Internal torque diagram

Internal torque diagram DEFAULT

Mechanics of Materials: Torsion


Torsional Deformation

Torque is a moment that twists a structure. Unlike axial loads which produce a uniform, or average, stress over the cross section of the object, a torque creates a distribution of stress over the cross section. To keep things simple, we're going to focus on structures with a circular cross section, often called rods or shafts. When a torque is applied to the structure, it will twist along the long axis of the rod, and its cross section remains circular. 

To visualize what I'm talking about, imagine that the cross section of the rod is a clock with just an hour hand. When no torque is applied, the hour hand sits at 12 o'clock. As a torque is applied to the rod, it will twist, and the hour hand will rotate clockwise to a new position (say, 2 o'clock). The angle between 2 o'clock and 12 o'clock is referred to as the angle of twist, and is commonly denoted by the Greek symbol phi. This angle lets us determine the shear strain at any point along the cross section.

Before we get into the details of this equation, it's important to note that because we're only discussing circular cross sections, we've switched from Cartesian coordinates to cylindrical coordinates. That's where the Greek symbol rho came from &#; it denotes the distance along the cross section, with rho=0 at the center and rho=c at the outer edge of the rod.

We can immediately learn a few things from this equation. The first thing might be obvious: the more angle of twist, the larger the shear strain (denoted by the Greek symbol gamma, as before). Second, and this is the big difference between axial-loaded structures and torque-loaded ones, the shear strain is not uniform along the cross section. It is zero at the center of the twisted rod, and is at a maximum value at the edge of the rod. Finally, the longer the rod, the smaller the shear strain.

So far, we've focused our attention on displacements and strain. To discuss the stress within a twisted rod we need to know how torque and stress relate. Since twist applies a shear strain, we expect that torque will apply a shear stress. The relationship between torque and shear stress is detailed in section of your textbook, and it results in the following relation:

In this equation, J denotes the second polar moment of area of the cross section. This is sometimes referred to as the "second moment of inertia", but since that already has a well-established meaning regarding the dynamic motion of objects, let's not confuse things here. We'll discuss moment's of area in more detail at a later point, but they take on a very simple form for circular cross sections:

(Note: those are both the same equation &#; solid rods have an inner radius of ci=0).

Now we have equations for our shear strain and our shear stress, all that is left to do is use Hooke's law in shear to see how they are related. Hooke's law lets us write down a nice equation for the angle of twist &#; a very convenient thing to measure in lab or our in the field.

And, just like we saw for axial displacements, we can use superposition for our shear deformations as well:

This final equation allows us to split up torques applied to different parts of the same structure. Let's work out a problem, and see if we understand what's going on for torsional deformations.

Power Transmission

One of the most common examples of torsion in engineering design is the power generated by transmission shafts. We can quickly understand how twist generates power just by doing a simple dimensional analysis. Power is measured in the unit of Watts [W], and 1 W = 1 N m s-1. At the outset of this section, we noted that torque was a twisting couple, which means that it has units of force times distance, or [N m]. So, by inspection, to generate power with a torque, we need something that occurs with a given frequency f, since frequency has the units of Hertz [Hz] or [s-1].  So, the power per rotation (2*pi) of a circular rod is equal to the applied torque times the frequency of rotation, or: 

On the far right hand side of the equation, we've used the relation that angular velocity, denoted by the Greek letter omega, is equal to 2pi times the frequency.

Statically Indeterminate Problems

One equation, two unknowns&#; we've been down this road before need something else. Although the type of loading and deformation are different, the statically indeterminate problems involving the torsion of rods are approached in the exact same manner as with axially loaded structures. We start with a free body diagram of twisted rod. Take, for example, the rod in the figure below, stuck between two walls.

Immediately upon inspection you should note that the rod is stuck to two walls, when only one would be necessary for static equilibrium.  More supports than is necessary: statically indeterminate.  And statically indeterminate means, draw a free body diagram, sum the forces in the x-direction, and you'll get one equations with two unknown reaction forces. So, we need to consider our deformations &#; for torsion, that means let's turn to our equation that describes the superposition of twist angles. For this equation, we should note that half the rod is solid, the other half is hollow, which affects how we calculate J for each half. Most importantly, we need to ask ourselves "what do we know about the deformation?" Well, since the rod is stuck to the wall at edge, the twist at A and must be equal to zero (just like the displacement in the last section). See if you can work the rest of this problem out on your own: What is the torque in each half of the rod?  

(Answer: Ta= lb ft & Tb= lb ft).

