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grandymaker [24]
3 years ago
14

a steal bar with 1 in diameter has been subjected to a tensile force of 3 tons find tensile stress in the bar

Engineering
1 answer:
Marysya12 [62]3 years ago
6 0

Answer:

stress = 8,556 Psi.   or  (stress = 59 Mpa)

Explanation:

stress = force / area

force P = 3 tons (convert to lbs. for units consistency)

1 ton = 2240 lbs.

P = 6,720 lbs.

steel bar Diameter D = 1 in. (convert to d

Area of steel bar = (π *  1²) / 4 = 0.785 in²

therefore, stress = 6720 lbs. / 0.785 in²

stress = 8,556 Psi.

in Mpa ----- 8556 Psi * 0.00689476 MPa/Psi = 59 Mpa

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Answer:

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Applying values we get

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\frac{T_{1}}{T_{o}}=\frac{[2\gamma M^{2}-(\gamma -1)][(\gamma -1)M^{2}+2]}{(\gamma +1)^{2}M^{2}}

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\frac{T_{1}}{423}=\frac{[2\times 1.4\times 1.8^{2}-(1.4-1)][(1.4-1)1.8^{2}+2]}{(1.4+1)^{2}\times 1.8^{2}}\\\\\therefore \frac{T_{1}}{423}=1.53\\\\\therefore T_{1}=647.85K=374.85^{o}C

The Mach number downstream is obtained by the relation

M_{d}^{2}=\frac{(\gamma -1)M^{2}+2}{2\gamma M^{2}-(\gamma -1)}\\\\\therefore M_{d}^{2}=\frac{(1.40-1)\times 1.8^{2}+2}{2\times1.4\times 1.8^{2}-(1.4-1)}\\\\\therefore M_{d}^{2}=0.38\\\\M_{d}=0.616

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