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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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Given :

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The vertical effective stress $=1500 \ lb/ft^2$

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Now,

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6 - 12 inch, $N_1=6 \left(\frac{350}{72+70} \right) $

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12 - 18 inch, $N_1=6 \left(\frac{350}{72+70} \right) $

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$N_{avg}=\frac{9.86+14.8+14.8}{3}$

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