Respuesta :
Answer:
1) 2304 kPa
2) B. 200 N/m
Explanation:
The internal pressure of the of the tank can be found from the following relations;
Resisting wall force F = p×(1/4·π·D²)
σ×A = p×(1/4·π·D²)
Where:
σ = Allowable stress of the tank
A = Area of the wall of the tank = π·D·t
t = Thickness of the tank = 15 mm. = 0.015 m
D = Diameter of the tank = 25 m
p = Maximum permissible internal pressure pressure
∴ σ×π·D·t = p×(1/4·π·D²)
p = 4×σ×t/D = 4 × 240 ×0.015/2.5 = 5.76 MPa
With a desired safety factor of 2.5, the permissible internal pressure = 5.76/2.5 = 2.304 MPa
2) The formula for average shear flow is given as follows;
[tex]q = \dfrac{T}{2 \times A_m}[/tex]
Where:
q = Average shear flow
T = Torque = 8 N·m
[tex]A_m[/tex] = Average area enclosed within tube
t = Thickness of tube = 6.35 mm = 0.00635 m
Side length of the square cross sectioned tube, s = 203 mm = 0.203 m
Average area enclosed within tube, [tex]A_m[/tex] = (s - t)² = (0.203 - 0.00635)² = 0.039 m²
[tex]\therefore q = \dfrac{8}{2 \times 0.039} = 206.9 \, N/m[/tex]
Hence the average shear flow is most nearly 200 N/m.
Following are the solution to the given question:
Calculating the allowable stress:
[tex]\to \sigma_{allow} = \frac{\sigma_y}{FS} \\\\[/tex]
[tex]= \frac{240}{2.5} \\\\= 96\\\\[/tex]
Calculating the Thickness:
[tex]\to t =15\ mm = \frac{15\ }{1000}= 0.015\ m\\\\[/tex]
The stress in a spherical tank is defined as
[tex]\to \sigma = \frac{pD}{4t}\\\\\to 96 = \frac{p(25)}{4(0.015)}\\\\\to p = 0.2304\;\;MPa\\\\\to p = 230.4\;\;kPa\\\\\to p \approx 230\;\;kPa\\\\[/tex]
[tex]\bold{\to A= 203^2= 41209\ mm^2} \\\\[/tex]
Calculating the shear flow:
[tex]\to q=\frac{T}{2A}[/tex]
[tex]=\frac{8}{2 \times 41209 \times 10^{-6}}\\\\=\frac{8}{0.082418}\\\\=97.066\\\\[/tex]
[tex]\to q=97 \approx 100 \ \frac{N}{m}\\[/tex]
Therefore, the final answer is "".
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