Same man. My professor never did examples with hinges. Also, it's not in the statics book or mechanics of materials book. I got an A in statics and they didn't even go over this.
Thank you for uploading these videos. I had a serious problem about imaginating situations in my mind, but now looks like i can get over for this, :) Thank you so much again..
Question1: When cutting the segments, why do you use always Ve of Vf and not Ey and Ex ? Question2: What if there is a couple moment of internal hinge/s, is it in same direction when cut the segment or it is unchangeble (the rotation of couple). Ty and more POWER!
We usually use the letter V to refer to a shear force. But of course we can use any other letter/symbol to refer to the internal shear force at the hinge when the beam is cut. Generally, there is no fast rule for labeling forces, there is only convention and/or preferences. Well, since the internal moment at the hinge is zero, there is no need to draw it at the hinge. But if we were to show the internal moment at a cut point in the beam, we would need to draw them at two opposite moments (similar to the way we draw two opposite shear forces). If we draw the moment at the left face in the counterclockwise direction, the moment at the right face of the cut needs to be drawn in the clockwise direction, and vice versa.
Thank you so much for the video! I just have one question, why can't we consider the beam as a whole system after we separate the two sides of the hinge?
The assumed direction of the force has no impact on the results of the analysis. If we assume a downward direction for an unknown force, and its magnitude comes out to be negative, then we know the actual direction of the force is upward. If we had assumed an upward direction for the force, we would have gotten a positive magnitude for the force. For example, a downward force of -10 kN is the same as an upward force with a magnitude of +10 kN. The same is true for bending moment. If we assume a clockwise direction for a bending moment, and its magnitude comes out to be negative, say -100 kN-m, then we know that the moment is actually acting in the counterclockwise direction with a magnitude of +100 kN-m. Here too it is important to keep in mind that a clockwise moment having a negative magnitude (say, +100 kN-m) is identical to a counterclockwise moment with a magnitude of -100 kN-m, and vice versa.
Great. I don't understand in university in own language. but this video helps me better than my teacher (professor) . even I don't understand all your words.
Why is it that in the first problem with 1 hinge, the moment of the fixed support is drawn clockwise, however in your first video when you were showing the types of support, the moment was drawn anti-clockwise. I drew the moment anti-clockwise and got M to be 16 kNM, whilst you got -16. Could you please explain to me how to determine the direction of moments?
When the force/moment is unknown, since we don’t yet know its direction, we need to make an assumption. In the case of a bending moment, we can assume it to be either in the clockwise or counterclockwise direction. If we guess correctly, we get a positive magnitude for the moment. If we guess incorrectly, the magnitude of the moment comes out negative, which tells us the correct direction. So, when the unknown moment becomes known, we have its correct direction and magnitude. That is, a negative moment magnitude tells us that the moment is actually acting in the opposite direction (with a positive magnitude).
So what's the use of the axial forces if their value is zero. Because both the problems you solved, their value comes =0 . So you can anyone kindly tell me this please?. Great video btw , got to learn the concept with ease
For the sake of completeness, we include the axial force(s) in the formulation. If the beam is not subjected to any horizontal load, the axial force(s) are always zero. But that would not be the case if an inclined load (with a horizontal component) is placed on the beam. Furthermore, if the structure is a frame (not a beam), axial forces may not necessarily be zero, even when the system carries gravity loads only. So, to be safe (not to make errors when solving the problem), we should include those forces in the formulation, even though for some problems they may end up to be zero.
The term fixed is mainly used for referring to supports, the ones that cannot move or rotate. A rigid joint, on the other hand, is a term we use for referring to connection between two members (often a beam and a column). A rigid connection simply means the beam and the column, at the joint, are going to move or rotate together; they act as a unit. Think of a welded connection between a beam and a column as a rigid joint. And think of a not overly bolted connection between the two members as a semi-rigid or even a non rigid connection where one member could rotate relative to the other one.
We say there are three equilibrium equations , FX,fy and sum of moments equal zero , however you say when we add a hinge the number of equilibrium equations increase to 4 or 5 , how is this possible? Does the hinge add another condition of equilibrium? My guess is that bending moment at hinge being equal to zero is the additional equilibrium equation that results when we add a hinge ? Please reply
For each beam segment we can write three static equilibrium equations. When the beam has an internal hinge, we can separate the beam at the hinge into two segments. For each segment we can write three equilibrium equations. That gives us a total of six equations. So we can solve for six unknown forces. The unknown forces being the support reactions and the shear force and axial force at the hinge (bending moment at the hinge = 0). Such a beam can be analyzed if it has a total of four unknown support reactions. The four support reactions plus the two unknown forces (shear force and axial force) at the hinge gives us a total of six unknowns. Such a beam is treated at a statically determinate system since we can determine all six unknowns using the six available equilibrium equations.
@@DrStructure my teacher has given me a problem that has a beam with one end fixed , the other end roller support with a hinge between the two supports , it says that the total number of reactions are four ,but the number of equilibrium equations are four , and the one extra equilibrium equation comes from the internal Hinge , can you please explain this ? Is there a difference between equilibrium conditions and equilibrium equations?
The number of unknown support reactions is 4 (three reactions at the fixed support and one reaction at the roller support). And if we separate the beam segments at the hinge, we introduce two additional unknowns (a shear force and an axial force at the hinge). So, we end up with 6 unknowns. Suppose we label the reactions at the fixed supports Fx, Fy, and M. And refer to the reaction at the roller as R. So, Fx, Fy, M, and R are our unknown forces. The three static equilibrium equations for the entire beam are going to be defined based on the four unknown forces mentioned above. Note that none of these unknows are related to the hinge. So, what does the additional equilibrium equation (for the hinge) look like? Does it use any of the four unknowns? The answer is no! The equation at the hinge references the internal moment at the hinge (the moment at the hinge = 0). But that moment is not a part of the support reactions (i.e., it is not one of the four unknowns referenced above). So that additional moment equation at the hinge is of no use for determining the support reactions. To solve the problem correctly, you need to write three equations for each segment of the beam (assuming we have separated the beam into two segments at the hinge). The four support reactions + the unknown shear and axial force at the hinge gives us a total of 6 unknowns which we can determine using the 6 equilibrium equations.
