Statics: Lesson 57 - Introduction to Internal Forces, M N V
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- เผยแพร่เมื่อ 30 ก.ค. 2024
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Professor Hanson, thank you for a positive introduction to Internal Forces M,N and V. This topic is very important for Mechanical and Civil Engineering students.
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I literally have an exam in an hour and realized that I have no clue what’s happening in this section, this just
Mr.Hanson ...I would say it would be good to see if you do a main topic such as this then you divided it into several calculation involving different shape such as triangle, right angle triangle , circle etc. We could see if there's any extra calculation when trying to calculate it and also different scenario like a triangle shape but it has extra length on the beam like in this video
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how did u identify the directions of the moments after you cut the beam??
Thank you!
When you get negative values for v does that mean it’s acting on the opposite direction? Are the directions the same even if you choose to go from the left/right?
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up stuff = down stuff didn't work for me and my problem. It looks exactly like yours but with an additional triangle distributed load. Is that why it's not working?
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Can someone explain why for when he was finding Ma for the -200 he didn't multiple by a distance
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Thank you
The value of the moments are the same regardless of whichever side of the beam you pick after cutting the beam but you will get a different direction (counterclockwise, clockwise,..) so when asked to express the direction of the moment at a point, which direction do we write ?
I have a very very good feeling that if it is negative, still write it as clockwise. If positive, then counterclockwise. I think when analyzing the beam you don't use the standard directions as the values of each side of the beam wont be equal (opposite signs (which makes sense since you are analyzing opposite sides of a point on a beam) ), hence the new direction rules. But when writing the final answer, you return back to using the conventional direction rules.
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Just to clarify, N is negative here because there's no force along it's direction?
Could someone explain why M and 200 is negative at 12:31? Much appreciated
M is negative because it is going clockwise. By convention, this is typically negative. 200 is negative for the same reason.
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deserves a sub
why at 12:01 and 12:26 that V is in negative sign and M and 200 all have a negative sign ?
how can i not uds this from my lecturer?
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how did he got the value of Ay?
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What changes Ay to 136.8 rather that it also being 113.2, I can solve for B but I don't know what is being done to it to make A, he says -250 but when I look I get minus 225, and I don't know where that would come form. 10:01 thanks
It's the equilibrium for fy, so the equation is Ay + By - 100 - 150 = 0 since those are all the variable for fy. From there, you just move the equation to Ay = 250 - By. And you should get Ay
Hope it helps.
Shouldn't the shear forces be same magnitude but different signs/ direction?
yes ı didn't understand that part too
can someone please explain how he found the 0.5 for that tiny distance between the cut?
in question,he asked to cut section at at 7' so remaining part is 0.5'
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if anyone here would love to explain it for me, why is V negative at 12:09 and why is it -M-200 at 12:36 😢 I don really get it the sign convention is confusing
there is a upward positive force By, so to balance it our V has to be equal and negative. And M is in the clockwise direction, clockwise moment is negative
this helped me fr lol 5:06
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we must have M N and V
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10:25 why did u cut the beam in hafl?
I think he cut it at the 7ft line. Around 7:00 he explained that he'll be cutting the beam and choosing a side like the method of sections. That's also why we need the positive sign conventions depending on which side we choose.
Hope this helped!
why is it (4.5) in the moment A in the concentrated force
4.5 is the distance between A and the concentrated force
3:25 Here's the replay button for the best moment 😁
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why is Ax=0 ?
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Sir why did you take negative M there 😔
Can someone help me understand why Ax = 0?
Thanks!
Hey, if you try to construct an equilibrium equation in the x direction ( Fx), you will only find Ax there, which means Ax=0, in other words there is no other forces in the x direction that Ax would balance and so there is no x-component for A.
Also, the reaction at B does not have an x component because it is a roller, refer back to Prof Hanson's video on supports and the reaction forces they give. So that leaves us with only Ax in the x direction, and due to equilibrium sum of forces in the x=0, Ax=0.
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3/2 = 1.5, not 2.5
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