So the two points over the axle will be 250 up and the two points on opposite sides of the axle will be 250 down.
This is what I officially have for the moment. http://imgur.com/a/NUGdm
For the right/left beam that is supporting the downward 250 lb load. I took the moment about point O. I...
The axle from earlier will be. Looking at the picture, http://imgur.com/a/fBvpE , it will be located on the dashed line, bisecting the two longer plates on the bottom of the assembly.
No, the 4 beams bolted together form the rectangular frame of a trailer. Something negligible will be placed on top of it to form a flat surface. Then a 1000lb load will be placed on it. The load has 4 legs that will all perfectly apply the loads to the 4 centers of the beams. The 4 squares with...
there is a 1000lb load standing on four legs. So there is a 250 lb force acting on the midpoints of all four beams.
Honestly, I tried my hardest searching through mechanics of materials, design, and Beam deflection books trying to assure myself of the effect the bolts have on my shear and...
Alright, I will review that.
I have another question if you are willing and able. Do bolts affect the shear and moment force diagrams for beam loading/deflection at all? I have dug through 4 of my textbooks and scoured the internet but haven't found anything to explain it.
See attached pic if...
Alright, no worries.
Ok, I suppose that makes sense then. A 1.5 diameter thick beam of 1018 steel sound like it should handle a 1000lb load.
Also, I just thought of something during my calculation. When i plug my torque into my shear stress equation to determine the torsion, I only need to...
Excellent. Thank you very much.
If you look at my diagram, the maximum moment i calculated was 5000 lb-in. Does that seem correct? When i calculate my sigma and tau and use the von mises criterion, I feel like the value seem awfully big. In the end I ended up with a diameter a little shorter...
Oh, ok. So then the force that needs to be overcome is 500(0.3) = 150 lb. Since each wheel gets its own share.
The resisting force acts between the wheel and the ground.
I suppose in this case, it would be 1000lb since there is two 500 lb forces acting on the axle, and it would be acting on the axis of the axle. So you would need to overcome 1000(0.3) = 300 lb?