Looks ok.
Insofar as 2 of the cables are in the horizontal plane, then the 500 pound vertical load must all be carried by the vertical comp of cable CD. And with that 3-4-5 triangle, 3 fourths of it or 375 pounds , must go horizontally to the other 2 cables, and their being symmetric, that’s...
Typical strain values for steel at yield might be in the order of
0.002 rather than .025. Something funny with the decimal point here. Factor of 10. Something is amiss.
It is best to find the resultant moment and use the radius of the circle as the c value. The angle of the resultant moment can also be calculated using vector analysis.
For finding end reactions, your treatment of the force at E in determining its moment about B is good. But you have signage errors in determining the sum of moments and you missed some loads and I don't know what is the 90.
I'm not into gears and such, but your load diagram shows the concentrated load of 1779 N practically right on top of the needle support B, implying B will take most all of the up reaction load, with small down reaction at A, and small up reaction at C, and there will be very little bending...
From symmetry, under vertical loads only, the horizontal reactions at the outside column pinned supports must be equal and opposite. That leaves no shear in the middle column, and thus no moment.
Yes this method is often used in the analysis of indeterminate beams and trusses. Often called the ‘unit load’ method, and considering a beam say supported on multiple pinned supports, the extra supports are removed and the beam analyzed as a determinate system with the equilibrium equations...
Newton’s 2nd law applies to acceleration of the center of mass of the system. Even though there is no rotational acceleration, you must sum torques about the COM.
There is a way around this by using the concept of inertial forces, called pseudo forces. Since F=ma, rewrite it as F - ma = 0...
What’s this? You have your moments about a point in (force x distance squared) units , which makes no sense. The moment of a couple about any point is the couple itself.
A couple is 2 equal and opposite forces F separated by a distance d. The magnitude of the couple is F(d). A free vector. Not the same as the moment of a force, as explained above.
A sea breeze sets up during the day when a large body of water is cooler than the land temperature and when the prevailing winds over land are light. A very localized area of low pressure sets up and the wind blows from the water during the day towards land, generally perpendicular to the...
M is the bending moment and at its maximum is equal to PL/4 from basic statics of a simple beam, along the short centerline of the plate for the simplified loading case I noted, where P is 200 pounds, L is 96 inches, and thus M-max is PL/4 or about 5000 in-lb. Then c is just half the board's...
If the foam board is simply supported at its short ends, and if your load is applied uniformly along the full width of the board but over a small length , then the max stress will occur when this load is applied at the center and the max stress at the center can be approximated using simple...
I'm not sure what energy method you are using, but since the problem is statically determinate, you don't need to know any rotation angles. B is a hinged joint, not a support, so there is no moment in the beam at B. The solution appears correct. Keep it simple.