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Amiraneli [1.4K]
3 years ago
5

A pair of spur gears with 20 degree pressure angle, full-depth, involute teeth transmits 65 hp. The pinion is mounted on a shaft

connected to a 4 cylinder diesel engine that operates at 1250 rpm in the clockwise direction. The pinion has 26 teeth and a diameteral pitch of 6. The gear has 48 teeth and drives a concrete mixer. The gears are manufactured to an AGMA quality number of A11. The face width of the gears is 3 inches and are machined from solid bar stock (blanks). The gears are enclosed in a commercial gear box.
1. What is the pitch line speed of the gear train in feet per minute?
2. What is the center distance of the gear pair in inches?
3. What is the torque on the pinion in Lbin?
4. What is the torque on the gear in lbin?
5. What is the output horsepower of the gearbox?
6. What is the tangential force acting on the gear teeth (Wt) in pounds?
7. What is the radial force acting on the gear teeth in pounds?
8. What is the normal force acting on the gear teeth in pounds?
9. What is the bending stress for the pinion teeth (St) in PSI ?
Engineering
1 answer:
vfiekz [6]3 years ago
4 0

Answer:

See explaination

Explanation:

a) The pitch circle diameter of pinion in inches is given by

Dp=Np/P

Where

Np= No. of teeth in pinion = 26

P =diametral pitch= 6

Hence

Dp= 26/6 = 4.333 in

Pinion angular speed\omega _{p} =1250 rpm = 130.9 rad/s

Therefore pitch line speed

V=Dp/2\omega _{p} = 4.333/2x130.9

= 283.62 in/s

V= 23.63 ft/s

V=1418 ft/min

b) The pitch circle diameter og gear is given by

Dg= Ng/P= 48/6 = 8 in

The center distance is given by

C=(Dp+Dg)/2

= (4.333+8)/2

C= 6.167 in

c) The torque on the pinion is given by

Tp= P/\omega _{p}

Where

P = transmitted power, =65 hp = 65x550= 35750 lt-lb/s

Tp= 35750/130.9

= 273.1 ft-lb

d) Speed ratio is given as

R=Ng/Np= 48/26 = 1.8461

Hence speed of gear is

\omega _{g}=\frac{\omega _{p}}{R}

= 130.9/1.8461

= 70.9 rad/s

Therefore torque on gear is given as

Tg= P/\omega _{g} = 35750/70.9

= 504.2 ft-lb

e) Assuming transmission eficiency of 100%

Output hp=input hp= 65 hp

f) Tangential force on gear teeth is given by

Fgt= Tg/(Dg/2)

= 504.2x2/8

= 126.05 lb

g) Radial force on ger teeth is given as

Fgr= Fgt tan\phi

Where

\phi is pressure angle = 200

Hence

Fgr= 126.05tan200

= 45.88 lb

h) The normal force on gear teeth is given as

F=Fgt/cos\phi

= 126.05/cos200

= 134.14 lb

You might be interested in
Determine the angular acceleration of the uniform disk if (a) the rotational inertia of the disk is ignored and (b) the inertia
lukranit [14]

Answer:

α = 7.848 rad/s^2  ... Without disk inertia

α = 6.278 rad/s^2  .... With disk inertia

Explanation:

Given:-

- The mass of the disk, M = 5 kg

- The right hanging mass, mb = 4 kg

- The left hanging mass, ma = 6 kg

- The radius of the disk, r = 0.25 m

Find:-

Determine the angular acceleration of the uniform disk without and with considering the inertia of disk

Solution:-

- Assuming the inertia of the disk is negligible. The two masses ( A & B )  are hung over the disk in a pulley system. The disk is supported by a fixed support with hinge at the center of the disk.

- We will make a Free body diagram for each end of the rope/string ties to the masses A and B.

- The tension in the left and right string is considered to be ( T ).

- Apply newton's second law of motion for mass A and mass B.

                      ma*g - T = ma*a

                      T - mb*g = mb*a

Where,

* The tangential linear acceleration ( a ) with which the system of two masses assumed to be particles move with combined constant acceleration.

