A brass alloy rod having a cross sectional area of 100 mm2 and a modulus of 110 GPa is subjected to a tensile load. Plastic deformation was observed to begin at a load of 39872 N. a. Determine the maximum stress that can be applied without plastic deformation. b. If the maximum length to which a specimen may be stretched without causing plastic deformation is 67.21 mm, what is the original specimen length

Answers

Answer 1

Answer:

a) the maximum stress that can be applied without plastic deformation is 398.72 N/mm²  

b) length of the specimen is 66.97 mm

Explanation:

Given the data in the question;

a) Determine the maximum stress that can be applied without plastic deformation

when know that; maximum stress σ[tex]_{max}[/tex]  = F / A

where F is the force in the rod ( 39872 N )

A is the cross-sectional area of the rod ( 100 mm² )

so we substitute;

σ[tex]_{max}[/tex]  = 39872 N / 100 mm²

σ[tex]_{max}[/tex]  = 398.72 N/mm²

Therefore, the maximum stress that can be applied without plastic deformation is 398.72 N/mm²  

b)  

strain in the members can be calculated using the expression

ε = σ / E

where σ is the stress in the rod

E is the module of elasticity (  110 GPa = 110000 N/mm² )

(Sl-L) / L = σ/E

where Sl-L is the change in length of the member

L is the original length of the specimen

so we substitute

(67.21 - L) / L = 398.72 / 110000

110000( 67.21 - L) = 398.72L

7393100 - 110000L = 398.72L

7393100 = 398.72L+ 110000L

7393100 = 110398.72L

L = 7393100 / 110398.72

L = 66.97 mm

Therefore; length of the specimen is 66.97 mm

 


Related Questions


Drum brakes are usually designed so that the condition of the lining can be checked even if the drum has not been
removed.

Answers

Answer:

no it has to be removed

Explanation:

It is completely inappropriate to mention that the drum brakes are usually designed so that the condition of the lining can be checked even if the drum has not been removed. Therefore, the statement given above is false.

What is the significance of drum brakes?

Drum brakes can be referred to or considered as the types of brakes that are useful in application of brakes to an object, such as wheels, in motion. To understand better, it can be stated that the system of braking under drum brakes is completely in contrast to that of disc brakes.

Drum brakes have a hydraulic pressure, which means that if the condition of lining is to be checked, removal of drums becomes essential. If the drums are not removed, correction or alignment of wheels cannot be performed.

Therefore, the significance regarding drum brakes has been aforementioned, and the statement given above with respect to their removal also holds false.

Learn more about drum brakes here:

https://brainly.com/question/14937026

#SPJ2

Please calculate the energy stored in a flying wheel. The steel flying wheel is 10 metric tons, in adiameter of 10m, and it rotates at a speed of 1000 rpm. The steel density is 8050 kg/m^3. When energyis fully released, the flying wheel stops its rotation.

Answers

Answer:

685.38 MJ

Explanation:

Given that:

mass = 10 tons = 1.0 × 10 ⁴ kg

diameter D = 10 m

radius R = 5 m

speed N = 1000 rpm

Using the formula for K.E = [tex]\dfrac{1}{2}I \omega^2[/tex] to calculate the energy stored

where;

[tex]= \dfrac{2 \pi \times N}{60}[/tex]

[tex]= \dfrac{2 \pi \times 1000}{60}[/tex]

= 104.719 rad/s

Hence, the energy stored is;

[tex]= \dfrac{1}{2}\times (\dfrac{MR^2}{2}) \times \omega^2[/tex]

[tex]= \dfrac{1}{2}\times (\dfrac{10^4\times 5^2}{2}) \times 104.719^2[/tex]

= 685379310.1

= 685.38 MJ

why you so mean to me? leave my questions please. answer them

Answers

Answer: Why is even here then.

Explanation:

PWM input and output signals are often converted to analog voltage signals using low-pass filters. Design and simulate the following: a PWM signal source with 1 kHz base frequency and adjustable pulse-width modulation(PWM source), an analog filter with time constant of 0.01 s. Simulate the input and output of the low pass filter for a PWM duty cycle of 25, 50, and 100%

Answers

Answer:

Attached below

Explanation:

PWM signal source has 1 KHz base frequency

Analog filter : with time constant = 0.01 s

low pass transfer function = [tex]\frac{1}{0.01s + 1 }[/tex]

PWM duty cycle is a constant block

Attached below is the design and simulation into Simulink at 25% , 50% and 100% respectively

64A geothermal pump is used to pump brine whose density is 1050 kg/m3at a rate of 0.3 m3/s from a depth of 200 m. For a pump efficiency of 74 percent, determine the required power input to the pump. Disregard frictional losses in the pipes, and assume the geothermal water at 200m depth to be exposed to the atmosphere.

