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Marrrta [24]
3 years ago
5

Write the definitions for engineering stress, true stress, engineering strain, and true strain for loading along a single axis.

A commercially pure copper wire originally 10.00 m long is pulled until its final length is 10.10 m. It is annealed, and then pulled again to a final length of 10.20 m. What is the engineering strain associated with each of the two steps in the process? What is the true strain for each step? What are the total engineering and true strains for the combined steps?
Engineering
1 answer:
shusha [124]3 years ago
5 0

Answer and Explanation:

Engineering Stress of a material is defined as the applied load or force divided by the original cross sectional Area of the material.

σ(engineering) = F/(Ao)

True Stress is defined as the applied load or force divided by the actual cross-sectional area (the changing area with respect to time) of the material at that point in time. It's an instantaneous stress.

σ(true) = F/A

Engineering strain is a measure of how much a material deforms under a particular load. It is the amount of deformation in the direction of the applied force divided by the initial length of the material.

ε(engineering) = Δl/lo

True Strain measures instantaneous deformation. It is obtained mathematically by integrating strain over small time periods and summing them up. Hence,

ε(true) = In (lf/lo)

The calculations,

First step, 10m to 10.1m, Δl = 0.1m, lf = 10.1m, lo = 10m

ε(engineering) = 0.1/10 = 0.01

ε(true) = In (10.1/10) = 0.00995

Second step, 10.1m to 10.2m, Δl = 0.1m, lf = 10.2m, lo = 10.1m

ε(engineering) = 0.1/10.1 = 0.0099

ε(true) = In (10.2/10.1) = 0.00985

Overall, 10m to 10.2m, Δl = 0.2m, lf = 10.2m, lo = 10m

ε(engineering) = 0.2/10 = 0.02

ε(true) = In (10.2/10) = 0.0198

QED!

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koban [17]

Answer:

26 lbf

Explanation:

The mass of the satellite is the same regardless of where it is.

The weight however, depends on the acceleration of gravity.

The universal gravitation equation:

g = G * M / d^2

Where

G: universal gravitation constant (6.67*10^-11 m^3/(kg*s))

M: mass of the body causing the gravitational field (mass of Earth = 6*10^24 kg)

d: distance to that body

15000 miles = 24140 km

The distance is to the center of Earth.

Earth radius = 6371 km

Then:

d = 24140 + 6371 = 30511 km

g = 6.67*10^-11 * 6*10^24 / 30511000^2 = 0.43 m/s^2

Then we calculate the weight:

w = m * a

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4 years ago
The base class Pet has attributes name and age. The derived class Dog inherits attributes from the base class Pet class and incl
Nonamiya [84]

Answer:

Explanation:

class Pet:

   def __init__(self):

       self.name = ''

       self.age = 0

   def print_info(self):

       print('Pet Information:')

       print('   Name:', self.name)

       print('   Age:', self.age)

class Dog(Pet):

   def __init__(self):

       Pet.__init__(self)

       self.breed = ''

def main():

   my_pet = Pet()

   my_dog = Dog()

   pet_name = input()

   pet_age = int(input())

   dog_name = input()

   dog_age = int(input())

   dog_breed = input()

   my_pet.name = pet_name

   my_pet.age = pet_age

   my_pet.print_info()

   my_dog.name = dog_name

   my_dog.age = dog_age

   my_dog.breed = dog_breed

   my_dog.print_info()

   print('   Breed:', my_dog.breed)

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Answer:

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In order to find the width of the address bus, we need first to know how many memory cells it is needed to address.

If the size memory is 64 KB, this means that the memory size, in bytes, is equal to the following quantity:

64 KB = 2⁶ * 2¹⁰ bytes = 2¹⁶ bytes.

In order to address this quantity of cell positions, the address bus must be able to address 2¹⁶ bytes, so it must have 16-bit wide.

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