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8_murik_8 [283]
3 years ago
13

A polyethylene rod exactly 10 inches long with a cross-sectional area of 0.04 in2 is used to suspend a weight of 358 lbs-f (poun

ds-force). Given the tensile modulus for this polymer is 25,000 psi and the viscosity is 1 × 109 psi-sec, calculate the length of the rod, in inches, 1 hour(s) after loading. Answer Format: X.XX Unit: inches
Engineering
1 answer:
Nadya [2.5K]3 years ago
7 0

Answer:

Final length of the rod = 13.90 in

Explanation:

Cross Sectional Area of the polythene rod, A = 0.04 in²

Original length of the polythene rod, l = 10 inches

Tensile modulus for the polymer, E = 25,000 psi

Viscosity, \eta = 1*10^{9} psi -sec

Weight = 358 lbs - f

time, t = 1 hr = 3600 sec

Stress is given by:

\sigma = \frac{Force}{Area} \\\sigma = \frac{358}{0.04} \\\sigma = 8950 psi

Based on Maxwell's equation, the strain is given by:

strain = \sigma ( \frac{1}{E} + \frac{t}{\eta} )\\Strain = 8950 ( \frac{1}{25000} + \frac{3600}{10^{9} } )\\Strain = 0.39022

Strain = Extension/(original Length)

0.39022 = Extension/10

Extension = 0.39022 * 10

Extension = 3.9022 in

Extension = Final length - Original length

3.9022 =  Final length - 10

Final length = 10 + 3.9022

Final length = 13.9022 in

Final length = 13.90 in

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3 years ago
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Determine the reactor volume (assume a CSTR activated sludge aerobic reactor at steady state) required to treat 5 MGD of domesti
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1.0MG

Explanation:

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a) 0.697*10³ lb.in

b) 6.352 ksi

Explanation:

a)

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Let Length of AB = 12 in

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\phi_B=\frac{T_{AB}L}{GJ}=\frac{T_{AB}*12}{3.3*10^6*0.1437}  =2.53*10^{-5}T_{AB}

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