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Lady_Fox [76]
3 years ago
6

Find the toughness (or energy to cause fracture) for a metal that experiences both elastic and plastic deformation. Assume Equat

ion 6.5 for elastic deformation, that the modulus of elasticity is 172 GPa (25 ´ 106 psi), and that elastic deformation terminates at a strain of 0.01. For plastic deformation, assume that the relationship between stress and strain is described by Equation 6.19, in which the values for K and n are 6900 MPa (1 ´ 106 psi) and 0.30, respectively. Furthermore, plastic deformation occurs between strain values of 0.01 and 0.75, at which point fracture occurs.

Engineering
1 answer:
PtichkaEL [24]3 years ago
7 0

Answer:

Detailed solution is attached below in three simple steps the problem is solved.

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A proposed embankment fill requires 7100 ft of compacted soil. The void ratio of the compacted fill is specified as 0.5. Four bo
podryga [215]

Answer:

1) the volume of solid (Vs) is 4733.33 ft³

2) the total cost to pay Pit A is $58220

3) the total cost to pay to Pit B is $53344.67

4) the total cost to pay to Pit C is $34269.32

5) the total cost to pay to Pit D is $49132.02

Explanation:  

given the data in the question;

Vs = ?, V = 7100 ft³, e = 0.5

1) Calculate the volume of solid (Vs)

Vs = V / ( 1 + e )

we substitute

Vs = 7100 / ( 1 + 0.5 )

Vs = 7100 / 1.5

Vs = 4733.33 ft³

Therefore, the volume of solid (Vs) is 4733.33 ft³

2) Calculate the total cost to pay to Pit A

VpitA / Vcomp = ( 1 + e_{PHA}) / ( 1 + e_{comp )

we substitute

VpitA / 7100 = ( 1 + 1.5) / ( 1 + 0.5)

VpitA / 7100 = 1.3666666

VpitA = 1.3666666 × 7100

VpitA  = 9703.33 ft³

Total cost = volume × unit cost = 9703.33 ft³ × 6 = $58220

Therefore, the total cost to pay Pit A is $58220

3) Calculate the total cost to pay to Pit B = Borrow Pit C: void ratio = 0.81 and the unit cost = $ 4 /ft3

Vpitb / Vcomp = ( 1 + e_{PHB}) / ( 1 + e_{comp )

we substitute

VpitB / 7100 = ( 1 + 0.61) / ( 1 + 0.5)

VpitB / 7100 = 1.0733333

VpitB = 1.0733333 × 7100

VpitB  = 7620.666 ft³

Total cost = volume × unit cost = 7620.666 × 7 = $53344.67

Therefore, the total cost to pay to Pit B is $53344.67

4) Calculate the total cost to pay to Pit C = Borrow Pit D: void ratio = 0.73 and the unit cost = $6 /ft3

VpitC / Vcomp = ( 1 + e_{PHC}) / ( 1 + e_{comp )

we substitute

VpitC / 7100 = ( 1 + 0.81) / ( 1 + 0.5)

VpitC / 7100 = 1.2066666

VpitC = 1.2066666× 7100

VpitC  = 8567.33 ft³

Total cost = volume × unit cost = 8567.33 × 4 = $34269.32

Therefore, the total cost to pay to Pit C is $34269.32

5) Calculate the total cost to pay to Pit D

VpitD / Vcomp = ( 1 + e_{PHD}) / ( 1 + e_{comp )

we substitute

VpitD / 7100 = ( 1 + 0.73) / ( 1 + 0.5)

VpitD / 7100 = 1.1533333

VpitD = 1.1533333 × 7100

VpitD  = 8188.67 ft³

Total cost = volume × unit cost = 8188.67 × 6 = $49132.02

Therefore, the total cost to pay to Pit D is $49132.02

3 0
3 years ago
In the following code, determine the values of the symbols here and there. Write the object code in hexadecimal. (Do not predict
allsm [11]

Answer:

Answer explained below

Explanation:

The value of here is 9

The value of there is hexadecimal value of DECO here, d = 0x39 aaaa (aaaa is the memory address of here )

We have the object code :-

let's take there address is 0x0007

0x0005 BR there :- 0x120020

0x0007 here: .WORD 9

310003 there: DECO here,d - 0x390007

310005 STOP

.END

4 0
3 years ago
Consider a system with two tasks, Task1 and Task2. Task1 has a period of 200 ms, and Task2 has a period of 300 ms. All tasks ini
Murrr4er [49]

<u>Explanation:</u>

Task 1 time period = 200ms, Task 2 time period = 300ms

Task ticked = \frac{1000ms}{200ms}= 5  →  5 times

Task 2 ticked =\frac{1000ms}{300ms} = 3.33 → 3 times

At 600 ms → 200ms 200ms 200ms

                     300ms → \frac{30ms}{60ms}

Largest time period = H.C.M of (200ms, 300ms)

                                 = 600ms

4 0
2 years ago
A 150-lbm astronaut took his bathroom scale (aspring scale) and a beam scale (compares masses) to themoon where the local gravit
Nonamiya [84]

Answer:

a) W = 25.5 lbf

b) W = 150 lbf

Explanation:

Given data:

Mass of astronaut = 150 lbm

local gravity = 5.48 ft/s^2

a) weight on spring scale

it can be calculated by measuring force against local gravitational force which is equal to weight of body

W = mg

W = (150 \times 5.48)\times \frac{1 lbm}{32.32 lbm. ft/s^2} = 25.5 lbf

b) As we know that beam scale calculated mass only therefore no change in mass due to variation in gravity

thus W= 150 lbf

7 0
3 years ago
Given a two-dimensional steady inviscid air flow field with no body forces described by the velocity field given below. Assuming
kolbaska11 [484]

Answer:

the pressure gradient in the x direction = -15.48Pa/m

Explanation:

  • The concept of partial differentiation was used in the determination of the expression for u and v.
  • each is partially differentiated with respect to x and the appropriate substitution was done to get the value of the pressure gradient as shown in the attached file.

4 0
3 years ago
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