Summary

We learned about torque and torsion in this lesson. This different type of loading creates an uneven stress distribution over the cross section of the rod &#; ranging from zero at the center to its largest value at the edge. From this analysis we can develop relations between the angle of twist at any a point along the rod and the shear strain within the entire rod. Using Hooke's law, we can relate this strain to the stress within the rod. We also used a method of dimensional analysis to determine the power generated by a transmission shaft (i.e. a rod) that spins with a given frequency under an applied torque. Finally, we showed that torsion problems are also often statically indeterminate, and even though the loading and deformation is different, the technique we established in the last section for solving problems with axial loading is the same technique for solving problems with torque loading.

This material is based upon work supported by the National Science Foundation under Grant No. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the National Science Foundation.

Sours: https://www.bu.edu/moss/mechanics-of-materials-torsion/

The torque diagram of a shaft is analogues to the shear force and bending moment diagram of a beam. It is an important engineering diagram from the pulley shaft design point of view. The steps required to draw it will be discussed with the help of the following example:

Problem: The below picture is showing a machine shop set up, where three machines namely: milling, drilling and lathe are drawing torques from the main shaft at the points B, C and D respectively. The main shaft is rotating by taking torque from the motor. The bearings E and F are assumed to be frictionless so not taking any torques from the torque, theoretically

The details of the input and output torques are shown below:

 

Milling M/C &#; 20 N-m

Drilling M/C &#; 30 N-m

Lathe M/C &#; 40 N-m

 

 

Solution:

StepDraw FBD of the entire shaft: The FBD of the entire shaft will look like below:

 

 

Please observe that the direction of the torque for the motor is opposite than the three machines. This is because the above FBD is showing the reaction torques and the direction of the reaction torques for the machines are opposite to that of the motor for obvious reason. The fact will be re-emphasised from the Step-2 below.

 

Step Calculate the input torque required from the motor: From the equation for equilibrium, we know that total torque must be equal to zero. So, we can write:

 

Σ T = 0

Or, TA+20+30+40=0

Or, TA= &#; 90 N-m

 

 

Step Calculate the torque @ section AB: Take a section anywhere between the point A and B and consider the portion of the shaft on the left of the section for equilibrium. The equilibrium equation becomes:

 

TA + TA-B=0

or, TA-B = &#; TA=90 N-m

 

Where,

TA-B – The torque at the section anywhere between A and B

 

 

Step Calculate the torque @ section BC: Take a section anywhere between the point B and C and consider the portion of the shaft on the left of the section for equilibrium. The equilibrium equation becomes:

 

TA + TB+TB-C=0

or, TB-C = &#; TA-TB=70 N-m

 

Where,

TB-C – The torque at the section anywhere between B and C

 

 

 

 

 

Step Calculate the torque @ section CD: Take a section anywhere between the point C and D and consider the portion of the shaft on the left of the section for equilibrium. The equilibrium equation becomes:

 

TA + TB+TC+TC-D=0

or, TC-D = &#; TA-TB-TC=40 N-m

 

Where,

TC-D – The torque at the section anywhere between C and D

 

 

Step Draw the torque diagram: Plot the length of the shaft in X-axis and the torque values (calculated in step-3,4,5) at the different sections in Y-axis. Like below:

 

The torque values obtained from the torque diagram is used as input for the pulley shaft design.

 

 

 

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Shibashis Ghosh

Hi, I am Shibashis, a blogger by passion and an engineer by profession. I have written most of the articles for mechGuru.com. For more than a decades i am closely associated with the engineering design/manufacturing simulation technologies. I am a self taught code hobbyist, presently in love with Python (Open CV / ML / Data Science /AWS + lines, + hrs. )

Tags: Torque

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Sours: https://mechguru.com/machine-design/draw-torque-diagram-in-six-steps/
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Axial Force Diagrams and Torque Diagrams

As an alternative to splitting a body in half and performing an equilibrium analysis to find the internal forces and moments, we can also use graphical approaches to plot out these internal forces and moments over the length of the body. Where equilibrium analysis is the most straightforward approach to finding the internal forces and moments at one cross section, the graphical approaches are the most straightforward approaches to find the internal forces or the internal moments across the entire length of a beam, shaft, or other body. This may be useful in complex loading scenarios where it may not be obvious where the maximum internal forces or internal moments exist. As a trade off however, we will need to plot out each type of internal load separately (one plot for internal axial forces, one for internal shear forces, one for internal torques, and one for internal bending moments).

