A three-point flexure test is conducted on a cylindrical specimen of aluminum oxide. The specimen radius is 5.0 mm and the dista
nce between the two supportsis 15.0 mm. If the specimen breaks when the applied force is 7500 N and the center deflection is 0.01 mm, what are the elastic modulus andthe flexural strength of the material?
1 answer:
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vₐ = 0.771 ft/s
vb = -1.54 ft/s
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<u>Explanation:</u>
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Block A:
F = ma
Nₐ = - 60 cos60° = 0
Nₐ = 30 lb
Fₐ = 0.1 X 30 = 3lb
Block B:
F = ma
Nb = - 40 cos30° = 0
Nb = 34.64lb
Fb = 0.1 X 34.64 = 3.464lb
T1 + ∑U = T2
(0 + 0) + 60 sin 60° |Δsₐ| - 40 sin 30° |Δsb| - 3|Δsₐ| - 3.464 |Δsb|
= 1/2 (60/32.2) vₐ² + 1/2 (40/32.2) vb²
and
2vₐ = - vb
On solving, we get
vₐ = 0.771 ft/s
vb = -1.54 ft/s
