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Ierofanga [76]
3 years ago
13

A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 5.0 mm (0.20 in

.); the specimen fractured at a load of 3000 N (675 lb) when the distance between the support points was 40 mm (1.6 in.). Another test is to be performed on a specimen of this same material, but one that has a square cross section of 15 mm (0.6 in.) length on each edge. At what load would you expect this specimen to fracture if the support point separation is maintained at 40 mm (1.6 in.)?
Engineering
1 answer:
Sonja [21]3 years ago
7 0

The load is 17156 N.

<u>Explanation:</u>

First compute the flexural strength from:  

σ = FL / πR^{3}

   = 3000 \times (40 \times 10^-3) / π (5 \times 10^-3)^3

σ = 305 \times 10^6 N / m^2.

We can now determine the load using:

F = 2σd^3 / 3L

  = 2(305 \times 10^6) (15 \times 10^-3)^3 / 3(40 \times 10^-3)

F = 17156 N.  

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1. Examine the following circuit. Find RT, I3, R1, R2, R3, V1, V2 and V3. Show all of your work clearly below.
Mkey [24]

Explanation:

Ohm's law is used here. V = IR, and variations. The voltage across all elements is the same in this parallel circuit. (V1 =V2 =V3)

The total supply current is the sum of the currents in each of the branches. (It = I1 +I2 +I3)

Rt = (8 V)/(8 A) = 1 Ω . . . . supply voltage divided by supply current

I3 = 8A -3A -4A = 1 A . . . . supply current not flowing through other branches

R1 = (8 V)/(3 A) = 8/3 Ω

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6 0
2 years ago
Consider a refrigerator that consumes 320 W of electric power when it is running. If the refrigerator runs only one quarterof th
ryzh [129]

Answer:

$5.184

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The cost can be calculated using the formula: Cost = Load \ factor \times Number \ of \ hours \ \\M_{month} = M_{units} \times W\\

Before using this, we require the following conversions:

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\frac {320}{1000} = 0.32

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Using the above stated formula:

M_{month} = 0.09 \times 0.32 \times \frac{1}{4} \times 720 = 5.184

4 0
2 years ago
A gas expands in a piston-cylinder assembly from p1 = 8 bar, V1 = 0.02 m3 to p2 = 2 bar. The relation between pressure and volum
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Answer:

The heat transfer is 29.75 kJ

Explanation:

The process is a polytropic expansion process

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Comparing PV^n = constant with PV^1.2 = constant

n = 1.2

(V2/V1)^n = P1/P2

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W = (P2V2 - P1V1)/1-n

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P2 = 2 bar = 2×100 = 200 kPa

V1 = 0.02 m^3

V2 = 0.064 m^3

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W = (200×0.064 - 800×0.02)/-0.2 = -3.2/-0.2 = 16 kJ

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Heat transfer (Q) = ∆U + W = 13.75 + 16 = 29.75 kJ

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3 years ago
where σ(t) and σ(0) represents the time-dependent and initial (i.e., time =0) stresses, respectively, and t and τ denote elapsed
artcher [175]

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Explanation:

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