Hey quick question, when finding the moment about A, wouldn't the force applied from Cy be negative since it is causing rotation in the clockwise sense? So then the member BE would be in tension instead of compression
I should have stuck with the convention in always labelling counterclockwise moments as negative, an written -T(0.5m) - 25kN(1m) = 0 but for some sloppy reason I decided to drop the negative signs. The result doesn't change, because the initial right hand side of the equation is zero, so either way, T will come out as -50kN.
Thanks for Solving my previous issue. My question is about member CD . Is the member CD a two force member? Eventhough there is a horizontal force of 10 kN acting on CD in its middle . And will it still be a two force member if I apply this force on pin joint D of member CD?
CD is considered a 3 force member. Each pin applies one force, and the horizontal 10kN in the middle is the third. If the 10kN horizontal one wasn’t there, and we still considered the vertical force to be acting on the lower Teal coloured member, then CD would be a 2 force member. If the horizontal force wasn’t there, and we considered the vertical force to be acting at D on CD, we could model CD and a 3 force member again. Sorry for the delay, hope that helps clear it up.
shouldn't the moment caused by the 10KN, Cx and Cy force be reversed since it causes it to rotate counterclockwise around D? Would that not mean that Cy has a force of -6.25KN?
Sorry I skipped a step. The initial expression for member CD sum of moments about D should be: -10kN(0.25) + Cy(0.25m) + 17.5kN(0.5m) = 0. This can be rearranged to 10kN(0.25m) = Cy(0.25m) + 17.5kN(0.5m) which is what I have. By moving the negative term to the other side of the equal sign, it becomes positive.
In trusses we assume force in member either to be tensile or compressive, but here in frames you treated every member forces only at joints not along ? Is this way we have to solve frames question
Exam questions be like If beam AB is subject to force 20kN and the structure is in equilibrium then how many apples does Susie have left? And how old was her mother in WW2?
Hi. Is there a way to tell if a connection between two parts (e.g. a pin joint) has a vertical and horizontal force acting on it? As in, why does B not have a horizontal force while C and D have both directions? I assumed after seeing a few examples that the general rule to follow is; if there is a pinned connection then there are equal, opposite and co-linear forces acting in both directions.
Hey Shoncay, there is a trick. You need to identify which members are two-force members. Watch this video on 2 force members if you haven't already: ruclips.net/video/Zc5zU8Piys4/видео.html the internal force of these members will be aligned along the axis that connects its two pins. So in this video, both teal coloured members are 2 force members. The vertical one (member BE) axis is vertical, so the Internal force has no horizontal component, only vertical. That's why the pin doesn't have any horizontal things going on. Member DE is on an angle, so it's internal force is aligned with that same angle. So 5he internal force has both x and y component. That's why you see both in the pin. Hope that's makes sense. You can also watch the video I did on 3 force members, as it's a good rule to know too: ruclips.net/video/1irV4NXRZJA/видео.html
Is it possible to take the moment about another arbitrary point which will include the Fy and Ay forces in the calculation and then to take that equation along with the summation of forces in the y direction equation and solve for Fy and Ay?
Yeah you can take moments about any point. It's common to pick a point though that has some lines of actions of forces passi g through it thiu because then those forces will not cause a moment about that chosen point and it will simplify the equation
What's an easy indication to know what to solve first?? I had trouble following along because of how we would skip to a new part and then come back. Thanks for the great video btw!
Look for members with the least amount of unknowns, and solve them first. As you solve them, other members will then also get less unknowns, and then can be candidates to solve next!
I'm struggling on sketching the correct FBD of internal forces on each member. Say for example in "Member ABC", on "Point B" I drew it having a Components "Bx" and "By" while on "Point C" is in Tension/Compression. Is this still correct?
These frame problems are messy. In this case, member BE is a straight 2 force member, which means it's only in pure tension or compression. Member CD however is a 3 force member, and has other internal forces (shear). This video and all introductory frame problems at the statics level will not even mention shear, and just talk about the overall force being transmitted across each pin. For now you should think of it like this: ABC and CD are both connected to a common pin. ABC pushes/pulls on that pin, and CD also pushes on that pin. ABC and BC don't really push and pull on each other, they push and pull on the pin. The pin is in static equilibrium, because its not accelerating in any direction. So the force balance on the pin must equal zero in the x and y direction. This means whatever push/pull of each x and y component from ABC on the pin must be equal and opposite to the push/pull on the pin from CD. That's the best way to deal with these IMO at the intro statics level.
With frame problems, if a force is applied on a joint, then only apply it once when you break apart the frame into multiple FBDs, otherwise it will be double counted and you won't get the right result.
@@Engineer4Free I tried to answer a different problem with the same scenario wherein an applied force is located at the pin. According to your explanation in the prey video, it doesn't matter where I decided to put the applied force at the joint as long as I don't duplicate it. My question is why is that the answer is different if I put the applied force on one of the member and got it right according to my book vs if I put it on the other member and got the wrong answer? Thank you.
