Thank for your courses :) I don't understand in the example at 17:30 why we don't consider the force F_B for calculating delta_AB. Indeed, the CB part applies a force F_B on the BA part and so we should consider that the tensile load applied on the BA part is F_A - F_B. Am i wrong ? To me, we should have delta_AB = (F_A-F_B) * L1/(E1 A1)
Thank you! I'm glad you liked it! You might also be interested in the playlist where I have all of the lectures for this course collected. ENGR 220: Statics and Mechanics of Materials: th-cam.com/play/PL1IHA35xY5H5sjfjibqn_XFFxk3-pFiaX.html Thanks for watching!
I don't really cover friction in a very focused way in this course. The elementary concepts of friction are covered in our Physics class that is a prereq for this class. I do work problems now and then involving friction though. You might check out these for example: th-cam.com/video/HHyTYgwfKw0/w-d-xo.html th-cam.com/video/vm98fhk2OqE/w-d-xo.html th-cam.com/video/6B0qDx-PnC0/w-d-xo.html th-cam.com/video/O8ZmcdvXeh4/w-d-xo.html th-cam.com/video/eEbCPIA4V9s/w-d-xo.html
Thank for your courses :) I don't understand in the example at 17:30 why we don't consider the force F_B for calculating delta_AB. Indeed, the CB part applies a force F_B on the BA part and so we should consider that the tensile load applied on the BA part is F_A - F_B. Am i wrong ? To me, we should have delta_AB = (F_A-F_B) * L1/(E1 A1)
I think I get it : the CB part does apply a force F_B on the BA part but not in the cross section in BA since the point A is free. Is this right ?
Very nice lecture!
Thank you! I'm glad you liked it! You might also be interested in the playlist where I have all of the lectures for this course collected.
ENGR 220: Statics and Mechanics of Materials: th-cam.com/play/PL1IHA35xY5H5sjfjibqn_XFFxk3-pFiaX.html
Thanks for watching!
You got any friction videos sir?
I don't really cover friction in a very focused way in this course. The elementary concepts of friction are covered in our Physics class that is a prereq for this class. I do work problems now and then involving friction though. You might check out these for example:
th-cam.com/video/HHyTYgwfKw0/w-d-xo.html
th-cam.com/video/vm98fhk2OqE/w-d-xo.html
th-cam.com/video/6B0qDx-PnC0/w-d-xo.html
th-cam.com/video/O8ZmcdvXeh4/w-d-xo.html
th-cam.com/video/eEbCPIA4V9s/w-d-xo.html
i didnt understand where does 90 kn come from.
if we are adding the forces . we should get 10 kn in the diagram
listen carefully at 33:45
the way the figure is labeled, two forces of 80kN each are applied at C
Hi Sir, just wonder if the DC section should experience 10KN because otherwise the rod is accelerating. Thank you.
listen carefully at 33:45
the way the figure is labeled, two forces of 80kN each are applied at C
@@TheBomPE got it. Thank you!