Write down the understanding of these designations of the materials (a)Ck45, (b)AISI 4140

The acrylic plastic rod is 200 mm long and 15 mm in diameter. If an axial load of 300 N is applied to it, determine the change i

ehidna [41]

Answer:

Change in length = 0.1257 mm

Change in diameter= -0.03771mm

Explanation:

Given

Diameter, d = 15 mm

Length of rod, L = 200mm

F = Force= 300N

d = 0.015m

Ep=2.70 GPa, np=0.4.

First, we have to calculate the normal stress using

σ = F/A where F = Force acting on the Cross-sectional area

A = Area

Area is calculated as πd²/4 where d = 0.015m

A = 22/7 * 0.015²/4

A = 0.000176785714285m²

A = 1.768E-4m²

So, stress. σ = 300N/1.768E-4m²

σ = 1696832.579185520Pa

σ = 1.697MPa

Calculating E(long)

E(long) = σ /Ep

E(long) = 1.697E-3/2.70

E(long) = 0.0006285

At this point, we fan now calculate the change in length of the element;

∆L = E(long) * L

∆L = 0.0006285 * 200mm

∆L = 0.1257mm

Calculating E(lat)

E(lat) = -np * E(long)

E(lat) = -4 * 0.0006285

E(lat) = -0.002514

At this point, we can now calculate the change in diameter of the element;

∆D = E(lat) * D

∆L = -0.002514 * 15mm

∆L = -0.03771mm

8 0

3 years ago

A cylindrical specimen of a metal alloy 48.8 mm long and 9.09 mm in diameter is stressed in tension. A true stress of 327 MPa ca

Gnom [1K]

Answer:

The true stress necessary to plastically elongate the specimen is 406.69MPa

Explanation:

The Explanation or Solving is on the attached document.

5 0

3 years ago

A vehicle of 1 200 kg is moving at a speed of 40 km/h on an incline of 1 in 50. The total constant rolling and wind resistance i

Readme [11.4K]

Answer:11.602 KW

Explanation:

mass of vehicle $\left ( m\right )=1200 kg$

speed=40Km/h

Resistance=600 N

$\eta =80%$

Gear ratio $\left ( G\right )=4:1$

$D_{effective}=500mm$

Net force to overcome by engine is

F=Resistance + sin component of weight

F=600+ $mgsin\theta$

Where $tan\theta =[tex]\frac{1}{50}$

$\theta =1.1457^{\circ}$

$F=600+1200\times 9.81\times sin\left ( 1.1457\right )$

F=600+235.38=835.38 N

power= $F.v=835.38\frac{100}{9}$

Engine Power= $\frac{835.\frac{100}{9}}{\eta }$ =11.602 KW

4 0

4 years ago

Technician A says that 18 gauge AWG wire can carry more current flow that 12 gauge AWG wire. Technician B says that metric wire

denpristay [2]

Answer:

Technician B

Explanation:

Both AWG and metric are sized by cross-sectional area.

Technician A is wrong: 12 gauge wire is larger diameter rated for 20 amps in free air. 18 awg is smaller diameter and typically used for speaker wiring, Class II or low voltage and sub-circuits within appliances.

6 0

4 years ago

A five legged intersection has safety issues. in 2018, the number of crashes reported was 48 and the average 24 hour volume ente

sashaice [31]

Answer:

The crash rate is 22 vehicles per 1 million vehicles

Explanation:

In this question, we are asked to determine the crash rate per million vehicles.

Crash rate is calculated using average crash frequency.

The crash rates are calculated based on the number of crashes per million vehicle miles travelled in a year.

Mathematically;

crash rate = (48 * 1,000,000)/ [(980 + 1560 + 1230 + 900 + 1435)* 365]

= 48,000,000/(6105*365)

= 21.54 approximately 22

4 0

4 years ago