Complete Question
Consider a single crystal of some hypothetical metal that has the BCC crystal structure and is oriented such that a tensile stress is applied along a [121] direction. If slip occurs on a (101) plane and in a [111] direction, compute the stress at which the crystal yields if its critical resolved shear stress is 2.9 MPa.
Answer:
The stress is 
Explanation:
From the question we are told that
The critical yield resolved shear stress is 
First we obtain the angle 
between the slip direction [121] and [111]
![\lambda = cos^{-1} [\frac{(u_1 u_2 + v_1 v_2 + w_1 w_2}{\sqrt{u_1^2 + v_1 ^2+ w_1^2})\sqrt{( u_2^2 + v_2^2 + w_2 ^2)} } ]](https://tex.z-dn.net/?f=%5Clambda%20%3D%20cos%5E%7B-1%7D%20%5B%5Cfrac%7B%28u_1%20u_2%20%2B%20v_1%20v_2%20%2B%20w_1%20w_2%7D%7B%5Csqrt%7Bu_1%5E2%20%2B%20v_1%20%5E2%2B%20w_1%5E2%7D%29%5Csqrt%7B%28%20u_2%5E2%20%2B%20v_2%5E2%20%2B%20w_2%20%5E2%29%7D%20%7D%20%20%5D)
Where 
are the directional indices
![\lambda = cos ^-[ \frac{(1) (-1) + (2) (1) + (1) (1)}{\sqrt{((1)^2 +(2)^2 + (1)^2)}\sqrt{((-1)^2 + (1)^2 + (1)^2 ) } } ]](https://tex.z-dn.net/?f=%5Clambda%20%20%3D%20%20cos%20%5E-%5B%20%5Cfrac%7B%281%29%20%28-1%29%20%2B%20%282%29%20%281%29%20%2B%20%281%29%20%281%29%7D%7B%5Csqrt%7B%28%281%29%5E2%20%2B%282%29%5E2%20%2B%20%281%29%5E2%29%7D%5Csqrt%7B%28%28-1%29%5E2%20%2B%20%281%29%5E2%20%2B%20%281%29%5E2%20%29%20%7D%20%20%7D%20%5D)
![= cos^{-1} [\frac{2}{\sqrt{6} \sqrt{3} } ]](https://tex.z-dn.net/?f=%3D%20cos%5E%7B-1%7D%20%5B%5Cfrac%7B2%7D%7B%5Csqrt%7B6%7D%20%5Csqrt%7B3%7D%20%20%7D%20%5D)
Next is to obtain the angle 
between the direction [121] and [101]
![\O = cos^{-1} [\frac{(u_1 u_3 + v_1 v_3 + w_1 w_3}{\sqrt{u_1^2 + v_1 ^2+ w_1^2})\sqrt{( u_3^2 + v_3^2 + w_3 ^2)} } ]](https://tex.z-dn.net/?f=%5CO%20%3D%20cos%5E%7B-1%7D%20%5B%5Cfrac%7B%28u_1%20u_3%20%2B%20v_1%20v_3%20%2B%20w_1%20w_3%7D%7B%5Csqrt%7Bu_1%5E2%20%2B%20v_1%20%5E2%2B%20w_1%5E2%7D%29%5Csqrt%7B%28%20u_3%5E2%20%2B%20v_3%5E2%20%2B%20w_3%20%5E2%29%7D%20%7D%20%20%5D)
Substituting 1 for 
, 2 for 
, 1 for 
, 1 for 
, 0 for 
, and 1 for 
![\O = cos^{-1} [\frac{1* 1 + 2*0 + 1*1 }{\sqrt{1^2 + 2^2 + 1^2 } \sqrt{(1^2 + 0^2 + 1^2 )} } ]](https://tex.z-dn.net/?f=%5CO%20%3D%20cos%5E%7B-1%7D%20%5B%5Cfrac%7B1%2A%201%20%2B%202%2A0%20%2B%201%2A1%20%7D%7B%5Csqrt%7B1%5E2%20%2B%202%5E2%20%2B%201%5E2%20%7D%20%5Csqrt%7B%281%5E2%20%2B%200%5E2%20%2B%201%5E2%20%29%7D%20%20%7D%20%5D)
![\O = cos^{-1} [\frac{2}{\sqrt{6} \sqrt{2} } ]](https://tex.z-dn.net/?f=%5CO%20%3D%20cos%5E%7B-1%7D%20%5B%5Cfrac%7B2%7D%7B%5Csqrt%7B6%7D%20%5Csqrt%7B2%7D%20%20%7D%20%5D)
The stress is mathematically represented as
Answer: quatrain
Explanation:
Reading the Shakespeare's "Sonnet 100", we can infer that the underlined section is referred to as a quatrain.
The quatrain simply refers to a type of stanza that is made up of four lines. For example, based on the information given, we can deduce that the rhyme scheme for the second quatrain is given as cdcd.
Answer:
Aggressive behavior
Explanation:
Alcohol consumption tends to cause more Aggressive behavior.
The consumption of alcohol plays a more role in our culture but drinking of too much alcohol can cause drowsiness, vomiting, Upset stomach, slurred speech, heart damage, infertile, numbness lung infections, and many more. Also too much alcohol can cause violence, anger and so on in the society.
Answer:
The limits of the hole size are;
The maximum limit of the hole diameter 0.255
The minimum limit of the hole diameter = 0.245
Explanation:
Tolerance is a standardized form of language that can be used to define the intended 'tightness' or 'clearance' degree between mating parts in a mechanical assembly process and in metal joining processes such as welding and brazing processes
In tolerancing, the size used in the description of a part is known as the nominal size while allowable variation of the nominal size that will still allow the part to function properly is known as the tolerance
A tolerance given in the form ±P is known as bilateral tolerancing, with the value being added to or subtracted from the nominal size to get the maximum and minimum allowable limits of the dimensions of the nominal size
Therefore;
The given nominal dimension of the hole diameter = 0.250
The bilateral tolerance of the dimension, = ±0.005
Therefore;
The maximum limit of the diameter of the hole = 0.250 + 0.005 = 0.255
The minimum limit of the diameter of the hole = 0.250 - 0.005 = 0.245
Answer:
See explanations for completed answers
Explanation:
Given that; The average grain diameter of an aluminum alloy is 14 mu m with a strength of 185 MPa. The same alloy with an average grain diameter of 50 mu m has a strength of 140 MPa. (a) Determine the constants for the Hall-Patch equation for this alloy, (b) How much more should you reduce the grain size if you desired a strength of 220 MPa
See attachment for completed solvings
