Answer:
A. True
Explanation:
MATLAB may be defined as a programming platform that is designed specifically for the engineers as well as the scientists to carry out different analysis and researches.
MATLAB makes use of a desktop environment which is tuned for certain iterative analysis and the design processes with a programming language which expresses matrix as well as array mathematics directly.
Thus the answer is true.
Answer:
with a square cross section and length L that can support an end load of F without yielding. You also wish to minimize the amount the beam deflects under load. What is the free variable(s) (other than the material) for this design problem?
a. End load, F.
b. Length, L.
c. Beam thickness, b
d. Deflection, δ
e. Answers b and c.
f. All of the above.
Answer:
The FEM is a general numerical method for solving partial differential equations in two or three space variables (i.e., some boundary value problems). To solve a problem, the FEM subdivides a large system into smaller, simpler parts that are called finite elements.
Answer:
zero off your workpiece so you can work from a datum, usually the centre of your work on a lathe. change your tool to a drill and drill a hole to a size smaller than your thread diameter, change out your tool for a threaded tap and away you go.
I'm not sure which part they want but I'd say ensure your tool is set to the right height, you have the tool lines up where you want to cut and that you have calculated the speed you need to cut at safety. Drill a hole before you tap though.
If you have a CNC lathe you just set the programme to do the processes and tool change for you.
Answer:
(a) Rate of heat transfer assuming there is no wind is 152.70J/s
(b) Rate of heat transfer assuming the stack is exposed to 20km/h winds is 151.49J/s
Explanation:
Q/t = KA(T2 - T1)/d
Range of heat transfer coefficient (K) for flue gases is 60-180W/mk
Assuming K=180W/mk, diameter (d) =0.6m, A=3.142×0.6^2/4 = 028278m^2, distance (d) = 10m, T2=40°C, T1=10°C
(a) Q/t = 180×0.28278×(40-10)/10 = 180×0.28278×30/10 = 152.70J/s
(b) V= 20km/h=20×1000/3600m/s= 5.6m/s, t=d/v=10/5.6=1.8s
Q = KA(T2 - T1)/V = 180×0.28278×(40 - 10)/5.6 = 180×0.28278×30/5.6 = 272.68J
Q/t = 272.68J/1.8s = 151.48J/s
