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Tom [10]
3 years ago
10

A three-point flexure test is conducted on a cylindrical specimen of aluminum oxide. The specimen radius is 5.0 mm and the dista

nce between the two supportsis 15.0 mm. If the specimen breaks when the applied force is 7500 N and the center deflection is 0.01 mm, what are the elastic modulus andthe flexural strength of the material?

Engineering
1 answer:
kondaur [170]3 years ago
7 0

Answer:

Detailed solution is given in the attached diagram

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The beam is supported by a pin at A and a roller at B which has negligible weight and a radius of 15 mm. If the coefficient of s
Anettt [7]

Answer:

33.4

Explanation:

Step 1:

\sumMo=0 (moment about the origin)

Fb(15)-Fc(15)=0

Fb=Fc

Step 2:

\sumFx=0

-Fb-Fccos\theta+Ncsin\theta=0

Fc=0.3Nc=Fb

-0.3Nc-0.3Nccos\theta+Ncsin\theta=0

(-0.3-cos\theta+sin\theta)Nc=0----(1)

\sumFy=0

Nccos\theta+Fcsin\theta-Nb=0

Nccos\theta+0.3Ncsin\theta-Nc=0

Nc[cos\theta+0.3sin\theta-1]=0--------(2)

Solving eq (1) and eq (2)

\theta=33.4

Step 3:

As the roller is a two force member

2(90-\phi)+\theta=180

\phi=\theta/2

\phi=Tan(\muN/N)-1

\phi=16.7

\theta=2x16.7=33.4

5 0
2 years ago
name the process by which mild steel can be converted into high carbon steel and explain it briefly ?​
Fudgin [204]

Answer:

please give me brainlist and follow

Explanation:

Mild steel can be converted into high carbons steel by which of the following heat treatment process? Explanation: Case hardening, also referred as carburizing increases carbon content of steel, thus, imparting hardness to steel.

6 0
2 years ago
A cross beam in a highway bridge experiences a stress of 14 ksi due to the dead weight of the bridge structure. When a fully loa
zlopas [31]

Answer:

a) 2.452

b) 1.256

Explanation:

Stress due to dead weight. = 14 Ksi

Stress due to fully loaded tractor-trailer = 45Ksi

ultimate tensile strength of beam = 76 Ksi

yield strength = 50 Ksi

endurance limit = 38 Ksi

Determine the safety factor for an infinite fatigue life

a) If mean stress on fatigue strength is ignored

β = ( 45 - 14 ) / 2

  = 15.5 Ksi

hence FOS ( factor of safety ) = endurance limit / β

                                                 = 38 / 15.5 = 2.452

b) When mean stress on fatigue strength is considered

β2 = 45 + 14 / 2

    = 29.5 Ksi

Ratio  = β / β2 = 15.5 / 29.5 = 0.5254

Next step: applying Goodman method

Sa =  [ ( 0.5254 * 38 *76 ) / ( 0.5254*76 + 38 ) ]

     = 19.47 Ksi

hence the FOS ( factor of safety ) = Sa / β

                                                      = 19.47 / 15.5 = 1.256

8 0
2 years ago
A 360 kg/min stream of steam enters a turbine at 40 bar pressure and 100 degrees of superheat. The steam exits the turbine as a
Brrunno [24]
Answer:
skskkdkdkfkgkgkgkkgkgkgigooigigi lol
Explanation:
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7 0
2 years ago
3–102 One of the common procedures in fitness programs is to determine the fat-to-muscle ratio of the body. This is based on the
gayaneshka [121]

Answer:

x_fat = [ 0.5*(Wsa + Wsw) -  p_muscle*V ] / V*( p_fat - p_muscle )

Explanation:

Given:

- The total volume of body = V

- The average density of the body = p_avg

- The density of muscle = p_muscle

- The density of fat = p_fat

Find:

Obtain a relation for the volume fraction of body fat x_fat

Solution:

- The volume of the fat is given by:

                          V_fat = x_fat*V

- The volume of the muscle is given by:

                          V_muscle = V - V_fat

                                            = V - x_fat*V

                                            =V*( 1 - x_fat )

- We will use the conservation of mass for the body related as:

                         mass_fat + mass_muscle = Total average mass

                         p_fat*V_fat + p_muscle*V_muscle = p_avg*V

                         p_fat*x_fat*V + p_muscle*V*( 1 - x_fat ) = p_avg*V

                         p_fat*x_fat + p_muscle*( 1 - x_fat ) = p_avg

- To determine p_1 we weigh the body in air:

                         Weight reading (Wsa) = m = p_1*V

                         p_1 = Wsa / V*g

- To determine p_2 we weigh the body in water:

                         Weight reading (Wsw) = m - p_w*V= p_1*V - p_w*V

                         Weight reading (Wsw) = V*(p_1 - p_w) = V*(p_2)

                         Where, p_2 = p_1 - p_water

                         p_2 = Wsw / V

- The average density p_avg:

                         p_avg = 0.5*(p_1 + p_2)  

                         p_avg = 0.5*(Wsa / V + Wsw / V)  

                         p_avg = 0.5*(Wsa + Wsw) / V                      

- Plug in the mass equation:

                         p_fat*x_fat + p_muscle*( 1 - x_fat ) = 0.5*(Wsa + Wsw) / V

                         x_fat*( p_fat - p_muscle ) = 0.5*(Wsa + Wsw) / V - p_muscle

                   x_fat = [ 0.5*(Wsa + Wsw) -  p_muscle*V ] / V*( p_fat - p_muscle )

                         

6 0
3 years ago
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