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Deffense [45]
3 years ago
9

To become familiar with the general equations of plane strain used for determining in-plane principal strain, maximum in-plane s

hear strain, and average normal strain. The state of strain at a point has components of ϵx=380.0×(10−6), ϵy=−250.0×(10−6), and γxy=180.0×(10−6).a. Equivalent in-plane strains on the oriented element Determine the equivalent in-plane strains on an element rotated counterclockwise at an angle of θ = 60.0 ∘ . Find ϵx',ϵy', γxy' Express your answers, separated by commas, to three significant figures.b. In-plane principal strains on the oriented element Determine the in-plane principal strains on the oriented element. Find ϵ1, ϵ2. Express your answers, separated by a comma, to three significant figures.c. Maximum in-plane shear strain and average normal strain on the oriented element Determine the maximum in-plane shear strain and the average normal strain on the oriented element. Express your answers, separated by a comma, to three significant figures.

Engineering
1 answer:
lukranit [14]3 years ago
3 0

Answer:

a) -1.46 x 10∧-5, 1.445x 10∧-4, -6.355 x 10∧-4

b) 3.926 x 10∧-4, -2.626 x 10∧-4

c) 6.552 x 10∧-4, 6.5 x 10∧-5

Explanation:

a) -1.46 x 10∧-5, 1.445x 10∧-4, -6.355 x 10∧-4

b) 3.926 x 10∧-4, -2.626 x 10∧-4

c) 6.552 x 10∧-4, 6.5 x 10∧-5

The explanation is shown in the attachment. I hope i have been able to help.

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The 10 foot wide circle quarter gate AB is articulated at A. Determine the contact force between the gate and the smooth surface
slamgirl [31]

Answer:

F = 641,771.52 \dfrac{lb-ft}{s^2}

Explanation:

Given that

R=8 ft

Width= 10 ft

We know that hydro statics force given as

  F=ρ g A X

ρ is the density of fluid

A projected area on vertical plane

X is distance of center mass of projected plane from free surface of water.

Here

X=8/2  ⇒X=4 ft

A=8 x 10=80  ft^2

So now putting the values

F=ρ g A X

F=62.4(32.14)(80)(4)

F = 641,771.52 \dfrac{lb-ft}{s^2}

   

4 0
3 years ago
Oil with a density of 800 kg/m3 is pumped from a pressure of 0.6 bar to a pressure of 1.4 bar, and the outlet is 3 m above the i
Naddik [55]

Answer:

23.3808 kW

20.7088 kW

Explanation:

ρ = Density of oil = 800 kg/m³

P₁ = Initial Pressure = 0.6 bar

P₂ = Final Pressure = 1.4 bar

Q = Volumetric flow rate = 0.2 m³/s

A₁ = Area of inlet = 0.06 m²

A₂ = Area of outlet = 0.03 m²

Velocity through inlet = V₁ = Q/A₁ = 0.2/0.06 = 3.33 m/s

Velocity through outlet = V₂ = Q/A₂ = 0.2/0.03 = 6.67 m/s

Height between inlet and outlet = z₂ - z₁ = 3m

Temperature to remains constant and neglecting any heat transfer we use Bernoulli's equation

\frac {P_1}{\rho g}+\frac{V_1^2}{2g}+z_1+h=\frac {P_2}{\rho g}+\frac{V_2^2}{2g}+z_2\\\Rightarrow h=\frac{P_2-P_1}{\rho g}+\frac{V_2^2-V_1^2}{2g}+z_2-z_1\\\Rightarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+\frac{6.67_2^2-3.33^2}{2\times 9.81}+3\\\Rightarrow h=14.896\ m

Work done by pump

W_{p}=\rho gQh\\\Rightarrow W_{p}=800\times 9.81\times 0.2\times 14.896\\\Rightarrow W_{p}=23380.8\ W

∴ Power input to the pump 23.3808 kW

Now neglecting kinetic energy

h=\frac{P_2-P_1}{\rho g}+z_2-z_1\\\Righarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+3\\\Righarrow h=13.19\ m\\

Work done by pump

W_{p}=\rho gQh\\\Rightarrow W_{p}=800\times 9.81\times 0.2\times 13.193\\\Rightarrow W_{p}=20708.8\ W

∴ Power input to the pump 20.7088 kW

6 0
3 years ago
Engineers are designing a cylindrical air tank and are trying to determine the dimensions of the tank. The proposed material for
lana66690 [7]

Answer:

The length of tank is found to be 0.6 m or 600 mm

Explanation:

In order to determine the length, we need to find a volume for the tank.

For this purpose, we use ideal gas equation:

PV  = nRT

n = no. of moles = m/M

Therefore,

PV = (m/M)(RT)

V = (mRT)/(MP)

where,

V = volume of air = volume of container

m = mass of air = 4.64 kg

R = General Gas Constant = 8.314 J/mol.k

T = temperature of air = 10°C + 273 = 283 K

M = molecular mass of air = 0.02897 kg/mol

P = Pressure of Air = 20 MPa = 20 x 10^6 N/m²

V = (4.64 kg)(8.314 J/mol.k)(283 k)/(0.02897 kg/mol)(20 x 10^6 N/m²)

V = 0.01884 m³

Now, the volume of cylindrical tank is given as:

V = 0.01884 m³ = π(Diameter/2)²(Length)

Length = (0.01884 m³)(4)/π(0.2 m)²

<u>Length = 0.6 m = 600 mm</u>

4 0
3 years ago
When measuring a Brake Drum, the Brake Micrometer is set to a Base Drum Diameter of 10 Inches plus four notches, and the dial re
kozerog [31]

Answer:

10.5

Explanation:

7 0
3 years ago
A screw is a simple machine that can be described as an inclined plane wrapped around a cylinder.
Phoenix [80]

Answer:

thanks thanks thanks thanks

Explanation:

7 0
3 years ago
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