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3 years ago

Which of the following structures is least likely to require a deep foundation?

Engineering

1 answer:

Ghella [55]3 years ago

3 0

Explanation:

a gas in a rigid container has a pressure of 632 torrs and a temperature of 45 celsius. The pressure has increased to 842 torrs. What is the new temperature of the gas

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simply supported beam is subjected to a linearly varying distributed load ( ) 0 q x x L 5 q with maximum intensity 0 q at B. The

Pavlova-9 [17]

Answer:

q₀ = 350,740.2885 N/m

Explanation:

Given

$q(x)=\frac{x}{L} q_{0}$

σ = 120 MPa = 120*10⁶ Pa

$L=4 m\\w=200 mm=0.2m\\h=300 mm=0.3m\\q_{0}=? \\$

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

$v(x)=\int\limits^L_0 {q(x)} \, dx\\v(x)=-\frac{q_{0}}{2L} x^{2}+\frac{q_{0} L}{6} \\M(x)=\int\limits^L_0 {v(x)} \, dx=-\frac{q_{0}}{6L} x^{3}+\frac{q_{0} L}{6}x$

Then, we can get the maximum bending moment as follows

$M'(x)=0\\ (-\frac{q_{0}}{6L} x^{3}+\frac{q_{0} L}{6}x)'=0\\ -\frac{q_{0}}{2L} x^{2}+\frac{q_{0} L}{6}=0\\x^{2} =\frac{L^{2}}{3}\\ x=\sqrt{\frac{L^{2}}{3}} =\frac{L}{\sqrt{3} }=\frac{4}{\sqrt{3} }m$

then we get

$M(\frac{4}{\sqrt{3} })=-\frac{q_{0}}{6*4} (\frac{4}{\sqrt{3} })^{3}+\frac{q_{0} *4}{6}(\frac{4}{\sqrt{3} })\\ M(\frac{4}{\sqrt{3} })=-\frac{8}{9\sqrt{3} } q_{0} +\frac{8}{3\sqrt{3} } q_{0}=\frac{16}{9\sqrt{3} } q_{0}m^{2}$

We get the inertia as follows

$I=\frac{w*h^{3} }{12} \\ I=\frac{0.2m*(0.3m)^{3} }{12}=4.5*10^{-4}m^{4}$

We use the formula

σ = M*y/I

⇒ M = σ*I/y

where

$y=\frac{h}{2} =\frac{0.3m}{2}=0.15m$

If M = Mmax, we have

$(\frac{16}{9\sqrt{3} }m^{2} ) q_{0}\leq \frac{120*10^{6}Pa*4.5*10^{-4}m^{4} }{0.15m}\\ q_{0}\leq 350,740.2885\frac{N}{m}$

8 0

4 years ago

9. Imagine that you're performing measurements on a circuit with a multimeter. You measure a total circuit

ratelena [41]

Answer:

Explanation:

the total resistance is equal to the total potential difference divided by the Current

3 0

2 years ago

Read 2 more answers

When watching your weight, you want to snack smart. To do that, you want a snack that is going to __________.

andreev551 [17]

Make u be healthy ...........

5 0

3 years ago

A certain piece of property is assessed at $150,000. If the tax rate is $2.50 per $100, what is the tax on this property?

stiks02 [169]

Answer:

The tax on this property is $3750$ dollars

Explanation:

Given

Tax on per $100 is $2.50

Tax on every $1 is $\frac{2.5}{100} = 0.025$ dollars

Tax on property of value $150,000 is

$150,000 * 0.025 = 3750$ dollars

The tax on this property is $3750$ dollars

7 0

3 years ago

For methyl chloride at 100°C the second and third virial coefficients are: B = −242.5 cm 3 ·mol −1 C = 25,200 cm 6 ·mol −2 Calcu

bogdanovich [222]

Answer:

a)W=12.62 kJ/mol

b)W=12.59 kJ/mol

Explanation:

At T = 100 °C the second and third virial coefficients are

B = -242.5 cm^3 mol^-1

C = 25200 cm^6 mo1^-2

Now according isothermal work of one mole methyl gas is

W=- $\int\limits^a_b {P} \, dV$

a= $v_2\\$

b= $v_1$

from virial equation

$\frac{PV}{RT}=z=1+\frac{B}{V}+\frac{C}{V^2}\\ \\P=RT(1+\frac{B}{V} +\frac{C}{V^2})\frac{1}{V}\\$

And

$W=-\int\limits^a_b {RT(1+\frac{B}{V} +\frac{C}{V^2}\frac{1}{V} } \, dV$

a= $v_2\\$

b= $v_1$

Now calculate V1 and V2 at given condition

$\frac{P1V1}{RT} = 1+\frac{B}{v_1} +\frac{C}{v_1^2}$

Substitute given values $P_1\\$ = 1 x 10^5 , T = 373.15 and given values of coefficients we get

$10^5(v_1)/8.314*373.15=1-242.5/v_1+25200/v_1^2$

Solve for V1 by iterative or alternative cubic equation solver we get

$v_1=30780 cm^3/mol$

Similarly solve for state 2 at P2 = 50 bar we get

$v_1=241.33 cm^3/mol$

Now

$W=-\int\limits^a_b {RT(1+\frac{B}{V} +\frac{C}{V^2}\frac{1}{V} } \, dV$

a=241.33

b=30780

After performing integration we get work done on the system is

W=12.62 kJ/mol

(b) for Z = 1 + B' P +C' P^2 = PV/RT by performing differential we get

dV=RT(-1/p^2+0+C')dP

Hence work done on the system is

$W=-\int\limits^a_b {P(RT(-1/p^2+0+C')} \, dP$

a= $v_2\\$

b= $v_1$

by substituting given limit and P = 1 bar , P2 = 50 bar and T = 373 K we get work

W=12.59 kJ/mol

The work by differ between a and b because the conversion of constant of virial coefficients are valid only for infinite series

8 0

3 years ago