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lesya [120]
3 years ago
11

The aluminum rod (E1 = 68 GPa) is reinforced with the firmly bonded steel tube (E2 = 201 GPa). The diameter of the aluminum rod

is d = 25 mm and the outside diameter of the steel tube is D= 45 mm. The length of the composite column is L = 761 mm. A force P = 88 kN is applied at the top surface, distributed across both the rod and tube.
Required:
Determine the normal stress σ in the steel tube.
Engineering
1 answer:
Vsevolod [243]3 years ago
8 0

Answer:

Explanation:

From the information given:

E_1 = 68 \ GPa \\ \\ E_2 = 201 \ GPa  \\ \\ d = 25 \ mm \  \\ \\ D = 45 \ mm \ \\ \\ L   = 761 \ mm  \\ \\ P = -88 kN

The total load is distributed across both the rod and tube:

P = P_1+P_2 --- (1)

Since this is a composite column; the elongation of both aluminum rod & steel tube is equal.

\delta_1=\delta_2

\dfrac{P_1L}{A_1E_1}= \dfrac{P_2L}{A_2E_2}

\dfrac{P_1 \times 0.761}{(\dfrac{\pi}{4}\times .0025^2 ) \times 68\times 10^4}= \dfrac{P_2\times 0.761}{(\dfrac{\pi}{4}\times (0.045^2-0.025^2))\times 201 \times 10^9}

P_1(2.27984775\times 10^{-8}) = P_2(3.44326686\times 10^{-9})

P_2 = \dfrac{ (2.27984775\times 10^{-8}) P_1}{(3.44326686\times 10^{-9})}

P_2 = 6.6212 \ P_1

Replace P_2 into equation (1)

P= P_1 + 6.6212 \ P_1\\ \\ P= 7.6212\ P_1 \\ \\  -88 = 7.6212 \ P_1  \\ \\ P_1 = \dfrac{-88}{7.6212} \\ \\  P_1 = -11.547 \ kN

Finally, to determine the normal stress in aluminum rod:

\sigma _1 = \dfrac{P_1}{A_1} \\ \\  \sigma _1 = \dfrac{-11.547 \times 10^3}{\dfrac{\pi}{4} \times 25^2}

\sigma_1 = - 23.523 \ MPa}

Thus, the normal stress = 23.523 MPa in compression.

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A water-filled manometer is used to measure the pressure in an air-filled tank. One leg of the manometer is open to atmosphere.
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Answer:

P = 150.335\,kPa (Option B)

Explanation:

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P = 101.3\,kPa + \left(1000\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{kg}{m^{3}} \right)\cdot (5\,m)\cdot \left(\frac{1\,kPa}{1000\,Pa} \right)

P = 150.335\,kPa

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Dust, dirt, or metal chips can pose a potential ____ injury risk in a shop.
Liono4ka [1.6K]

Answer: Eye injury

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7 0
2 years ago
Which of the following are some of the problems found in city streets?
storchak [24]

Answer: drugs and rushing cars

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4 0
2 years ago
I don’t get this it’s hella hard
qwelly [4]

Answer:

V₂ = 20 V

Vt = 20 V

V₁ = 20 V

V₃ = 20 V

I₁ = 10 mA

I₃ = 3.33 mA

It = 18.33 mA

Rt = 1090.91 Ω

Pt = 0.367 W

P₁ = 0.2 W

P₂ = 0.1 W

P₃ = 0.067 W

Explanation:

Part of the picture is cut off.  I assume there is a voltage source Vt there?

First, use Ohm's law to find V₂.

V = IR

V₂ = (0.005 A) (4000 Ω)

V₂ = 20 V

R₁ and R₃ are in parallel with R₂ and the voltage source Vt.  That means V₁ = V₂ = V₃ = Vt.

V₁ = 20 V

V₃ = 20 V

Vt = 20 V

Now we can use Ohm's law again to find I₁ and I₃.

V = IR

I = V/R

I₁ = (20 V) / (2000 Ω)

I₁ = 0.01 A = 10 mA

I₃ = (20 V) / (6000 Ω)

I₃ = 0.00333 A = 3.33 mA

The current It passing through Vt is the sum of the currents in each branch.

It = I₁ + I₂ + I₃

It = 10 mA + 5 mA + 3.33 mA

It = 18.33 mA

The total resistance is the resistance of the parallel resistors:

1/Rt = 1/R₁ + 1/R₂ + 1/R₃

1/Rt = 1/2000 + 1/4000 + 1/6000

Rt = 1090.91 Ω

Finally, the power is simply each voltage times the corresponding current.

P = IV

Pt = (0.01833 A) (20 V)

Pt = 0.367 W

P₁ = (0.010 A) (20 V)

P₁ = 0.2 W

P₂ = (0.005 A) (20 V)

P₂ = 0.1 W

P₃ = (0.00333 A) (20 V)

P₃ = 0.067 W

7 0
3 years ago
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