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Naddika [18.5K]
3 years ago
5

Using the AASHTO 1993 Flexible Pavement Design Procedure, design a pavement cross section that will provide 10 years service. Th

e initial PSI is 4.2 and the final PSI is determined to be 2.5. The subgrade has a soil resilient modulus of 18,000 psi. Reliability is 95% with an overall standard deviation of 0.4. For design, the daily truck traffic consists of 400 passes of trucks with two single axles and 350 passes of semitrailer truck with tandem axles. The axle weights are:
Single-Unit truck = 8000-lb steering, single axle

= 22,000-lb drive, single axle

Semiunit truck = 10,000-lb steering, single axle

= 16,000-lb drive, tandem axle

= 34,000-lbs trailer, tandem axle

M2 and M3 are equal to 1.0 for the materials in the pavement structure.

Engineering
1 answer:
natali 33 [55]3 years ago
5 0

Answer:

Check the explanation

Explanation:

Single Unit Truck ESAL = 43.38 + 5.16 = 48.54

Semi Unit Truck ESAL = 43.38+ 6.00+7.4 = 56.78

So total ESAL's during design life = (400*48.54 + 350*56.78)*365*10/18000 = (19416+19873)*3650= 3939

Kindly check the attached image

Here

Reliability = 95% = 0.95, therefore ZR = -1.645, S0 = 0.4, MR = 18

Delta PSI = 4.2-2.5= 1.7

Resilient Modulus = 18000 psi, So MR = 18

Assume SN = 3.0 for flexible pavements

There W18 calculates to 0.26807

So

log10 (3939) = 9.36*log10(SN+1) -.2/(.4+1094/(SN+1)5.19)) -6.01

Structural Number SN = a1*d1 + a2*d2 *m2 +a2*d3 *m3

= a1*d1 + a2*d2 +a2*d3

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4 0
3 years ago
Which of the following impulse responses correspond(s) to stable LTI systems? (a) h1(t) = e-(1-2j)ru(t) (b) h2(t) = e-r cos(2t)u
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LTI system is stable if Impulse response is finite.

so the correct answer is "b"

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4 years ago
Which of the following is not a relationship set between elements in a sketch​
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6 0
2 years ago
A solid circular shaft has a uniform diameter of 5 cm and is 4 m long. At its midpoint 65 hp is delivered to the shaft by means
AlekseyPX

Answer:

A) τ_max = 59.139 x 10^(6) Pa

B) θ = 0.0228 rad.

Explanation:

A) In the left half of the shaft we have 25 hp which corresponds to a torque T1 given by;

P = Tω

Where P is power and ω is angular speed.

Power = 25 HP = 25 x 746 W = 18650W

ω = 200 rev/min = 200 x 0.10472 rad/s = 20.944 rad/s

P = T1•ω

T1 = P/ω = 18650/20.944

T1 = 890.47 N.m

Similarly, in the right half we have 40 hp corresponding to a torque T2

given by;

P = T2•ω

T2 = P/ω

Where P = 40 x 760 = 30,400W

T2 = 30400/20.944 = 1451.49 N.m

The maximum shearing stress consequently occurs in the outer fibers in the right half and is given by;

τ_max = Tρ/J

Where J is polar moment of inertia and has the formula ;J = πd⁴/32

d = 5cm = 0.05m

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ρ = 0.05/2 = 0.025m

T will be T2 = 1451.49 N.m

Thus,

τ_max = Tρ/J

τ_max = 1451.49 x 0.025/6.136 x 10^(-7)

τ_max = 59139022.94 N/m² = 59.139 x 10^(6) Pa

B) The angles of twist of the left and right ends relative to the center are, respectively, using θ = TL/GJ

G = 80 Gpa = 80 x 10^(9) Pa

θ1 = (890.47 x 2)/(80 x 10^(9) x 6.136 x 10^(-7)) = 0.0363 rad

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θ2 = (1451.49 x 2)/(80 x 10^(9) x 6.136 x 10^(-7)) = 0.0591 rad

Since θ1 and θ2 are in the same direction, the relative angle of twist between the two ends of the shaft is

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Answer:

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Explanation:

The step by step explanation is as shown in the attachment

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