Torsion Repaired)

8/4/2019 Torsion Repaired)

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STRENGTH

OF

MATERIALSII TORSION


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Objective:

To understand and appreciate the torsion phenomenon

Specific objectives

By the end of the lecture you should be able to:

Define torsion Describe some applications which make use of torsion Develop formula associated with torsion Solve problems

Pre-requisite

Aki Ola : fractions

Aki Ola : change of subject

Maths I from last semester

SOM I : shear stresses and strains

Mini topics

Introduction Torsion - What is it?

Applications of torsion Assumptions Angle of twist - what is it? Shear Stresses and strains within the shaft Definition of terms


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: Angle of twist (radians degrees) Polar second moment of area, J - What is it? Section modulus Torisional rigidity

Power transmitted by shafts Summary

Evaluation

It is important to note that at the end of the topic you would be examined on how you have

understood the text with a set of 10 true or false questions and 1 application question. This would

be recorded.


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Introduction

Last semester we discussed shearing stresses and strains. Fortunately for us these two

phenomena happen (play a part) during torsion. That is to say that during torsion it is shearing

stress that helps with the twisting. It would be important to revise the notes on shear stresses.

Actually, we would realize that without shear there would virtually be no torsion.

Torsion - What is it?

Torsion is the twisting of an object due to an applied torque. In circular sections, the resultant

shearing stress is perpendicular to the radius.

Another definition from Gere and Timoshenko defines it as:

The twisting of a structural member when it is loaded by couples that produce rotation about its

longitudinal axis. Couples mentioned here refer to the twisting moments or torques caused by the

shear forces.
http://en.wikipedia.org/wiki/Torquehttp://en.wikipedia.org/wiki/Shear_stresshttp://en.wikipedia.org/wiki/Shear_stresshttp://en.wikipedia.org/wiki/Torque


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Applications of torsion

Heavy trucks

Heavy trucks that ply our roads need huge torques to be transmitted from the engine to the

wheel. This is achieved using the universal joint. Hence the universal joint undergoes torsion at

that instant where the gear is pushed to the first and the clutch is released.

Picture of the truck with the universal joint

Ships and submarines

These also use propeller shafts to move them forward

Others:

Drive shafts in other machinery Drill rods

Torsional pendulums Screw drivers Steering rods

In reality whenever torsion is applied an engineering material (in our case circular shafts) we

must bear some few things in mind.

1. Every section of the shaft is in a state of pure shear. This means that no compressive or tensile stresses exist.

2. The moment of resistance is the same everywhere. This means that the value of the moment of resistance is the same across the

length of the shaft

If the torque is being applied at one end that is positive and becomes fully

negative at the receiving end.


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Assumptions

To enable us develop formulae to help us solve problems associated with torsion, it is expedient

to also keep these assumptions at the back of our minds. Note that without these assumptions the

formula developed will not hold.

1. Material (under consideration) must be homogeneous That is to say that the material under consideration must be of uniform properties

throughout.

2. Material must be elastic This means that the material under consideration must obey Hookes law Stresses should be proportional to the strains

3. Stresses applied should not exceed the elastic limit or the limit of proportionality same as two because if one crosses the elastic limit he has ventured into the

plastic range and that area is out of our range. Some other equations cater for that.

4. Circular sections must remain circular The diameter of the shaft should the same throughout its length

5. Cross sections must remain plane

Torsion must occur in only one plane, for example in the x-y plane or the y-zplane

6. Cross section of the material must rotate as if rigid This means that at least one side of the shaft must be rigid and the shaft must

behave as a solid, not a gel and certainly not liquid.


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With these behind us we can now determine the angle of twist.

Angle of twist - what is it?

Angle subtended on the cross section when a point A is gradually shifted to B due to an applied

torque.

Usually this is considered so small that it will not cause a change in the length of the bar or its

radius.

From the diagram, one can see how the shaft behaves like a solid. Due to its solid nature, as

torsion is applied at one end of the shaft, say end A, the particles there have less resistance. As

one moves along the length of the shaft the resistance to the torsion increases. And so if we place

dots showing the resistance to the torsion along the shaft we would see the resistance of theparticles right up to the last particle which do not move at all due to full resistance at that point ,

B.

It is important to note that the greater the torque / torsion the greater the angle of twist.

A

B

OR


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From a simplified diagram of the cross-section of the shaft, assuming we apply torsion to theshaft in the clockwise direction all particles at A will gradually shift to B (thanks to shear). Themotion will subtend an angle, which we can determine.

Mathematically we know that:

The length of an arc = radius angle subtended

So in our case,

Length of arc AB = radius (R) angle subtended ( ) (1)

From the earlier diagram, one would agree that theta is causing the shearing strain.

