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Ratling [72]
3 years ago
9

A crystalline grain of aluminum in a metal plate is situated so that a tensile load is oriented along the [1 1 1] direction. Wha

t tensile stress is required (to 3 significant figures) to produce a critical resolved shear stress of 0.242 MPa along [1 0 1] in the (1 1 -1) plane
Engineering
1 answer:
Ivenika [448]3 years ago
3 0

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

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The velocity of a particle which moves along the s-axis is given by = 40 − 3 2/ , ℎ t is in seconds. Calculate the displacement
scoundrel [369]

The displacement ∆S of the particle during the interval from t = 2sec to 4sec is; 210 sec

<h3>How to find the displacement?</h3>

We are given the velocity equation as;

s' = 40 - 3t²

Thus, the speed equation will be gotten by integration of the velocity equation to get;

s = ∫40 - 3t²

s = 40t - ¹/₂t³

Thus, the displacement between times of t = 2 sec and t = 4 sec is;

∆S = [40(4) - ¹/₂(4)³] - [40(2) - ¹/₂(2)³]

∆S = 210 m

Read more about Displacement at; brainly.com/question/4931057

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8 0
2 years ago
For surface-mounted and pendant-hung luminaires, support rods should be placed so that they extend about ____
7nadin3 [17]

Answer:

One

For surface-mounted and pendant-hung luminaires, support rods should be placed so that they extend about _one___

<h3>what is supported mounted?</h3>
  • A structure that holds up or serves as a foundation for something else. Support is a synonym for mounting.

To learn more about it, refer

to brainly.com/question/25689052

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4 0
2 years ago
Q4. (20 points) For a bronze alloy, the stress at which plastic deformation begins is 271 MPa and the modulus of elasticity is 1
babunello [35]

Answer:

a) P = 86720 N

b) L = 131.2983 mm

Explanation:

σ = 271 MPa = 271*10⁶ Pa

E = 119 GPa = 119*10⁹ Pa

A = 320 mm² = (320 mm²)(1 m² / 10⁶ mm²) = 3.2*10⁻⁴ m²

a) P = ?

We can apply the equation

σ = P / A     ⇒    P = σ*A = (271*10⁶ Pa)(3.2*10⁻⁴ m²) = 86720 N

b) L₀ = 131 mm = 0.131 m

We can get ΔL applying the following formula (Hooke's Law):

ΔL = (P*L₀) / (A*E)    ⇒  ΔL = (86720 N*0.131 m) / (3.2*10⁻⁴ m²*119*10⁹ Pa)

⇒  ΔL = 2.9832*10⁻⁴ m = 0.2983 mm

Finally we obtain

L = L₀ + ΔL = 131 mm + 0.2983 mm = 131.2983 mm

3 0
3 years ago
A race car is travelling around a circular track. The velocity of the car is 20 m/s, the radius of the track is 300 m, and the m
Zielflug [23.3K]

Answer:

μ = 0.136

Explanation:

given,

velocity of the car = 20 m/s

radius of the track = 300 m

mass of the car = 2000 kg

centrifugal force

F_c = \dfrac{mv^2}{r}

F_c = \dfrac{2000\times 20^2}{300}

F c = 2666. 67 N

F f= μ N

F f = μ m g

2666.67  =  μ × 2000 × 9.8

μ = 0.136

so, the minimum coefficient of friction between road surface and car tyre is equal to μ = 0.136

5 0
3 years ago
What are the 3 dimensions that used in isometric sketches?
noname [10]

Answer:

The three dimensions shown in an isometric drawing are the height, H, the length, L, and the depth, D

Explanation:

An isometric drawing of an object in presents a pictorial projection of the object in which the three dimension, views of the object's height, length, and depth, are combined in one view such that the dimensions of the isometric projection drawing are accurate and can be measured (by proportion of scale) to draw the different views of the object or by scaling, for actual construction of the object.

5 0
3 years ago
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