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3 years ago

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The assembly consists of two red brass C83400 copper alloy rods AB and CD of diameter 30 mm, a stainless 304 steel alloy rod EF

of diameter 40 mm, and a rigid cap G. If the supports at A, C and F are rigid, determine the average normal stress developed in rods AB, CD, and EF.

1 answer:

dezoksy [38]3 years ago

6 0

Answer:

Normal Stress = 26.5MPa and 33.8MPa

Explanation:

See explanation attached

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An alloy is evaluated for potential creep deformation in a short-term laboratory experiment. The creep rate is found to be 1% pe

LenaWriter [7]

Answer:

a) Q = 251.758 kJ/mol

b) creep rate is $= 1.751 \times 10^{-5} \% per hr$

Explanation:

we know Arrhenius expression is given as

$\dot \epsilon =Ce^{\frac{-Q}{RT}$

where

Q is activation energy

C is pre- exponential constant

At 700 degree C creep rate is $\dot \epsilon = 5.5\times 10^{-2}$ % per hr

At 800 degree C creep rate is $\dot \epsilon = 1$ % per hr

activation energy for creep is $\frac{\epsilon_{800}}{\epsilon_{700}}$ = $= \frac{C\times e^{\frac{-Q}{R(800+273)}}}{C\times e^{\frac{-Q}{R(700+273)}}}$

$\frac{1\%}{5.5 \times 10^{-2}\%} = e^{[\frac{-Q}{R(800+273)}] -[\frac{-Q}{R(800+273)}]}$

$\frac{0.01}{5.5\times 10^{-4}} = ln [e^{\frac{Q}{8.314}[\frac{1}{1073} - \frac{1}{973}]}]$

solving for Q we get

Q = 251.758 kJ/mol

b) creep rate at 500 degree C

we know

$C = \epsilon e^{\frac{Q}{RT}}$

$=- 1\% e{\frac{251758}{8.314(500+273}} = 1.804 \times 10^{12} \% per hr$

$\epsilon_{500} = C e^{\frac{Q}{RT}}$

$= 1.804 \times 10^{12} e{\frac{251758}{8.314(500+273}}$

$= 1.751 \times 10^{-5} \% per hr$

4 0

2 years ago

Sun of first 1 nayural numbers

marin [14]

Answer:

the answer is

$n(n + 1) \div 2$

n(n+1)/2

4 0

2 years ago

Read 2 more answers

A power plant operates on a regenerative vapor power cycle with one open feedwater heater. Steam enters the first turbine stage

faltersainse [42]

Answer:

a) 0.489

b) 54.42 kg/s

c) 247.36 kW/s

Explanation:

Note that all the initial enthalpy and entropy values were gotten from the tables.

See the attachment for calculations

4 0

3 years ago

A traffic flow has density 61 veh/km when the speed is 59 veh/hr. If a flow has a jam density of 122 veh/km, what is the maximum

antoniya [11.8K]

Since this traffic flow has a jam density of 122 veh/km, the maximum flow is equal to 3,599 veh/hr.

<u>Given the following data:</u>

Density = 61 veh/km.

Speed = 59 km/hr.

Jam density = 122 veh/km.

<h3>How to calculate the maximum flow.</h3>

According to Greenshield Model, maximum flow is given by this formula:

$q_{max}=\frac{V_f \times K_i}{4}$

<u>Where:</u>

$V_f$ is the free flow speed.

$K_i$ is the Jam density.

In order to calculate the free flow speed, we would use this formula:

$V_f =2 V\\\\V_f =2\times 59\\\\V_f=118\;km/hr$

Substituting the parameters into the model, we have:

$q_{max}=\frac{118 \times 122}{4}\\\\q_{max}=\frac{14396}{4}$

Max flow = 3,599 veh/hr.

Read more on traffic flow here: brainly.com/question/15236911

6 0

1 year ago

A specimen of commercially pure copper has a strength of 240 MPa. Estimate its average grain diameter using the Hall-Petch equat

romanna [79]

Answer:

3.115× $10^{-3}$ meter

Explanation:

hall-petch constant for copper is given by

$S_0$ =25 MPa

k=0.12 for copper

now according to hall-petch equation

$S_Y$ = $S_0$ + $\frac{K}{\sqrt{D}}$

240=25+ $\frac{0.12}{\sqrt{D}}$

D=3.115× $10^{-3}$ meter

so the grain diameter using the hall-petch equation=3.115× $10^{-3}$ meter

5 0

2 years ago