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Introduction to internal forces
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- čas přidán 3. 05. 2017
- This engineering statics tutorial introduces internal forces in members. The internal forces that we look at here are axial force, shear force, and internal bending moment. If you “virtually cut” a member and remove one side, the part that remains must still be in static equilibrium. It’s the internal forces that are transferred across the virtual cut that keep each side in static equilibrium.
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Thanks for watching, I hope it helps!
As a mostly uninterested Mechanical Engineering student I was failing my classes and did not have the motivation to study. Your videos are a life saver and the clarity and simplicity of them gives me more reason to study than the classes I have failed ever could. Thank you.
Thanks for letting me know. I'm glad I can help you, and I hope you come to appreciate and crush your classes!!
tysm for this playlist. It's 3am and I've been watching these as a refresher for tomorrow's final.
I gochu! Good luck 💪💪
Thank you for making this video. It really helped me understand everything I needed to know.
Glad it was helpful! Make sure to check out the rest of the playlist here: engineer4free.com/statics
Thank you for saving my statics😂😂
Any time 😂
Can you explain how you got 6m for By and 4m for 15kN?
I am appreciated thanks
@Engineer4Free Could you explain why the the moment at left in anti-clockwise at 6:24
Hey Tabris. The internal shear and bending moment there are just drawn as unknowns in their positive sign convention. But you will find out that the shear must be oriented downward for the sum of forces in vertical direction to net to zero. Because the reaction pushes up on the left side of the member, and the internal shear pushes down on the right side, those cause that section of the member to want to rotate clockwise. It’s in static equilibrium, and not rotating, so the internal bending moment has to counter-act it. You will find that the internal moment there is actually how we have drawn it, as counter clockwise, and will have the same magnitude as the moment caused by the force couple that’s just mentioned. Hope that makes sense.
Thank you for the interesting video. Could you be so nice and tell me what software did you use to make the video? Is it some kind of whiteboard soft? Tnx in advance.
Yeah the full list is at engineer4free.com/tools 👌
Thanks for the video. When question asks shear force at point C , should we say +5 or -5 ?
You're welcome =). The shear at point C is +5kN. Check out this video: I go over this problem in more detail and draw the SFD and BMD: www.engineer4free.com/4/internal-force-sign-convention
@@Engineer4Free After i wrote this comment i found your video about sign convention. Thanks four your reply.
What is the physical meaning of +5KN shear force at point C... ???
Is it internal shear force at point C should be +5 to neutralize the shear force applied by external load at point C?
why didn't you include the 15 kn in the EFy on the right side
Can you be more clear. The right side of what?
at 8:15 you sum the forces in the y direction, but don't include the 15kn @@Engineer4Free
Colton McDorman it’s actually included, instead of writing 10+v-15=0 he wrote 10+v=15
Hey Tolga thanks for helping out! I missed the notification after Colton replied so thanks for jumping in :) :)
Engineer4Free thank you, for being an amazing teacher...
Also 2020 anyone?
Me 🤣
@@Engineer4Free Lol
@@tsegahagos1440 Me in 2021.
I think this is a faulty setup. A won't resist any moment and B won't resist lateral movement. There is no additional support in between A and B. There is a hinge and we are loading this span.
The span will just fold at the hinge.
Hi, point C is not a hinge, sorry for the confusion. It is just a point of interest. This is a simply supported beam that is stable. The pin at A resists lateral movement, and the roller at B ensures no rotation or translation with the loading.
Mechanical lesson should be easy not complicated for that reason iam out 😂
huh.... he posted this on may 4th and he's talking about forces....