The external load/moment needs to be placed either just to the left of the hinge (at the right end of segment AC) or just to the right of the hinge at the left end of segment CB. The load/moment cannot be placed exactly at the hinge, since it is not an actual structural member but a means of connecting two members. Place the applied bending moment on the free-body diagram when the two segments are separated. So, the moment appears at the end of one of the two segments (AC or CB). Then proceed with the analysis as usual.
Thank you so much for this video. It was explained well and easily understandable. I have a small doubt in the final problem for which Moment at E was given 2(8)(4)+8Vf- 3Cy=0.I am not sure why 2(8)(4) is coming...Isnt it be 2(8).?
Dinesh Kumar Two times eight is not bending moment, rather it is the magnitude of the total load, the equivalent concentrated load for the uniformly distributed one. That is, if we were to replace the distributed load with a concentrated load we would get a load magnitude of 2(8). This equivalent concentrated load is placed at the center of the distributed load (at center of the rectangle representing the uniform load). This center is located 4 meters away from point E. But what is the moment of this load about point E? The total load is 2(8), the moment arm is the 4 m, so we get 2(8)(4).
Hi, @ 3:00, for the left part. if Mc = 0, then FAy × 8m = 0, then FAy = 0. i know it's wrong, but I couldn't figure out why. Could you explain it to me?
Hi Sir, will the hinge reduce the moment in a beam ? or will it increase the moment ? for example at video 2:50, the structure is stable without adding the hinge, but it will be statically indeterminate ( we can't count ) but it is still stable, assuming if the above structure at 2:50 has no hinge, will the moment suffer in the beam be greater or less ?
Yes, if we get rid of the hinge, we end up with an indeterminate beam. That beam would have a smaller bending moment at A, but a non-zero moment at C. The beam with the hinge has a zero moment at C but a larger moment at A. So, when the hinge is removed, the moment gets redistributed resulting in a moment reduction in one part of the beam (at A) and a moment increase in another part of the beam (at C).
Do you mean a beam with a circular cross-section or one that has a curved (circular) geometry? If the former, yes, a beam with a circular cross-section could have a splice that acts as an internal hinge. If the latter, the analysis of curved beams is beyond the scope of introductory structural analysis, but they too could be designed to include internal hinges.
SF @ A = 2KN SF is constant from A to point LOAD as there is no load in between in the segment. SF @ pint load = 2KN - 4KN= -2KN THEREFORE SHEAR VALUE CHANGES FROM POSITIVE 2KN TO NEGATIVE 2KN. AGAIN SF -2KN IS CONSTANT FROM POINT LOAD TO SUPPORT B. THEREFORE SF @B = -2KN IS IT CORRECT SIR??
By definition, a hinge allows rotation. Think of two popsicle sticks connected together using a push pin. We can easily rotate one of the sticks around the pin relative to the other. This simply means the pin cannot resist bending moment, that is why the rotation takes place. A hinge behaves the same way. It permits rotation, therefore no bending moment develops at the connection.
If a force passes through a point, its moment about the point is zero. But that is not what we have here. At the fixed support, there is a vertical reaction force and a bending moment. These are called the support reactions. The bending moment (M) does not act like a force. We don't find its moment by multiplying its magnitude by a distance. The moment of M taken about any point on the beam is M.
Hi, thanks for the tutorial, very informative. However I have encountered a problem that is what if there is a point force acting right at C, the hinge? How should I deal with such force? Thanks !
Fottonela1 Wys For analysis purposes, a hinge is considered to be very small (more like a dimensionless point). It does not contribute to the overall length of the beam. As such the load can be paced either to the right of it or to the left of it. Alternatively, you can divide the load in half. Place half the load to the right of the hinge and place the other half-load to the left of the point.
hey ! @2.26, it is said that , although the force Cx is not acting on the beam, for completeness we have to take this forces too. Why so ? What is that completeness meant there ?
In 2D space, there are three equilibrium equations: (1) Sum of the forces in the x direction must be zero. (2) Sum of the moments in the y direction must be zero. (3) Sum of the moments about any point (often one of the support points) must be zero. In most cases, if there is no applied force in the x direction, the first equation above is automatically satisfied. So, when we are solving the problem, we can ignore (1) and just use (2) and (3) to analyze the beam. By completeness, it meant to indicate that we are also using (1) in our formulation. That is, we are using the complete set of equilibrium equations.
It causes a redistribution of the internal forces in the structure. That does not necessarily mean a weak structure. Generally speaking, we design a structure for a specific purpose; to safely carry a specific set of forces. Of course, we can over-design the structure. We can design it to carry more loads than it will be subjected to during its life cycle. In a sense, the over-designed structure can be said to be “stronger” than the optimized structure. But that does not mean the optimized system is weak. If we add redundancies to the system and make the members heavier and stronger, we end up with an unnecessarily stronger structure which renders the fitting (optimized) design a weaker option.
Yes, our lectures can be used for educational purposes. You may want to direct your students to our (free) online course which can be used as a supplementary resource for your teaching. The link is provided in the video description field.
Thank you so much for the video.But , i have a small doubt regarding this beam(2), it is statically determinate i.e 5 unknowns,3 EOQ remaining two equations are obtained from two hinges ok .My doubt is how many members,joints in this beam & at support is it joint there or not ,so that how can we determine(from formulas or through Free body diagrams) beam (2) is statically determinate.
+pavan kumar Not sure if I understand your question correctly. As you said, there are five unknown forces (support reactions). The three equilibrium equations + two hinge equations give us a total of five equations, hence we have a statically determinate system. Suppose the beam had another segment, resting on the roller, attached to the exiting system using a hinge. So now, the number of unknowns is increased to 6 (because of the additional roller), but so is the number of equations (we get one additional equation due to the additional hinge), so the structure remains statically determinate. As long as the number of equations matches the number of unknowns, the system remains statically determinate.
