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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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ω=314.15 rad/s.

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Explanation:

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You are about to perform PMCS on your M1114? What resource should you use for the procedures and instructions for performing PMC
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The resources and instructions that should be used for the procedures of performing PMCS are:

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<h3>What is PMCS?</h3>

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Read more on PMCS here: brainly.com/question/15720250

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An object of mass 521 kg, initially having a velocity of 90 m/s, decelerates to a final velocity of 14 m/s. What is the change i
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Change in kinetic energy=\frac{1}{2}\times m\left ( V_0^{2}-V^2\right )

Change in kinetic energy=\frac{1}{2}\times521\left ( 90^{2}-14^2\right )

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