex] :)

    However as far as I can see, the problem wasn't with the direction of F1, but that of Fw2. Of course it is my interpretation, and not the only valid one.

    I am trying to figure out where that 0.008 came from.
    I guess it is related to shear modulus, and the contraction of the rod due to the fact that a line paralell to its axis on its surface will be bent.
    My try:
    From the definition of Young modulus:
    [tex] E = \frac{\frac{F}{A}}{\frac{\delta x}{x}} = \frac{\frac{F}{x^2 2 \pi }}{\frac{\delta x}{x}}[/tex]
    I get [tex] \delta x [/tex] from the above:
    [tex] \delta x = \frac{F}{2 \pi E x} [/tex]
    Integrating [tex] \delta x [/tex] along the length of the rod should give the overall contraction:
    [tex] \Delta l = \int_0^l \delta x dx = \int_0^l \frac{F}{2 \pi E x} dx = \left [ \frac{F \operatorname{log}\left(x\right)}{2 \pi E} \right ]_0^l = \frac{F \operatorname{log}\left(l\right)}{2 \pi E} + \frac{F}{0}[/tex]
    But is is infinity. I guess I should have combined in the moment somehow...
     
  14. Feb 16, 2010 #13

    nvn

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    Yes, Fw2 is backwards in your last diagram. I forgot to mention that. Reverse the direction of Fw2 in your last diagram. F1 is backwards in the perspective view, but is correct in the profile view. Reverse the direction of F1 in the perspective view of your last diagram.

    I would recommend to not worry about the 0.8 %. It is too small to spend time on; almost negligible for axle 1. Just multiply your answer by 1.008. It is not computed in any way like what you have imagined and posted.
     
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