During mounting of the coupling, the parts will have to deform. This is not easy in your design. If you make slots in the female part, this will be easier.
See as an example this feston plug that I designed.
This is a statically determinate structure (four reactions minus one release equals three static equilibrium equations).
So we should be able to do this using simple statics:
\sum F_x &= A_x + C_x = 0 \\
\therefore A_x &= -C_x \\
\sum F_y &= A_y + C_y - 300\cdot6 = 0 \\
\therefore C_y &= 300\cdot6 - A_y \\
\sum M_A &= -300\...
I am 68 kilograms and you could never convince me to sit on this!
let's say the table is 90cm high and the legs can extend out, by just eyeballing and correct me please, 50cm. and your lege ar 2cm pipe and the screws are on a 3cm diameter circle.
One can not carefully place the load at the center of the table every time, also because of the fact the legs are ...
If you are a beginner and you already use solidworks I would suggest sticking with it. It has a decent FEA/CFD addon that you can use in a single piece of software. That alone will save you loads of time. The main drawback is that you might not have the material models and customizability of other pure bred FEA Tools.
Ansys, (and a lot others) is/are more ...
I am not sure which end you are writing the equation from (support at the left end or the right/cut)
However I think:
I think you are forgetting the 120 kNm
Also the uniform load creates a counterclockwise rotation about the cut/section, so it has the same sign with 120 kNm, which is the opposite of the 21 and the 18*(4+x)
also note that x is defined from ...
I haven't had to do this for years so please confirm my analysis and calculations.
Take the left wheel as a pivot point.
The left deck has a mass 67 kg so will have a force of 67 × g = 670 N approximately through its centre, 0.6 m from the wheel. The resultant turning moment will be M = F × d = 670 × 0.6 = 402 Nm clockwise.
If the ram is below the top of ...
the way your figure shows the two leafs are locked in place by not allowing the pump to change its angle.
If you have your two leaves hinged on the bottom and the hydraulic piston on top with a space or bracket giving it freedom to push the lieves edges out to rotate, so they could swing each 90 degrees to come in full contact.
then the maximum for required ...