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lisabon 2012 [21]
3 years ago
13

A contractor needs to excavate 50,000 yd3 of silty clay and haul it with Caterpillar 69C dump trucks. Each truck can carry 30.9

yd3 of soil per load, and operates on a 15-minute cycle. The job must be completed in five working days with the trucks working two 8-hour shifts per day. Using a bulking factor of 30%, how many trucks will be required
Engineering
1 answer:
zalisa [80]3 years ago
5 0

Answer:

Number of trucks required = 203

Explanation:

Calculate the amount of soil need to be excavated from the borrow site by using the following equation:

VF = Vf + (◇V/◇Vf) × Vf

Here,  ◇V is change in volume during grading, Vf  is the Volume of fill, and ◇V/◇Vf  is the bulkage factor

Substitute in the equation.

VF = 50000 + (30/100) 50000

VF = 50000 + 15000

VF = 65000yd^3

Number of trucks needed = (Truck cycle time / Excavator time ) × VF

Number of truck heeded= (15 / (16× 5 ×60) × 65000

Number of trucks = 975000/ 4800

Number of trucks = 203

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3 years ago
Electric current originates from which part of an atom? *
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3 0
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Problem 34.3 The elevation of the end of the steel beam supported by a concrete floor is adjusted by means of the steel wedges E
Natasha2012 [34]

The image is missing, so i have attached it.

Answer:

A) P = 65.11 KN

B) Q = 30 KN

Explanation:

We are given;

The end reaction of the beam; F = 100KN

Coefficient of static friction between two steel surfaces;μ_ss = 0.3

Coefficient of static friction between steel and concrete;μ_sc = 0.6

So, F1 = μ_ss•F =0.3 x 100 = 30 KN

F2 = μ_ss•N_EF = 0.3N_EF

From the screen shot, we see that the angle is 12°

Sum of forces in the Y-direction gives;

F2•sin12 - N_EF•cos12 + 100 = 0

Rearranging gives;

N_EF•cos12 - F2•sin12 = 100

Let's put 0.3N_EF for F2 to give;

N_EF•cos12 - 0.3N_EF•sin12 = 100

Thus;

N_EF(0.9158) - 0.1247 = 100

N_EF(0.9781) = 100 + 0.1247

N_EF = 100.1247/0.9158

N_EF = 109.33 KN

Thus, F2 = 0.3N_EF = 0.3 x 109.33 = 32.8 KN

Wedge will move if;

P = (F1 + F2cos12 + N_EFsin12)

Thus;

P = 10 + (32.8 x 0.9781) + (109.33 x 0.2079)

P ≥ 65.11 KN

B) For static equilibrium, Q = F1

Thus, Q = 30 KN

3 0
3 years ago
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