Answer:
q₀ = 350,740.2885 N/m
Explanation:
Given
![q(x)=\frac{x}{L} q_{0}](https://tex.z-dn.net/?f=q%28x%29%3D%5Cfrac%7Bx%7D%7BL%7D%20q_%7B0%7D)
σ = 120 MPa = 120*10⁶ Pa
![L=4 m\\w=200 mm=0.2m\\h=300 mm=0.3m\\q_{0}=? \\](https://tex.z-dn.net/?f=L%3D4%20m%5C%5Cw%3D200%20mm%3D0.2m%5C%5Ch%3D300%20mm%3D0.3m%5C%5Cq_%7B0%7D%3D%3F%20%5C%5C)
We can see the pic shown in order to understand the question.
We apply
∑MB = 0 (Counterclockwise is the positive rotation direction)
⇒ - Av*L + (q₀*L/2)*(L/3) = 0
⇒ Av = q₀*L/6 (↑)
Then, we apply
![v(x)=\int\limits^L_0 {q(x)} \, dx\\v(x)=-\frac{q_{0}}{2L} x^{2}+\frac{q_{0} L}{6} \\M(x)=\int\limits^L_0 {v(x)} \, dx=-\frac{q_{0}}{6L} x^{3}+\frac{q_{0} L}{6}x](https://tex.z-dn.net/?f=v%28x%29%3D%5Cint%5Climits%5EL_0%20%7Bq%28x%29%7D%20%5C%2C%20dx%5C%5Cv%28x%29%3D-%5Cfrac%7Bq_%7B0%7D%7D%7B2L%7D%20x%5E%7B2%7D%2B%5Cfrac%7Bq_%7B0%7D%20L%7D%7B6%7D%20%5C%5CM%28x%29%3D%5Cint%5Climits%5EL_0%20%7Bv%28x%29%7D%20%5C%2C%20dx%3D-%5Cfrac%7Bq_%7B0%7D%7D%7B6L%7D%20x%5E%7B3%7D%2B%5Cfrac%7Bq_%7B0%7D%20L%7D%7B6%7Dx)
Then, we can get the maximum bending moment as follows
![M'(x)=0\\ (-\frac{q_{0}}{6L} x^{3}+\frac{q_{0} L}{6}x)'=0\\ -\frac{q_{0}}{2L} x^{2}+\frac{q_{0} L}{6}=0\\x^{2} =\frac{L^{2}}{3}\\ x=\sqrt{\frac{L^{2}}{3}} =\frac{L}{\sqrt{3} }=\frac{4}{\sqrt{3} }m](https://tex.z-dn.net/?f=M%27%28x%29%3D0%5C%5C%20%28-%5Cfrac%7Bq_%7B0%7D%7D%7B6L%7D%20x%5E%7B3%7D%2B%5Cfrac%7Bq_%7B0%7D%20L%7D%7B6%7Dx%29%27%3D0%5C%5C%20-%5Cfrac%7Bq_%7B0%7D%7D%7B2L%7D%20x%5E%7B2%7D%2B%5Cfrac%7Bq_%7B0%7D%20L%7D%7B6%7D%3D0%5C%5Cx%5E%7B2%7D%20%3D%5Cfrac%7BL%5E%7B2%7D%7D%7B3%7D%5C%5C%20%20x%3D%5Csqrt%7B%5Cfrac%7BL%5E%7B2%7D%7D%7B3%7D%7D%20%3D%5Cfrac%7BL%7D%7B%5Csqrt%7B3%7D%20%7D%3D%5Cfrac%7B4%7D%7B%5Csqrt%7B3%7D%20%7Dm)
then we get
![M(\frac{4}{\sqrt{3} })=-\frac{q_{0}}{6*4} (\frac{4}{\sqrt{3} })^{3}+\frac{q_{0} *4}{6}(\frac{4}{\sqrt{3} })\\ M(\frac{4}{\sqrt{3} })=-\frac{8}{9\sqrt{3} } q_{0} +\frac{8}{3\sqrt{3} } q_{0}=\frac{16}{9\sqrt{3} } q_{0}m^{2}](https://tex.z-dn.net/?f=M%28%5Cfrac%7B4%7D%7B%5Csqrt%7B3%7D%20%7D%29%3D-%5Cfrac%7Bq_%7B0%7D%7D%7B6%2A4%7D%20%28%5Cfrac%7B4%7D%7B%5Csqrt%7B3%7D%20%7D%29%5E%7B3%7D%2B%5Cfrac%7Bq_%7B0%7D%20%2A4%7D%7B6%7D%28%5Cfrac%7B4%7D%7B%5Csqrt%7B3%7D%20%7D%29%5C%5C%20M%28%5Cfrac%7B4%7D%7B%5Csqrt%7B3%7D%20%7D%29%3D-%5Cfrac%7B8%7D%7B9%5Csqrt%7B3%7D%20%7D%20q_%7B0%7D%20%2B%5Cfrac%7B8%7D%7B3%5Csqrt%7B3%7D%20%7D%20q_%7B0%7D%3D%5Cfrac%7B16%7D%7B9%5Csqrt%7B3%7D%20%7D%20q_%7B0%7Dm%5E%7B2%7D)
