Recent content by shawn87411

I am curious if anyone here is familiar with ABAQUS and using it for crack propagation. I am doing a project on fatigue life and need to use a computer program (ABAQUS is our FEA) to predict fatigue life under specific loading conditions. I am planning to use crack propagation to do this and...

Ahhh I see so the screws only have the force up but because the distance the torque is equated for. Now is it correct that the screws A and C won't compensate for the 30 N-m torque and the screws B and D don't compensate for the 20 N-m torque because of their locations?

Sorry I attached a more specific picture. Distance between D and B is 300mm and the distance between A and C is 450mm. If you draw something in just the XY plane you have something like the second picture attached, correct (the 20 is into the page)? I don't see how you can find the force on...

Well the forces aren't known, this is a general design problem. All that is given is the torques on the shafts and I need to find the torques on the screw. Also the screws are on the edge of the box for each one with the length of the box being 300mm and the width 450mm so those are considered...

This isn't a specific problem, I just can't remember how to solve problems like these. I attached a brief picture. The dots on the top are screws that hold the box in place. There's torques on both the shafts in the clockwise direction (shaft coming out towards you is 30 and the one going...

Alright, I'm really confused. You said I know the values for E(4), E(X), Var(X) and E(X^2) and those need to be plugged in. You also said E(4+4x+X^2) = E(4)+4E(X)+E(X^2) = E(4)+4E(X)+Var(X)+E(X)^2 Its given that E(X)=2, Var(X)=3. This gives E(X)^2=4 and I assumed E(4)=4? Plugging those...

E(4)=4 E(X)=2 Var(X)=3 E(X^2)=4 E(4)+4E(X)+4E(X^2)=13?

Your just trying to find the expected value of 4+4x+x^2, so all I have to say is E(4+4x+x6x^2)=E(4)+4E(x)+Var(X)+E(X^2)? Thats it? You don't say Var(X)=3 and E(X^2)=4 and plug that in?

E(X^2) = Var(X) + E(X)^2, with Var(X)=3 and E(X)^2=4 so 4+3 gives the 7. So do you not solve anything and it stays at 7=E(4)+4E(X)+E(X^2)?

So I have 7=E(4)+E(4x)+E(x^2) but I'm confused what i do now. Do I plug in E(x)=2 so E(4x)=8 and E(x^2)=E(4)? if so where do you go from there, I'm just confused what I'm solving for

Var(X)+E(X)^2=E(X^2) 3+4=7=E(x^2) Then use the fact that E(4+4X+X^2)=E(4)+E(4X)+E(X^2).

Right but what's the difference between E(X^2) and E(X)^2? Is E(X^2)=E(4^2)+E(4x^2)+E((x^2)^2) with x=2?

I think I'm mixing up the terms E(X^2) and e(X)^2. Which one is E[X]=2? So E(X^2)=E(4)^2+E(4x)^2+(x^2)^2?

So Var(x) = 3 = E(2^2-(E(4)+E(4x)+E(x^2))? Sorry I had a bad flu the past week so I missed my monday and wednesday lecture and our book is pretty lackluster

Is the E(X^2-E(X)^2) = E(2^2-(4+4x+X^2)^2)?