Mechanics of Materials: Lesson 5 - Bearing Stress Explained, Example Problem
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- čas přidán 4. 09. 2024
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Your videos have helped me through Statics, Strength of Materials, Structural Analysis, and Indeterminate Structures.
I am an engineer now, but still in school. I can't thank you enough for all the times you've helped me over the years!
Great comment! Thank you!
@@1234jhanson Can u explain angle FAB?
@@joeking3688 Angle FAB in the FBD Dr. Hanson draws comes from the initial 30 degree angle, PLUS the 20 degrees he rotates the coordinate system. Lmk if this makes sense.
Keep it up professor 💪💪 you’re doing wonders for us mechanical engineering students
I wish you do fluid mechanics course
Can you make a video explaining why projection is used as opposed to cylindrical contact area.
ohh may daysss, finally the best professor is back.
In this course right now, perfect timing
the professor is back!
Wonderful solids lessons!! really taking the time to explain the problems along with a good sense of humor :))
Miss you dr Hanson! Hope all is going well at tech!
This is the best explanation I've heard!! Absolute life saver. Sending love from new zealand
I have a question. At 10:20 he assumed to measure Fab as Fab(cos50) but wouldn’t it be cos40 bc you measure the angle from the horizontal? Someone please clarify thank you!!
20+30=50
Need to keep up the videos. Once per week would be epic if you can sir. You were thinking of moving onto other topics once upon a time 😊
@9:24 why did you make F_ABsin50 on the horizontal and F_ABcos50 on the vertical. Initially I paused the video and went ahead and did my own calculation but with the moment at point B. I Got stumped not realizing you had your sin and cos mixed up. If I recalculated my moment to have it at point C, I realize I wouldve got the same answers as before. So im just lost on why x is sin and y is cos.
absolute hero
So how do we find the normal stress here? What should be the area for Normal Stress?
Hi, professor. Why the area in the bearing stress isn't 3.14(4.5)(15)? because it's half the perimeter times the thickness, that the surface that the bold hold the force with it.
Hello. I believe it is because if you look at it from the "bottom" it looks like a rectangle and as mentioned in the problem the diameter of the circle was 9 mm which becomes the height if you are still looking it from the bottom. And the thickness becomes the "length".
Glad you’re back!
Panutanku dan Guruku!
Can someone explain how he arrived at the angle 20 at F(AB) to get the overall angle of 50 deg?😪😮💨☹🥴😖😫
This confused me too, drawing a picture of the axes helped me.
Draw a normal x-y axis grid, and put a line above the +x-axis at 20° for the beam and another in the third quadrant 30° from the -y-axis for your AB force.
Now, plot the NEW x'-y' axis with your 20° beam. Since the angle between the x'-y' axes must be 90°, you should be able to visually see that the "remaining angle" in the new plane is 70°.
Looking back at the original x-y axis, you now have a 70° in the fourth quadrant below the +x-axis, and since all the angles in one quadrant must equal 90°, that leaves us with the 20° that Prof Henderson shows in his drawing.
From there you simply add the given 30° and found 20°.
hello, here in this video, we didn't transform from mm to m because we are dividing in the end and it will simplify true??
Correct.
Well come back sir
Absolute legend!
I'm still struggling to understand why you have to find the resultant force at pin A and then divide by the area. I thought the shear force was only on the "y" direction
At pin C? Because he adjusted the neutral axis to make it easier to solve for the real reaction at point C. If u flip it back to the original axis the Y component is not perpendicular to the axis( it will also adjust by 20degress back) so it doesn't make sense to use Y component for it but u need y and x to compute the real load at point C. Hope it helps
It is average shear stress, so makes sense to take the resultant of the two forces for the total shear stress
@@jaysonvillabito1511 How did he get 50 deg at FAB?
@@joeking3688 make the BCD your neutral axis x. or make the bcd straight line in your free body diagram so you can picture out that the horizontal force(tension) and the vertical force is not 30 degress angle but 50 degress 30+20 degress(inclined since you adjusted it for better picture of the diagram) i hope you understand my english this is not my native language
Can’t relate to metric…dang. Pounds I can relate! Great lesson though! Hilarious! I am a structural guy in Austin…we do this stuff a lot…gotta love statics and mechanics of materials.
sir, thank you!
Come on, man. I just graduated. Where were you 7 months ago??
Congratulations, I should’ve been graduating this school year in 2023 but typical engineering rookie mistakes, I’ll be longer.
I graduated last August. Time to brush up for the FE exam!
For A I got 702.47Pa, since I converted the mm diameter to m when calculating area. How is my answer wildly different from yours?
b/c you're dealing with area, a mm^2 would be 1/10^6 of a m^2, not 1/10^3. So you need to divide by 10^3 again to get the right answer
how to get 50 degree?
Did you get it
@@Ahmad-os3si Can you just explain and stop being cheeky?
We need machine elements course sir..
how are you man long time
Thx
Where did 250N come from?
Given. It is a part of the problem statement.
Petition to rename MEs to Statics Party People
منووو عربي ما فاهم ع استاذ المادة وجاي هنااا 😍
No one could see he reversed cos and sin 50 for x and y.