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worty [1.4K]
3 years ago
7

At a point on the free surface of a stressed body, the normal stresses are 20 ksi (T) on a vertical plane and 30 ksi (C) on a ho

rizontal plane. An unknown negative shear stress exists on the vertical plane. The absolute maximum shear stress at the point has a magnitude of 32 ksi. Determine the principal stresses and the shear stress on the vertical plane at the point.
Engineering
1 answer:
victus00 [196]3 years ago
4 0

Answer:

The principal stresses are σp1 = 27 ksi, σp2 = -37 ksi and the shear stress is zero

Explanation:

The expression for the maximum shear stress is given:

\tau _{M} =\sqrt{(\frac{\sigma _{x}^{2}-\sigma _{y}^{2}  }{2})^{2}+\tau _{xy}^{2}    }

Where

σx = stress in vertical plane = 20 ksi

σy = stress in horizontal plane = -30 ksi

τM = 32 ksi

Replacing:

32=\sqrt{(\frac{20-(-30)}{2} )^{2} +\tau _{xy}^{2}  }

Solving for τxy:

τxy = ±19.98 ksi

The principal stress is:

\sigma _{x}+\sigma _{y} =\sigma _{p1}+\sigma _{p2}

Where

σp1 = 20 ksi

σp2 = -30 ksi

\sigma _{p1}  +\sigma _{p2}=-10 ksi (equation 1)

\tau _{M} =\frac{\sigma _{p1}-\sigma _{p2}}{2} \\\sigma _{p1}-\sigma _{p2}=2\tau _{M}\\\sigma _{p1}-\sigma _{p2}=32*2=64ksi equation 2

Solving both equations:

σp1 = 27 ksi

σp2 = -37 ksi

The shear stress on the vertical plane is zero

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The four strokes in a four stroke cycle engine in proper order.
Morgarella [4.7K]

Answer:

A four-stroke cycle engine is an internal combustion engine that utilizes four distinct piston strokes (intake, compression, power, and exhaust) to complete one operating cycle. The piston make two complete passes in the cylinder to complete one operating cycle.

Explanation:

6 0
2 years ago
(a) A non-cold-worked 1040 steel cylindrical rod has an initial length of 100 mm and initial diameter of 7.50 mm. is to be defor
serg [7]

Answer:

A) 1040 steel is not a possible candidate for this application

B) 35.94%

Explanation:

Initial length = 100 mm =  0.1 m

Initial diameter ( d ) = 7.5 mm = 0.0075 m

Tensile load ( p ) = 18,000 N

Condition : The 1040 steel must not experience plastic deformation or a diameter reduction of more than 1.5 * 10^-5 m

<u>A) would the 1040 steel be a possible candidate for this application</u>

<em>Yield strength of 1040 steel < stress  ( in order to be a possible candidate )</em>

stress = p / A0 = ( 18000 ) / ( \frac{\pi }{4} ) * 0.0075^2

                      = 18,000 / (4.418 * 10^-5 )   =  407.424 MPa

Yield strength of 1040 steel = 450 MPa

stress = 407.424 MPa

∴ Yield strength ( 450 MPa ) > stress ( 407.424 MPa )  

Therefore 1040 steel is not a possible candidate for this application

<u>B) Determine How much cold work would be required to reduce the diameter of the steel to 6.0 mm</u>

Area1 = ( \frac{\pi }{4} ) ( 0.006 )^2 = 2.83 * 10^-5 m^2

therefore % of cold work done = ( A0 - A1 ) / A0  * 100 = 35.94%

6 0
2 years ago
One kilogram of water fills a 150-L rigid container at an initial pressure of 2 MPa. The container is then cooled to 40∘C. Deter
lukranit [14]

The pressure of water is 7.3851 kPa

<u>Explanation:</u>

Given data,

V = 150×10^{-3} m^{3}

m = 1 Kg

P_{1} = 2 MPa

T_{2}  = 40°C

The waters specific volume is calculated:

v_{1} = V/m

Here, the waters specific volume at initial condition is v_{1}, the containers volume is V, waters mass is m.

v_{1} = 150×10^{-3} m^{3}/1

v_{1} = 0.15 m^{3}/ Kg

The temperature from super heated water tables used in interpolation method between the lower and upper limit for the specific volume corresponds 0.15 m^{3}/ Kg and 0.13 m^{3}/ Kg.

T_{1}= 350+(400-350) \frac{0.15-0.13}{0.1522-0.1386}

T_{1} = 395.17°C

Hence, the initial temperature is 395.17°C.

The volume is constant in the rigid container.

v_{2} = v_{1}= 0.15 m^{3}/ Kg

In saturated water labels for T_{2}  = 40°C.

v_{f} = 0.001008 m^{3}/ Kg

v_{g} = 19.515 m^{3}/ Kg

The final state is two phase region v_{f} < v_{2} < v_{g}.

In saturated water labels for T_{2}  = 40°C.

P_{2} = P_{Sat} = 7.3851 kPa

P_{2} = 7.3851 kPa

7 0
2 years ago
A labor-intensive process to manufacture a product has a fixed cost of $338,000 and a variable cost of $143 per unit. An automat
ozzi

Answer:

no of unit is 17941

Explanation:

given data

fixed cost = $338,000

variable cost = $143 per unit

fixed cost = $1,244,000  

variable cost = $92.50 per unit

solution

we consider here no of unit is = n

so here total cost of labor will be sum of fix and variable cost i.e

total cost of labor = $33800 + $143 n  ..........1

and

total cost of capital intensive  = $1,244,000 + $92.5 n   ..........2

so here in both we prefer cost of capital if cost of capital intensive less than cost of labor

$1,244,000 + $92.5 n  <  $33800 + $143 n

solve we get

n > \frac{906000}{50.5}

n > 17941

and

cost of producing less than selling cost so here

$1,244,000 + $92.5 n < 197 n

solve it we get

n > \frac{1244000}{104.5}  

n > 11904

so in both we get greatest no is 17941

so no of unit is 17941

3 0
2 years ago
Given a force of 72 lbs at a distance of 15 ft, calculate the moment produced.​
Elis [28]

Answer:

1425.78 N.m

Explanation:

Moments of force is calculated as ;

Moments= Force * distance

M= F*d

The S.I unit for moment of force is Newton-meter (N.m)

Given in the question;

Force = 72 lbs

1 pound = 4.45 N

72 lbs = 4.45 * 72=320.4 N

Distance= 15 ft

1ft= 0.3048 m

15 ft = 15*0.3048 = 4.57 m

d= 4.57 m

M= F*d

M=320.4*4.57 =1425.78 N.m

5 0
3 years ago
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