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3 years ago

8

A cylindrical 2014-t6 aluminum alloy bar is subjected to compression-tension stress cycling along its axis; results of these tes

ts are shown in fig. 1. if the bar diameter is 12.0 mm, calculate the maximum allowable load amplitude (in newtons) to ensure that fatigue failure will not occur at 107 cycles.

2 answers:

rosijanka [135]3 years ago

6 0

Answer:

The maximum allowable load amplitude is 6408.85 Newtons

Step-by-step explanation:

Assume a factor of safety N = 3.0

From the graph attached, the maximum stress the cylindrical 2014-t6 aluminum alloy bar is subjected to is S = 170MPa

diameter d = 12.0 mm = 12*10^-3 m

area A

S = F/A

But A = (Π* d²)/4

S = 4F/(Π*d²)

By incorporating factor of safety N

S/N = 4F/(Π*d²)

F = SΠd²/4N

F = {170*10^6 *Π *(12*10^-3)}/4*3.0

F = 76906.19/12 = 6408.85 Newtons

Pls find the solution in the attached file for more explanation.

Pani-rosa [81]3 years ago

3 0

Complete part of Question:

Assume a factor of safety of 3.0

The Maximum stress versus logarithm of the number of cycles to fatigue failure curve for 2014-T6 Aluminium bar is attached to this solution

Answer:

The maximum allowable load amplitude is 6403.33 N

Step-by-step explanation:

diameter of the bar, d = 12 mm

d = 0.012 m

Cross sectional area of the bar, A = πd²/4

A = π*0.012²/4

A = 0.000113 m²

From the Maximum stress versus logarithm of the number of cycles to fatigue failure for 2014-T6 Aluminium bar attached to this solution;

At 10⁷ cycles, Maximum stress, S = 170 MPa

S = 170 * 10⁶ Pa

Factor of safety, N = 3.0

The tensile stress is given by the formula:

$\sigma = S/N\\\sigma = \frac{170 * 10^{6} }{3} \\\sigma = 56.67 * 10^{6} N/m^{2}$

The tensile stress is also given by:

$\sigma = F/A\\\sigma = F/0.000113$

F/0.000113 = 56.67 * 10⁶

F = 56.67 * 10⁶ * 0.000113

F = 6403.33 N

The maximum allowable load amplitude is 6403.33 N

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