Hey no it is Cy(0.25) because the horizontal distance between C and D is 0.25m, you can see at the very beginning of the video when the distances are labelled on the diagram.
Fy is not included because there are too many unknowns. Once the frame is broken into its individual parts, there will be enough equations to solve for all of the unknowns.
This problem is a mess. The members are not in pure tension/compression, but there is also an element of internal shear force, and ignoring its analysis for simplified statics problems like these can lead to real-world errors. But this is still how frame problems are generally introduced in introductory statics classes. Because CD is a 3 force member, and not a 2 force member, we cant just talk about it just being in pure tension or compression. What we do with this simplified version of frame analysis is consider that the pin/joint at C is in equilibrium. That pin that connects ABC and CD is like a reaction force from the perspective of each member. The pin pulls/pushes ABC to the right. The pin pulls/pushes Cx to the left. The pin is in equilibrium, so it doesn't accelerate in the x direction, and therefor the net x forces on it must sum to zero so the magnitude of the x component of it's force balance is zero. Same in the y direction. With frame problems it's helpful to think of how members are interacting with each pin, rather than other members directly.
Thank you for taking the time out to make these videos. They are easier to understand than the lectures in class
Great videos! I found your channel a few days ago and your statics and linear algebra videos are super helpful, much better than my teachers!
Hey thanks for letting me know! Glad to hear it =)
thanks have a midterm tomorrow! you helped me a lot brother :]
Hope it went well!!!!
Hey quick question, when finding the moment about A, wouldn't the force applied from Cy be negative since it is causing rotation in the clockwise sense? So then the member BE would be in tension instead of compression
I should have stuck with the convention in always labelling counterclockwise moments as negative, an written -T(0.5m) - 25kN(1m) = 0 but for some sloppy reason I decided to drop the negative signs. The result doesn't change, because the initial right hand side of the equation is zero, so either way, T will come out as -50kN.
Thanks for Solving my previous issue. My question is about member CD . Is the member CD a two force member? Eventhough there is a horizontal force of 10 kN acting on CD in its middle . And will it still be a two force member if I apply this force on pin joint D of member CD?
CD is considered a 3 force member. Each pin applies one force, and the horizontal 10kN in the middle is the third. If the 10kN horizontal one wasn’t there, and we still considered the vertical force to be acting on the lower Teal coloured member, then CD would be a 2 force member. If the horizontal force wasn’t there, and we considered the vertical force to be acting at D on CD, we could model CD and a 3 force member again. Sorry for the delay, hope that helps clear it up.
@@Engineer4Free Thank you for solving my issue👍♥️
This is just cleared up so much, thanks a bunch!
Glad it helped! Check out the full playlist here: engineer4free.com/statics the frame videos are # 52-57 =)
shouldn't the moment caused by the 10KN, Cx and Cy force be reversed since it causes it to rotate counterclockwise around D? Would that not mean that Cy has a force of -6.25KN?
Sorry I skipped a step. The initial expression for member CD sum of moments about D should be: -10kN(0.25) + Cy(0.25m) + 17.5kN(0.5m) = 0. This can be rearranged to 10kN(0.25m) = Cy(0.25m) + 17.5kN(0.5m) which is what I have. By moving the negative term to the other side of the equal sign, it becomes positive.
In trusses we assume force in member either to be tensile or compressive, but here in frames you treated every member forces only at joints not along ? Is this way we have to solve frames question
I wish the exam questions were that simple
Life would be too easy
Exam questions be like
If beam AB is subject to force 20kN and the structure is in equilibrium then how many apples does Susie have left? And how old was her mother in WW2?
Hi. Is there a way to tell if a connection between two parts (e.g. a pin joint) has a vertical and horizontal force acting on it? As in, why does B not have a horizontal force while C and D have both directions? I assumed after seeing a few examples that the general rule to follow is; if there is a pinned connection then there are equal, opposite and co-linear forces acting in both directions.
Hey Shoncay, there is a trick. You need to identify which members are two-force members. Watch this video on 2 force members if you haven't already: ruclips.net/video/Zc5zU8Piys4/видео.html the internal force of these members will be aligned along the axis that connects its two pins. So in this video, both teal coloured members are 2 force members. The vertical one (member BE) axis is vertical, so the Internal force has no horizontal component, only vertical. That's why the pin doesn't have any horizontal things going on. Member DE is on an angle, so it's internal force is aligned with that same angle. So 5he internal force has both x and y component. That's why you see both in the pin. Hope that's makes sense. You can also watch the video I did on 3 force members, as it's a good rule to know too: ruclips.net/video/1irV4NXRZJA/видео.html
Question, why wouldn't member "BE" be a zero force member?
