1 kip = 1000 pounds = 4.45 kN (approximately). It’s actually a standard US unit for force in civil engineering, no joke. The only surprising thing is that it’s a factor of 1000 in a unit system that typically eschews powers of 10!
Hello Professor H, for a falsework bent supporting a sloping concrete slab at 10% cross slope, do the vertical posts have to be braced to resist the horizontal force due to the sloping bent cap beam?? Some on the team think that yes, others think that friction will transfer the dead load vertically into the posts regardless of the slope, so there is no lateral force to consider. What does Prof. H say? Thank you.
At 5:57, I am looking to find the internal forces in Beam BC. The downward (axial) internal force at the top of AB must be applied as an equal and opposite upward (shear) force at the left end of BC. It's not technically accurate to call this upward force a "reaction"... it's actually the internal shear force at the left end of BC. Again, we know it acts up because it must be equal and opposite that on the top of AB. As for computing shear forces, the shear diagram (which is just a plot of the internal shear forces at all locations throughout the structures) "follows" the applied loads. Shear is always perpendicular to the direction of the structures, so shear in the columns goes right-and-left, while shear in the beam goes up-and-down. Because of this change in direction, we have to re-orient ourselves with the internal forces whenever rounding a corner. Hence as described above, the axial force in the column becomes the shear force in the beam when going from AB to BC.
The moment at the upper right corner (location C) is 27 k-ft in the beam and column. It doesn't change around the column. Honestly, I drew a vertical line there just out of habit... the moment doesn't go from 27 k-ft down to zero and then back up to 27 k-ft. It just is 27 k-ft at that point.
The 1.25 k is an external (applied) force at point C, not the shear, which is an internal force at that same point. The external force moves the shear diagram down by 1.25 k, in the direction of the force.
thanks for the video, why is the 1.8k positive ? would it be negative if it is applied to point B or if it was reversed at point C and what would happen if it would be below point A in a mirror like picture ?
I believe in the video, I call this 1.8-kip force at C a negative force, not positive. But, to be honest, positive and negative signs for forces are completely relative. It depends on the coordinate system you are using. The more important consideration is how the load changes the axial, shear, and moment diagrams. So for the 1.8-kip load at C, we are looking at the shear diagram from C to D. The positive x-direction goes down from C to D because I defined it that way, and the positive y-direction follows the right-hand rule for coordinate systems (x points down, so y must point to the right). The 1.8-kip force points left and is therefore "negative," but all you need to remember is that the shear diagram moves in the same direction as the applied load. So because the load points left, the shear diagram from C and D also moves to the left, taking me from 0 to -1.8 kips in shear. Note that flipping the coordinate system turns this around, but we can always use the same shear diagram principle. So if I instead define the positive x-direction as UP from D to C, now the positive y-direction points to the left... my leftward 1.8-kip load is now suddenly positive! In this case, we would draw the shear diagram starting at D (you always need to draw diagrams in the direction of positive x, never backwards) and going to C. When you do this, though the shear diagram may look flipped, you will find that the shear demand is still -1.8 kips along the entire length of the column from D to C... exactly the same as in the video where the shear diagram was drawn from C to D.
Hey sir at 10:04 can you explain why you did not connect the shear and moment diagram to zero at the end? Since you solved the for the missing shear and moment i thought u would complete the diagrams?
Thanks for your question! At point C, there is still an internal shear (4.33 k) and moment (27 k-ft). I recognize that some people have different conventions for this, but I like to leave the diagrams "open" when I'm still in the middle of the structure, and only "close" it back to zero when I get to the end. You will notice I did the same thing at point B at 5:38. I do this because the shear and moment diagrams physically represent the internal forces or moments at that point in the structure. Closing the diagrams back to zero at corners B or C seems to imply that the internal forces jump to zero at those points, which is clearly not the case. Anyway, regardless of my opinion on this topic, I've see both ways done, so I'd say it mainly comes down to preference.
The full answer might be a full video, but in the meantime, I'd probably take slices of the free body diagram at some angle theta (sort of like polar coordinates) rather than at some position x. For a semi-circular frame, the axial force will be tangential, the shear force will be radial, and the bending moment will still just be a moment. Even for a concentrated force at the peak of frame, the axial and shear force values will not be constants - they will vary by the angle theta. From there, you can apply equilibrium (sum of forces and sum of moments) to find the internal forces.
The beam is rigidly connected to the columns and the brace has hinge connections. The columns have pin supports so how to draw SFD BMD and axial force?
The moment arm for Dx about point A equals zero, so it doesn't show up in the sum of moments about A. The moment arm is always measured perpendicular to the line of action of the force. Dx acts horizontally, so the moment arm is the vertical distance from A to D... which is zero because A and D are at the same height. Said another way that may be more familiar to some: the line of action of Dx passes through point A, so it has no moment about point A.