+pavan kumar :Thank you for answering.I understand 3 Equilibrium equations+2 hinge Equations,hence it is statically determinate. But, my doubt is how many members, joints in this beam & at support(A or D) is it joint there or not?And also how can we determine(from formulas ) beam (2) is statically determinate.
+pavan kumar for analysis purposes, there are three segments in the beam: AE, EF and FD. There is no internal hinge at A, but the point rests on a pin support. There is no internal hinge at D either, but the point rests on a roller support. To determine if the beam is statically determinate, separate the beam into the three segments mentioned above. For each segment write the equilibrium equations. Since we have three equilibrium equations per segment, we end up with a total of nine equations. Now, count the number of unknowns. If there are exactly nine unknown forces, the structure is said to be statically determinate. The unknowns are: Vertical and horizontal reactions at A, vertical reaction at B, C and D, shear and axial force at E, and shear and axial force at F. This gives a total of 9 unknown forces. So, the beam is statically determinate.
sir,The purpose of providing internal hinge is : 1. to connect different members on different Materials. 2. to transfer the load from one part to another part. 3. to convert indeterminacy of structure into determinate structure. right... But my question is how would it transfer the loads from one part to another part.. next question in your first example you would have just separated the given beam into two segments in which there were five unknowns in the first segments but we have only 3 equilibrium equations then how could you say it is statically determinate?? you have not explained about the force releases concept here?? what does it mean force releases??
The hinge transfers the load via a shear force, and possibly an axial force, if the member is subjected to such a force. Since in theory, the hinge allows rotation, there would be no resistance to moment at that point, meaning bending moment becomes zero at the hinge, so no transfer of moment takes place there. With regard to determinacy, we say a structure is statically determinate if it can be analyzed using the static equilibrium conditions (equations). This is different than saying a structure is determinate if it only has three unknowns. As the example illustrates, sometimes more than three unknowns appear in the formulation, but as long as we can determine all the forces using the three equilibrium equations, the structure is considered determinate. If the entire structure was that of Segment AE in the example, and there was no other parts to the beam, then we would have had an indeterminate system.
this is the best video... but i have small confusion what is the degree of static indeterminancy if more than 2 members are connected in internal hinge.? plz i am looking forward for answer?
+Rajendra Kunwar Say, we have a simply supported beam, with degree of indeterminacy of zero. If we attach another beam segment that rests on a roller to the original beam using a hinge, the degree of indeterminacy remains zero. We can continue adding beam segments (each resting on a roller) using a hinge without increasing the degree of indeterminacy of the entire system. The beam remains statically determinate.
No, it does not change the nature of the pin support. A hinge located at a (short) distance from a pin support makes the segment between them act as a truss element. Meaning, it can carry an axial force only, no shear force or bending moment develops in the segment. Assuming that the resulting system is still statically determinate, it can be analyzed just like a beam with an internal hinge.
@@DrStructure you mean the support reaction which is in the direction of shear at hinge will be zero by making the equilibrium at the segment. Right? And then the only force at the support will be the axial force balanced at support?
Sir Please solve one problem in beam with internal hinge because I can able to find the support reaction but even I can calculate the shear force and bending moment values but having a confusions in calculating shear force. Because according to positive shear force sign convention the sfd should come at the support A is positive 2kN but in books it has shown in downward direction there I am getting confusions.
The first example, which involves an internal hinge, shows an upward reaction of 2 kN at A. This results in a positive shear in the left segment of the beam. If you have a counter example, printed in a book, that is causing confusion, feel free to email an image of the solution to the address below for additional feedback. Dr.Structure@EducativeTechnologies.net
You just need to place the load either a tiny bit to the right of the hinge, or a bit to the left of the hinge (either would work, and give the same results) and proceed as usual.
An internal hinge can carry (transfer) a shear force and an axial force. If the beam is not subjected to any axial force, then no internal axial force develops in the member. In such a case, when drawing the free-boy diagram, knowing that the axial force at the hinge is zero, we often omit drawing it.
That is a design consideration which can be decided on case by case bases. For example, if the splice is near the point of zero moment in the beam, we can design it as a shear splice. If however the splice is to be placed at a point with a high internal moment, a moment splice would be more fitting.
@@DrStructureThanks indeed for the reply. 👍 Let's us say for example, that in a continuous beam, one of the spans has zero moment at one point and a maximum at the other one. Is it permitted and practical to have at one point a shear connection and the other a moment? Is it really happening in the real-world?
Permitted? Yes! Practical? That depends on the cost. If the cost of building such splice combinations for that beam is deemed reasonable, then yes, it is practical.
The process would be the same. If the load is at the hinge, for analysis purposes, move it a tiny bit, say, 0.0001 mm either to the right or to the left of the hinge (so that you can separate the two segments at the hinge) then proceed as usual.
That equation is written using the free-body diagram of the right segment of the beam (Segment BC). There is no M in in the free-body diagram of Segment BC. M is located at Point A and not a part of Segment BC.
That is the moment of the uniformly distributed load about E. The load is rectangular, the area of the rectangle is (2)(8). The rectangular load can be replaced by a concentrated load having a magnitude of (2)(8). That is the magnitude of the concentrated load that replaces the distributed load equals to the area of the rectangle. But where on the beam would we place this concentrated load? At the center of the rectangle which is 4 m away from Point E. So what is the moment of the distributed load about E? It equals to the moment of the concentrated load about E. What is the moment of the concentrated load about E? It equals to the load magnitude (2)(8) times the moment arm (4). So, we get (2)(8)(4).
@@DrStructure as bdm at inflection point was zero before adding the hinge, so it won't change the bending moment diagram, but only changes the elastic curve, right? it is like changing the boundary condition.
