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3 years ago

Chaplets are used to support a sand core inside a sand mold cavity. The design of the

Engineering

1 answer:

baherus [9]3 years ago

3 0

Answer:

I think the answer is option 1.54

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A vacuum gage connected to a chamber reads 35 kPa at a location where the atmospheric pressure is 92 kPa. The absolute pressure

Elodia [21]

Answer:

Absolute pressure= 57 KPa

Explanation:

Given that

Vacuum gauge pressure = 35 KPa

Atmospheric pressure = 92 KPa

We know that

Absolute pressure=Atmospheric pressure + gauge pressure

But we should remember that Vacuum gauge pressure is also called negative gauge pressure.So when given that pressure is vacuum gauge then subtract gauge pressure from atmospheric pressure instead of addition.

So now by putting the values

Absolute pressure=Atmospheric pressure - Vacuum gauge pressure

Absolute pressure=92 - 35 KPa

Absolute pressure= 57 KPa

7 0

3 years ago

What is the stress concentration factor of a shaft in torsion, where D=1.25 in. and d=1 in. and the fillet radius is, r=0.2 in.a

klio [65]

Answer:

Concentration factor will be 1.2

So option (C) will be correct answer

Explanation:

We have given outer diameter D = 1.25 in

And inner diameter d = 1 in and fillet ratio r = 0.2 in

So $\frac{r}{d}$ ratio will be $=\frac{0.2}{1}=0.2$

And $\frac{D}{d}$ ratio will be $=\frac{1.25}{1}=1.25$

Now from the graph in shaft vs torsion the value of concentration factor will be 1.2

So concentration factor will be 1.2

So option (C) will be correct answer.

6 0

3 years ago

A 20cm-long rod with a diameter of 0.250 cm is loaded with a 5000 N weight. If the diameter of the bar is 0.490 at this load, de

Margaret [11]

If the diameter of the bar is 0.490 at this load, determine I. the engineering stress and strain, and [2] II. the true stress and strain is 1561. 84 MPa.

<h3>What is strain?</h3>

Strain is a unitless degree of ways a great deal an item receives larger or smaller from an implemented load. Normal stress happens while the elongation of an item is in reaction to an everyday pressure (i.e. perpendicular to a surface), and is denoted via way of means of the Greek letter epsilon.

L = 20 cm d x 1 = 0.21 cm
dx 2 = 0.25 cmF=5500 a) σ= F/A1= 5000/(π/4x(0.0025)^2)= 1018.5916 MPa lateral stress= Ad/d1= (0.0021-0.0025)/0.0025 = - 0.1 longitudinal stress (ɛ_l)= -lateral stress/v = -(-0.16)/0.3
(assuming a poisson's ration of 0.3) ε_l=0.16/0.3 = 0.5333
b) σ_true= σ(1+ ɛ_I)= 1018.5916(1+0.5333
= 1561.84 MPa.