Piston-Rod-Crank Stress - Axial Loading in 90 Seconds!

[@okaytree7711]

Thank you.

[@joshuasirgany9617]

sum of forces formula looks a little wierd to me can you explain if steps were skipped or anything?

[@EngineerHarris]

I'm not sure what your confusion is but I'll take a guess and if you are still unsure just reply back. He basically compartmentalized the FBD. The way you can think about it is if you imagine what the internal force is in member BC near joint C. This "internal force" is simply Fbc. P is a load that is holding the structure in static state. So then the only vertical force that is "fighting" against P is Fbcsin(theta) and the only horizontal force fighting against reaction Rx is Fbccos(theta). When he does the summation of Moment about point A. He goes back to the entire structure. Point B is free because it is not constrained. It is free to move in X, Y and in rotation which is why there are no Reaction forces about point B. So then the only force that makes a moment couple at point A is Rx multiplied by perpendicular distance.

[@lavis4448]

@@EngineerHarris why was 450 x10^6? where did he get the 6 from? isnt mm supposed to be 10^-3?

[@EngineerHarris]

@@lavis4448 Abc is 450 mm^2. But in his stress equation he is converting the area to m^2. It's the different between Square Milimeters and Square Meters. 1 m^2 = 1e+6 mm^2

[@yannicknelva9299]

CovidImages need to be invested more than half19