Which bar will support the most weight if supported on both ends with the weight hanging in the middle....
2'' diameter solid square bar
2'' diameter hollow square bar with 1/2'' wall thickness
2'' diameter solid round bar
2'' diameter hollow round bar with 1/2'' wall thickness
....all bars are 6 feet in length. If you could help me with that question. Thanks
2006-07-01 10:02:35 · 12 answers · asked by nrolater 1 in Science & Mathematics ➔ Engineering
If the 2" diameter solid square bar is in fact 2" square then it should have the greatest section modulus (sp?) and therefore the greatest resistance to bending.
2006-07-01 12:38:15 · answer #1 · answered by oil field trash 7 · 0⤊ 0⤋
Not enough information.
Material information: What are the bars made of?
Application information: what are the orientations of the square section? How is the weight to be applied? Can you make a truss, or a triangular support?
The material props work with something called the "moment of inertia". The section with the largest moment of inertia, for the same material, will hold the larger load.
If you are looking for load per bar weight, then the hollow bar will work better. You might seriously consider U-channel or I-beam stock. They can be TONS cheaper than rod or pipe stock.
If you are looking for pure maximum load, then get a non-hollow bar.
The moment of inertia for a rectangular section is (1/12)*b*h^3
For the 2" square thats 4/3
The moment of intertia for a circular section is (1/4)*pi*R^4
For the 2" circle thats 0.785
The square has a moment of inertia that is 70% larger, so its stronger.
If you have a choice of materials, steels tend to be very strong, and very cheap. Aluminum costs more, and is ductile (soft) and not as strong.
If you can tell me the material, I will tell you how much it should hold, and I can use the Von-Mises failure model to do so.
The subject you are asking about is called "strength of materials" or "deformable solids" or "continuum mechanics". Solving specific, very simple, problems like this requires algebra. Truly understanding the subject, and being able to use it for unique & challenging problems requires understanding vectors, calculus, and differential equations.
2006-07-01 13:27:41 · answer #2 · answered by Curly 6 · 0⤊ 0⤋
If I understand you, the 2" solid square bar (2inch by 2inch) would be the heaviest and also hold the most weight.
f you are thinking of different shapes, all 6 feet long and weighing the same, I would GUESS the hollow square tilted to a diamond, that is; with the corners up/down/left/right.
2006-07-01 18:22:36 · answer #3 · answered by science_curious 2 · 0⤊ 0⤋
Well my guess would be,(and this is only a guess) the solid circle.My mind says the circles will bend easier than the squares.The hollow circle would collaspe in the center (because the metal has space to collaspe into the hollow) at some point before the solid circle.
2006-07-01 10:25:27 · answer #4 · answered by Anonymous · 0⤊ 0⤋
2" diameter solid round bar
2006-07-01 10:08:55 · answer #5 · answered by James B 1 · 0⤊ 0⤋
2" hollow square bar with 1/2 " wall thickness....guarenteed
2006-07-01 10:07:56 · answer #6 · answered by da_hammerhead 3 · 0⤊ 0⤋
2'' diameter solid square bar
Most mass, most material
2006-07-02 18:39:09 · answer #7 · answered by Lee J 4 · 0⤊ 0⤋
I'm a bit puzzled as a square doesn't have a diameter!
2006-07-01 12:00:15 · answer #8 · answered by Robert A 5 · 0⤊ 0⤋
It depends on the strength of the material compared to it's weight... If the material is very heavy and relatively weak, you will get a different answer than if the material is relatively light and strong.
2006-07-01 12:35:05 · answer #9 · answered by Paul B 1 · 0⤊ 0⤋
The solid square. The material is immaterial.
Ha ha ha!
2006-07-04 16:33:49 · answer #10 · answered by Prof. Frink 3 · 0⤊ 0⤋