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 plot of the function production rate m(t) (in kg/min) against time t (in min) is attached to this answer.
The production rate function M(t) is:
![m(t)=[H(t)\cdot10+H(t-20)\cdot5-H(t-80)\cdot14+H(t-81)\cdot9]kg/min](https://tex.z-dn.net/?f=m%28t%29%3D%5BH%28t%29%5Ccdot10%2BH%28t-20%29%5Ccdot5-H%28t-80%29%5Ccdot14%2BH%28t-81%29%5Ccdot9%5Dkg%2Fmin)
(1)
The Laplace transform of this function is:
![\displaystyle m(s)=[\frac{10+5e^{-20s}-14e^{-80s}+9e^{-81s}}{s}]kg/min](https://tex.z-dn.net/?f=%5Cdisplaystyle%20m%28s%29%3D%5B%5Cfrac%7B10%2B5e%5E%7B-20s%7D-14e%5E%7B-80s%7D%2B9e%5E%7B-81s%7D%7D%7Bs%7D%5Dkg%2Fmin)
(2)
Explanation:
The function of the production rate can be considered as constant functions by parts in the domain of time. To make it a continuous function, we can use the function Heaviside (as seen in equation (1)). To join all the constant functions, we consider at which time the step for each one of them appears and sum each function multiply by the function Heaviside.
For the Laplace transform we use the following rules:
![\mathcal{L}[f(x)+g(x)]=\mathcal{L}[f(x)]+\mathcal{L}[g(x)]=F(s)+G(s)](https://tex.z-dn.net/?f=%5Cmathcal%7BL%7D%5Bf%28x%29%2Bg%28x%29%5D%3D%5Cmathcal%7BL%7D%5Bf%28x%29%5D%2B%5Cmathcal%7BL%7D%5Bg%28x%29%5D%3DF%28s%29%2BG%28s%29)
(3)
![\mathcal{L}[aH(x-b)]=\displaystyle\frac{ae^{-bs}}{s}](https://tex.z-dn.net/?f=%5Cmathcal%7BL%7D%5BaH%28x-b%29%5D%3D%5Cdisplaystyle%5Cfrac%7Bae%5E%7B-bs%7D%7D%7Bs%7D)
(4)
Answer:
you never mentioned the options
Answer:
Input power = 465.63 W
current = 1.94 A
Explanation:
we have the following data to answer this question
V = 9130
i = 0.051
the input power = VI
I = 51.0 mA = 0.051
= 9130 * 0.051
= 465.63 watts
the current = 465.63/240
= 1.94A
therefore the input power is 465.63 wwatts
while the current is 1.94A
the input power is the same thing as the output power.
Answer:
Question 1 A 1020 Cold-Drawn steel shaft is to transmit 20 hp while rotating at 1750 rpm. Calculate the transmitted torque in lbs. in. Ignore the effect of friction. Answer with three decimal points. 60.024 Question 2 Based on the maximum-shear-stress theory, determine the minimum diameter in inches for the shaft in Q1 to provide a safety factor of 3. Assume Sy = 57 Kpsi. Answer with three decimal points. 0.728 Question 3 If the shaft in Q2 was made of ASTM 30 cast iron, what would be the factor of safety? Assume Sut = 31 Kpsi, Suc = 109 Kpsi 0 2.1 O 2.0 O 2.5 0 2.4 2.3 O 2.2
Explanation:
hope it helps