Complex Loading of a Beam
Internal forces in 3D

In this section, we will be focusing on the methods used to generate the plots for the internal axial forces, and the internal torques. This will be the force and moment acting along the length of the beam or shaft. The axial force plot is used primarily for vertical columns or cables supporting multiple loads along its length. The torque diagram is used primarily for shafts supporting multiple inputs and outputs. Each of these plots will have a different practical application, but we have grouped them together here because the process used to generate each of the two plots is very similar.

Creating the Axial Force Diagram

The axial force diagram will plot out the internal axial (normal) forces within a beam, column, or cable that is supporting multiple forces along the length of the beam itself. This can be thought of as the internal tension or compression forces. The most relevant practical scenarios that match this description will be main support columns in a multi-story building, or hanging cables that are used to support multiple loads.

Main support collumns in a multi-story building

To create the axial force plot for a body, we will use the following process.

  1. Solve for all external forces acting on the body.
  2. Draw out a free body diagram of the body horizontally. In the case of vertical structures, rotate the body so that it sits horizontally and all the forces act horizontally
  3. Lined up below the free body diagram, draw a set of axes. The x-axis will represent the location (lined up with the free body diagram above), and the y-axis will represent the internal axial forces, with positive numbers indicating tension and negative numbers indicating compression.
  4. Starting at zero at the right side of the plot, you will move to the right, pay attention to forces in the free body diagram above. As you move right in your plot, keep steady except
    • Jump upwards by the magnitude of the force for any forces in our free body diagram to the left.
    • Jump downwards by the magnitude of the force for any forces to the right.
    • You can ignore any moments or vertical forces applied to the body.
    By the time you get to the left end of the plot, you should always wind up coming back to zero. If you don't wind up back at zero, go back and check your previous work.
A free body diagram with the axial force diagram.

To read the plot, you simply need to find the location of interest from the free body diagram above, and read the corresponding value on the y-axis from your plot. Again, positive numbers represent an internal tension at that location and negative numbers represent an internal compression at that location.

Creating the Torque Diagram

The torque diagram will plot out the internal torsional moment within a shaft that is supporting multiple inputs and/or outputs along its length. The most relevant practical scenarios that match this description are shafts within complex gear or pulley driven systems.

A line shaft delivering power via multiple pulleys

To create the torque diagram for a shaft, we will use the following process.

  1. Solve for all external moments acting on the shaft.
  2. Draw out a free body diagram of the shaft horizontally, rotating the shaft if necessary, so that all torques act around the horizontal axis.
  3. Lined up below the free body diagram, draw a set of axes. The x-axis will represent the location (lined up with the free body diagram above), and the y-axis will represent the internal torsional moment, with positive numbers indicating an internal torsional moment vector to the right and negative numbers indicating an internal torsional moment to the left.
  4. Starting at zero at the right side of the plot, you will move to the right, pay attention to moments in the free body diagram above. As you move right in your plot, keep steady except
    • Jump upwards by the magnitude of the moment for any torques in our free body diagram where the moment vector would point left.
    • Jump downwards by the magnitude of the moment for any torques in our free body diagram where the moment vector would point right.
    • You can ignore any forces in the free body diagram or moments not about the x axis.
    By the time you get to the left end of the plot, you should always wind up coming back to zero. If you don't wind up back at zero, go back and check your previous work.
A free body diagram of a shaft with the torque diagram.

To read the plot, you simply need to take the find the location of interest from the free body diagram above, and read the corresponding value on the y-axis from your plot.

Worked Problems:

Question 1:

A wooden beam is subjected to the forces shown below (forces applied at base of vector). Draw the axial force diagram for the beam.

Problem 1 Diagram

Question 2:

A steel shaft is subjected to the torques shown below. Draw the torque diagram for this shaft.

Problem 2 Diagram
Sours: http://mechanicsmap.psu.edu/websites/6_internal_forces/_axial_torque_diagrams/axial_torque_diagrams.html
Understanding Torsion

Axial Force Diagrams and Torque Diagrams

As an alternative to splitting a body in half and performing an equilibrium analysis to find the internal forces and moments, we can also use graphical approaches to plot out these internal forces and moments over the length of the body. Where equilibrium analysis is the most straightforward approach to finding the internal forces and moments at one cross section, the graphical approaches are the most straightforward approaches to find the internal forces or the internal moments across the entire length of a beam, shaft, or other body. This may be useful in complex loading scenarios where it may not be obvious where the maximum internal forces or internal moments exist. As a trade off however, we will need to plot out each type of internal load separately (one plot for internal axial forces, one for internal shear forces, one for internal torques, and one for internal bending moments).

