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

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

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:

The tensile stress is also given by:

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

ok

step-by-step explanation:

7! = 7×6×5×4×3×2

3! = 3× 2

as 7! is multiple of 3! because 7! contains 3! that is 3× 2 as factor

so 7! /(3! ) = 7× 6× 5× 4

= 42 × 20

= 840

its b) 840✌✌✌

dhruv⭐