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2 years ago

Determine the critical load if the bottom is fixed and the top is pinned. ewew = 1. 6 ×(10)3ksi×(10)3ksi ,σyσy = 5 ksiksi

Engineering

1 answer:

Katen [24]2 years ago

4 0

<h3>What is a Critical Load?</h3>

Critical load Fcr or buckling load is the value of load that causes the phenomenon of change from stable to unstable equilibrium state.

With that beign said, first it is neessary to calculate the moment of inercia about the x-axis:

$Ix= \frac{db^3}{12}\\ Ix = \frac{2.(4)^3}{12} = 10.667in$

Then it is necessary to calculate the moment of inercia about the y-axis:

$Iy = \frac{db^3}{12}\\ Iy = \frac{4.(2)^3}{12} = 2.662in$

Comparing both moments of inercia it is possible to assume that the minimun moment of inercia is the y-axis, so the minimun moment of inercia is 2662in.

And so, it is possible to calculate the critical load:

$Pc\gamma = \frac{2046\pi ^2E.I}{L^2} \\Pc\gamma= \frac{2046.\pi ^2.(1,6.10^3.10^3).2662}{(10.12)^2} \\Pc\gamma= 5983,9db$

See more about critical load at: brainly.com/question/22020642

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Which option identifies the concept represented in the following scenario?

dlinn [17]

Answer:

project object

Explanation:

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3 years ago

An ideal gas undergoes two processes: one frictionless and the other not. In both the cases, the gas is initially at 200 ℉ and 1

Zarrin [17]

Answer:

The process which has friction

Explanation:

The entropy is simply the change in the state of the things or the molecules in the system. It is simply the change in the energy of the system with a focus on the atoms in the system. This is also known as the internal energy of the system and is given the symbol, G. The friction contributes to the change in the energy of the system. This is because friction generates another form of energy - that is heat energy. This energy causes the internal temperature id the system to increase. Hence the greater change in the temperature.

6 0

3 years ago

Read 2 more answers

Water is stored in a tank which has vent open to the atmosphere. The water level is 1.0 m below the top the tank and the water i

sp2606 [1]

Answer:

6.99 x 10⁻³ m³ / s

Explanation:

Th e pressure difference at the two ends of the delivery pipe due to atmospheric pressure and water column will cause flow of water.

h = difference in the height of water column at two ends of delivery pipe

6 - 1 = 5 m

Velocity of flow of water

v = √2gh

= √ (2 x 9.8 x 5)

= 9.9 m /s

Volume of water flowing per unit time

velocity x cross sectional area

= 9.9 x 3.14 x .015²

= 6.99 x 10⁻³ m³ / s

7 0

3 years ago

A rigid tank whose volume is 2 m3, initially containing air at 1 bar, 295 K, is connected by a valve to a large vessel holding a

bazaltina [42]

Answer:

$Q_{cv}=-339.347kJ$

Explanation:

First we calculate the mass of the aire inside the rigid tank in the initial and end moments.

$P_iV_i=m_iRT_i$ (i could be 1 for initial and 2 for the end)

State1

$1bar*|\frac{100kPa}{1}|*2=m_1*0.287*295$

$m_1=232kg$

State2

$8bar*|\frac{100kPa}{1bar}|*2=m_2*0.287*350$

$m_2=11.946$

So, the total mass of the aire entered is

$m_v=m_2-m_1\\m_v=11.946-2.362\\m_v=9.584kg$

At this point we need to obtain the properties through the tables, so

For Specific Internal energy,

$u_1=210.49kJ/kg$

For Specific enthalpy

$h_1=295.17kJ/kg$

For the second state the Specific internal Energy (6bar, 350K)

$u_2=250.02kJ/kg$

At the end we make a Energy balance, so

$U_{cv}(t)-U_{cv}(t)=Q_{cv}-W{cv}+\sum_i m_ih_i - \sum_e m_eh_e$

No work done there is here, so clearing the equation for Q

$Q_{cv} = m_2u_2-m_1u_1-h_1(m_v)$

$Q_{cv} = (11.946*250.02)-(2.362*210.49)-(295.17*9.584)$

$Q_{cv}=-339.347kJ$

The sign indicates that the tank transferred heat<em> to</em> the surroundings.

8 0

3 years ago

There are two identical oil tanks. The level of oil in Tank A is 12 ft and is drained at the rate of 0.5 ft/min. Tank B contains

Luba_88 [7]

Answer:

16 minutes

Explanation:

This is an example of a class of problems in which two quantities start with different initial values and change at different rates. In such problems, the rates of change are generally ones that cause the values to converge.

The question usually asks when the values will be the same. The generic answer is, "when the difference in rates makes up the difference in initial values."

Here the tanks differ in initial fill height by 12 -8 = 4 ft. The rates of change differ by 0.5 -0.25 = 0.25 ft/min. The more filled tank is draining faster (important), so the fill heights will converge after ...

(4 ft)/(0.25 ft/min) = 16 min

The level in the two tanks will be the same after 16 minutes.

<em>Additional comment</em>

The oil levels at that time will be 4 ft.

You can write two equations for height:

y = 12 -0.5x . . . . . . . height in feet after x minutes (tank A)

y = 8 -0.25x . . . . . . height in feet after x minutes (tank B)

These will be equal when ...

y = y

12 -0.5x = 8 -0.25x

4 = 0.25x . . . . . . . . . . add 0.5x -8

16 = x . . . . . . . . . . . . multiply by 4 . . . . time to equal height

The graph shows when the tanks will have equal heights and when they will be drained.

4 0

2 years ago