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Solution video - H27.A
Solution video - H27.B
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HOMEWORK 27 - Fall 24
70 thoughts on “HOMEWORK 27 - Fall 24”
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Should we cut the sections of the beam right at C and D? Or do we cut at some point before them?
I would recommend watching the video solution for 9B2 he explains of he cuts the segment. I approached this problem similarly and cut the segments between AC to solve for M and v over the first segement.
The segments should be cut between each point, however your answer (for M(x)) should be a function that also allows you to see the moment at C and D.
I believe you technically need to cut right before B and C in order to not include the reactions to the right which would cause more unknowns than equations which is unsolvable.
You cut right before the point in order to not include the reactions at that point in your equations.
Similar to lecture this week. It is less so about "cutting" at a point and more so about setting up equations that represent the summation of forces between the applied forces/moments such that every point in between at a distance x is in equilibrium.
You should cut before them; its less about where you cut and more about making a general equation for the sections that are outlined.
You will want to cut the beam at some point right before C and D, this allows you to see the moment at these points and guarantee getting a value unaffected by the reaction forces occuring at these points.
I cut at a point between A and C for my first cut, and then cut between C and B, and then between B and D while still keeping my starting point at A in each of these Free body diagrams. Doing this should give you three separate equations that you'll be able to graph in your shear force and bending moment diagrams.
It would be best if you cut before them like some other people are saying. I also found the video solution for 9B2 to be beneficial as someone else mentioned.
You can cut it anywhere technically as long as you make sure to account for all forces/moments acting on the system being investigated. However, I would suggest cutting it at B because it is being supported at this segment so only this cut will be essential to solve everything else. Make sure to account for the left side isolated (AB) and then the right side as a whole system without the force P acting on beam (AC). This way you will be able to graph the Moment in respect to distance as well as the sheer in respect to distance.
Prof. Jones posted a video in week 11 that explains this very well!
I would cut at B and do two sections
It is best to cut the segments right before and after B, this way you avoid the forces acting at point B. With the equilibrium equations, there are also not as many unknowns.
You can choose to cut it at points C and B, resulting in your beam being segmented into 3 sections. However, you will get the same result if you only make a cut at B. This is because there are no reaction forces occurring at C to make a difference in the shear or moment equations. It is beneficial to identity these patterns prior to solving the problem to simplify the calculations. If you segment the beam into 3, the first two segments from the left will have the same moment and shear force equations.
When splitting the segments up we must cut right before the joint. The best way I would describe this is to imagine the length as a number and we will represent the new length of the cut as x. We want the length of this new found variable x to be less than but not equal to our real length. This eliminates any forces, or moments acting on the segment at the end of the cut.
If you mean where you should consider the moment at, you really need to only do it where a new force is present, otherwise you can continue the trend otherwise present. If you're asking if you need to consider the forces at some given point or not, you definitely consider them at that point as if they happen directly there.
I cut off the sections at C and D.
Prof. Jones has an example very similar to this under week 11 on bright space
I also found Prof Jones example to be helpful in working this problem.
I made two cuts and analyzed three sections of the beam. I also used the entire beam to solve for some reaction forces.
Does the presence of a moment force affect how we would approach making the graph?
The moment forces play a part in your moment equilibrium equations, which you graph on the M(x) graph, so they impact the graph we make.
Yes they could eliminate some forces which impact the graph that you create
I approach the graph around point B as a calculus discontinuity problem. So, to find the two values at B, you break the question down into looking at how the graph approaches point B from the left and right. For how the graph approaches from the left you use the moment equation you found when you split the FBD at B, when approaching from the right, you can use the equation from when you split the FBD before D. This gives you the two values to graph at B on the M(x) graph.
Yes, the presence of the couples does affect making the graph. The moment forces are added to the M(X) summation when doing the analysis for the corresponding segments of the beam.
This affects the final value of the M that we are solving for. If we didn't add these couples it would result in a much different answer for M(X).
I think to like actually do this right, you need to find the actual piecewise functions related to each section. To solve for shear it isn't too bad (sum of forces in y), and then to solve for bending moment, just don't include any extra moment past the cut you make.
Because the moments (M and 2M) are pointing clockwise in the diagram, should we treat them as negative in our moment equations even though it lists the value of M as positive?
I treated them as negative since in the image they are going clockwise.
Yes, they should be treated as negative in the moment equations, as the given value is the magnitude of M and the arrows on the diagram determine whether it will be positive or negative in the equations.
When graphing this Piecewise function, would the line connecting them need to be dotted. Or is a solid line acceptable?
My lines are solid, because in the examples the lines are solid.
Are the graphs always directly correlated such that the zero point of the shear force is at the same distance as the peak of the bending moment?
Yes, this is because the integral of the shear force gives you the bending moment, therefore the bending moment will be at its maximum when the shear force is 0.
I have noticed that the graph for the shear force looks like the derivative of the graph for the moment. Is this always true and if it is, why is that?
Yes it is. The shear force is actually the derivative of the moment. It is a useful way to check your answers by checking if the your sheer force is the derivative of the moment.