No quite. At the inflection point the of curvature of the elastic curve changes direction (i.e., going from concave up/down to concave down/up). That is not a point at which bending moment is zero. So, the bending moment diagram for such a beam is not going to show zero at the inflection point. But if we introduce an internal hinge at the point, we end up with a zero moment at that point. A good example for this would be a two-span continuous beam where only one of the spans is subjected to loads. The loaded span deflects downward (having a concave up deflection) while a significant part of the other span deflects upward with a concave down deflection. The point of inflection is in the unloaded span close to the middle support where concave up turns into concave down. Now, if we place a hinge at that point, we turn the beam into a statically determine one in which the entire load is going to be carried by the loaded span (and the two supports that the span rests on) only. The other span carries no part of the load. This results in a moment diagram, and an elastic curve that are completely different than those for the (indeterminate) continuous beam with no internal hinge.
@@DrStructure thank you for your detailed reply, but I feel confused. maybe I didn't express it clearly. I mean adding a hinge on the inflection point of the elastic curve, not the curvature diagram. the inflection point of elastic curve is where the curvature changes sign. As curvature (not slope) is in direct proportion to bending moment, so it is also the point where bending moment equals zero.
I think it would be best if we put this discussion in the context of an actual problem. You can work out a problem as you have envisioned it, save it as a PDF (still images works too) and email it to Dr.Structure@EducativeTechnologies.net We'll take a look at your solution and comment on it accordingly.
The beam is divided into three segments (free-body diagrams). Start from the right-most free-body diagram. The equilibrium equations for that free-body diagram is given @5:00. Solving those equations yields: Dy = 0, Vf = 0, Fx = 0. Knowing Fx and Vf, now consider the middle free-body diagram. The only remaining unknowns on that free-body diagram are: Ex, Ve, and Cy. Solving the three equilibrium equations for the middle segments yields: Ex = 0, Cy = 21.33 kN, and Ve = -5.33 kN. Knowing Ex and Ve, now consider the left-most free-body diagram where Ay resides. Solving the three equilibrium equations for the free-body diagram yields: Ax = 0, By = -7.46 and Ay = 2.14 kN.
+Dr. Structure ... sir I have gone through your profile .. thank u sir.. This is my humble request if possible please do some videos on design of reinforced concrete and steel structures so that we could learn parallely about the design along with the analysis. sir it is just a request pleas dont mind.. I always respect you sir.. in India we too have some excellent professors but they are very less in numbers. Every one can teach with the help of the board but only few professors like you touch the heart of the students you are one among.. in India number one institution is IIT even professor's who were working in this institution can't teach like you.. finally sorry for my English there might be some grammatical errors, since it is not our mother tongue.
Thank you for your feedback and kind words. Yes, our plan is to cover the design topics as well. Time permitting, hopefully we will get to those sooner rather than later.
Examine the free-body diagram and the equilibrium equations for the right segment. The third equation gives us: Dy = 0. Then, from the second equation we get: Vf = 0. Now, examine the middle segment where Vf has already been determined to be zero. There are three equilibrium equations associated with this segment. From the third equation, since Vf = 0, we can determine Cy. It equals (2)(8)(4)/3. Or, Cy = 21.33 kN. The second equation equation for this segment is: Ve + Cy - Vf - 2(8) = 0. Since Vf = 0 and Cy = 21.33, then Ve must be equal to -5.33 kN.
this channel is a digital treasure
Perfect explanation!!! Frames were one of the hardest topics to me, even more difficult than trusses and cables! Thank you for this tutorial
You’re welcome!
I spent hours trying to figure out a question with the hinge....... This video helped me a lot!
Same man. My professor never did examples with hinges. Also, it's not in the statics book or mechanics of materials book. I got an A in statics and they didn't even go over this.
Thanks a lot lot lot as this example gave me 4 points in upda test in qatar. Thank you vvvv much.
Thank you for uploading these videos. I had a serious problem about imaginating situations in my mind, but now looks like i can get over for this, :) Thank you so much again..
same here!:)
Thank you...I solved problems by this video...FBD I confused but now I get it.... awesome video..
Thank you sooooo much you're a life saver! I don't know how to say thank you but, you really helped me
So well explained… really helped me a lot… thanks
Question1: When cutting the segments, why do you use always Ve of Vf and not Ey and Ex ?
Question2: What if there is a couple moment of internal hinge/s, is it in same direction when cut the segment or it is unchangeble (the rotation of couple).
Ty and more POWER!
We usually use the letter V to refer to a shear force. But of course we can use any other letter/symbol to refer to the internal shear force at the hinge when the beam is cut. Generally, there is no fast rule for labeling forces, there is only convention and/or preferences.
Well, since the internal moment at the hinge is zero, there is no need to draw it at the hinge. But if we were to show the internal moment at a cut point in the beam, we would need to draw them at two opposite moments (similar to the way we draw two opposite shear forces). If we draw the moment at the left face in the counterclockwise direction, the moment at the right face of the cut needs to be drawn in the clockwise direction, and vice versa.
I get what exactly I was looking for
Thank you
Thank you Thank you thank you thank you thank you thank you thank you
Great work keep it up.
thanks very for this wonderful video,,,,,,,,,,,you are AWESOME
Brilliant video
Can u pls make more videos on hinged comnections and also continuous beam
Thank you so much for the video! I just have one question, why can't we consider the beam as a whole system after we separate the two sides of the hinge?
We can only write three equilibrium equations for the beam as a whole. Three equations are not enough to determine all the unknowns.
Nicely explained
Does the direction of the force that we assumed effect on the reactions or the moment
Thank you very much dr structure
The assumed direction of the force has no impact on the results of the analysis.
If we assume a downward direction for an unknown force, and its magnitude comes out to be negative, then we know the actual direction of the force is upward. If we had assumed an upward direction for the force, we would have gotten a positive magnitude for the force. For example, a downward force of -10 kN is the same as an upward force with a magnitude of +10 kN.
