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BabaBlast [244]
4 years ago
14

The vertical force P acts on the bottom of the plate having a negligible weight. Determine the shortest distance d to the edge o

f the plate at which it can be applied so that it produces no compressive stresses on the plate at section a-a. The plate has a thickness of 10 mm and P acts along the center line of this thickness.
200mm is the width and d is the distance from the right edge to the force P.

Engineering
1 answer:
Greeley [361]4 years ago
6 0

Answer:

The shortest distance d to the edge of the plate is 66.67 mm

Concepts and reason

Moment of a force:

Moment of a force refers to the propensity of the force to cause rotation on the body it acts upon. The magnitude of the moment can be determined from the product of force’s magnitude and the perpendicular distance to the force.

Moment(M) = Force(F)×distance(d)

Moment of inertia ( I )

It is the product of area and the square of the moment arm for a section about a reference. It is also called as second moment of inertia.

First prepare the free body diagram of sectioned plate and apply moment equilibrium condition and also obtain area and moment of inertia of rectangular cross section. Finally, calculate the shortest distance using the formula of compressive stress (σ) in combination of axial and bending stress

Solution and Explanation:

[Find the given attachments]

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(b) the volumetric flow rates (Umin) of the feed air  is 256 x 10³ 1/m

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Explanation:

a)   Molecular weight of pentane = 72.15 g/mol

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Flow rate of feed liquid nitrogen = 50.4 l/min

                                                  = 626*50.4*10-3

                                                   = 31.55 kg/min

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b)

C₂H₅ + 8O₂ ----------->   5CO₂ + 6H₂O

From the Stoichiometric reaction,

8 mol of O2 are used for combustion of 1 mol of pentane

for 0.4372 kmol/min of pentane = 8 * 0.4372 kmol/min of Oxygen will be required

                                                  = 3.49 kmol/min of O2

amount of air will be = 3.49/0.21 = 16.62 kmol/min

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Amount of O2 fed by air = 0.21*19.12 = 4.0215 kmol/mim

unused O2 left = 4.0215 - 3.146

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Carbon Dioxide:

1 mol of pentane = 5 mol of CO2

0.3932 kmol/min of pentane = 5*0.3932 kmol/min of CO2..................(from (2)

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