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Grace [21]
3 years ago
7

A tension test was conducted on a specimen of AISI 1020 hot rolled steel having an initial diameter of 9.11 mm. The load at the

0.2% plastic strain offset was 17.21 kN, the highest load reached was 25.75 kN and the load at fracture was 17.39 kN. After fracture, the broken halves were assembled and the following measurements were made: (1) Marks on opposite sides of the necked region that were originally 25 mm apart had stretched to 38 mm apart. Similar marks originally 50 mm apart had stretched to 68.5 mm apart. (3) The final diameter was 5.28 mm. Determine the following properties from this test: yield stress, ultimate strength, percent elongation and percent reduction in area. Estimate the toughness.
Engineering
1 answer:
Ad libitum [116K]3 years ago
5 0

Answer:

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Explanation:

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At what times should you use your headlights?
SVETLANKA909090 [29]

Answer:

Headlights are required to be used 1/2 hour after sunset to 1/2 hour before sunrise, when windshield wipers are being used, when visibility is less than 1000 feet, or when there is insufficient light or adverse weather.

Explanation:

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8 0
3 years ago
Read 2 more answers
A cylindrical specimen of steel has an original diameter of 12.8 mm. It is tested in tension its engineering fracture strength i
Mama L [17]

Answer:

a) The ductility = -30.12%

the negative sign means reduction

Therefore, there is 30.12% reduction

b) the true stress at fracture is 658.26 Mpa

Explanation:

Given that;

Original diameter d_{o} = 12.8 mm

Final diameter d_{f} = 10.7

Engineering stress  \alpha _{E} = 460 Mpa

a) determine The ductility in terms of percent reduction in area;

Ai = π/4(d_{o} )²  ; Ag = π/4(d_{f} )²

% = π/4 [ ( (d_{f} )² - (d_{o} )²) / ( π/4  (d_{o} )²) ]

= ( (d_{f} )² - (d_{o} )²) / (d_{o} )² × 100

we substitute

= [( (10.7)² - (12.8)²) / (12.8)² ] × 100

= [(114.49 - 163.84) / 163.84 ] × 100

= - 0.3012 × 100

= -30.12%

the negative sign means reduction

Therefore, there is 30.12% reduction

b) The true stress at fracture;

True stress  \alpha _{T} = \alpha _{E} ( 1 +  E_{E} )

E_{E}  is engineering strain

E_{E}  = dL / Lo

= (do² - df²) / df² = (12.8² - 10.7²) / 10.7² = (163.84 - 114.49) / 114.49

= 49.35 / 114.49  

E_{E} = 0.431

so we substitute the value of E_{E}  into our initial equation;

True stress  \alpha _{T} = 460 ( 1 +  0.431)

True stress  \alpha _{T} = 460 (1.431)

True stress  \alpha _{T} = 658.26 Mpa

Therefore, the true stress at fracture is 658.26 Mpa

6 0
2 years ago
Write short notes on: (any four) a) Suspended ground floor b) Soil exploration c) Baulking of sand d) Bearing capacity of soil e
vredina [299]

Answer:

a) A suspended floor is a ground floor with a void underneath the structure. The floor can be formed in various ways, using timber joists, precast concrete panels, block and beam system or cast in-situ with reinforced concrete. However, the floor structure is supported by external and internal walls.

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Explanation:

<h2>please follow me</h2>
6 0
3 years ago
What is soil texture in civil engineering​
Bond [772]

Answer:

It will indicate the percentage of sand, silt, and clay present in the soil.

Explanation:

4 0
2 years ago
Develop a chase plan that matches the forecast and compute the total cost of your plan. (Negative amounts should be indicated by
vladimir2022 [97]

Answer:

The answer for the question :

"Develop a chase plan that matches the forecast and compute the total cost of your plan. (Negative amounts should be indicated by a minus sign. Leave no cells blank - be certain to enter "0" wherever required. Omit the "$" sign in your response.)"

is explained in the attachment.

Explanation:

4 0
2 years ago
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