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kondaur [170]
2 years ago
8

A series of experiments is conducted in which a thin plate is subjected to biaxial tension/compression, σ1, σ2 , the plane surfa

ces of the plate being traction-free (i.e. σ 3-O) Unbeknown to the experimenter, the material contains microscopic defects which can be idealized as a sparse distribution of small circular holes through the thickness of the plate. The hoop stress around the circumference of one of these holes when the plate is loaded in uniaxial tension σ is known to be σ 0-o(1-cos 2θ), where the angle θ is measured from the direction of the applied stress. Show graphically the relation that will hold at yield between the stresses ' σ2 applied to the defective plate if the Tresca criterion applies for the undamaged material (Hint: The hoop stress due to biaxial stress can be constructed by superposition. The maximum must occur at either θ = 0 or θ-m2, depending on the relative magnitude of σ, σ2 .)
Engineering
1 answer:
Andrew [12]2 years ago
3 0

Answer:

                                               

Explanation:

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(a) A non-cold-worked 1040 steel cylindrical rod has an initial length of 100 mm and initial diameter of 7.50 mm. is to be defor
serg [7]

Answer:

A) 1040 steel is not a possible candidate for this application

B) 35.94%

Explanation:

Initial length = 100 mm =  0.1 m

Initial diameter ( d ) = 7.5 mm = 0.0075 m

Tensile load ( p ) = 18,000 N

Condition : The 1040 steel must not experience plastic deformation or a diameter reduction of more than 1.5 * 10^-5 m

<u>A) would the 1040 steel be a possible candidate for this application</u>

<em>Yield strength of 1040 steel < stress  ( in order to be a possible candidate )</em>

stress = p / A0 = ( 18000 ) / ( \frac{\pi }{4} ) * 0.0075^2

                      = 18,000 / (4.418 * 10^-5 )   =  407.424 MPa

Yield strength of 1040 steel = 450 MPa

stress = 407.424 MPa

∴ Yield strength ( 450 MPa ) > stress ( 407.424 MPa )  

Therefore 1040 steel is not a possible candidate for this application

<u>B) Determine How much cold work would be required to reduce the diameter of the steel to 6.0 mm</u>

Area1 = ( \frac{\pi }{4} ) ( 0.006 )^2 = 2.83 * 10^-5 m^2

therefore % of cold work done = ( A0 - A1 ) / A0  * 100 = 35.94%

6 0
2 years ago
Question 2 (Multiple Choice Worth 3 points)
ololo11 [35]

Answer:

the car to the right

Explanation:

its in the name the RIGHT of way hope it helps good luck

6 0
3 years ago
Steam enters a two-stage adiabatic turbine at 8 MPa and 5008C. It expands in the first stage to a state of 2 MPa and 3508C. Stea
Nataly [62]

Answer:

1) The exergy of destruction is approximately 456.93 kW

2) The reversible power output is approximately 5456.93 kW

Explanation:

1) The given parameters are;

P₁ = 8 MPa

T₁ = 500°C

From which we have;

s₁ = 6.727 kJ/(kg·K)

h₁ = 3399 kJ/kg

P₂ = 2 MPa

T₂ = 350°C

From which we have;

s₂ = 6.958 kJ/(kg·K)

h₂ = 3138 kJ/kg

P₃ = 2 MPa

T₃ = 500°C

From which we have;

s₃ = 7.434 kJ/(kg·K)

h₃ = 3468 kJ/kg

P₄ = 30 KPa

T₄ = 69.09 C (saturation temperature)

From which we have;

h₄ = h_{f4} + x₄×h_{fg} = 289.229 + 0.97*2335.32 = 2554.49 kJ/kg

s₄ =  s_{f4} + x₄×s_{fg} = 0.94394 + 0.97*6.8235 ≈ 7.563 kJ/(kg·K)

The exergy of destruction, \dot X_{dest}, is given as follows;

\dot X_{dest} = T₀ × \dot S_{gen} = T₀ × \dot m × (s₄ + s₂ - s₁ - s₃)

\dot X_{dest} = T₀ × \dot W×(s₄ + s₂ - s₁ - s₃)/(h₁ + h₃ - h₂ - h₄)

∴ \dot X_{dest} = 298.15 × 5000 × (7.563 + 6.958 - 6.727 - 7.434)/(3399 + 3468 - 3138  - 2554.49) ≈ 456.93 kW

The exergy of destruction ≈ 456.93 kW

2) The reversible power output, \dot W_{rev} = \dot W_{} + \dot X_{dest} ≈ 5000 + 456.93 kW = 5456.93 kW

The reversible power output ≈ 5456.93 kW.

6 0
3 years ago
What are the advantages of studying a scheduled course over a more freeform one?
anyanavicka [17]

Answer:

Taking responsibility for your own learning makes it easier to identify your strengths and weaknesses. Once these have been identified you can work on a learning plan that focuses on the areas that you need most help with, increasing the speed of your learning, and build the skills you have been trying to perfect.

Explanation:

8 0
2 years ago
Read 2 more answers
Steam enters a radiator at 16 psia and 0.97 quality. The steam flows through the radiator, is con- densed, and leaves as liquid
AnnZ [28]

Answer:

5.328Ibm/hr

Explanation:

Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)  

through prior knowledge of two other properties such as pressure and temperature.  

for this case we can define the following equation for mass flow using the first law of thermodynamics

m=\frac{Q}{h1-h2}

where

Q=capacity of the radiator =5000btu/hr

m = mass flow

then using thermodynamic tables we found entalpy in state 1 and 2

h1(x=0.97, p=16psia)=1123btu/lbm

h2(x=0, p=16psia)=184.5btu/lbm

solving

Q=\frac{5000}{1123-184.5} =5.328Ibm/hr

3 0
3 years ago
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