Where the total shearing strain = length (L) x shearing strain ( ) (2)

This means that we can conclude that

Length of arc AB = radius ( ) angle subtended ( ) = length (L) x shearing strain ( )

Arc AB = = L (3)

Shearing strain, = (4)

Recall from the definition of modulus of rigidity, G

G =

= (5a)

From (5a), shear strain can be determined as = (5b)

A combination of (5a) and (5b) would yield

= = = (6)

Rewriting (6)

= (7)

Deductions from (6)

1. Shear stress is directly proportional to modulus of rigidity and angle of twist. This means

that as shear stress increases one should expect that angle of twist should also increase.


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2. Shear stress is inversely proportional to length. This means that the smaller the length of

the circular shaft the greater the shear stress and vice versa.

3. Rewriting (7) into this form, = , one would realize that shear stress increases with

increasing radius, from R = 0 to R = R max . At R max , shear stress is also maximum and isusually written as

(7) can further be written as

= (8)

Shear Stresses within the shaft

Recall we said earlier that the twisting and eventual failure of the shaft is facilitated by shear

stresses. It should therefore be possible to determine the shear stress occurring at any point

within the shaft. To do that would break the cross-section of the shaft into elements and take one

as shown in the diagram below. Then when we have generated expressions for the element we

can sum it up for all the elements.

In the diagram, our element is shown white.

Recall that Force set up in each element, F e = shear stress area

: shear stress

t: thickness = dr

r

O


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Shear stress = (because it is not at the maximum r = R max )

Area of the given element = (approx)

Therefore, F e = (9)

Due to the twisting about the axis (axis is the pivot), moment will occur. We can call thismoment elemental moment, M e because it is occurring on the element and not on the wholeshaft. Moment is also known as torque. So we can find the elemental torque, T e.

Recall

Moment / Torque = force x perpendicular distance

The force acting is what we have generated in (9) = F e =

The perpendicular distance = r, the radius

So M e = T e = F e

Te =

Te =

Remember that all this while we have working on just one element. To find the total torque forall the elements in the shaft, we sum up the equation for all the elements. To make things easier

on ourselves we apply the principle of integration (if you have forgotten, revise get aSTROUD!)

Total torque, T = a. We need to replace with a term which contains r that will make it integrable. We know

from (8) that it can be rewritten as = . So substituting this term, we would get

T = T =

b. At this particular moment / torque, all the parameters which make up are all constant

and are therefore taken outside the summation sign.

T =


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This step brings us to an important parameter known as the polar second moment of area, J

Therefore T = (10)

(10) can also be rewritten as

(11) One would realize that (11) and (8) can be equated as follows

(12) (12) is what is known as the torsion theory in mathematical formDefinition of terms

: Shear stress at any radius, r, (N/m 2)

R: Maximum radius (equal to R max ) (m)

: Maximum shear stress occurring at R max

G: modulus of rigidity/shear modulus (GPa)

L: Length (m)

: Angle of twist (radians)

T: torque (Nm)

J: polar second moment of area (m 4)


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: Angle of twist (radians degrees)

Radians is the usual unit applied in this context but sometimes to give us a better picture of howmuch twisting has occurred we convert it to degrees using the following formula.

a. Radians to degrees b. Degrees to radians

Polar second moment of area, J

What is it?

J is the torsion constant for the section. It is identical to the polar moment of inertia for a roundshaft or concentric tube only. It can be determined for other shapes.

For solid shafts

This is a parameter which must be calculated for before applied in the torsion theory equation.First, we have to derive it. We know that

is a constant and therefore has to be taken outside the summation
http://en.wikipedia.org/wiki/Torsion_constanthttp://en.wikipedia.org/wiki/Polar_moment_of_inertiahttp://en.wikipedia.org/wiki/Polar_moment_of_inertiahttp://en.wikipedia.org/wiki/Torsion_constant


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By substituting the limits

*+ (13)

Knowing the R and substituting J will become

(14)

For hollow shafts

There is an internal and external radius as shown in the diagram below.

In the diagram:

r: internal radius

R: external radius

These two are the limits for calculation

Note that maximum shear stress still occurs at R = R max

Substituting into the J equation

rR


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By substituting the limits

(15) Knowing the R and substituting J will become

(16) D: external diameter

d: internal diameter

At this juncture we are almost at the end of the topic. To gain insight it would be expedient to trysome few examples. It is recommended that you not only read the examples as a means tolearning it but make a conscious effort to work through it with pencil, paper, and calculator,making sure all the values are correct. This method helps you to train your mental faculty toapply what you have learnt when called upon.

