It has to be shear stress doesn't it? The gage pressure doesn't cause any shear stress, right?
Does it cause another kind of stress too? Or am I just wrong? Can I get a multiple choice? :smile:
I think I may have figured it out. I just used Mc/I where c is the distance from the neutral axis.
Beam is 18.06 in deep, minus 0.625in per each flange, is 16.85, divided by two is 8.425. I used that as C, and 712 for I. Its about 22 ksi. seems reasonable to me.
Homework Statement
Homework Equations
σ=pr/t (hoop)
σ=pr/2t (longitudinal)
τ=VQ/It
The Attempt at a Solution
I have longitudinal stress and hoop stress. I'm trying to find shear stress, which should come only as a result of the 40kN from the collar correct? I'm confused...
Homework Statement
Homework Equations
σ=P/A
σ=My/I
τ=VQ/ItThe Attempt at a Solution
I'm on part C, pretty sure I have A and B, so I'm using a W18x46 Beam. I have drawn shear and moment diagrams. I just want to be clear on what part C is asking...
I'm assuming this point is at the end of...
i think that's right, because coulombs is the current coming in over an amount of time. (amps times however many seconds) so derivative would give instantaneous current at a given time.
q(t) = 7.5C · (1 − e(+t/τ)).
In this equation, the final output is in C, is it not? then would the units of tau be seconds also, so then to have this find current, can I multiply the whole thing by 1/t leaving me with amps as the resultant unit?
Homework Statement
Over time (0s ≤ t < ∞), charge enters an element according to q(t) = 7.5C · (1 − e(+t/τ)).
1. What is the current into the element as a function of time? (Find a symbolic answer!)
2. What is the unit of τ?
3. Prove that the unit of the result indeed is A.
The...
ok that's what i thought. so since the larger mass is larger and moving faster, how does it stopping make sense? obviously the math shows it but it just seems like it would have some motion left. I guess that it transfers all it's energy to the smaller box and the rest (the extra 320) is...
Homework Statement
A 20kg cart traveling to the right at 8 m/s collides head-on with a 10 kg car traveling 6 m/s to the left. After the collision, the 10 kg cart is traveling 10 m/s to the right
a) What is the velocity of the 20 kg cart after the collision?
b) determine whether or not the...
so according to that...
1/2*49pi*7^2 [I_outer] - (1/2*16pi*4^2+16*1)[I_inner]=3319
3319/(33pi)=32
sqrt(32)=5.66
And that was right...
thanks guys. I understand it a little better now..