Problem statements
Solution video - H26.A
Solution video - H26.B
DISCUSSION THREAD
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HOMEWORK 26 - Fall 24
65 thoughts on “HOMEWORK 26 - Fall 24”
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How can angle R and 20 degrees help me find the height of the force of 2F?
you can make a triangle at points O, C, and D. The hypotenuse is R (3ft) in length and you are given the x value. Use trig to find the y value of the triangle. You don't need to 20 degree angle to find it.
I was wrong. The triangle is at O, D and just below C so it is a right triangle. The hypotenuse is from O to D and is 3ft long and you have the x value of the triangle.
But if the triangle is right below C is that still okay to be used to approximate the height of the force
To find the height at point D, you don't need to know the height of C. If you draw the triangle, you have the hypotenuse and the x distance between OC and point D
I used the Pythagorean theorem to find the side lengths of the triangle from O to D being the hypotenuse of 3 and the bottom side length being r/2 (from O to halfway to B). This forms a right triangle. The vertical distance to 2F is the third side.
Since the bar is a semicircle, you know that any point on the bar is at a distance R from point O. So point D and O are two points of a triangle and the third point, what I'm going to call X, is located directly above point O and in a location such at that the line between X and D is parallel to the line AB. This will give you a right triangle with a hypotenuse of R and one leg of R/2. The third leg of that triangle is the height of point D. You can use trig functions to find this value.
you need R to find the moments around C for the forces acting on C from E
To find the height of 2F, I looked at the semicircle in relation to the unit circle. Since you consider R = 1 with the unit circle, R/2(along the x-axis) is equal to a 60 degree, or pi/3, angle. If you recall the unit circle, the y of 60 degrees is equal to sqrt(3)/2. Relate this back to R and you should have the correct height of y.
The 20 degrees isn't really needed for this portion I took the square root of the rum of a variable squared and R/2 squared and set that equal to R and solved for that variable. I then used this found length to find ay in the moment equations in my first step of completing the problem. I hope this helps.
We can use trigonometry to find the height (y-coordinate) of the triangle. We can calculate the height by knowing the hypotenuse (R) and the base (x-coordinate).
Even if the member is a curved surface (as shown above), can we approximate it as a straight member for calculations, i.e. does it being curved change our calculations?
I don't think the member being curved changes the way you calculate the forces but it will affect distances when solving the moment equations.
The curve won't affect the calculations but you will have to use the triangle with twenty degrees and another triangle for the force of member AC to calculate the necessary distances for your force and moment equations.
I am stuck on how to find the distance D (specifically 2F) away from A when taking the moment at A. Can someone help me?
I'm assuming you have to find the angles on the other side, then take the circumference of the whole semicircle bar and find the arc length. This process is assuming that point D and E are symmetrical. I hope this helps let me know if you find any other info!
I don't think that point D and E are symmetrical about the y-axis. Using trig, you'll find that the angle of imaginary line OD with respect to the y-axis is 30 degrees, whereas the angle of OE is 20 degrees with respect to the y-axis. Hope this helps.
For taking the moment, are there multiple ways to find the vertical length? What is the easiest way?
Because it is a circle with a center at O, the heights would all be the radius, or r.
For the moment of the forces F and 2F at point D, you can use the Pythagorean Theorem knowing that the distance from the origin to the point D is R, and making a right triangle. For taking the moment of the forces at point E, you can use the same method, but now you know the angle so you can find the components knowing that the length from O to E is R.
Using the provided angle you can use trig and the Pythagorean theorem to get the needed distances, and the radius, which greatly helped in solving for some of the needed reactions.
When drawing the FBD, is it okay to consider the curved members as straight members?
they would have to be curved members as you have to use the curved distance for moment calculations
for this homework, would I split the semicircular bar into 2 curved members for the FBD breakdown? or would it just be one whole semicircular bar?
I broke the bar into 2 curved members for my FBDs.
Do we need to use the arc length equation when we're summing the forces? Since it is a semicircle and not a triangle
You shouldn't need to use arc length in this problem. When calculating the moment, you should use the perpendicular distance from the point to the force. Because of this it doesn't matter that the members are curved.
You shouldn't need to, since for the moment you just use the perpendicular length, which isn't curved.
Someone told me that we were solving for four unknowns. Is this true? If so, it would be the moment, shear, the force related to the cut, and what else?
To my knowledge this is correct. I believe because we are only dealing with two dimensions and the questions is asking for the internal resultants this means we need to solve for all four. I think that the four we need to solve for are axial force, shear force, the torsional moment, and the bending moment.
Actually I think I am wrong about what I said before. You might only need to solve for three things, Ve, Ne, and Me. Again I could be wrong.