It is also important to note that hollow bars are more efficient in resisting torsional loads than aresolid bars. The shear stresses for a solid circular shaft are at the maximum at the outer boundary

of the cross section and zero at the centre. Therefore most of the material in the solid shaft isstressed significantly below the maximum shear stress. If weight reduction weight reduction andsavings in material are important then it is advisable to use hollow shafts.

What this statement simply means it is that instead of wasting material creating solid shafts andstill adding weight, if savings in material and weight reduction is important to you, then youshould use hollow materials because they can resist the same amount of shear stress.


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Example 1

What torque, applied to hollow circular shaft of 25cm outside diameter and 17.5cm insidediameter will produce maximum shearing stress of 75MN/m 2 in the material?

Given dataExternal diameter, D = 25cm = 0.25m

Internal diameter, d = 17.5cm = 0.175m

Maximum shear stress, = 75MN/m 2

Required

Torque

Calculation and Solution

External radius, R = = 0.125mInternal radius, r = = 0.0875mThe torsion theory formula is given by

Terms are defined above and are expected in any assignment

The appropriate equation to use would be

The values for the other parameters have been provided but J would have to be calculated

For a hollow shaft

Use of either formula would yield the same result


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Using the radius and substituting the values

= = Therefore substituting values to find torque

= 174850.8Nm

Therefore the torque required is 174.8kNm

All the parameters can be demanded from questions given so one must practice how tomanipulate them.


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Example 3

Research shows that especially in industries where weight is of essence, use of hollow shaft,being of less mass and weight than a solid shaft, can transmit the same amount of torque andpower. Aircraft industries take this very serious and this next example is aimed drawing your

mind to the reality.

A solid circular shaft of 25cm diameter is to be replaced by a hollow shaft; the ratio of theexternal to internal diameters being 2:1. Find the size of the hollow shaft if the maximumshearing stress is to be the same as for the solid shaft. What % economy in mass with this changeeffect?

Data given

External diameter, D = 25cm = 0.25m

For hollow shaft, external diameter: internal diameter = 2:1

This implies that if the internal radius is r then the external radius R = 2r

Required

a. Size of hollow shaft, which means dimensions for the internal and external diametersb. % economy the change would effect


External radius, R = = 0.125mThe torsion theory formula is given by

For a solid shaft,

*+ Since we know the radius we can substitute the values and calculate for J


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*+= For the hollow shaft:

But we inferred from the question that R = 2r. Therefore substituting

factorizing = So J solid shaft = whilst J hollow shaft = .

These shafts will experience the same maximum shearing stress (inference from question) andtherefore we can determine the dimensions

[ ] [ ] Note that the r in the hollow shaft which must generate maximum shearing stress = R whichfrom the question = 2r

Substituting values and expressions

[ ] Simplifying

r = 0.02977m

a. dimensions of hollow shaftHence the internal radius, r = 29.77mm ;

Internal diameter, d = 59.54mm

The external radius, R = 2r = 2 (29.77) = 59.54mm ;

External diameter, D = 119.09mm


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b. % economy in mass

% economy in mass =

= 23.8%

This means that by using the hollow shaft 23.8% of mass which would have gone waste doingnothing has been saved.


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Section Modulus

It is sometimes convenient to rewrite part of the torsion theory formula to obtain the maximumshear stress in the shaft as follows:

From this equation, we can define a parameter known as polar section modulus, Z

Z = Substituting this into the equation

Without developing the formula because the basics have already been developed (try it),

Z for solid shafts =

Z for hollow shafts =

Torsional Rigidity

Angle of twist per unit length of shafts is given by the torsion theory as

The GJ is known as torsional rigidity, a property of the material and is calculated as

Remember that angle of twist per unit length means that the length has a value of 1

Therefore = =


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Composite shafts

Sometimes shafts are not just made of one material. They can be of different materials andconnected differently depending on what torque is needed and hence get work done.

Types of connectionsThere are basically two types of connections; series and parallel connections

Series connection

This is where 2 or more shafts of same ordifferent materials, diameter or basic form areconnected in such a way that each carries thesame torque. Here composite shaft strength istreated separately applying the torsion theory

in turn.

Mathematically,Torsion theory states that

Designating different materials as 1 and 2

Since torque is the same for both materials, it means that

Conditions

1. There are real situations where it is convenient that the angles of twist are equal. Thismeans that

The resultant equation by this condition would be

2. Where the material used is said to be the same, it means



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Parallel connection

Shafts are said to be connected in parallel if two or more materials are rigidly fixedtogether such that the applied torque is

shared between them and angle of twistis equal in each portion.

Mathematically, if torque is shared

T = T 1 + T 2

From the torsion theory

If angle of twist is equal in each portion

Conditions

1. Where the lengths are the same, it would mean that




Maximum shearing stresses can be found in each part as follows

It is up to the student to read the question and deduce the conditions from it.


