Answer:
Explanation:
Small-angle grain boundaries are not as effective in interfering with the slip process as are high-angle grain boundaries because there is not as much crystallographic misalignment in the grain boundary region for small-angle, and therefore not as much change in slip direction.
Low angle grain boundaries (quasi-coherent) are formed by the dislocation network positioned along the geometric plane with small tilt angle differences between successive peers that is tilt boundary made up edge dislocations therefore it may only divert the slip direction of the incoming gliding dislocation with very little frictional stresses. And on the other hand, a high angle grain boundary region because of their disordered almost liquid like structure which acts as a strong barrier against dislocation slip motion and causes actually formation of dislocations file-up against it by arresting their motion unless that the stress concentration at the leading dislocation becomes high enough to go though the barrier.
Explanation:
Step1
Factor of safety is the number that is taken for the safe design of any component. It is the ratio of failure stress to the maximum allowable stress for the material.
Step2
It is an important parameter for design of any component. This factor of safety is taken according to the environment condition, type of material, strength, type of component etc.
Step3
Different material has different failure stress. So, ductile material fails under shear force. Ductile material’s FOS is based on yield stress as failure stress as after yield point ductile material tends to yield. Brittle material’s FOS is based on ultimate stress as failure stress.
The expression for factor of safety for ductile material is given as follows:

Here,
is yield stress and
is allowable stress.
The expression for factor of safety for brittle material is given as follows:

Here,
is ultimate stress and
is allowable stress.
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Answer:
13.4 mm
Explanation:
Given data :
Load amplitude ( F ) = 22,000 N
factor of safety ( N )= 2.0
Take ( Fatigue limit stress amplitude for this alloy ) б = 310 MPa
<u>calculate the minimum allowable bar diameter to ensure that fatigue failure will not occur</u>
minimum allowable bar diameter = 13.4 * 10^-3 m ≈ 13.4 mm
<em>attached below is a detailed solution</em>