The same is true for bending moment. If we assume a clockwise direction for a bending moment, and its magnitude comes out to be negative, say -100 kN-m, then we know that the moment is actually acting in the counterclockwise direction with a magnitude of +100 kN-m. Here too it is important to keep in mind that a clockwise moment having a negative magnitude (say, +100 kN-m) is identical to a counterclockwise moment with a magnitude of -100 kN-m, and vice versa.
@@DrStructure
Got it 💚
Thank you it's not enough for you 💚
Fantastic. Short but illustrative . like it.
Great. I don't understand in university in own language. but this video helps me better than my teacher (professor) . even I don't understand all your words.
Why is it that in the first problem with 1 hinge, the moment of the fixed support is drawn clockwise, however in your first video when you were showing the types of support, the moment was drawn anti-clockwise. I drew the moment anti-clockwise and got M to be 16 kNM, whilst you got -16. Could you please explain to me how to determine the direction of moments?
When the force/moment is unknown, since we don’t yet know its direction, we need to make an assumption. In the case of a bending moment, we can assume it to be either in the clockwise or counterclockwise direction. If we guess correctly, we get a positive magnitude for the moment. If we guess incorrectly, the magnitude of the moment comes out negative, which tells us the correct direction.
So, when the unknown moment becomes known, we have its correct direction and magnitude. That is, a negative moment magnitude tells us that the moment is actually acting in the opposite direction (with a positive magnitude).
Great video! Thanks!
Watching your videos really helps me a lot! I’m having this subject’s exam today. I’m so scared i cant sleep lmao. Thank you!
Are you a civil engineer by now? I really need motivation...! 😫
Like if you think our teacher should only show us these video in lecture time then their sleepy lecture.
So what's the use of the axial forces if their value is zero. Because both the problems you solved, their value comes =0 .
So you can anyone kindly tell me this please?.
Great video btw , got to learn the concept with ease
For the sake of completeness, we include the axial force(s) in the formulation.
If the beam is not subjected to any horizontal load, the axial force(s) are always zero. But that would not be the case if an inclined load (with a horizontal component) is placed on the beam.
Furthermore, if the structure is a frame (not a beam), axial forces may not necessarily be zero, even when the system carries gravity loads only.
So, to be safe (not to make errors when solving the problem), we should include those forces in the formulation, even though for some problems they may end up to be zero.
Good evening Dr. What is the difference between a fixed beam and a rigidly jointed beam as we we see in the analysis of frames?
The term fixed is mainly used for referring to supports, the ones that cannot move or rotate. A rigid joint, on the other hand, is a term we use for referring to connection between two members (often a beam and a column). A rigid connection simply means the beam and the column, at the joint, are going to move or rotate together; they act as a unit. Think of a welded connection between a beam and a column as a rigid joint. And think of a not overly bolted connection between the two members as a semi-rigid or even a non rigid connection where one member could rotate relative to the other one.
I find it very useful.. Thank you
We say there are three equilibrium equations , FX,fy and sum of moments equal zero , however you say when we add a hinge the number of equilibrium equations increase to 4 or 5 , how is this possible? Does the hinge add another condition of equilibrium? My guess is that bending moment at hinge being equal to zero is the additional equilibrium equation that results when we add a hinge ? Please reply
For each beam segment we can write three static equilibrium equations.
When the beam has an internal hinge, we can separate the beam at the hinge into two segments.
For each segment we can write three equilibrium equations. That gives us a total of six equations. So we can solve for six unknown forces.
The unknown forces being the support reactions and the shear force and axial force at the hinge (bending moment at the hinge = 0).
Such a beam can be analyzed if it has a total of four unknown support reactions. The four support reactions plus the two unknown forces (shear force and axial force) at the hinge gives us a total of six unknowns. Such a beam is treated at a statically determinate system since we can determine all six unknowns using the six available equilibrium equations.
@@DrStructure my teacher has given me a problem that has a beam with one end fixed , the other end roller support with a hinge between the two supports , it says that the total number of reactions are four ,but the number of equilibrium equations are four , and the one extra equilibrium equation comes from the internal Hinge , can you please explain this ? Is there a difference between equilibrium conditions and equilibrium equations?
The number of unknown support reactions is 4 (three reactions at the fixed support and one reaction at the roller support). And if we separate the beam segments at the hinge, we introduce two additional unknowns (a shear force and an axial force at the hinge). So, we end up with 6 unknowns.
Suppose we label the reactions at the fixed supports Fx, Fy, and M. And refer to the reaction at the roller as R. So, Fx, Fy, M, and R are our unknown forces.
The three static equilibrium equations for the entire beam are going to be defined based on the four unknown forces mentioned above. Note that none of these unknows are related to the hinge. So, what does the additional equilibrium equation (for the hinge) look like? Does it use any of the four unknowns? The answer is no! The equation at the hinge references the internal moment at the hinge (the moment at the hinge = 0). But that moment is not a part of the support reactions (i.e., it is not one of the four unknowns referenced above). So that additional moment equation at the hinge is of no use for determining the support reactions.
To solve the problem correctly, you need to write three equations for each segment of the beam (assuming we have separated the beam into two segments at the hinge). The four support reactions + the unknown shear and axial force at the hinge gives us a total of 6 unknowns which we can determine using the 6 equilibrium equations.
Hello Dr Structure. How is the beam in the first example any different if we have an external applied moment at the hinge? Thank you :)
The external load/moment needs to be placed either just to the left of the hinge (at the right end of segment AC) or just to the right of the hinge at the left end of segment CB. The load/moment cannot be placed exactly at the hinge, since it is not an actual structural member but a means of connecting two members.
Place the applied bending moment on the free-body diagram when the two segments are separated. So, the moment appears at the end of one of the two segments (AC or CB). Then proceed with the analysis as usual.
I am taking this in statics!
Thank you so much for this video. It was explained well and easily understandable. I have a small doubt in the final problem for which Moment at E was given 2(8)(4)+8Vf- 3Cy=0.I am not sure why 2(8)(4) is coming...Isnt it be 2(8).?
