# simply supported beam with a midspan load

First off non-engineer apologies,

A plate of 5251 Aluminium is used to support the load of a centrally seated human I estimate a load of 1 kilonewton.

The beam measures a length of 1000mm, 100mm wide.

For a thickness 6mm I calculate the Moment of Inertia to be 1800mm4 and given a Elastic modulus of 70000MPa the maximum deflection to be about 14cm.

Assuming the beam will fail under tension of 130-240 N/mm2 will it break or permanently deform ?

How thick would the plate have to be before the process would be elastic ?

(6mm is the thickest plate though I could stack them; I can get 5mm being much cheaper)

• I have reveiwed engineering.stackexchange.com/questions/5808/… but I am no clearer Aug 12 '20 at 14:59
• A single flat plate, or a stack of plates, is a very inefficient way to support the load. For example a T-section beam would be much lighter and stiffer. Aug 12 '20 at 22:56

Using the properties you mentioned, the equation for bending stress is

$$\sigma = \frac{M}{I_{xx}}\cdot y_{max}$$

where :

• $$M=F\frac{l}{2}=0.5kNm$$ (that's a lot)
• $$I_{xx}$$= 1800mm4
• $$y_{max} = \frac{thickness}{2}=3[mm]$$

The results is a whopping 833.[MPa], so it will fail.

Assuming you use a solid beam (same material), you can estimate the minimum thickness from the following equation:

$$h_{min}= \sqrt{\frac{6 M}{b\cdot \sigma_{max}}}$$

The only thing new thing here is that $$\sigma$$ is substituted for $$\sigma_{max}$$. Al does not have a clear yield area like steel has. Therefore "elastic region" is a bit vague (normally you'd settle for proof stress). However, given the ultimate tensile properties you posted and the intended use, I'd settle for at least 100[MPa], which yields a minimum thickness of 17.5[mm].

OK, now for the tricky part. If you plan to stack them, then you'd need a lot more. A rough estimation is between 8 or 9 plates of 6[mm]. For why that happens you might want to read the following question