Sorry, no. Fortunately for structural analysis we only really need length and force units, so 1 Kip = 4.5 kN and 1 ft = 0.3 meters (approximate conversions) will get you pretty close. The process is otherwise the same regardless of your unit system.
For the point loads at connection C, you can pick either free-body diagram (FBD) and get the same results. So we could apply the 1.8-kip x-force to either BC or CD (but not both at once), and we could apply the 1.25-kip y-force to either BC or CD (but again, not both at once). In this video, I chose to apply the vertical force to (3) BC and the horizontal force to (4) CD because I find it more intuitive to modify shear diagrams rather than axial diagrams, but that's just a matter of personal preference. At any rate, each external force should only be placed on one FBD; for example, if the 1.8-kip x-force were placed on both segments (3) BC and (4) CD, then the force would be double counted.
The M-diagram is indeed the integral of the V-diagram, but being an integral, you will have a constant of integration (the internal moment at the beginning of the beam). That starting point will have an impact on the sign of the M-diagram. But yes, positive shear means moment moves in the positive direction, and likewise for negative shear. The other way to keep this is mind: the moment diagram is always drawn on the side of the beam or column that is being compressed. This can be used as a quick check if the signs make sense by sketching the deflected shape of the structure.
The column from A to B is supported by a roller at the base. This means it only has a vertical reaction at the base, and therefore it only carries axial force. The moment is determined by integrating the shear (that is, taking the area under the shear diagram), but the shear is zero for the entire column. Therefore, the moment is also zero for that column, including up to point B at the top left corner.
what in the american si units is this
Exactly 😭 Foot is alright but what on earth is Kips
1 kip = 1000 pounds = 4.45 kN (approximately). It’s actually a standard US unit for force in civil engineering, no joke. The only surprising thing is that it’s a factor of 1000 in a unit system that typically eschews powers of 10!
Thank you so much for this video it explains it perfectly.
Hello Professor H, for a falsework bent supporting a sloping concrete slab at 10% cross slope, do the vertical posts have to be braced to resist the horizontal force due to the sloping bent cap beam?? Some on the team think that yes, others think that friction will transfer the dead load vertically into the posts regardless of the slope, so there is no lateral force to consider. What does Prof. H say? Thank you.
I really get it so easily thanks a lot...
sir, may i know in 5:57 , why do we use the reaction force(already given: 12.92, upward) instead of the internal foce(downward)
also,, how do you calculate the shear forces sir,, sorry if i happen to ask you lots of quest. ,,
At 5:57, I am looking to find the internal forces in Beam BC. The downward (axial) internal force at the top of AB must be applied as an equal and opposite upward (shear) force at the left end of BC. It's not technically accurate to call this upward force a "reaction"... it's actually the internal shear force at the left end of BC. Again, we know it acts up because it must be equal and opposite that on the top of AB.
As for computing shear forces, the shear diagram (which is just a plot of the internal shear forces at all locations throughout the structures) "follows" the applied loads. Shear is always perpendicular to the direction of the structures, so shear in the columns goes right-and-left, while shear in the beam goes up-and-down. Because of this change in direction, we have to re-orient ourselves with the internal forces whenever rounding a corner. Hence as described above, the axial force in the column becomes the shear force in the beam when going from AB to BC.
@@StructuresProfH thank you very much sir for the explanation 🙂
What if we have pinned connections at B, C and D and fixed at A..
Link to any video with these conditions? It will be very helpful
The final moment diagram, why is there a small vertical line going down at 27k/ft to the top corner of the frame? Just before the triangle?
The moment at the upper right corner (location C) is 27 k-ft in the beam and column. It doesn't change around the column. Honestly, I drew a vertical line there just out of habit... the moment doesn't go from 27 k-ft down to zero and then back up to 27 k-ft. It just is 27 k-ft at that point.
Why here 10:08 didn’t take the total Vc=3.08k? (4.33-1.25)
THANKS!!!!YOU saved my staying in collage
Happy to help!
Interesting video. Keep it up.
7:45 i dont get it,isnt the shear at point c equal to -1,25 kips?
The 1.25 k is an external (applied) force at point C, not the shear, which is an internal force at that same point. The external force moves the shear diagram down by 1.25 k, in the direction of the force.
thanks for the video, why is the 1.8k positive ? would it be negative if it is applied to point B or if it was reversed at point C and what would happen if it would be below point A in a mirror like picture ?
I believe in the video, I call this 1.8-kip force at C a negative force, not positive. But, to be honest, positive and negative signs for forces are completely relative. It depends on the coordinate system you are using. The more important consideration is how the load changes the axial, shear, and moment diagrams.