Answers

Answer:

835,175.68W

Explanation:

Calculation to determine the required power input to the pump

First step is to calculate the power needed

Using this formula

P=V*p*g*h

Where,

P represent power

V represent Volume flow rate =0.3 m³/s

p represent brine density=1050 kg/m³

g represent gravity=9.81m/s²

h represent height=200m

Let plug in the formula

P=0.3 m³/s *1050 kg/m³*9.81m/s² *200m

P=618,030 W

Now let calculate the required power input to the pump

Using this formula

Required power input=P/μ

Where,

P represent power=618,030 W

μ represent pump efficiency=74%

Let plug in the formula

Required power input=618,030W/0.74

Required power input=835,175.68W

Therefore the required power input to the pump will be 835,175.68W

A river has an average rate of water flow of 59.6 M3/s. This river has three tributaries, tributary A, B and C, which account for 36%, 47% and 17% of water flow respectively. How much water is discharged in 30 minutes from tributary B?

Answers

Answer:

50421.6 m³

Explanation:

The river has an average rate of water flow of 59.6 m³/s.

Tributary B accounts for 47% of the rate of water flow. Therefore the rate of water flow through tributary B is:

Flow rate of water through tributary B = 47% of 59.6 m³/s = 0.47 * 59.6 m³/s = 28.012 m³/s

The volume of water that has been discharged through tributary B = Flow rate of water through tributary B * time taken

time = 30 minutes = 30 minutes * 60 seconds / minute = 1800 seconds

The volume of water that has been discharged through tributary B in 30 seconds = 28.012 m³/s * 1800 seconds = 50421.6 m³

A horizontal curve of a two-lane undivided highway (12-foot lanes) has a radius of 678 feet to the center line of the roadway. An old building (sight obstruction) is located 30 feet from the edge of the innermost lane. The road is level and the superelevation is 0.06. Please determine the maximum speed for safe vehicle operation on this horizontal curve.

Answers

Answer:

maximum speed for safe vehicle operation = 55mph

Explanation:

Given data :

radius ( R ) = 678 ft

old building located ( m )= 30 ft

super elevation = 0.06

Determine the maximum speed for safe vehicle operation

firstly calculate the stopping sight distance

m = R ( 1 - cos [tex]\frac{28.655*S}{R}[/tex] )  ----  ( 1 )

R = 678  

m ( horizontal sightline ) = 30 ft

back to equation 1

30 = 678 ( 1 - cos (28.655 *s / 678 ) )

( 1 - cos (28.655 *s / 678 ) )  = 30 / 678 = 0.044

cos [tex]\frac{28.65 *s }{678}[/tex]  = 1.044

hence ; 28.65 * s = 678 * 0.2956

s = 6.99 ≈ 7 ft

next we will calculate the design speed ( u ) using the formula below

S = 1.47 ut  + [tex]\frac{u^2}{30(\frac{a}{3.2} )-G1}[/tex]  ----  ( 2 )

t = reaction time,  a = vehicle acceleration, G1 = grade percentage

assuming ; t = 2.5 sec , a = 11.2 ft/sec^2, G1 = 0

back to equation 2

6.99 = 1.47 * u * 2.5 + [tex]\frac{u^2}{30[(11.2/32.2)-0 ]}[/tex]

3.675 u  + 0.0958 u^2 - 6.99 = 0

u ( 3.675 + 0.0958 u ) = 6.99

3/4 + 1/2
Ashskfnrjcisj

Answers

1.25, because change the fractions into decimals and then add it from there so it would be add like, 0.75+0.5=1.25
5/4 or 1 1/4. Hdhdjvdvfjkbxb

Write the heat equation for each of the following cases:

a. A wall, steady state, stationary, one-dimensional, incompressible and no energy generation.
b. A wall, transient, stationary, one-dimensional, incompressible, constant k with energy generation.
c. A cylinder, steady state, stationary, two-dimensional (radial and axial), constant k, incompressible, with no energy generation.
d. A wire moving through a furnace with constant velocity, steady state, one-dimensional (axial), incompressible, constant k and no energy generation.
e. A sphere, transient, stationary, one-dimensional (radial), incompressible, constant k with energy generation.