Dinesh Kumar Two times eight is not bending moment, rather it is the magnitude of the total load, the equivalent concentrated load for the uniformly distributed one. That is, if we were to replace the distributed load with a concentrated load we would get a load magnitude of 2(8). This equivalent concentrated load is placed at the center of the distributed load (at center of the rectangle representing the uniform load). This center is located 4 meters away from point E.
But what is the moment of this load about point E? The total load is 2(8), the moment arm is the 4 m, so we get 2(8)(4).
Hi, @ 3:00, for the left part. if Mc = 0, then FAy × 8m = 0, then FAy = 0. i know it's wrong, but I couldn't figure out why. Could you explain it to me?
First of all take only one segment. If taken the left one,consider the forces on the left one creating moments and the moment at A too.
Hi Sir, will the hinge reduce the moment in a beam ? or will it increase the moment ?
for example at video 2:50, the structure is stable without adding the hinge, but it will be statically indeterminate ( we can't count ) but it is still stable, assuming if the above structure at 2:50 has no hinge, will the moment suffer in the beam be greater or less ?
Yes, if we get rid of the hinge, we end up with an indeterminate beam. That beam would have a smaller bending moment at A, but a non-zero moment at C. The beam with the hinge has a zero moment at C but a larger moment at A. So, when the hinge is removed, the moment gets redistributed resulting in a moment reduction in one part of the beam (at A) and a moment increase in another part of the beam (at C).
great video but what do i do if i have a circular beam can i use those hinges ????
Do you mean a beam with a circular cross-section or one that has a curved (circular) geometry? If the former, yes, a beam with a circular cross-section could have a splice that acts as an internal hinge. If the latter, the analysis of curved beams is beyond the scope of introductory structural analysis, but they too could be designed to include internal hinges.
@@DrStructure thanks a lot for replying
SF @ A = 2KN
SF is constant from A to point LOAD as there is no load in between in the segment.
SF @ pint load = 2KN - 4KN= -2KN
THEREFORE SHEAR VALUE CHANGES FROM POSITIVE 2KN TO NEGATIVE 2KN.
AGAIN SF -2KN IS CONSTANT FROM POINT LOAD TO SUPPORT B.
THEREFORE SF @B = -2KN
IS IT CORRECT SIR??
Yes, your analysis is correct.
@@DrStructure thank u sir for your time and replying me.🎓🎓🎓
Thank you for the video...one question..why we don't have moment at the C hinged point in the free body diagram?
By definition, a hinge allows rotation. Think of two popsicle sticks connected together using a push pin. We can easily rotate one of the sticks around the pin relative to the other. This simply means the pin cannot resist bending moment, that is why the rotation takes place.
A hinge behaves the same way. It permits rotation, therefore no bending moment develops at the connection.
Thank you!!
You are welcome!
I am wondering where did you get the M in the summation of moment at point A. Isnt that supposed to be zero because it passes in the point?
If a force passes through a point, its moment about the point is zero. But that is not what we have here. At the fixed support, there is a vertical reaction force and a bending moment. These are called the support reactions. The bending moment (M) does not act like a force. We don't find its moment by multiplying its magnitude by a distance. The moment of M taken about any point on the beam is M.
@@DrStructure thank you for being responsive!! I understand it now 💗
Hi, thanks for the tutorial, very informative. However I have encountered a problem that is what if there is a point force acting right at C, the hinge? How should I deal with such force? Thanks !
Fottonela1 Wys For analysis purposes, a hinge is considered to be very small (more like a dimensionless point). It does not contribute to the overall length of the beam. As such the load can be paced either to the right of it or to the left of it.
Alternatively, you can divide the load in half. Place half the load to the right of the hinge and place the other half-load to the left of the point.
hey !
@2.26, it is said that , although the force Cx is not acting on the beam, for completeness we have to take this forces too. Why so ? What is that completeness meant there ?
In 2D space, there are three equilibrium equations:
(1) Sum of the forces in the x direction must be zero.
(2) Sum of the moments in the y direction must be zero.
(3) Sum of the moments about any point (often one of the support points) must be zero.
In most cases, if there is no applied force in the x direction, the first equation above is automatically satisfied. So, when we are solving the problem, we can ignore (1) and just use (2) and (3) to analyze the beam.
By completeness, it meant to indicate that we are also using (1) in our formulation. That is, we are using the complete set of equilibrium equations.
Does the presence of internal hinge make structure weak
It causes a redistribution of the internal forces in the structure. That does not necessarily mean a weak structure.
Generally speaking, we design a structure for a specific purpose; to safely carry a specific set of forces. Of course, we can over-design the structure. We can design it to carry more loads than it will be subjected to during its life cycle. In a sense, the over-designed structure can be said to be “stronger” than the optimized structure. But that does not mean the optimized system is weak.
If we add redundancies to the system and make the members heavier and stronger, we end up with an unnecessarily stronger structure which renders the fitting (optimized) design a weaker option.
Hello. Can I use your video as a supplement for my teaching?
Yes, our lectures can be used for educational purposes. You may want to direct your students to our (free) online course which can be used as a supplementary resource for your teaching. The link is provided in the video description field.
You have anyhing on shear center and shear flow? Your lectures are awesome helps me a lot
Eventually we will get into mechanics topics, but don't have anything as of now.
I have a similar problem like this, but I need to draw sfd and bmd diagram out of this.. and I don't know how
Thank you so much for the video.But , i have a small doubt regarding this beam(2), it is statically determinate i.e 5 unknowns,3 EOQ remaining two equations are obtained from two hinges ok .My doubt is how many members,joints in this beam & at support is it joint there or not ,so that how can we determine(from formulas or through Free body diagrams) beam (2) is statically determinate.
+pavan kumar Not sure if I understand your question correctly. As you said, there are five unknown forces (support reactions). The three equilibrium equations + two hinge equations give us a total of five equations, hence we have a statically determinate system.