So for the 1.8-kip load at C, we are looking at the shear diagram from C to D. The positive x-direction goes down from C to D because I defined it that way, and the positive y-direction follows the right-hand rule for coordinate systems (x points down, so y must point to the right). The 1.8-kip force points left and is therefore "negative," but all you need to remember is that the shear diagram moves in the same direction as the applied load. So because the load points left, the shear diagram from C and D also moves to the left, taking me from 0 to -1.8 kips in shear.
Note that flipping the coordinate system turns this around, but we can always use the same shear diagram principle. So if I instead define the positive x-direction as UP from D to C, now the positive y-direction points to the left... my leftward 1.8-kip load is now suddenly positive! In this case, we would draw the shear diagram starting at D (you always need to draw diagrams in the direction of positive x, never backwards) and going to C. When you do this, though the shear diagram may look flipped, you will find that the shear demand is still -1.8 kips along the entire length of the column from D to C... exactly the same as in the video where the shear diagram was drawn from C to D.
Hey sir at 10:04 can you explain why you did not connect the shear and moment diagram to zero at the end? Since you solved the for the missing shear and moment i thought u would complete the diagrams?
Thanks for your question! At point C, there is still an internal shear (4.33 k) and moment (27 k-ft). I recognize that some people have different conventions for this, but I like to leave the diagrams "open" when I'm still in the middle of the structure, and only "close" it back to zero when I get to the end. You will notice I did the same thing at point B at 5:38.
I do this because the shear and moment diagrams physically represent the internal forces or moments at that point in the structure. Closing the diagrams back to zero at corners B or C seems to imply that the internal forces jump to zero at those points, which is clearly not the case.
Anyway, regardless of my opinion on this topic, I've see both ways done, so I'd say it mainly comes down to preference.
Hey what’s the reason behind separating those 2 point loads in the beginning
What if I have a curved (semi-circular in my instance) frame? How do I implement NQM there?
The full answer might be a full video, but in the meantime, I'd probably take slices of the free body diagram at some angle theta (sort of like polar coordinates) rather than at some position x. For a semi-circular frame, the axial force will be tangential, the shear force will be radial, and the bending moment will still just be a moment. Even for a concentrated force at the peak of frame, the axial and shear force values will not be constants - they will vary by the angle theta. From there, you can apply equilibrium (sum of forces and sum of moments) to find the internal forces.
The beam is rigidly connected to the columns and the brace has hinge connections. The columns have pin supports so how to draw SFD BMD and axial force?
when taking the moment about a, why dont you consider dx?
The moment arm for Dx about point A equals zero, so it doesn't show up in the sum of moments about A. The moment arm is always measured perpendicular to the line of action of the force. Dx acts horizontally, so the moment arm is the vertical distance from A to D... which is zero because A and D are at the same height.
Said another way that may be more familiar to some: the line of action of Dx passes through point A, so it has no moment about point A.
Is there a version where you use SI units?
Sorry, no. Fortunately for structural analysis we only really need length and force units, so 1 Kip = 4.5 kN and 1 ft = 0.3 meters (approximate conversions) will get you pretty close. The process is otherwise the same regardless of your unit system.
should 1.8K be included as a force acting in x direction in 3-BC ?
For the point loads at connection C, you can pick either free-body diagram (FBD) and get the same results. So we could apply the 1.8-kip x-force to either BC or CD (but not both at once), and we could apply the 1.25-kip y-force to either BC or CD (but again, not both at once). In this video, I chose to apply the vertical force to (3) BC and the horizontal force to (4) CD because I find it more intuitive to modify shear diagrams rather than axial diagrams, but that's just a matter of personal preference.
At any rate, each external force should only be placed on one FBD; for example, if the 1.8-kip x-force were placed on both segments (3) BC and (4) CD, then the force would be double counted.
@@StructuresProfH Thanks Professor.
isn't the M-diagram the integral of the V-diagram? and therefore the M-diagram should have been drawn on the other side of the last beam?
The M-diagram is indeed the integral of the V-diagram, but being an integral, you will have a constant of integration (the internal moment at the beginning of the beam). That starting point will have an impact on the sign of the M-diagram. But yes, positive shear means moment moves in the positive direction, and likewise for negative shear.
The other way to keep this is mind: the moment diagram is always drawn on the side of the beam or column that is being compressed. This can be used as a quick check if the signs make sense by sketching the deflected shape of the structure.
why is there no moment at top left corner, at B?
The column from A to B is supported by a roller at the base. This means it only has a vertical reaction at the base, and therefore it only carries axial force. The moment is determined by integrating the shear (that is, taking the area under the shear diagram), but the shear is zero for the entire column. Therefore, the moment is also zero for that column, including up to point B at the top left corner.
Sway analyisis of frame to be explain