Complex Loading of a Beam
Internal forces in 3D

In this section, we will be focusing on the methods used to generate the plots for the internal axial forces, and the internal torques. This will be the force and moment acting along the length of the beam or shaft. The axial force plot is used primarily for vertical columns or cables supporting multiple loads along its length. The torque diagram is used primarily for shafts supporting multiple inputs and outputs. Each of these plots will have a different practical application, but we have grouped them together here because the process used to generate each of the two plots is very similar.

Creating the Axial Force Diagram

The axial force diagram will plot out the internal axial (normal) forces within a beam, column, or cable that is supporting multiple forces along the length of the beam itself. This can be thought of as the internal tension or compression forces. The most relevant practical scenarios that match this description will be main support columns in a multi-story building, or hanging cables that are used to support multiple loads.

Main support collumns in a multi-story building

To create the axial force plot for a body, we will use the following process.

  1. Solve for all external forces acting on the body.
  2. Draw out a free body diagram of the body horizontally. In the case of vertical structures, rotate the body so that it sits horizontally and all the forces act horizontally
  3. Lined up below the free body diagram, draw a set of axes. The x-axis will represent the location (lined up with the free body diagram above), and the y-axis will represent the internal axial forces, with positive numbers indicating tension and negative numbers indicating compression.
  4. Starting at zero at the right side of the plot, you will move to the right, pay attention to forces in the free body diagram above. As you move right in your plot, keep steady except...
    • Jump upwards by the magnitude of the force for any forces in our free body diagram to the left.
    • Jump downwards by the magnitude of the force for any forces to the right.
    • You can ignore any moments or vertical forces applied to the body.
    By the time you get to the left end of the plot, you should always wind up coming back to zero. If you don't wind up back at zero, go back and check your previous work.
A free body diagram with the axial force diagram.

To read the plot, you simply need to find the location of interest from the free body diagram above, and read the corresponding value on the y-axis from your plot. Again, positive numbers represent an internal tension at that location and negative numbers represent an internal compression at that location.

Creating the Torque Diagram

The torque diagram will plot out the internal torsional moment within a shaft that is supporting multiple inputs and/or outputs along its length. The most relevant practical scenarios that match this description are shafts within complex gear or pulley driven systems.

A line shaft delivering power via multiple pulleys

To create the torque diagram for a shaft, we will use the following process.

  1. Solve for all external moments acting on the shaft.
  2. Draw out a free body diagram of the shaft horizontally, rotating the shaft if necessary, so that all torques act around the horizontal axis.
  3. Lined up below the free body diagram, draw a set of axes. The x-axis will represent the location (lined up with the free body diagram above), and the y-axis will represent the internal torsional moment, with positive numbers indicating an internal torsional moment vector to the right and negative numbers indicating an internal torsional moment to the left.
  4. Starting at zero at the right side of the plot, you will move to the right, pay attention to moments in the free body diagram above. As you move right in your plot, keep steady except...
    • Jump upwards by the magnitude of the moment for any torques in our free body diagram where the moment vector would point left.
    • Jump downwards by the magnitude of the moment for any torques in our free body diagram where the moment vector would point right.
    • You can ignore any forces in the free body diagram or moments not about the x axis.
    By the time you get to the left end of the plot, you should always wind up coming back to zero. If you don't wind up back at zero, go back and check your previous work.
A free body diagram of a shaft with the torque diagram.

To read the plot, you simply need to take the find the location of interest from the free body diagram above, and read the corresponding value on the y-axis from your plot.

Worked Problems:

Question 1:

A wooden beam is subjected to the forces shown below (forces applied at base of vector). Draw the axial force diagram for the beam.

Problem 1 Diagram

Question 2:

A steel shaft is subjected to the torques shown below. Draw the torque diagram for this shaft.

Problem 2 Diagram
Sours: http://mechanicsmap.psu.edu/websites/6_internal_forces/6-3_axial_torque_diagrams/axial_torque_diagrams.html

Torque diagram internal

Mechanics of Materials: Torsion


Torsional Deformation

Torque is a moment that twists a structure. Unlike axial loads which produce a uniform, or average, stress over the cross section of the object, a torque creates a distribution of stress over the cross section. To keep things simple, we're going to focus on structures with a circular cross section, often called rods or shafts. When a torque is applied to the structure, it will twist along the long axis of the rod, and its cross section remains circular. 

To visualize what I'm talking about, imagine that the cross section of the rod is a clock with just an hour hand. When no torque is applied, the hour hand sits at 12 o'clock. As a torque is applied to the rod, it will twist, and the hour hand will rotate clockwise to a new position (say, 2 o'clock). The angle between 2 o'clock and 12 o'clock is referred to as the angle of twist, and is commonly denoted by the Greek symbol phi. This angle lets us determine the shear strain at any point along the cross section.

