Answer:
Mining would go under Industry organization.
Answer: required tensile stress is 0.889 MPa
Explanation:
Given that;
tensile load is oriented along the [1 1 1] direction
shear stress is 0.242 MPa along [1 0 1] in the (1 1 -1) plane
first we determine
λ which is Angle between [1 1 1] and [1 0 1]
so
cosλ = [ 1(1) + 1(0) + 1(1) ] / [ √(1² + 1² + 1²) √(1² + 0² + 1²)]
= 2 / √3√2 = 2/√6
Next, we determine ∅ which is angle between [1 1 1] and [1 1 -1]
so,
cos∅ = [ 1(1) + 1(1) + 1(-1) ] / [ √(1² + 1² + 1²) √(1² + 1² + (-1)²)]
cos∅ = [ 2-1] / [√3√3 ]
cos∅ = 1/3
Now, we know that;
σ = T_stress/cosλcosθ
so we substitute
σ = 0.242 / ( 2/√6 × 1/3 )
σ = 0.242 / 0.2721
σ = 0.889 MPa
Therefore, required tensile stress is 0.889 MPa
Answer:
R = 148.346 N
M₀ = - 237.2792 N-m
Explanation:
Point O is selected as a convenient reference point for the force-couple system which is to represent the given system
We can apply
∑Fx = Rx = - 60N*Cos 45° + 40N + 80*Cos 30° = 66.8556 N
∑Fy = Ry = 60N*Sin 45° + 50N + 80*Sin 30° = 132.4264 N
Then
R = √(Rx²+Ry²) ⇒ R = √((66.8556 N)²+(132.4264 N)²)
⇒ R = 148.346 N
Now, we obtain the moment about the origin as follows
M₀ = (0 m*40 N)-(7 m*60 N*Sin 45°)+(4 m*60 N*Cos 45°)-(5 m*50 N)+ 140 N-m + (0 m*80 N*Cos 30°) + (0 m*80 N*Sin 30°) = - 237.2792 N-m (clockwise)
We can see the pic shown in order to understand the question.
Answer:
rounding
Explanation:
each of the digits of a number that are used to express it to the required degree of accuracy, starting from the first non-zero digit.
Answer:
(a) Precipitation hardening
(1) The strengthening mechanism involves the hindering of dislocation motion by precipitates/particles.
(2) The hardening/strengthening effect is not retained at elevated temperatures for this process.
(4) The strength is developed by a heat treatment.
(b) Dispersion strengthening
(1) The strengthening mechanism involves the hindering of dislocation motion by precipitates/particles.
(3) The hardening/strengthening effect is retained at elevated temperatures for this process.
(5) The strength is developed without a heat treatment.
