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I am looking for the formula allowing to calculate the deflection of a rectangular plate with a "transverse loading in its center" (maybe the kind of loading will be more clear with this picture : enter image description here

I already found this formula for the case of a rectangular plate with a concentrated load at center (and I know the one I'm looking for is very close). https://www.engineersedge.com/material_science/rectangular_plate_concentrated_load_13645.htm

But I can't find the coefficients for my precise case with the load uniformly distributed across the width (as in the first picture)

Thank you

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  • $\begingroup$ A quick google search gives engineersedge.com/calculators/flat-plate-deflection.htm, but what did you find? $\endgroup$ – Solar Mike Apr 20 at 15:40
  • $\begingroup$ @SolarMike Thanks but I actually want the formula which could match to my case to have an expression of the deflection as a function of E (young modulus).I actually will fix a deflection to find my E and I need to justify the caclulation of this E with a formula (because I do this on a project basis) $\endgroup$ – Gey Apr 20 at 16:06
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The behavior of a plate supported on only two opposing sides with loads that are uniform along the width of the plate is identical to that of a beam, so the standard beam deflection equations can be used. The deflection of a beam with a concentrated load at its midspan is

$$\delta = \dfrac{F\ell^3}{48EI}$$

where $F$ is the total concentrated load at midspan, $\ell$ is the beam's span length, $E$ is the material's elastic modulus, and $I$ is its moment of inertia. Considering the plate's cross-section of $w \times t$, we have

$$I = \dfrac{wt^3}{12}$$

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  • $\begingroup$ Ok thank you very much. Could you just please explain why here the width w is not taken into account at all? We actually had an answer from our professor who said that we were supposed to find δ=Fw^3/48EI so we don't know If there is not a another way to calculate It with formulas for the rectangulare plates? $\endgroup$ – Gey Apr 20 at 16:34
  • $\begingroup$ @Gey I think that might've been a typo on your professor's end, getting $w$ and $\ell$ mixed up. And you'll notice that $w$ and $t$ are taken into account: they are the inputs to calculate the moment of inertia, which is then used for the deflection. $\endgroup$ – Wasabi Apr 20 at 17:48
  • $\begingroup$ Sorry actually the formula that has been given to us by our professor is δ=Fw^3/48Et with t the thickness and w the width, that's what we don't understand. Maybe using the beams formula here is a kind of simplification and the formula given by our professor is more precise? Or is there really no other way to calculate the deflection than the one you gave us? $\endgroup$ – Gey Apr 20 at 18:18
  • $\begingroup$ @Gey I can tell you without a doubt that equation is wrong. Just think about it: you have $w$ in the numerator. So the thicker the slab, the greater the deflection? And the length of the span doesn't appear? So a slab a mile long will deflect as much as one an inch long? That makes no sense. Or do dimensional analysis on that equation. You have $\delta = Fw^3/48Et$, which in dimensions is $\delta = [F] \cdot [L]^3/48 [F / L^2] \cdot [L] = [L]^4$. Obviously the dimension of deflection is length, not length to the 4th power. $\endgroup$ – Wasabi Apr 20 at 19:20
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    $\begingroup$ @Gey now, it may be that simply-supported slabs as in your case (often called "one-way slabs") show some transversal bending moment, but it is insignificant. I very much doubt your teacher is discussing that, since one-way slabs are a standard part of early engineering lessons (from your question, I infer that's your current level) and are taught as "just pretend they're beams". Two-way slabs (where supports exist on adjacent sides) are completely different, but one-way slabs are just very wide beams. $\endgroup$ – Wasabi Apr 20 at 19:25
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Roark’s Formulas for Stress and Strain by WARREN C. YOUNG RICHARD G. BUDYNAS On pp 435 has a reference to this question.

A very wide beam of rectangular cross section can be treated as a beam if E is replaced by $\frac{E}{1-\nu^2}$ and $I$ by $t^3/12$ (see Sec. 8.11). It can also be treated as a plate with two opposite edges free as shown in Figs. 8.16 and 11.2. For details see Ref. 88.

If you don't have access to it I suppose I can take picture of the page and give it to you. Because another conditions apply too.

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