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NNADVOKAT [17]
3 years ago
12

A 0.25" diameter A36 steel rivet connects two 1" wide by .25" thick 6061-T6 Al strips in a single lap shear joint. The shear str

ength of steel is 0.58oy. What is the maximum load before yielding? A. 2313 psi B. 1025 psi C. 3207 psi D. 1654 psi
Engineering
1 answer:
just olya [345]3 years ago
3 0

Answer:

Option B

1025 psi

Explanation:

In a single shear, the shear area is \frac {\pi d^{2}}{4}=\frac {\pi 0.25^{2}}{4}

The shear strength=0.58\sigma_y and in this case \sigma_y=36 000 psi

Shear strength=\frac {Load}{Shear area} hence making load the subject then

Load=Shear area X Shear strength

Load=\frac {\pi 0.25^{2}}{4} \times 0.58\times 36000\approx 1025 psi

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You want to plate a steel part having a surface area of 160 with a 0.002--thick layer of lead. The atomic mass of lead is 207.19
Pepsi [2]

Answer:

<u><em>To answer this question we assumed that the area units and the thickness units are given in inches.</em></u>

The number of atoms of lead required is 1.73x10²³.    

Explanation:

To find the number of atoms of lead we need to find first the volume of the plate:

V = A*t

<u>Where</u>:

A: is the surface area = 160

t: is the thickness = 0.002

<u><em>Assuming that the units given above are in inches we proceed to calculate the volume: </em></u>

V = A*t = 160 in^{2}*0.002 in = 0.32 in^{3}*(\frac{2.54 cm}{1 in})^{3} = 5.24 cm^{3}    

Now, using the density we can find the mass:

m = d*V = 11.36 g/cm^{3}*5.24 cm^{3} = 59.5 g

Finally, with the Avogadros number (N_{A}) and with the atomic mass (A) we can find the number of atoms (N):

N = \frac{m*N_{A}}{A} = \frac{59.5 g*6.022 \cdot 10^{23} atoms/mol}{207.19 g/mol} = 1.73 \cdot 10^{23} atoms    

Hence, the number of atoms of lead required is 1.73x10²³.

I hope it helps you!

3 0
3 years ago
Marks
Lina20 [59]

Answer:

i think C

Explanation:

if not really sorry

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What is the term RF exiciter?
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