# Maximum deflection of a beam with both ends fixed and distributed load

I'm designing a part, and I'm trying to figure out what distance a slot should be from the bottom of the part. For simplicity in analysis, I'm approximating the region between the bottom of the slot and the bottom of the part as a beam that is fixed on both sides. This beam will have a uniform load over the entire span.

I've found this equation to find the maximum deflection of the beam $y_{max}$, given the load $w$, beam length $l$, elasticity modulus $E$, thickness $b$, and distance between the slot and the bottom of the part (or beam height) $h$.

$$y_{max}=-\frac{w l^4}{32bh^3}$$

I'm trying to figure out what $h$ ought to be, so I rewrote it in terms of $h$.

$$h=-\frac{1}{2}\left(\frac{wl^4}{4Eby_{max}}\right)^{1/3}$$

I already know what $l$, $E$, and $b$ are, and my plan is to write $h$ as a function of $w$, but I need to eliminate $y_{max}$. I've tried finding it using the formula for bending stress ($\sigma=My/I$), but that resulted me in an identity for $y_{max}$. I tried finding $y_{max}$ by solving the ODE $\ddot{y}=M/(EI)$, but that got me the first formula that's above.

It would be awesome to have another formula for $y_{max}$, but a better/different method of determining what $h$ should be would be just as good.

• I don't understand what you are asking. You seem to be asking for some way to define y_max. This will be a limit that is set by what your part can tolerate. Figure out what the maximum deflection is that you can allow and use that. Or add some more background to your question. – hazzey Mar 3 '16 at 22:06
• The problem above lies where to get the value of ymax. Maximum deflection are normally given and decided by which code you follow for certain design criteria. Example, Philippine codes specify deflection limits up to 1/360 of the span length, such that: ymax = L / 360 – Jem Eripol Aug 19 '17 at 1:58

## 2 Answers

You need to decide whether the design limitation which determines h is max stress or max deflection.

If you decide deflection is more important insert the value for y which is the maximum allowable deflection for this part of your component.

If stress is more important (ie you don't care how much it moves but you don't want it to break) then you need to use the stress equation not the deflection one and solve for the yield stress (or acceptable working stress) your material.

If you're not sure which is more important do it both ways and go with the most conservative result.

Note in this situation for a solid beam with fixed ends and a uniform load, the max tress is just the basic shear stress of [Force/cross sectional area] of the beam.

In order to have a successful design, the beam needs to satisfy the following conditions:

\begin{align} f_v &\leq f_v' \\ f_b &\leq f_b' \\ y_{max} &\leq y_{permissible} \text{ (deflection)} \end{align}

$$f_v'$$ and $$f_b'$$ are allowable stress of shear and bending, respectively. For ductile metal, $$f_v'$$ is usually limited to $$0.4F_y$$, and $$f_b'$$ is in the range of $$0.6F_y - 0.66F_y$$. The permissible deflection is usually limited by the need of your application.

Now we can setup equations for these conditions. For fixed end beam with uniformly distributed load ($$w$$),

1. Shear stress: $$f_v = \dfrac{wL}{2A} = \dfrac{wL}{2bh} \leq 0.4F_y$$
2. Bending stress: $$M_{max} = \dfrac{wL^2}{12},\quad f_b = \dfrac{6M}{bh^2} = \dfrac{wL^2}{2bh^2} \leq 0.6F_y$$ (conservative)
3. Deflection: $$I = \dfrac{bh^3}{12},\quad y_{max} = \dfrac{wL^4}{384EI} = \dfrac{wL^4}{32bh^3} \leq y_{permissible}$$

From equations above, you can workout $$h$$ that satisfies each respective limit, and finally, the largest $$h$$ will yield a satisfactory design.