I'm working on a project where I know that force will be applied by a bolt pulled through a hole in the center of an aluminum plate for load testing purposes (2200 lbs, actual worst case max force around 1500 with a 1.5 times safety factor), the dimensions (27"x6.25"x?), and the material (5052 Aluminum -- I thought about 6061 but I'm afraid it is probably too brittle). We will be pulling on the plate with a static 2200 lbs for 1 minute for real world load testing. I was hoping to get an equation that related plate thickness to max load without permanent deformation. I would prefer to do the work myself(might come back for confirmation), but if needed I can provide more details! Thanks a lot for any help!
You first need to specify the support conditions of the plate...is it supported on all 4 sides along the edges...fixed or simple, ,,,or just at the ends, fixed or simple.
There are 14 bolts around the edges of the plate which affix it to a structure of very heavy aluminum. 7 (equidistance -- 4" apart) along each 27" side. I'm not sure what you mean by fixed vs simple. Thanks for your help.
A fixed edge support means that there is no plate rotation at that edge. A simply supported edge means that there is plate rotation at the edge. Plate bending stresses and deformations depend on the support conditions. With so many bolts along the long edges, you can probably consider the long edges fixed, and the short edges free, although in reality, the 'fixed' edge is not quite fully fixed . You would need a plate handbook or computer program to solve for these stresses and deformations. I don't have one handy. EDIT: In the absence of plate tables or a computer software program, you can conservatively assume that the plate is a simply supported beam 4 inches wide by 6 inches long with a concentrated load at the center, calculate the max moment, and then the section modulus required, and find the plate thickness required. You can try this yourself if you are familar with stress/deflection beam equations. I know what thickness I would use, but if I tell you that, I'd have to put a disclaimer on it, for obvious reasons. __________________
You have piqued my curiosity. Why are you using 4"x6" for the beam dimensions? The equation I was looking at was Deflection[meters]= ((Force[newtons]*(Length[meters]/2))/(48 * Elastic Modulus[Newtons/Meter^2] * Moment of Inertia[meters^4])) * (2 Length[meters]^2) Is this the correct equation for a single point load on a beam? I'm getting insanely small deflection (10^-8 m). I'm really not familiar with deflection of materials, so maybe this is correct? Thanks.
cntchds: Your formula in post 7 is correct for deflection. However, post 1 indicates you need stress, not deflection. The formula for stress on the beam in post 4 is, sigma = 1.5*P*L/(b*t^2), where b = beam width. In post 4, PhanthomJay is giving you dimensions of an equivalent beam that will give you a stress close to the maximum stress on your given plate. Here is an arbitrary example of the above stress formula. sigma = 1.5*P*L/(b*t^2) = 1.5(10 000 N)(152.4 mm)/[(101.6 mm)(7.20 mm)^2] = 434 MPa. Here is an arbitrary example of your deflection formula. y = P*(L^3)/(48*E*I) = (10 000 N)[(152.4 mm)^3]/{48(69 000 MPa)[(1/12)(101.6 mm)(7.20 mm)^3]} = 3.38 mm.
In the case of the stress formula that you have provided is there a way to know the max stress possible with a given material? I can see that for any value of t as it goes up stress decreases exponentially, and the opposite as it decreases. There must be a point at which the material fails from being too thin.
For a given material, maximum failure or yield stresses are a property of the molecular structure of the material. Steel for example will yield at somewhere between 30,000 to 100,000 psi depending on its chemical content and grade etc. Wood pine or fir species might fail at 8000 psi or so. These are axial or bending stress maximums. Shear failures occur at lower stresses. Concrete unreinforced might rupture at 4000 psi compression or just say less than 1000 psi in tension, depending on the mix (primarily water/cement ratio). These values are found in tables and often determined by testing. One would hesitate to come up with a plate thickness in your example without looking at all possible failure parameters. In nvn's example, you are looking at about 5/16 inch thick plate for the assumed max stress noted. But as I mentioned, check other factors...like punching shear or bearing stresses.
Yes, I currently would say, probably use the tensile yield strength (Sty) for your material, instead of tensile ultimate strength (Stu). You will need to know the alloy name and number, including heat-treatment condition (or temper) number. What is the alloy name, number, and heat-treatment condition number, for your material? Do you have Stu and Sty for your specific material? E.g., for Al 5052-O (i.e., annealed), Stu and Sty might be Stu = 182 MPa, and Sty = 78 MPa (?). Is your plate at room temperature?