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kakasveta [241]

3 years ago

13

Electrical strain gauges were applied to a notched specimen to determine the stresses in the notch. The results were εx=0.0019 a

nd εy=0.00072. Find σx and σy if the material is made of carbon steel where E=200GPa, Sy=360 MPa.

1 answer:

pickupchik [31]3 years ago

6 0

Answer:

$\sigma_x=451.8MPa$

$\sigma_y=265.99MPa$

Explanation:

$\varepsilon _x=0.0019$

$\varepsilon _y=0.00072$

Poission's ratio for steel =0.28

We know that

$\varepsilon _x=\dfrac{\sigma _x}{E}-\mu\dfrac{\sigma _x}{E}$

$\varepsilon _y=\dfrac{\sigma _y}{E}-\mu\dfrac{\sigma _x}{E}$

Now by putting the values

$0.0019=\dfrac{\sigma _x}{200\times 1000}-0.28\dfrac{\sigma _y}{200\times 1000}$

$0.00072=\dfrac{\sigma _y}{200\times 1000}-0.28\dfrac{\sigma _x}{200\times 1000}$

So $\sigma_x=451.8MPa$

$\sigma_y=265.99MPa$

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3 years ago

The roof of a refrigerated truck compartment consists of a layer of foamed urethane insulation (t2 = 21 mm, ki = 0.026 W/m K) be

lakkis [162]

Answer:

Tso = 28.15°C

Explanation:

given data

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t1 = 9 mm

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solution

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Therefore, the equation of a single straight vessel is given by

$F_{f_1}=\frac{8flQ_1^2}{\pi^2gd_1^5}$ ......................(i)

So there are 100 similar parallel pipes of the same cross section. Therefore, the equation for the area is

$\frac{\pi d_1^2}{4}=1000 \times\frac{\pi d_2^2}{4}$

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Now for parallel pipes

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Solving the equations (i) and (ii),

$\frac{H_{f_1}}{H_{f_2}}=\frac{\frac{8flQ_1^2}{\pi^2 gd_1^5}}{\frac{8flQ_2^2}{\pi^2 gd_2^5}}$

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$=\frac{10^6}{10^7}$

Therefore,

$\frac{H_{f_1}}{H_{f_2}}=\frac{1}{10}$

or $H_{f_2}=10 \ H_{f_1}$

Thus the answer is option A). 10

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