Before we get into the details of this equation, it's important to note that because we're only discussing circular cross sections, we've switched from Cartesian coordinates to cylindrical coordinates. That's where the Greek symbol rho came from – it denotes the distance along the cross section, with rho=0 at the center and rho=c at the outer edge of the rod.

We can immediately learn a few things from this equation. The first thing might be obvious: the more angle of twist, the larger the shear strain (denoted by the Greek symbol gamma, as before). Second, and this is the big difference between axial-loaded structures and torque-loaded ones, the shear strain is not uniform along the cross section. It is zero at the center of the twisted rod, and is at a maximum value at the edge of the rod. Finally, the longer the rod, the smaller the shear strain.

So far, we've focused our attention on displacements and strain. To discuss the stress within a twisted rod we need to know how torque and stress relate. Since twist applies a shear strain, we expect that torque will apply a shear stress. The relationship between torque and shear stress is detailed in section 5.2 of your textbook, and it results in the following relation:

In this equation, J denotes the second polar moment of area of the cross section. This is sometimes referred to as the "second moment of inertia", but since that already has a well-established meaning regarding the dynamic motion of objects, let's not confuse things here. We'll discuss moment's of area in more detail at a later point, but they take on a very simple form for circular cross sections:

(Note: those are both the same equation – solid rods have an inner radius of ci=0).

Now we have equations for our shear strain and our shear stress, all that is left to do is use Hooke's law in shear to see how they are related. Hooke's law lets us write down a nice equation for the angle of twist – a very convenient thing to measure in lab or our in the field.

And, just like we saw for axial displacements, we can use superposition for our shear deformations as well:

This final equation allows us to split up torques applied to different parts of the same structure. Let's work out a problem, and see if we understand what's going on for torsional deformations.

Power Transmission

One of the most common examples of torsion in engineering design is the power generated by transmission shafts. We can quickly understand how twist generates power just by doing a simple dimensional analysis. Power is measured in the unit of Watts [W], and 1 W = 1 N m s-1. At the outset of this section, we noted that torque was a twisting couple, which means that it has units of force times distance, or [N m]. So, by inspection, to generate power with a torque, we need something that occurs with a given frequency f, since frequency has the units of Hertz [Hz] or [s-1].  So, the power per rotation (2*pi) of a circular rod is equal to the applied torque times the frequency of rotation, or: 

On the far right hand side of the equation, we've used the relation that angular velocity, denoted by the Greek letter omega, is equal to 2pi times the frequency.

Statically Indeterminate Problems

One equation, two unknowns… we've been down this road before need something else. Although the type of loading and deformation are different, the statically indeterminate problems involving the torsion of rods are approached in the exact same manner as with axially loaded structures. We start with a free body diagram of twisted rod. Take, for example, the rod in the figure below, stuck between two walls.

Immediately upon inspection you should note that the rod is stuck to two walls, when only one would be necessary for static equilibrium.  More supports than is necessary: statically indeterminate.  And statically indeterminate means, draw a free body diagram, sum the forces in the x-direction, and you'll get one equations with two unknown reaction forces. So, we need to consider our deformations – for torsion, that means let's turn to our equation that describes the superposition of twist angles. For this equation, we should note that half the rod is solid, the other half is hollow, which affects how we calculate J for each half. Most importantly, we need to ask ourselves "what do we know about the deformation?" Well, since the rod is stuck to the wall at edge, the twist at A and must be equal to zero (just like the displacement in the last section). See if you can work the rest of this problem out on your own: What is the torque in each half of the rod?  

(Answer: Ta=51.7 lb ft & Tb=38.3 lb ft).

Summary

We learned about torque and torsion in this lesson. This different type of loading creates an uneven stress distribution over the cross section of the rod – ranging from zero at the center to its largest value at the edge. From this analysis we can develop relations between the angle of twist at any a point along the rod and the shear strain within the entire rod. Using Hooke's law, we can relate this strain to the stress within the rod. We also used a method of dimensional analysis to determine the power generated by a transmission shaft (i.e. a rod) that spins with a given frequency under an applied torque. Finally, we showed that torsion problems are also often statically indeterminate, and even though the loading and deformation is different, the technique we established in the last section for solving problems with axial loading is the same technique for solving problems with torque loading.

This material is based upon work supported by the National Science Foundation under Grant No. 1454153. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the National Science Foundation.

Sours: https://www.bu.edu/moss/mechanics-of-materials-torsion/
How to Draw a Torque Diagram Using Equilibrium Equations

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