Is it possible to take the moment about another arbitrary point which will include the Fy and Ay forces in the calculation and then to take that equation along with the summation of forces in the y direction equation and solve for Fy and Ay?
Yeah you can take moments about any point. It's common to pick a point though that has some lines of actions of forces passi g through it thiu because then those forces will not cause a moment about that chosen point and it will simplify the equation
If it was asked force in member AB or any other, how will be find it out?
Thanks for these videos!
You're welcome, thanks for watching!!
What's an easy indication to know what to solve first?? I had trouble following along because of how we would skip to a new part and then come back. Thanks for the great video btw!
Look for members with the least amount of unknowns, and solve them first. As you solve them, other members will then also get less unknowns, and then can be candidates to solve next!
I'm struggling on sketching the correct FBD of internal forces on each member. Say for example in "Member ABC", on "Point B" I drew it having a Components "Bx" and "By" while on "Point C" is in Tension/Compression. Is this still correct?
These frame problems are messy. In this case, member BE is a straight 2 force member, which means it's only in pure tension or compression. Member CD however is a 3 force member, and has other internal forces (shear). This video and all introductory frame problems at the statics level will not even mention shear, and just talk about the overall force being transmitted across each pin. For now you should think of it like this: ABC and CD are both connected to a common pin. ABC pushes/pulls on that pin, and CD also pushes on that pin. ABC and BC don't really push and pull on each other, they push and pull on the pin. The pin is in static equilibrium, because its not accelerating in any direction. So the force balance on the pin must equal zero in the x and y direction. This means whatever push/pull of each x and y component from ABC on the pin must be equal and opposite to the push/pull on the pin from CD. That's the best way to deal with these IMO at the intro statics level.
Thanks a ton ❤
Very helpful
Glad to hear it =)
At 7:00, shouldn't the value of Cx be -17.5kN bse its direction is reversed?
Yes, but he transposed it to the other side making it a positive value.
Super.. Thank you very much
You’re welcome!!! 🙂
Why is the 5 kN not applied on member CD?
With frame problems, if a force is applied on a joint, then only apply it once when you break apart the frame into multiple FBDs, otherwise it will be double counted and you won't get the right result.
@@Engineer4Free I tried to answer a different problem with the same scenario wherein an applied force is located at the pin. According to your explanation in the prey video, it doesn't matter where I decided to put the applied force at the joint as long as I don't duplicate it. My question is why is that the answer is different if I put the applied force on one of the member and got it right according to my book vs if I put it on the other member and got the wrong answer? Thank you.
Couldn't you have gotten the Fy and Ay by doing the moment about point D? Or can you not because there's an external force?
Nevermind, there would still be 2 unknowns I see now
wow thank you sir!!!
You're welcome peanuts! 🥜😉
God damn that textbook was spitting straight gibberish until I saw your video.
Math text books are really the worst. Thanks boss.
sum{Md}=isn't it Cy(0.5) bro let me know I'm really confused
Hey no it is Cy(0.25) because the horizontal distance between C and D is 0.25m, you can see at the very beginning of the video when the distances are labelled on the diagram.
Hi, can you take the moment in F to find the force Ax
It works both ways, I solved with the moment but I think it's just easier to solve with sum of forces
Will I get the same answer if I apply 5kN at point D of section FED?
I’m not sure what you mean, the 5kN force is indicated there
@@Engineer4Free I'm sorry I framed the question wrong. What I meant was if I assume 5kN at FBD CD instead of FBD at FED, will I get the same answer?
Great Video i got it right
Awesome, glad you gave it a try yourself =). Best way to use the videos IMO.
Really helpfull !
Thanks for letting me know! =)
clutch video
well you guys, this is a lesson, where we learn from others' mistakes.
xoxo
boom boom
What about Fy? Why didn't we consider that?
Fy is not included because there are too many unknowns. Once the frame is broken into its individual parts, there will be enough equations to solve for all of the unknowns.
why would the Cy for member ABC be in tension but in compression for member CD lol bro you messed this whole thing up
This problem is a mess. The members are not in pure tension/compression, but there is also an element of internal shear force, and ignoring its analysis for simplified statics problems like these can lead to real-world errors. But this is still how frame problems are generally introduced in introductory statics classes. Because CD is a 3 force member, and not a 2 force member, we cant just talk about it just being in pure tension or compression. What we do with this simplified version of frame analysis is consider that the pin/joint at C is in equilibrium. That pin that connects ABC and CD is like a reaction force from the perspective of each member. The pin pulls/pushes ABC to the right. The pin pulls/pushes Cx to the left. The pin is in equilibrium, so it doesn't accelerate in the x direction, and therefor the net x forces on it must sum to zero so the magnitude of the x component of it's force balance is zero. Same in the y direction. With frame problems it's helpful to think of how members are interacting with each pin, rather than other members directly.