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Slav-nsk [51]
4 years ago
14

Starting with the column in Problem 1, perform enough additional calculations to determine the effects of increasing fc from 500

0 to 8000 psi on column capacity at both high and low axial loads. Assuming that a compressive strength of 8000 psi is appropriate for the lower stories of a high-rise structure, would you recommend using concrete with fc = 8000 psi for the columns supporting all stories within the building? Use your analysis to support your answer.

Engineering
1 answer:
Hoochie [10]4 years ago
5 0

Answer:

Explanation:

Find attached the solution to the question

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Find an expectation value in the n th state of the harmonic oscillator.Find an expectation value in the n th state of the harmon
s344n2d4d5 [400]

The classical motion for an oscillator that starts from rest at location x₀ is

                                           x(t) = x₀ cos(ωt)

The probability that the particle is at a particular x at a particular time t

is given by ρ(x, t) = δ(x − x(t)), and we can perform the temporal average

to get the spatial density. Our natural time scale for the averaging is a half

cycle, take t = 0 → π/ ω

Thus,

ρ =   \frac{1}{\pi / w} \int\limits^\pi_0 {d(x - x_o cos(wt))} \, dt

Limit is 0 to π/ω

We perform the change of variables to allow access to the δ, let y = x₀ cos(ωt)  so that

ρ(x) = -\frac{w}{\pi } \int\limits^x_x {\frac{d ( x - y)}{x_ow sin(wt)} } \, dy

Limit is x₀ to -x₀

\frac{1}{\pi } \int\limits^x_x {\frac{d (x-y)}{x_o\sqrt{1 - cos^2(wt)} } } \, dy

Limit is -x₀ to x₀

= \frac{1}{\pi } \int\limits^x_x {\frac{d(x-y)}{\sqrt{x_o^2 - y^2} } } \, dy\\ \\= \frac{1}{\pi\sqrt{x_o^2 - x^2}  }

This has \int\limits^x_x {p(x)} \, dx  = 1 as expected. Here the limit is -x₀ to x₀

The expectation value is 0 when the ρ(x) is symmetric, x ρ(x) is asymmetric and the limits of integration are asymmetric.

6 0
4 years ago
Two stepped bar is supported at both ends.At the join point of two segments,the force F is applied(downwards).Calculate reactive
nlexa [21]

Answer:

F=200kN

Explanation:

8 0
3 years ago
A 900 kg car is accelerated from a speed of 10 m/s to 30 m/s. An estimated heat loss of 20 BTU's occurs during the acceleration.
Strike441 [17]

Answer:

Work = 651,1011 kJ

Explanation:

Let´s take the car as a system in order to apply the first law of thermodynamics as follows:

E_{in}- E_{out}=E_{system,final}- E_{system,initial}

Where

E_{in}- E_{out}=(Q_{in}-Q_{out})_{heat}+(W_{in}-W_{out})_{work}+(Em_{in}-Em_{out})_{mass}

And considering that there is no mass transfer and that the only energy flows that interact with the system are the heat losses and the work needed to move the car we have:

E_{in}- E_{out}=-Q_{out}+W_{in}

Regarding the energy system we have the following:

E_{system,final}- E_{system,initial}=(U_{f}-U_{i})_{internal}+(1/2m(V^2_{f}-V^2_{i}))_{kinetic}+(mg(h_{f}-h_{i}))_{potential}

By doing the calculations we have:

E_{system,final}- E_{system,initial}=[0,1*900]_{internal}+[0,5*900(30^2-10^2)/1000)_{kinetic}+(900*10*(20-0)/1000)_{potential}\\E_{system,final}- E_{system,initial}=90+360+180=630kJ

Consider that in the previous calculation, the kinetic and potential energy terms were divided by 1.000 to change the units from J to kJ.

Finally, the work needed to move the car under the required conditions is calculated as follows:

W_{in}=Q_{out}+E_{system,final}- E_{system,initial}\\W_{in}=21,1011+630=651,1011kJ

Consider that in the previous calculation, the heat loss was changed previously from BTU to kJ.

4 0
3 years ago
can you translate this me gusta el queso :sorry i would have put 300 points buut i used them all for my last question
Irina-Kira [14]
Translate in Spanish: lo siento, hubiera puesto 300 puntos pero los usé todos para mi última pregunta



If you meant a different language lmk
4 0
3 years ago
Software piracy occurs when
mario62 [17]

Answer:

Someone steals software

3 0
3 years ago
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