This is correct, the integral of the shear force (plus an initial condition) will give you the bending moment. This is because when you take the sum of moments around the middle of the general case (with M(x) clockwise, M(x+dx) counterclockwise, V(x) in +y, and V(x+dx) in -y), you can solve for M(x) as the fundamental theorem of calculus with the shear force as the function of x.
Is the best way to do this problem to split the beam up in two different locations, one at B and one at around E?
It will not be split exactly at B and E. It will be slightly before it. That is, you should split one section just before B (will include reactions at A, the applied moment at A, and the internal shear, normal, and moment) and the other section just before E (does not include P or internal reactions that you found from the first section). This will allow you to make separate equilibrium equations that solve for M and V in section 1 and 2.
In order to solve for the shear and moment forces, I split once before C, once before B, and once before B. You do not need to make a split before C because there are no additional forces acting between C and B.
*D
I would split it into two locations. That will be split similarly to B and E.
Should the shear forces from AB and CD counter each other?
Depending on where the segments are cut, the shear forces should be in opposite directions of each other. The shear force should always cause clockwise rotation.
For this question do the moments from the previous sections impact the moment of the following sections like section BC?
The applied moments are a part of the overall equilibrium equation, so you have to include them at every individual section's moment equation to solve for the unknown moment.
Are you supposed to include the moments at points A and B in the FBDs of the sections?
Yes, because they are pure moments they will still have an effect on that section, so include them in the FBD.
Of course! Similar to most problems, include Ax and Ay in A since it is fixed and then include By for B since it is roller. With that included you can include them in the equilibrium equations for moment and that should be included since you have to consider them in the problem. Hope that helps!
Yes, the moments at A and B exist at the sections created to solve the problem, so they would be included in the Free Body Diagram
I solved for the moments at the points in-between A and B, and A and D like we were taught in class, but when I go to graph my Moment graph, there's a jump in moment around point B, I was wondering if that normal, or if I did something wrong.
Yes, there is a jump as you suddenly have entirely new forces contributing to the moment when you get to point B. It is not continuous B doesn’t contribute right up until you hit it in your cut
This is normal because the forces are not continuous. The forces solved separately in the AB and BD segments will not be the same at point B.
This is normal because there is an applied moment at point B. In that case, this problem has a negative applied moment at point B which will result in a positive jump when you graph the moment plot. I believe you are on the right track.
I also found there to be a jump in the moment at point B. The equation for segment AB considers the moment right up until point B. There is a moment applied at point B so it will be seen in the equation for segment BD.
For a 9-foot simply supported beam divided into three equal 3-foot segments at points labeled A, C, B, and D, I need to construct the shear force and bending moment diagrams. The beam has a 50 kip downward point load applied at D (the far right end), a 50 kip-ft counterclockwise moment applied at A (the far left end), and a 100 kip-ft clockwise moment applied at B, which is located 3 feet from the right end (one-third of the beam length). Given these loads, I need to understand the correct method to set up the shear and moment equations for each segment, specifically how to incorporate the effect of the counterclockwise moment at A and the clockwise moment at B in these equations. Additionally, I want to confirm at which points along the beam—such as immediately to the left or right of points A, B, and D—I should expect to see discontinuities or sudden changes in the values of the shear force and bending moment in the diagrams.
I cut the beam at AB, did anyone else take this approach, or should I reconsider?
I cut the beam directly before and after B, meaning you would have to solve for the unknown values of two different segments. One segment being A to before B, and A to sometime before the force on the far right side. Hope this helps!
Is moment at B needed at all in the calculation for the region on the right?
Yes, the moment about B will be used when you find the moment and shear for the section from B to D. When finding the moment and shear force, you will make a cut just to the left of where a force is applied. To find how the shear and moment varies from B to D you will cut between B and D and include points A to B, meaning that you will include the moment at B in your calculation for the moment between B and D.
Why do we not include the forces at the ends of the cut sections in the local FBDs, I saw some examples where they weren't included and was curious why not?
If the reaction force at Ay is negative, do we show V acting upwards or still show V acting downwards?
As far as I can tell, if Ay is negative, V is also negative at that point.
I should rephrase, I meant negative slope no negative, so downward slope.
There are three dotted lines on the graphs, however can we choose to have two lines for the forces?
Yes, you can because there's no external reaction working on point C, the first dotted line, so there won't be anything that can change the graphs between section AB. You only need to pay attention on point A, B, and D because there's external force working on it and there's force from pin joint and roller which result reaction at point A and B.
Are the graphs always directly correlated so that the zero point of the shear force coincides with the peak of the bending moment?
since the moment is the integral of the shear force, a zero point for the shear force will be either a peak or a trough of the bending moment (assuming the function is continuous around that point)
When cutting the beam to analyze segments, should we consider making cuts right before points B and D to avoid including the reactions at those points, or is it better to cut at an arbitrary point within each segment?
For the one not assigned for homework, the diagram does not say there are moments at the joints. Does that just mean it is normal and the moment is equal to zero.