- g: The gravitational acceleration constant = 9.81 m/s^2

- Sum the two equations for both masses A and B:

                      g* ( ma - mb ) = ( ma + mb )*a

                      a =  g* ( ma - mb ) / ( ma + mb )

                      a = 9.81* ( 6 - 4 ) / ( 6 + 4 ) = 9.81 * ( 2 / 10 )

                      a = 1.962 m/s^2  

- The rope/string moves with linear acceleration of ( a ) which rotates the disk counter-clockwise in the direction of massive object A.

- The linear acceleration always acts tangent to the disk at a distance radius ( r ).

- For no slip conditions, the linear acceleration can be equated to tangential acceleration ( at ). The correlation between linear-rotational kinematics is given below :

                     a = at = 1.962 m/s^2

                     at = r*α      

Where,

           α: The angular acceleration of the object ( disk )

                    α = at / r

                    α = 1.962 / 0.25

                    α = 7.848 rad/s^2                                

- Take moments about the pivot O of the disk. Apply rotational dynamics conditions:

             

                Sum of moments ∑M = Iα

                 ( Ta - Tb )*r = Iα

- The moment about the pivots are due to masses A and B.

 

               Ta: The force in string due to mass A

               Tb: The force in string due to mass B

                I: The moment of inertia of disk = 0.5*M*r^2

                   ( ma*a - mb*a )*r = 0.5*M*r^2*α

                   α = ( ma*a - mb*a ) / ( 0.5*M*r )

                   α = ( 6*1.962 - 4*1.962 ) / ( 0.5*5*0.25 )

                   α = ( 3.924 ) / ( 0.625 )

                   α = 6.278 rad/s^2

6 0
3 years ago
A cylindrical specimen of a titanium alloy having an elastic modulus of 107 GPa (15.5 × 106 psi) and an original diameter of 3.7
Keith_Richards [23]

Answer:

the maximum length of specimen before deformation is found to be 235.6 mm

Explanation:

First, we need to find the stress on the cylinder.

Stress = σ = P/A

where,

P = Load = 2000 N

A = Cross-sectional area = πd²/4 = π(0.0037 m)²/4

A = 1.0752 x 10^-5 m²

σ = 2000 N/1.0752 x 10^-5 m²

σ = 186 MPa

Now, we find the strain (∈):

Elastic Modulus = Stress / Strain

E = σ / ∈

∈ = σ / E

∈ = 186 x 10^6 Pa/107 x 10^9 Pa

∈ = 1.74 x 10^-3 mm/mm

Now, we find the original length.

∈ = Elongation/Original Length

Original Length = Elongation/∈

Original Length = 0.41 mm/1.74 x 10^-3

<u>Original Length = 235.6 mm</u>

5 0
3 years ago
3. Low-voltage conductors rarely cause<br> injuries.
hram777 [196]
True
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8 0
2 years ago
8.19 - Airline Reservations System (Project Name: Airline) - A small airline has just purchased a computer for its new automated
e-lub [12.9K]

Answer:

The App is written in C++ language using dev C++.

Explanation:

/******************************************************************************

You can run this program in any C++ compiler like dev C++ or any online C++ compiler

*******************************************************************************/

#include <iostream>

using namespace std;

class bookingSeat// class for airline reservation system

{

  private:

   

   

  bool reserveSeat[10];// 10 seats (1-5) for first class and 6-10 for economy class

  int firstClassCounter=0;//count first class seat

  int economyClassCounter=5;//count economy class seat

  char seatPlacement;/* switch between economy and first clas seat----- a variable for making decision based on user input*/

  public:  

  void setFirstClassSeat()//

  {

      if(firstClassCounter<5)// first class seat should be in range of 1 to 5

      {

          reserveSeat[firstClassCounter]=1; /*set first class seat..... change index value to 1 meaning that it now it is reserved*/

          cout<<"Your First Class seat is booked and your seat no is "<<firstClassCounter+1; //display seat number reserved

          firstClassCounter++; //increament counter

      }

      else//in case seats are ful

      {

          cout<<"\nSeats are full";

          if(economyClassCounter==10 && firstClassCounter==5)

          {

              cout<<"\n Next flight leaves in 3 hours.";

