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2 years ago

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The fatigue data for a brass alloy are given as follows: Stress Amplitude (MPa) Cycles to Failure 170 3.7 × 104 148 1.0 × 105 13

0 3.0 × 105 114 1.0 × 106 92 1.0 × 107 80 1.0 × 108 74 1.0 × 109 i. Make an S–N plot (stress amplitude versus logarithm of cycles to failure) using these data. ii. Determine the fatigue strength at 4 × 106 cycles. iii. Determine the fatigue life for 120 MPa.

1 answer:

prohojiy [21]2 years ago

8 0

Answer:

i) S–N plot is attached

ii) fatigue strength = 100 MPa

iii) fatigue life = 5.62 x 10^(5) cycles

Explanation:

i) I have attached the S–N plot (stress amplitude versus logarithm of cycles to failure)

ii) The question says we should find the fatigue strength at 4 × 10^(6) cycles.

So let's find the log of this and trace it on the graph attached.

Log(4 × 10^(6)) = 6.6

From the graph attached, at log of cycle value of 6.6, the fatigue strength is approximately 100 MPa

iii) The question says we should find the fatigue life for 120 MPa.

Thus, from the graph, at stress amplitude of 120 MPa, the log of cycles is approximately 5.75.

Thus,the fatigue life will be the inverse log of 5.75.

Thus, fatigue life = 10^(5.75)

Fatigue life = 5.62 x 10^(5)

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Answer:

25 - $\sqrt[4]{26.66*10^{-8} }$ mm

Explanation:

Given data

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next we calculate the inner diameter using the relation

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c2 = (50/2) = 25 mm

$\frac{J}{c_{2} }$ = $\frac{\pi }{2c_{2} } ( c^{4} _{2} - c^{4} _{1} )$ = $\frac{\pi }{0.050} [ ( 0.025^{4} - c^{4} _{1} ) ]$

therefore; 0.025^4 - $c^{4} _{1}$ = 0.050 / $\pi$ (7.8 *10^-6)

$c^{4} _{1}$ = 39.06 * 10 ^-8 - ( 1.59*10^-2 * 7.8*10^-6)

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THE TUBE THICKNESS

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Answer:

a.) -147V

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c.) 51V

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a.) Equation for potential difference is the integral of the electrical field from a to b for the voltage V_ba = V(b)-V(a).

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Complete question is;

Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a 100 mm thick plate (ρ = 7830 kg/m3, Cp = 550 J/kg K, k = 48 W/m K). The plate initially is at 200 °C and is to be heated to a minimum temperature of 550 °C. Heating is effected in a gas-fired furnace where the products of combustion at T∞ = 800 °C maintain a convection heat transfer coefficient of h = 250 W/m.K on both surfaces of the plate. How long should the plate be left in the furnace?

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860 seconds

Explanation:

We are given;

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Let's find the biot number from the formula;

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Now, lowest temperature in the slab is given as;

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θ_o = 0.4167

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Thus, root of the biot number Bi = 0.2604 is;

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Answer:

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Answer:

See explanations for step by step procedures to get answer.

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