Answers

Answer:

Explanation:

a) the steady-state, 1-D incompressible and no energy generation equation can be expressed as follows:

[tex]\dfrac{\partial^2T}{\partial x^2}= \ 0 \ ; \ if \ T = f(x) \\ \\ \dfrac{\partial^2T}{\partial y^2}= \ 0 \ ; \ if \ T = f(y) \\ \\ \dfrac{\partial^2T}{\partial z^2}= \ 0 \ ; \ if \ T = f(z)[/tex]

b) For a transient, 1-D, constant with energy generation

suppose T = f(x)

Then; the equation can be expressed as:

[tex]\dfrac{\partial^2T}{\partial x^2} + \dfrac{Q_g}{k} = \dfrac{1}{\alpha} \dfrac{dT}{dC}[/tex]

where;

[tex]Q_g[/tex] = heat generated per unit volume

[tex]\alpha[/tex] = Thermal diffusivity

c) The heat equation for a cylinder steady-state with 2-D constant and no compressible energy generation is:

[tex]\dfrac{1}{r}\times \dfrac{\partial}{\partial r }( r* \dfrac{\partial \ T }{\partial \ r}) + \dfrac{\partial^2 T}{\partial z^2 }= 0[/tex]

where;

The radial directional term = [tex]\dfrac{1}{r}\times \dfrac{\partial}{\partial r }( r* \dfrac{\partial \ T }{\partial \ r})[/tex] and the axial directional term is [tex]\dfrac{\partial^2 T}{\partial z^2 }[/tex]

d) The heat equation for a wire going through a furnace is:

[tex]\dfrac{\partial ^2 T}{\partial z^2} = \dfrac{1}{\alpha}\Big [\dfrac{\partial ^2 T}{\partial ^2 t}+ V_z \dfrac{\partial ^2T}{\partial ^2z} \Big ][/tex]

since;

the steady-state is zero, Then:

[tex]\dfrac{\partial ^2 T}{\partial z^2} = \dfrac{1}{\alpha}\Big [ V_z \dfrac{\partial ^2T}{\partial ^2z} \Big ][/tex]'

e) The heat equation for a sphere that is transient, 1-D, and incompressible with energy generation is:

[tex]\dfrac{1}{r} \times \dfrac{\partial}{\partial r} \Big ( r^2 \times \dfrac{\partial T}{\partial r} \Big ) + \dfrac{Q_q}{K} = \dfrac{1}{\alpha}\times \dfrac{\partial T}{\partial t}[/tex]

A freezer compartment consists of a cubical cavity that is 2 m on a side. Assume the bottom to be perfectly Problems 49 CH001.qxd 2/24/11 12:03 PM Page 49 insulated. What is the minimum thickness of styrofoam insulation (k 0.030 W/m K) that must be applied to the top and side walls to ensure a heat load of less than 500 W, when the inner and outer surfaces are 10 and 35 C

Answers

Answer:

30 mm is the minimum thickness that must be applied.

Explanation:

Given the data in the question;

Using Fourier's equation. the heat rate is  

q = kA(ΔT/Δx)

where

A is the surface area, we must consider all surfaces through which the heat can dissipate through

i.e 2×2 for one wall gives you 4m²,

there are 5 walls, so we will  have 20m² for surface area.

k is thermal conductivity of the styrofoam ( 0.030 W/m K)    

q is the heat loss (500 W  )

ΔT is the Temperature difference ( 35 - 10) = 25°C

Δx  = ?

So we substitute

500 = (0.030)(20)(25/Δx)

500 = 0.6 (25/Δx)

500 = 15 / Δx

Δx = 15 / 500

Δx = 0.03 m = 30 mm

Therefore, 30 mm is the minimum thickness that must be applied.

Determine the voltage drop from the top terminal to the bottom terminal, vab, in the right hand branch and, vcd, in the left hand branch of the circuit. Determine each voltage drop based on the elements in the corresponding branch.

Answers

Answer:

Hello your question is incomplete attached below is the missing part of the question

answer ;

voltage drop in the Vcd branch = 30 V

Voltage drop in the middle branch = 40v - 30v = 10 volts

voltage drop in AB = 60 + ( -600 * 0.05 ) = 60 - 30 = 30 volts

Explanation:

Determine voltage drop from top terminal to bottom terminal ( Vab ) in the right hand branch and Vcd in left hand branch

40v and 50mA are in series hence;  Ix = 50mA

also Vcd = 30V

CD is parallel to AB hence; Vcd = Vab = 30 V

Vab = ∝*Ix + 60 v

 30v  = ∝ ( 50mA ) + 60

therefore ∝ = -600

voltage drop in the Vcd branch = 30 V

Voltage drop in the middle branch = 40v - 30v = 10 volts

voltage drop in AB = 60 + ( -600 * 0.05 ) = 60 - 30 = 30 volts

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