We get the inertia as follows
![I=\frac{w*h^{3} }{12} \\ I=\frac{0.2m*(0.3m)^{3} }{12}=4.5*10^{-4}m^{4}](https://tex.z-dn.net/?f=I%3D%5Cfrac%7Bw%2Ah%5E%7B3%7D%20%7D%7B12%7D%20%5C%5C%20I%3D%5Cfrac%7B0.2m%2A%280.3m%29%5E%7B3%7D%20%7D%7B12%7D%3D4.5%2A10%5E%7B-4%7Dm%5E%7B4%7D)
We use the formula
σ = M*y/I
⇒ M = σ*I/y
where
![y=\frac{h}{2} =\frac{0.3m}{2}=0.15m](https://tex.z-dn.net/?f=y%3D%5Cfrac%7Bh%7D%7B2%7D%20%3D%5Cfrac%7B0.3m%7D%7B2%7D%3D0.15m)
If M = Mmax, we have
![(\frac{16}{9\sqrt{3} }m^{2} ) q_{0}\leq \frac{120*10^{6}Pa*4.5*10^{-4}m^{4} }{0.15m}\\ q_{0}\leq 350,740.2885\frac{N}{m}](https://tex.z-dn.net/?f=%28%5Cfrac%7B16%7D%7B9%5Csqrt%7B3%7D%20%7Dm%5E%7B2%7D%20%29%20q_%7B0%7D%5Cleq%20%5Cfrac%7B120%2A10%5E%7B6%7DPa%2A4.5%2A10%5E%7B-4%7Dm%5E%7B4%7D%20%20%20%7D%7B0.15m%7D%5C%5C%20q_%7B0%7D%5Cleq%20350%2C740.2885%5Cfrac%7BN%7D%7Bm%7D)
Answer:
Fully Automated
Periodic Maintenance Activities
Answer: For #1 I'm going to go with A because that has to do with biology
For #2 I'm going to go with B oceans because that has to do with plant life (and life in general).
For #3 I'll say marine/maritime engineer (you can just say marine)
Hope it helps!
Answer:
Explanation:
Mean temperature is given by
![T_mean = \frac{T_i + T_ \infinity}{2}\\\\T_mean = \frac{200 + 15}{2}](https://tex.z-dn.net/?f=T_mean%20%3D%20%5Cfrac%7BT_i%20%2B%20T_%20%5Cinfinity%7D%7B2%7D%5C%5C%5C%5CT_mean%20%3D%20%5Cfrac%7B200%20%2B%2015%7D%7B2%7D)
Tmean = (Ti + T∞)/2
![T_mean = 107.5^{0}](https://tex.z-dn.net/?f=T_mean%20%3D%20107.5%5E%7B0%7D)
Tmean = 107.5⁰C
Tmean = 107.5 + 273 = 380.5K
Properties of air at mean temperature
v = 24.2689 × 10⁻⁶m²/s
α = 35.024 × 10⁻⁶m²/s
= 221.6 × 10⁻⁷N.s/m²
= 0.0323 W/m.K
Cp = 1012 J/kg.K
Pr = v/α = 24.2689 × 10⁻⁶/35.024 × 10⁻⁶
= 0.693
Reynold's number, Re
Pv = 4m/πD²
where Re = (Pv * D)/![\mu](https://tex.z-dn.net/?f=%5Cmu)
Substituting for Pv
Re = 4m/(πD
)
= (4 x 0.003)/( π × 6 ×10⁻³ × 221.6 × 10⁻⁷)
= 28728.3
Since Re > 2000, the flow is turbulent
For turbulent flows, Use
Dittus - Doeltr correlation with n = 0.03
Nu = 0.023Re⁰⁸Pr⁰³ = (h₁D)/k
(h₁ × 0.006)/0.0323 = 0.023(28728.3)⁰⁸(0.693)⁰³
(h₁ × 0.006)/0.0323 = 75.962
h₁ = (75.962 × 0.0323)/0.006
h₁ = 408.93 W/m².K
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