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.
I’m not sure what the textbook method was for the load applied at D. You somehow recognized that member ACF was a 2-force member. Otherwise the x and y components of the pin reactions at C and F would not be trigonometrically related . You should have used tangents instead of sin and cos, but...
what sort of full time engineering job do you have? You might be better off staying in engineering. Please explain why you want to leave engineering besides fulfilling your dream . Your dream might become true, or perhaps never become true.
It is best to forget about sign conventions for clockwise and counterclockwise. Rather, show the couple in its vector form using the right hand rule , and the moment vector will point in the direction of your thumb. Then you should be able to more easily find the projection and signage of the...
That’s for sure. Your answer for torsional shear stress is off by a factor of one million. Don’t feel bad, I’m from the US and I seldom use metric , so it confuses the living daylights out of me. Once I ordered 1/4 inch thick plate from China and they offered me 6mm instead. It took me awhile...
It is not shown in the diagram, but the 225 N is a force, not a torque, and it acts tangent to the circumference of the shaft at its free end. What is the Torque caused by it? Then watch your conversion factors, there's a lot of zeros when using SI units.
I am not sure I understand your concern. The shear and bending moment diagrams for the UDL case look good. For a point load at the center of the beam, 1/2 of the point load goes to each reaction . For the bending moments at the ends and center, you look them up in a table like you did for the...
Your static equilibrium equation for summed forces in the y direction is wrong. You are leaving out some forces. Watch your subscripts also for mb vs. md. And check your algebra.
I always felt that a one hour STEM lecture was too short, and a 2 hour lecture was too long. 13 hours? Students would be tuned out after hour 2, and even if not, absolutely no learning would take place. Nothing at all.
You did not solve the problem correctly. You noted that the components of P1 and P3 cancel out when looking at the forces on the member along its length axis. correct. But what about the other components?
In the first, F and d are the magnitudes of the vectors F and d, respectively, and work is a scalar.in the second, it is W = F.s, the dot product of 2 vectors, which is fscostheta, the same result.
scalars do not have direction. They can be plus or minus, but the plus or minus signs do not...
Yes kinetic energy is always positive and it is a scalar quantity. Work is also a scalar, and can be negative or positive. You can’t differentiate at a discontinuity in the curve, but in actuality, it is not really a discontinuity because you can’t have an instantaneous change in acceleration...