          }

          else

          {

              cout<<"\nIt’s acceptable to be placed to you in the first-class section  y/n ";//take input from user

              cin>>seatPlacement;//user input

              if(seatPlacement=='y')//if customer want to reserve seat in first class

              {

                  setEconomyClassSeat();// then reserve first class seat

              }

              else

              {

                  cout<<"\n Next flight leaves in 3 hours.";

               }

               

          }

      }

       

  }

  void setEconomyClassSeat()//set economy class seat

  {

    if(economyClassCounter<10)//seat ranges between 6 and 10

      {

          reserveSeat[economyClassCounter]=1;// reserve economy class seat

          cout<<"Your Economy class seat is booked and your seat no is "<<economyClassCounter+1;//display reservation message about seat

          economyClassCounter++;//increament counter

      }

      else// if economy class seats are fulled

      {

          cout<<"\nSeats are full";

          if(economyClassCounter==10 && firstClassCounter==5)//check if all seats are booked in both classes

          {

              cout<<"\n Next flight leaves in 3 hours.";

          }

          else

          {

              cout<<"\nIt’s acceptable to be placed to you in the first-class section  y/n ";//take input from user

              cin>>seatPlacement;//user input

              if(seatPlacement=='y')//if customer want to reserve seat in first class

              {

                  setFirstClassSeat();// then reserve first class seat

              }

              else

              {

                  cout<<"\n Next flight leaves in 3 hours.";

               }

               

          }

      }

  }

   

   

};

int main()

{   int checkseat=10;// check seat

   int classType;//class type economy or first class

   bookingSeat bookseat;//object declaration of class bookingSeat

   while(checkseat<=10)//run the application until seats are fulled in both classes

   {

       cout<<"\nEnter 1 for First Class and 2 for Economy Class ";

       cin>>classType;//what user entered

       switch (classType)//decide which seat class to be reserved  

       {

           case 1://if user enter 1 then reserve first class seat

           bookseat.setFirstClassSeat();

           break;

           case 2://if user enter 2 then reserve the economy class seat

           bookseat.setEconomyClassSeat();

           

       }

       

   }

   

   return 0;

}

8 0
3 years ago
A large class with 1,000 students took a quiz consisting of ten questions. To get an A, students needed to get 9 or 10 questions
VMariaS [17]

Answer:

a. 0.11

b. 110 students

c. 50 students

d. 0.46

e. 460 students

f. 540 students

g. 0.96

Explanation:

(See attachment below)

a. Probability that a student got an A

To get an A, the student needs to get 9 or 10 questions right.

That means we want P(X≥9);

P(X>9) = P(9)+P(10)

= 0.06+0.05=0.11

b. How many students got an A on the quiz

Total students = 1000

Probability of getting A = 0.11 ---- Calculated from (a)

Number of students = 0.11 * 1000

Number of students = 110 students

So,the number of students that got A is 110

c. How many students did not miss a single question

For a student not to miss a single question, then that student scores a total of 10 out of possible 10

P(10) = 0.05

Total Students = 1000

Number of Students = 0.05 * 1000

Number of Students = 50 students

We see that 5

d. Probability that a student pass the quiz

To pass, a student needed to get at least 6 questions right.

So we want P(X>=6);

P(X>=) =P(6)+P(7)+P(8)+P(9)+P(10)

=0.08+0.12+0.15+0.06+0.05=0.46

So, the probability of a student passing the quiz is 0.46

e. Number of students that pass the quiz

Total students = 1000

Probability of passing the quiz = 0.46 ----- Calculated from (d)

Number of students = 0.46 * 1000

Number of students = 460 students

So,the number of students that passed the test is 460

f. Number of students that failed the quiz

Total students = 1000

Total students that passed = 460 ----- Calculated from (e)

Number of students that failed = 1000 - 460

Number of students that failed = 540

So,the number of students that failed is 540

g. Probability that a student got at least one question right

This means that we want to solve for P(X>=1)

Using the complement rule,

P(X>=1) = 1 - P(X<1)

P(X>=1) = 1 - P(X=0)

P(X>=1) = 1 - 0.04

P(X>=1) = 0.96

7 0
3 years ago
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