Suppose the beam had another segment, resting on the roller, attached to the exiting system using a hinge. So now, the number of unknowns is increased to 6 (because of the additional roller), but so is the number of equations (we get one additional equation due to the additional hinge), so the structure remains statically determinate.
As long as the number of equations matches the number of unknowns, the system remains statically determinate.
+pavan kumar :Thank you for answering.I understand 3 Equilibrium equations+2 hinge Equations,hence it is statically determinate. But, my doubt is how many members, joints in this beam & at support(A or D) is it joint there or not?And also how can we determine(from formulas ) beam (2) is statically determinate.
+pavan kumar for analysis purposes, there are three segments in the beam: AE, EF and FD.
There is no internal hinge at A, but the point rests on a pin support.
There is no internal hinge at D either, but the point rests on a roller support.
To determine if the beam is statically determinate, separate the beam into the three segments mentioned above. For each segment write the equilibrium equations. Since we have three equilibrium equations per segment, we end up with a total of nine equations.
Now, count the number of unknowns. If there are exactly nine unknown forces, the structure is said to be statically determinate.
The unknowns are: Vertical and horizontal reactions at A, vertical reaction at B, C and D, shear and axial force at E, and shear and axial force at F. This gives a total of 9 unknown forces.
So, the beam is statically determinate.
thank you this vedio maker
sir,The purpose of providing internal hinge is :
1. to connect different members on different Materials.
2. to transfer the load from one part to another part.
3. to convert indeterminacy of structure into determinate structure.
right... But my question is how would it transfer the loads from one part to another part..
next question in your first example you would have just separated the given beam into two segments in which there were five unknowns in the first segments but we have only 3 equilibrium equations then how could you say it is statically determinate??
you have not explained about the force releases concept here??
what does it mean force releases??
The hinge transfers the load via a shear force, and possibly an axial force, if the member is subjected to such a force. Since in theory, the hinge allows rotation, there would be no resistance to moment at that point, meaning bending moment becomes zero at the hinge, so no transfer of moment takes place there.
With regard to determinacy, we say a structure is statically determinate if it can be analyzed using the static equilibrium conditions (equations). This is different than saying a structure is determinate if it only has three unknowns. As the example illustrates, sometimes more than three unknowns appear in the formulation, but as long as we can determine all the forces using the three equilibrium equations, the structure is considered determinate. If the entire structure was that of Segment AE in the example, and there was no other parts to the beam, then we would have had an indeterminate system.
+Dr. Structure thank u so much sir.
this is the best video... but i have small confusion
what is the degree of static indeterminancy if more than 2 members are connected in internal hinge.? plz i am looking forward for answer?
+Rajendra Kunwar Say, we have a simply supported beam, with degree of indeterminacy of zero. If we attach another beam segment that rests on a roller to the original beam using a hinge, the degree of indeterminacy remains zero. We can continue adding beam segments (each resting on a roller) using a hinge without increasing the degree of indeterminacy of the entire system. The beam remains statically determinate.
+Dr. Structure thank u
What if a column (in a frame) has an hinge right at top of its bottom pin support? Does it make the pin support to a roller support?
No, it does not change the nature of the pin support. A hinge located at a (short) distance from a pin support makes the segment between them act as a truss element. Meaning, it can carry an axial force only, no shear force or bending moment develops in the segment. Assuming that the resulting system is still statically determinate, it can be analyzed just like a beam with an internal hinge.
@@DrStructure thank you for the prompt response! You are the best...
@@DrStructure you mean the support reaction which is in the direction of shear at hinge will be zero by making the equilibrium at the segment. Right? And then the only force at the support will be the axial force balanced at support?
Correct!
Sir Please solve one problem in beam with internal hinge because I can able to find the support reaction but even I can calculate the shear force and bending moment values but having a confusions in calculating shear force. Because according to positive shear force sign convention the sfd should come at the support A is positive 2kN but in books it has shown in downward direction there I am getting confusions.
The first example, which involves an internal hinge, shows an upward reaction of 2 kN at A. This results in a positive shear in the left segment of the beam.
If you have a counter example, printed in a book, that is causing confusion, feel free to email an image of the solution to the address below for additional feedback.
Dr.Structure@EducativeTechnologies.net
you are the best
What if the point load(S.F.) is acting on the internal hinge itself ?
You just need to place the load either a tiny bit to the right of the hinge, or a bit to the left of the hinge (either would work, and give the same results) and proceed as usual.
at the internal hinge does it have two reactions or only one reaction??
An internal hinge can carry (transfer) a shear force and an axial force. If the beam is not subjected to any axial force, then no internal axial force develops in the member. In such a case, when drawing the free-boy diagram, knowing that the axial force at the hinge is zero, we often omit drawing it.
@@DrStructure at 02:40 considering the left segment, why is Cx Axial force not considered? Shouldn't this equation be
Ax+Cx=0? 🤔🤔🤔
Yes, the equation should have been written as Ax+Cx=0. Although the result still is: Ax = Cx = 0
well done
How one decides if the splice be a moment or shear?
That is a design consideration which can be decided on case by case bases. For example, if the splice is near the point of zero moment in the beam, we can design it as a shear splice. If however the splice is to be placed at a point with a high internal moment, a moment splice would be more fitting.
@@DrStructureThanks indeed for the reply. 👍
Let's us say for example, that in a continuous beam, one of the spans has zero moment at one point and a maximum at the other one. Is it permitted and practical to have at one point a shear connection and the other a moment? Is it really happening in the real-world?
Permitted? Yes!
Practical? That depends on the cost. If the cost of building such splice combinations for that beam is deemed reasonable, then yes, it is practical.
@@DrStructure 👍 Again, many thanks.
Why is that in summation of Fx is not Ax+Cx=0? (In the first example)
There is an annotation at 2:41 showing the correct equation: Ax + Cx = 0.
