A series of lectures and examples for mechanical engineering topics from statics, dynamics to finite element analysis. It is most suitable for mechanical engineering students
In the previous video, you used a fixed beam with UDL to calculate the equivalent load for an evenly distributed load. My question is, how does this work for determining the equivalent load for the second element of this problem?
You put it in the matrix form with all elements, all you need to know is what reactions you have at each support, if it is fixed you will have a moment and for pinned supports, you set zero for the moment value
That is the harder case as there are moments at the two end involved make the system statically undermined, the simply supported case is an easier system where reaction forces can be determined using equilibrium equations
Great questions! Not normally, that is part of the job! Unless you have an accepted grant, then you can pay yourself summer salary or supplement your academic year salary up to 20%.
I have a question about compensating the toe region. A continuation of the linear part should be generated in order to find the corrected zero strain point. Am I right? after that, should stress start from zero too? after this procedure, at the beginning, before the linear part, still there will be an inclined part. Is it right?
At the zero strain, the force is not necessarily zero due to tightening the grip! Stress-strain curve is not a good indication of initial conditions and it is hard to find the force at zero strain unless you read the machine recording at the time of experimentation! Extrapolation of stress strain curve will give you inaccurate beginning point. You can shift the stress-strain curve to give you more reasonable stress and strain values but do not extend it beyond its limit of recording
That’s our agreed convention to call a clockwise moment negative and counter clockwise moment positive, but in general if you use the right hand rule, a clock wise motion is towards negative Z which is the axes perpendicular to the page and counter clock wise would be towards positive z
In practice, there are forces, but in theory if only external vertical forces are applied and in absence of external horizontal forces, there is no reaction forces to oppose any forces
if you analyse the joint A, you can notice an Y force (as well as on joint B). then, you do: Mo= 6 times YB + 900 times 4 + 600 times 8. YB= -1400 N ΣFY= YA+YB= 0 so YA= 1400N afterwards, we can find the angle of the ABE triangle: 4/3 (half of the base)= 1.333333 1.333= tg 53.13 if Fea= Ya times sin 53.13 then Fea equals 1750N you can find Feb easily after that.
Using similarities of triangles, before it was cut the bigger triangle had 200N/m, after it was cut the smaller triangle was exactly half of the bigger one, finding the ratio(i.e 200:x=6:3) , x= 100. Note that 6 and 3 are just the base distances of a bigger and smaller triangles respectively.
Good question! For a rectangle, it is in the middle and for a triangular load, the resultant force is acting 1/3 of the base from the right angle of the triangle.
