Atensile test was made on a tensile specimen, with a cylindrical gage section which had a diameter of 10 mm, and a length of 40 mm. after fracture, the length of the gage section was found to be 50 mm, the reduction in area 90 percent, and the load at fracture 1000 n. compute: a) b) c) d) the specimen elongation. the engineering fracture stress. the true fracture stress. the true strain at the neck. note: the reduction in area is defined as the initial area, minus the final area, divided by the initial area

Answers

Half of the burn victims die when the burn area exceeds 70 percent of the body

Find the complete solution in the given attachments


A cylindrical metal specimen having an original diameter of 12.8 mm (0.505 in.) and gauge length of
A cylindrical metal specimen having an original diameter of 12.8 mm (0.505 in.) and gauge length of

Beagle = 494,080

Brash = 295,229.7

Earle = 53,093.4

Heroina = 455,416

Platter = 100,809

Step-by-step explanation:

Beagle

38.60% × 1,280,000

= 494,080

Brash

44.13% × 669,000

= 295,229.7

Earle

24.81% × 214,000

= 53,093.4

Heroina

43.79% × 1,040,000

= 455,416

Platter

48.7% × 207,000

= 100,809

hope this helps

a. The percent of change in perimeter is 100% increase

b. The percent of change in Area in 300% increase.

Step-by-step explanation:

The diagram was missing we have attached for your reference.

Given:

Length of the sandbox = 10 ft

Width of the sandbox = 6 ft

We need to find the change in percentage of perimeter and area when the dimensions are doubled.

Solution:

First we will find the Perimeter and area of the sand box with current dimension.

Now we know that;

Perimeter of rectangle is twice the sum of length and width.

Perimeter of rectangle = 2(6+10)=2\times16 =32\ ft

Area of rectangle is length times width.

Perimeter of rectangle = 6\times 10 = 60\ ft^2

Now When the dimension are doubled.

Length of the sand box = 20 ft

width of the sandbox = 12 ft

Perimeter of sandbox when dimensions are doubled = 2(20+12)=2\times 32 =64 \ ft

Area of the sand box when dimension are doubled = 20\times 12= 240 \ ft^2

Now we need to find the Percent of change in Perimeter and Area.

Percent of change can be calculated by new value minus original value divided by original value multiplied by 100.

Percent of change in Perimeter = \frac{64-32}{32} \times 100 = 100\%

Percent of change in area = \frac{240-60}{60}\times100 = \frac{180}{60}\times 100 = 300\%

Hence When the dimension are doubled then the percent of change in perimeter is 100% increase and the percent of change in Area in 300% increase.


Suppose the length and the width of the sandbox are doubled. a. Find the percent of change in the pe

494,080

295,229.7

53,093.4

455,416

100,809

Step-by-step explanation:

Beagle:

38.6/100 × 1280000

494,080

Brash:

44.13/100 × 669000

295,229.7

Earle:

24.81/100 × 214000

53,093.4

Heroina:

43.79/100 × 1040000

455,416

Platter:

48.8/100 × 207000

100,809

%Reduction in area = 73.41%

%Reduction in elongation = 42.20%

Explanation:

Given

Original diameter = 12.8 mm

Gauge length = 50.80mm

Diameter at the point of fracture = 6.60 mm (0.260 in.)

Fractured gauge length = 72.14 mm.

%Reduction in Area is given as:

((do/2)² - (d1/2)²)/(do/2)²

Calculating percent reduction in area

do = 12.8mm, d1 = 6.6mm

So,

%RA = ((12.8/2)² - 6.6/2)²)/(12.8/2)²

%RA = 0.734130859375

%RA = 73.41%

Calculating percent reduction in elongation

%Reduction in elongation is given as:

((do) - (d1))/(d1)

do = 72.14mm, d1 = 50.80mm

So,

%RA = ((72.24) - (50.80))/(50.80)

%RA = 0.422047244094488

%RA = 42.20%

% reduction in area==PR=0.734=73.4%

% elongation=EL=0.42=42%

Explanation:

given do=12.8 mm

df=6.60

Lf=72.4 mm

Lo=50.8 mm

% reduction in area=((\pi*(do/2)^2)-(\pi*(df/2)^2)))/\pi*(do/2)^2

substitute values

% reduction in area=73.4%

% elongation=EL=((Lf-Lo)/Lo))*100

% elongation=((72.4-50. 8)/50.8)*100=42%

B. temperature, acidity, mineral composition and surface area

Independent variables are those which can be changed and manipulated in the experiment the effect of which can be observed on the changes that occur in the dependent variable. For example the effect of quantity of manure and water on the height of the plant. Here, quantity of manure and water are the independent variables and the height is the dependent variable.

The temperature is the independent variable because the level of temperature can be increased or lowered down manually.

The acidity is the independent variable because the acidity can be manipulated by addition of chemicals that exhibit more hydrogen ions.

Mineral composition can be manipulated and it is an independent variable by treating a mineral with a suitable chemical this will bring a change in the chemical composition of the mineral.

Surface area is the independent variable because  it is the total area required for the conduction of the experiment . The surface area can be increased or decreased.

1. Parallel Execution
3. Hope that a company calls you and offers you a position
4. Intranet
5. Percent

The answers are as follow:

a) 10 mm

b) 12.730 N/mm^{2}

c) 127.307 N/mm^{2}

d) 0.25

Explanation:

d1 = 10mm , L1 = 40 mm, L2 = 50 mm, reduction in area = 90% = 0.9

Force = F =1000 N

let us find initial area first, A1 = pi*r^{2} = 78.55 mm^{2}

using reduction in area formula : 0.9 = (A1 - A2 ) / A1

solving it will give,  A2 = 0.1 A1 = 7.855  mm^{2}

a) The specimen elongation is final length - initial length

50 - 40 = 10 mm

b) Engineering stress uses the original area for all stress calculations,

Engineering stress = force / original area  = F / A1 = 1000 / 78.55  

Engineering stress = 12.730 N / mm^{2}

c) True stress uses instantaneous area during stress calculations,

True fracture stress = force / final  area  = F / A2 = 1000 / 7.855

True Fracture stress = 127.30 N / mm^{2}

e) strain = change in length / original length

strain = 10 / 40  = 0.25



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