huangyhg

help solving an easy stress and displacement problem
hi,
i'm an entry level engineer fresh out of college. i'm still having problems learning how to apply the basics to real world problems because they did not do a good job teaching us that in school.
i was hoping someone could tell me how i go about solving the problem below by using hand calculations. i need to find the max stress and max displacement and am not sure how to go about solving this. i don't need the exact solution but i provided a picture in hopes someone could teach me how to set it up and solve it myself.
i realize this is probably a really easy problem but i do appreciate any help i can get. thanks.
it's impossible to tell what you are asking from the picture. describe the structure and the loading in words, and make some sketches of plan, elevation and section as appropriate. that will be the first step in preparing your hand calculation, and it will help us understand your problem.
this is a small sheet metal support made out of al 6061-0 that is 0.045 inches thick throughout. there is a load that is evenly distributed on the top surface totaling 50 pounds. the bottom of the two supports are resting on a firm surface. the part is 1 inch high...from the bottom surface of the leg to the top surface holding the load.
thank you for your help and i'm sorry that i wasn't very clear in the first post. if this still isn't enough info i'm very sorry, but please let me know and i can tell you whatever i'm missing.
thanks for your help!
this can be treated as a frame analysis with the ends pinned. the midspan moment can be approximated by wl^2/8 to wl^2/12 (8 would be conservative, 12 not). the corner moments would be somewhere in the range of wl^2/18 for the pinned condition, going to 0 at the support. you then need to check buckling of the turn-down. 0.045 aluminum is fairly stiff, but your temper is t0, so your allowable stresses are relatively low, as compared to a t6 condition.

i'm getting 37 ksi with a 1/2" deflection or so.
step 1 - free body diagram. i simplified it as a simple beam spanning 6" with an 8.333 lb/in dist load
step 2 - shear diagram = 25-8.33x; moment diagram = 25x-4.167x^2; rotation diagram = ct+12.5x^2-1.388x^3; deflection diagram = ct x + 4.17x^3-0.3472x^4
step 3 - calculate max stress due to max moment = mxx y/ixx
step 4 - calculate max deflection (x=3)
step 5 - check assumptions - bc of step 1 simplification, the rotation at each end amplifys the deflection at the center so add a little on to the max deflection calculated (ie cut with axe).

teguci.......i tried it your method and got the same answers you did. i have also ran this through pro/e mechanica to compare the answers. i know i won't get the exact answers but i was wondering, generally, how much error is acceptable. in this case mechanica found a 0.155" deflection and a 28 ksi von mises stress.
i would model it with a roller at one end
did i miss something? what the heck are the support conditions? pin...pin, pin...roller or what? in any event, whatever they are, the problem is easily solved by hand methods.
ba
in a normal beam, the assumption is made that the flanges are able to freely expand in width and thickness. with a wide flat bar, the tension flange and compression flange are attached together, and no longer free to expand/contract laterally. if i re
something else to consider is that if it's a 1-off item, it may be cheaper to add $2 worth of steel that you don't need than to do $50 worth of extra analysis on it.
pin-pin means that a horizontal reaction can be generated, enabling a horizontal coupling force to be generated as the vertical deflection increases that will counter the vertical deflection. the greater the deflection seen, the higher the coupling force and resiulting moment generated.
such is not the case with a pinned - roller model where no horizontal force can be generated. the model should be pinn - roller as tegici suggested.
i am assuming the shape is just sitting on a flat surface here with no special attachment to the next shape. however, if the legs are attached together with horizontal tek screws, this becomes a different problem with more of a fixed condition in the short legs. not true fixity, but partial.
mike mccann
mmc engineering

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