In 1st question When 4KN load is applied at hinge then how will we calculate support reaction...Please reply
The process would be the same. If the load is at the hinge, for analysis purposes, move it a tiny bit, say, 0.0001 mm either to the right or to the left of the hinge (so that you can separate the two segments at the hinge) then proceed as usual.
@@DrStructure Thanks!
In 3:23 , Why is there no M in the equation of 4 (5) - 10 By = 0?
That equation is written using the free-body diagram of the right segment of the beam (Segment BC). There is no M in in the free-body diagram of Segment BC. M is located at Point A and not a part of Segment BC.
@@DrStructure Internal hinge doesn't have M?
Correct. There is no bending moment at the internal hinge.
where did you get 2(8)(4) for the moment about e? im lost
That is the moment of the uniformly distributed load about E.
The load is rectangular, the area of the rectangle is (2)(8).
The rectangular load can be replaced by a concentrated load having a magnitude of (2)(8). That is the magnitude of the concentrated load that replaces the distributed load equals to the area of the rectangle.
But where on the beam would we place this concentrated load? At the center of the rectangle which is 4 m away from Point E.
So what is the moment of the distributed load about E?
It equals to the moment of the concentrated load about E.
What is the moment of the concentrated load about E?
It equals to the load magnitude (2)(8) times the moment arm (4).
So, we get (2)(8)(4).
Dr. Structure thank you very much I figured since it was distributed that it would be half, but wasn't 100% sure
You are welcome.
Where does "M" came from? tia
Hey! The subtitles covered the whole thing😩
Thanks a lot 😄
Thanx!
Thank you very much. I have a question: adding a hinge on the inflection point, will the hinge affect the bending moment diagram and elastic curve?
Yes, it does. Bending moment at the hinge becomes zero causing the slope of the elastic curve to become discontinuous.
@@DrStructure as bdm at inflection point was zero before adding the hinge, so it won't change the bending moment diagram, but only changes the elastic curve, right? it is like changing the boundary condition.
No quite. At the inflection point the of curvature of the elastic curve changes direction (i.e., going from concave up/down to concave down/up). That is not a point at which bending moment is zero. So, the bending moment diagram for such a beam is not going to show zero at the inflection point. But if we introduce an internal hinge at the point, we end up with a zero moment at that point.
A good example for this would be a two-span continuous beam where only one of the spans is subjected to loads. The loaded span deflects downward (having a concave up deflection) while a significant part of the other span deflects upward with a concave down deflection. The point of inflection is in the unloaded span close to the middle support where concave up turns into concave down. Now, if we place a hinge at that point, we turn the beam into a statically determine one in which the entire load is going to be carried by the loaded span (and the two supports that the span rests on) only. The other span carries no part of the load. This results in a moment diagram, and an elastic curve that are completely different than those for the (indeterminate) continuous beam with no internal hinge.
@@DrStructure thank you for your detailed reply, but I feel confused. maybe I didn't express it clearly. I mean adding a hinge on the inflection point of the elastic curve, not the curvature diagram. the inflection point of elastic curve is where the curvature changes sign. As curvature (not slope) is in direct proportion to bending moment, so it is also the point where bending moment equals zero.
I think it would be best if we put this discussion in the context of an actual problem. You can work out a problem as you have envisioned it, save it as a PDF (still images works too) and email it to Dr.Structure@EducativeTechnologies.net
We'll take a look at your solution and comment on it accordingly.
thanks alot
how to get Ay= 2.14 please help
The beam is divided into three segments (free-body diagrams).
Start from the right-most free-body diagram. The equilibrium equations for that free-body diagram is given @5:00. Solving those equations yields: Dy = 0, Vf = 0, Fx = 0.
Knowing Fx and Vf, now consider the middle free-body diagram. The only remaining unknowns on that free-body diagram are: Ex, Ve, and Cy. Solving the three equilibrium equations for the middle segments yields: Ex = 0, Cy = 21.33 kN, and Ve = -5.33 kN.
Knowing Ex and Ve, now consider the left-most free-body diagram where Ay resides. Solving the three equilibrium equations for the free-body diagram yields: Ax = 0, By = -7.46 and Ay = 2.14 kN.
sir, are you an Indian professor??
visit Lab101.space, then the LinkedIn page to learn more about me.
+Dr. Structure ... sir I have gone through your profile .. thank u sir.. This is my humble request if possible please do some videos on design of reinforced concrete and steel structures so that we could learn parallely about the design along with the analysis.
sir it is just a request pleas dont mind..
I always respect you sir.. in India we too have some excellent professors but they are very less in numbers. Every one can teach with the help of the board but only few professors like you touch the heart of the students you are one among.. in India number one institution is IIT even professor's who were working in this institution can't teach like you..
finally sorry for my English there might be some grammatical errors, since it is not our mother tongue.
Thank you for your feedback and kind words.
Yes, our plan is to cover the design topics as well. Time permitting, hopefully we will get to those sooner rather than later.
+Dr. Structure thank u sir..
@@venkatesh2285 You're welcome.
I am sorry but i can't find the value of Ve. May u help me plz ?
Please be more specific. What Ve? Where?
Ahh i'm sorry again. I was talking about the equation of the beam AE of last example, and I couldn't figure out the value of Ve.
Examine the free-body diagram and the equilibrium equations for the right segment.
The third equation gives us: Dy = 0.
Then, from the second equation we get: Vf = 0.
Now, examine the middle segment where Vf has already been determined to be zero. There are three equilibrium equations associated with this segment. From the third equation, since Vf = 0, we can determine Cy. It equals (2)(8)(4)/3. Or, Cy = 21.33 kN.
The second equation equation for this segment is: Ve + Cy - Vf - 2(8) = 0.
Since Vf = 0 and Cy = 21.33, then Ve must be equal to -5.33 kN.
Oh, thank a lot. Such an informative and helpful tutorial. I like the way you quickly responded.
You are welcome.
talk a little louder it's hard to hear you