Recent content by cd19

1. A decision has been made to make a plate out of an advanced engineering material ceramic that has better high temperature and wear properties than alloys. The plate must withstand a tensile load of 177.92kN. The part will be non-destructively tested to ensure that no flaws are present that...

Can anyone help me out this is the answer for the closed loop function

1. Homework Statement 2. Homework Equations I need to reduce the above block diagram to find the closed loop transfer function This is my solution to the reduction part of the problem: i) Add the feed forward paths; (10 + 10) ii) cascade G(s)2 x G(s)3 , {G(s)2 = 1/1+12, G(s)3= 20/s}...

max shear 7 kn, 6.45mpa... thanks

ah yeah silly mistake, so its y=25 giving me an answer of 9.19 Mpa. I've went over my calculations and they all look correct but not getting an exact value. Thanks for the help btw. The answers; (A - 12.86), (B - 6.43), (C - 4.34), (D - 3.76), (E - 9.86) so its E, I'm preparing for a...

Thanks for clarifying, definitely over complicated it! What I got using the bottom area; A\Upsilon= 20*50*20 with y being 20\tau= \frac{10000*(20*50*20)}{1.36*10^6 *(20)}= 6.87 N/mm^2 or 6.87Mpa The question is multiple choice and one of the answers is 6.43 Mpa so...

Yeah true, so if I use the bottom area the first moment of area = a*y = 60*20 + 20*[] in the brackets I should have 'y' the centroid co-ordinates I do not know how to calculate this... Then all I need to do is sub into equation above, How do I find 'y' **looking at something similar...

1. \tau = (V*Q)/(I*t) Q=first moment of area, t=width, I= Ixx and V=shear force 3. First I found the greatest shear for to be 10Kn at point b Then I found second moment of inertia = 1360000 Now I need to calculate for Q (first moment of area) = A*y...

I've just noticed something, My Re equation is wrong it should be (u) Mean velocity instead of \mu Viscosity, And to find u = Q/A , Now I'm completely confused.

Ah ok, If its a laminar flow no need to account for friction. Just seems a bit too easy, was one of the questions with the most marks

Its used to estimate the coefficient of friciton, In the problem it states ignore all loses other than friciton, this is where I'm having difficulty, I don't know how to incorporate the ficiton loss in the final equation http://www.engineeringtoolbox.com/moody-diagram-d_618.html

Anyone have any help?

1. The pressure at the start of a 180km stainless steel natural gas pipeline is 4Mpa and at the end is 2Mpa. If the diameter of the pipe is 0.4m, what will be the gas flow rate through the system? (ignore losses other than friciton) (for natural gas, \rho= 40kg/m3, \mu= 11x10-6 Pas2...

Homework Statement Air in an automobile tire is initially at -10oC and 190kPa. After the automobile is driven while the temperature rises to +10oC. Estimate the new pressure in the tire. State all assumptions that you make in providing this estimate. Homework Equations Pv=nRt...

Right so someone tell me if I'm correct here: p=f/a therefore f = pa and the mass will be equal to pa/g saturated water pressure at 120C is .2 Mpa atmospheric pressure is 101.3 Kpa m = (200000-101300)(.005)/9.8 = 50.36g Am I correct?