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LenKa [72]
3 years ago
9

A simply supported wood roof beam is loaded with single point dead and roof live loads applied at midspan (PD = 400 lb, PLr = 16

00 lb) of an 8 ft span. It is located in an industrial situation where the moisture content in-service is greater than 19%. The controlling load combinations are D + Lr (ASD) and 1.2D + 1.6Lr (LRFD). A No. 1 DF-L 4 x 8 is used for the beam and it is continuously braced (lu = 0). The allowable live load deflection is Δ ≤ L/360. Determine whether the beam is adequate for ASD and LRFD by determining: a. Size Category (Dimension lumber, B&S, P&T) b. Reference design values: Fv, and E c. Adjusted ASD values: F’v, and E’ d. Actual stresses (ASD) and deflection: fv, and Δ e. Compare the actual stresses and adjusted design values, and determine if the member is adequate for shear and deflection (ASD) f. Nominal LRFD values Fvn, and E g. Adjusted LRFD values F’vn, and E’ h. Adjusted LRFD shear resistance: V’n i. Factored shear (LRFD), and actual deflection: Mu, Vu, and Δ j. Determine if the member is adequate for shear and deflection using LRFD
Engineering
1 answer:
Lynna [10]3 years ago
3 0

Answer:mold i belive

Explanation:

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A hypothetical A-B alloy of composition 57 wt% B-43 wt% A at some temperature is found to consist of mass fractions of 0.5 for b
Dennis_Churaev [7]

Answer:

composition of alpha phase is 27% B

Explanation:

given data

mass fractions  = 0.5 for both

composition = 57 wt% B-43 wt% A

composition = 87 wt% B-13 wt% A

solution

as by total composition Co = 57 and by beta phase composition  Cβ = 87  

we use here lever rule that is

Wα = Wβ   ...............1

Wα = Wβ = 0.5

now we take here left side of equation

we will get

\frac{C_\beta - Co}{C_\beta - Ca}   = 0.5

\frac{87 - 57}{87 - Ca} = 0.5  

solve it we get

Ca = 27

so composition of alpha phase is 27% B

8 0
3 years ago
A monatomic ideal gas undergoes a quasi-static process that is described by the function p(????)=p1+3(????−????1) , where the st
Alenkasestr [34]

A pure gas made up only of atoms. The noble gases argon, krypton, and xenon are some examples.

Concepts:

Perfect gas law: Work performed on the system: PV = nRT W = -∫PdV

Energy preservation formula: U = Q + W

Reasoning:

W = nRT ln(Vi/Vf) when the process is isothermal.

The temperature is said to be constant, and we are given n, Pfinal, and Vfinal.

Calculation information:

(A) A process that is isothermal has a constant temperature.

PV = nRT, and hence, constant

nRT = PV = 101000 Pa*25*10-3 m3

For a process that is isothermal, W = nRT ln(Vi/Vf).

W/(nRT)=3000 J/(101000 Pa*25*10-3 m3)=-1.19

(The gas produces -W of labor.)

Vi = (25*10-3 m3)/3.28 = 7.62*10-3 m3 = 7.62 L where Vf/Vi = exp(1.19) = 3.28 Vi (b) for a perfect gas PV = nRT. 101000 Pa*25*10-3 m3 = (8.31 J/K) T. T = 303.85 K.

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brainly.com/question/29310303

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5 0
10 months ago
Information or signals entered into a computer system is
Virty [35]

<em>Logs.</em>

<em>Like data logs. Sometimes people make these logs to keep tabs on other people or to get important information put down somewhere that way it is saved and can be looked back upon later. Anytime someone makes an action on the computer, it makes a TMP file representing a log of what you want it to do before the computer quickly get's rid of the file.</em>

<em>-Ɽ3₮Ɽ0 Ⱬ3Ɽ0</em>

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2 years ago
PLZZ HELP
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Answer:

Could ask a family member to help

Explanation:

5 0
3 years ago
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Un mol de gas ideal realiza un trabajo de 3000 J sobre su entorno, cuando se expande de manera isotermica a una temperatura de 5
Shkiper50 [21]

Answer:

74,4 litros

Explanation:

Dado que

W = nRT ln (Vf / Vi)

W = 3000J

R = 8,314 JK-1mol-1

T = 58 + 273 = 331 K

Vf = desconocido

Vi = 25 L

W / nRT = ln (Vf / Vi)

W / nRT = 2.303 log (Vf / Vi)

W / nRT * 1 / 2.303 = log (Vf / Vi)

Vf / Vi = Antilog (W / nRT * 1 / 2.303)

Vf = Antilog (W / nRT * 1 / 2.303) * Vi

Vf = Antilog (3000/1 * 8,314 * 331 * 1 / 2,303) * 25

Vf = 74,4 litros

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3 years ago
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