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4 years ago

Which statement best describes a hybrid design?

Engineering

2 answers:

mario62 [17]4 years ago

4 0

Answer:

B. Hybrid

Explanation:

A hybrid desing implicate combining two different elements/methods, such a hybrid car, which works using both mechanic and electric energy.

mash [69]4 years ago

3 0

<u>Answer:</u>

Hybrid design - (B) It is a design based on multiple methodologies.

<u>Explanation:</u>

Hybrid simply means more than one. Certain times one thing can't be able to sustain the whole process for that multiple ideas are taken together and packed into one and a new design is made called hybrid design which includes properties of more than one designs.

It is simple as a child have abilities of parents as well as grandparents inherited. It is said as hybrid because it is a combination of more than one design. It helps to carry forward over what is already there and take it to the needed next level and helps to get desired successful results."

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3 years ago

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4 years ago

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natta225 [31]

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3 years ago

A square plate of titanium is 12cm along the top, 12cm on the right side, and 5mm thick. A normal tensile force of 15kN is appli

lukranit [14]

Answer:

For the Top Side

- Strain ε = 0.00021739

- Elongation is 0.00260868 cm

For The Right side

- Strain ε = 0.00021739

-Elongation is 0.00347826 cm

Explanation:

Given the data in the question;

Length of the squared titanium plate = 12 cm by 12 cm = 0.12 m by 0.12 m

Thickness = 5 mm = 0.005 m

Force to the Top F $_t$ = 15 kN = 15000 Newton

Force to the right F $_r$ = 20 kN = 20000 Newton

elastic modulus, E = 115 GPa = 115 × 10⁹ pascal

Now, For the Top Side;

- Strain = σ/E = F $_t$ / AE

we substitute

= 15000 / ( 0.12 × 0.005 × (115 × 10⁹) )

= 15000 / 69000000

Strain ε = 0.00021739

- Elongation

Δl = ε × l

we substitute

Δl = 0.00021739 × 12 cm

Δl = 0.00260868 cm

Hence, Elongation is 0.00260868 cm

For The Right side

- Strain = σ/E = F $_r$ / AE

we substitute

Strain = 20000 / ( 0.12 × 0.005 × (115 × 10⁹) )

= 20000 / 69000000

Strain ε = 0.000289855

- Elongation

Δl = ε × l

we substitute

Δl = 0.000289855× 12 cm

Δl = 0.00347826 cm

Hence, Elongation is 0.00347826 cm

5 0

3 years ago

Three identical fatigue specimens (denoted A, B, and C) are fabricated from a nonferrous alloy. Each is subjected to one of the

Law Incorporation [45]

Answer:

B A and C

Explanation:

Given:

Specimen σ $_{max}$ σ $_{min}$

A +450 -150

B +300 -300

C +500 -200

Solution:

Compute the mean stress

σ $_{m}$ = (σ $_{max}$ + σ $_{min}$ )/2

σ $_{mA}$ = (450 + (-150)) / 2

= (450 - 150) / 2

= 300/2

σ $_{mA}$ = 150 MPa

σ $_{mB}$ = (300 + (-300))/2

= (300 - 300) / 2

= 0/2

σ $_{mB}$ = 0 MPa

σ $_{mC}$ = (500 + (-200))/2

= (500 - 200) / 2

= 300/2

σ $_{mC}$ = 150 MPa

Compute stress amplitude:

σ $_{a}$ = (σ $_{max}$ - σ $_{min}$ )/2

σ $_{aA}$ = (450 - (-150)) / 2

= (450 + 150) / 2

= 600/2

σ $_{aA}$ = 300 MPa

σ $_{aB}$ = (300- (-300)) / 2

= (300 + 300) / 2

= 600/2

σ $_{aB}$ = 300 MPa

σ $_{aC}$ = (500 - (-200))/2

= (500 + 200) / 2

= 700 / 2

σ $_{aC}$ = 350 MPa

From the above results it is concluded that the longest fatigue lifetime is of specimen B because it has the minimum mean stress.

Next, the specimen A has the fatigue lifetime which is shorter than B but longer than specimen C.

In the last comes specimen C which has the shortest fatigue lifetime because it has the higher mean stress and highest stress amplitude.

7 0

4 years ago