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The first back-of-the envelope blah blah calculation that we will make is for the total deformation of the crank under the specified applied load. A list of different cantilevered beam loading cases along with their closed-form maximum deflections formulas can be accessed on this website. Because our beam is loaded at the second hole instead of at the tip, our loading is best represented by the case 2 presented (i.e for a cantilevered beam with a concentrated load, P, at any point). The appropriate formula for the maximum deflection is therefore

{latex}
\begin{equation*}
\delta_{max} = -\frac{Pa^{2}}{6EI}(3L-a)
\end{equation*}
{latex}

where "P" is the load, "a" is the distance from the support to the load, "L" is the distance from the support to the end of the beam, "E" is Young's Modulus and "I" is the moment of inertia.

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We also know that "P" is 100 lbs and "E" is 10,000 ksi. The tricky part is to determine the moment of inertia. Recall that for a rectangular cross-section of height h and depth b, 

{latex}
\begin{equation*}
I = \frac{1}{12}bh^{3}
\end{equation*}
{latex}

where "h" is the height and "b" is the depth. In this case we know that the depth is 0.375 in but what should we do about this varying height? Since height varies as a function of x, the moment of inertia also varies with x. Finding the maximum deflection for a varying moment of inertia is actually very complex. The goal here is not necessarily to get the exact answer but to get a reasonable idea of what we should expect our ANSYS solution to be.

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We now have all variables needed to find the maximum deflection. Using the equations above, we estimate the maximum deflection to be 0.039 inches.

{latex}$\sigma_x${latex}

along the height of the cross-section

Now we wish to do hand-calculations to predict the stress in the x-component across the height of the cross-section at the thinnest part of the crank (i.e the middle). Because

{latex}$\sigma_x${latex}

is linear along the height of the cross-section, and because from symmetry, the value at the top of the cross-section is the same as the value at the bottom of the cross-section, one can simply calculate the

{latex}$\sigma_x${latex}

at the top to compare with ANSYS. This value is the only real unknown because stress at other heights are directly dependent on that value and can be found quite easily by interpolation. So let's calculate

{latex}$\sigma_x${latex}

at the top of the cross section. Dividing the minimum height found above by two, we find the coordinate of this point to be (3.345", 0.306") with the origin at the left hole center.

{latex}
\begin{equation*}
M=-P*(a-x)=-(100)*(6.69-3.345) = -334.5 lb*inches
\end{equation*}

\begin{equation*}
I = \frac{1}{12}bh^{3} = \frac{1}{12}(0.375)(0.612)^{3} = 0.00716 in^{4}
\end{equation*}

\begin{equation*}
\sigma_{x} = \frac{M*y}{I} = \frac{-334.5*0.306}{0.00716} = -14295.7 psi
\end{equation*}

{latex}

We now have a good idea of what to expect from our ANSYS results. We will come back to these hand-calculation results in the verification and validation step. Let's now solve this problem using ANSYS!

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