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Ratling [72]
2 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]2 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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6 0
1 year ago
Using the Distortion-Energy failure theory: 8. (5 pts) Calculate the hydrostatic and distortional components of the stress 9. (1
WITCHER [35]

Answer:

Detailed solution is given below:

7 0
3 years ago
A slender rod AB, of weight W, is attached to blocks A and B, which move freely in the guides shown. The blocks are connected by
gregori [183]

Answer:

(a) T = W/2(1-tanθ)  (b) 39.81°

Explanation:

(a) The equation for tension (T) can be derived by considering the summation of moment in the clockwise direction. Thus:

Summation of moment in clockwise direction is equivalent to zero. Therefore,

T*l*(sinθ) + W*(l/2)*cosθ - T*l*cosθ = 0

T*l*(cosθ - sinθ) = W*(l/2)*cosθ

T = W*cosθ/2(cosθ - sinθ)

Dividing both the numerator and denominator by cosθ, we have:

T = [W*cosθ/cosθ]/2[(cosθ - sinθ)/cosθ] = W/2(1-tanθ)

(b) If T = 3W, then:

3W = W/2(1-tanθ),

Further simplification and rearrangement lead to:

1 - tanθ = 1/6

tanθ = 1 - (1/6) = 5/6

θ = tan^(-1) 5/6 = 39.81°

8 0
2 years ago
When block C is in position xC = 0.8 m, its speed is 1.5 m/s to the right. Find the velocity of block A at this instant. Note th
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Answer:

The answer is "2 m/s".

Explanation:

The triangle from of the right angle:

\to (x_c-0.8)+(1.5+y_4) +\sqrt{x_c^2 + 1.5^2}= constant

Differentiating the above equation:

\to V_c +V_A+ \frac{X_cV_c}{\sqrt{x_c^2 +1}}=0\\\\\to 1-V_A+ \frac{0.8 \times 1.5}{\sqrt{ 0.8^2+1.5}}=0\\\\

\to V_A=  \frac{1.2}{\sqrt{ 0.64+1.5}}+1\\\\

        = \frac{1.2}{ 1.46}+1\\\\= \frac{1.2+ 1.46}{ 1.46}\\\\ = \frac{2.66}{1.46}\\\\= 1.82 \ \frac{m}{s}\\\\= 2 \ \frac{m}{s}

3 0
2 years ago
Technician A says that a way to prevent galvanic corrosion is to duplicate the original installation method. Technician B says t
sertanlavr [38]

Answer:

Technician A

Explanation:

Galvanic corrosion is not on only one metal alone but caused when two metals are interacting. Thus, Duplicating the original installation method is a better option because re-using a coated bolt doesn't prevent galvanic corrosion because both materials must be coated and not just the bolt and in technician B's case he is coating just the bolt. Thus, technician B's method will not achieve prevention of galvanic corrosion but technician A's method will achieve it.

8 0
2 years ago
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