For finding the sum of forces, would we use the entire body or just the cut body from A to E?
Both, you need to use the entire body to solve for a couple unknowns, than the cut body from A to E to solve for a few more unknowns using your sum of forces. Than you can use the unknowns you solved for to find the 3 unknowns asked for in the problem. Hope this Helps!
When dividing the frame into sections, would we have a section for A to E, A to C, and C to B, assuming that CB is part of the whole frame?
Exactly! I Used those 3 different FBDs to calculate for the unknowns within each of those different FBDs to calculate the unknowns forces that are on point E, using the summation of forces. Hope this Helps!
Can Member AC be considered a 2 force member or not since there is the force F at C?
Member AC is not a two-force member because it has an additional external force, F, acting at point C that is not aligned along the line that connects A and C. For a member to qualify as a two-force member, only two forces can act on it, and they must be equal, opposite, and collinear. Hope this helps!
Member AC is not a two-force member because it has an additional external force, F, acting at point C that is not aligned along the line that connects A and C. For a member to qualify as a two-force member, only two forces can act on it, and they must be equal, opposite, and collinear. Hope this helps!
What direction would the final components of the shear vector be? I got that the magnitude of my vector is negative, but I'm not sure whether the x-component or the y-component would be the one that is negative in the full vector.
Depending on the direction of the arrow you drew, it will depend. For instance, if you assumed the arrow to be pointing right along the surface of the cut, and you got a negative value, you would flip the direction when sorting out your signs. I use trig to determine exact components, and then look at my FBD to determine sign conventions. Happy Halloween!
To solve for the final vector, are you going to have to solve for the force vectors of both points a and b to solve for e or is there a way to do it while just directly solving for e? The only way I can think of getting the e vector involves solving for a and b as well.
Its just Pythagorean theorem for positions x^2 +y^2 = r^ 2, I don't think You should need Trig for it
Ignore this I was being a silly goose
In full honestly, I don't know where to start with this problem. Any recommendations?
What should I find the moment around to start after dividing up the different members?
I found the moment around A so that you don't have to worry about the reaction forces at A and instead can look at the forces at C and D.
Would the final answer be the vector of the reaction forces at E?
I started by finding the reaction forces on points A and B. You could view the system as a whole to find one component and then you will need to use method of sections to find the other component of the forces on those two points. Also use method of sections and the known values to find the resultants on section AE of AC. Hope this helps.
Yeah I think so. You would have the V, N, and moment about point E. The way I solved for this was finding the Fac on point A (you can assume it's a two force bar/system) and then you can take the moment about point B and sum forces to find the final answers. I believe you can also find reaction forces at point A or B to solve for the answers, but that may be more work.
You do not need the 20 degree angle since you are already given the 3ft hypotenuse of OCD.
When solving these types of problems, what is the difference between the axial force and the internal shear? I am confused on this as they both are forces, but aren't the same.
The axial force is a force that acts in the same direction as your FBD and the internal shear is a force that acts tangent to your FBD. When doing an internal analysis you have three parts at the portion you split at; the axial force, the shear force and a moment about the point you split at. If you are still confused about this your professor probably provided a diagram on what this looks like.
Do we need to find the shear force for this problem?
Yes, you need to find the shear force.
Yes the shear force should be one of your results and is the y components when writing in vector form for your final answer.
How would you solve for the height between D and A for taking the moment around A?
We know the horizontal distance between O and D, and the length of the segment OD. If you consider the angle between OD and OB, you can use trigonometry to solve for the length of the perpendicular dropped from D on OB, and this length will be the distance between A and D.
Do we exclude Bx from the moment around A to solve for By?
By taking the moment about A, it should automatically get rid of the Bx component, because it is in line with the point A where you are taking the moment around. However, I found it easier to take the moment about B first, then to use solving for the shear and normal internal forces using some trig knowledge and sum of forces. Then finally, for the moment, it is most helpful to do it around E, because it gets rid of some forces to were we don't have to do so much work.
When you take the moment about A, bx is excluded since it goes through point A
Yes. Since Bx acts directly through Point A, there is no perpendicular "lever arm" that could be used to solve for the moment Bx causes around Point A, so therefore it is excluded.
Is the only way to solve this problem involve splitting the half circle into three different FBD's?
There's also another way by cutting into two different FBD's, which is section AC's FBD to find the reaction at point A in x direction (Ax) and section AE's FBD to find the reaction at point E (Ex and Ey) along with its moment (ME). From the main FBD, the half circle diagram, we can find the reaction at B (Bx and By) and reaction at A in y direction (Ay).
Why is there moment on this given problem?
On problem B, would the force at point W have any actual impact on the problem? I'm not sure since to my understanding the resultant shouldn't change due to the position of the force alone.