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Power transmitted by shafts

Usually shafts in huge trucks are connected from the engine to the drive shafts at the back axlesand in small vehicles shafts are used to transmit power from the engine to the front axles to drivethem. Power generated, Torque needed or applied can be determined with appropriate equations.

If a shaft carries a torque, (Nm) and rotates (rad/s), then the work done per second (Power, P Joules, Watts) is given by:

Power, P = (17)

: Angular / rotational speed =

Therefore power,

P = (18)

N: number of revolutions per second (usually given in rev/min so we divide by 60 to convert it torev/sec)


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Example 2

A ships propeller shaft has external and internal diameters of 25cm and 15cm respectively.What power can be transmitted at 110 rev/min with a maximum shearing stress of 75MN/m 2.What then must be the twist in degrees of a 10m length of the shaft? G = 80GN/m 2

Given data


Internal diameter, d = 15cm = 0.15m

Maximum shear stress, = 75MN/m 2 = N/m 2

G, Shear modulus = 80GN/m 2 = N/m 2

Length, L = 10m

N, no. of revolutions = 110 rev/min

Required

Angle of twist,


External radius, R = = 0.125mInternal radius, r =

= 0.075m

The torsion theory formula is given by

We need the full torsion theory, so we have to use the parameters given to get the others. Firstwe need this part of the equation:

To get the power, we need torque. We need to calculate for J after which we can substitutebecause the others have been provided

= =


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Therefore substituting values to find torque

= 200.3kNm

Therefore the torque required is 200.3kNm

a. Power, P =

Substituting values

P = = = =

Therefore power transmitted =

b. Angle of twist

Now to find the angle of twist by employing , therefore = 0.075 radians

This value we can convert to degrees to give us a better picture using either formula

Therefore angle of twist, =


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Example 4

a. A solid shaft 100mm diameter transmits 75kW at 150 rev/min. Determine the value of the maximum shear stress set up in the shaft and the angle of twist per meter of the shaftlength if G = 80GN/m 2

b. If the shaft were now bored in order to reduce the weight to produce a tube of 100mmoutside diameter and 60mm inside diameter, what torque could be carried if the samemaximum shear stress is not to be exceeded? What is the percentage increase in power / weight ratio effected by this modification

Given data


Length, L = 1m

Power, P =75kW = W

G, Shear modulus = 80GN/m 2 = N/m 2

N, no. of revolutions = 150 rev/min

Required

a. Maximum shear stress, b. Angle of twist per meter of shaft


The torsion theory formula is given by

T and J are unknown that have to be determined. Hence, we need this part of the equation:

Substituting values

= =

The value for power, P has been provided so with its equation we can determine the torque, T

= therefore

=


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Therefore

remember that R = Rewriting the equation

N/m 2= MN/m 2

Therefore maximum shear stress in the shaft is MN/m 2.

Angle of twist per meter of shaft

Converting to degrees

a. Therefore

b. When the shaft is bored, we must note that the polar second moment of area willdefinitely change. J will become:

= = The torque carried by the modified shaft will be given by

= = Nm = Nm

Let us have a short discussion in order to understand this weight per meter issue

Recall that force, F = mass ( m ) x gravitational constant, ( g) this gives us the weight

But mass, m =


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Substituting mass into the first formula,

Weight =

Again, volume, v =

Therefore, Weight =

Finally Area, A of shaft is given by, A = and per meter means L = 1 m

Therefore substituting everything

The weight/meter of the original shaft is given by:

Weight / meter = = =

The weight/meter of the modified shaft is given by:

Weight / meter = =

=

=

Power, P = T

Power / weight ratio for original shaft is given by:

Power / weight =

Power / weight ratio for modified shaft is given by:

Power / weight =


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Notice that even though the weight of the modified shaft is less than that of the original shaft, the

power it can generate is almost 1.4x what the original shaft can generate. The percentage

increase in power / weight ratio can be calculated as follows:

% increase in power / weight = = 36%

We can also explain it t his way that for the same length of shaft material, a hollow shaft

generated 36% more power than the solid shaft whilst reducing its weight.


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Example 5

Determine the dimensions of a hollow shaft with a diameter ratio of 3:4 which is to transmit

60kW at 200rev/min. the maximum shear stress in the shaft is limited to 70MN/m 2 and the angle

of twist is 3.8o

in a length of 4m. For the shaft material, G = 80GN/m2

.

Data given

2 limiting conditions; that determined by:

a. Maximum shear stress = 70MN/m 2

b. Angle of twist = 3.8 o

Please understand that a. and b. will lead to two different answers for the required diameter. The

larger shaft however must be chosen as the one for which neither condition will be exceeded.

Proof of this statement will be shown.

Ratio of internal diameter: external diameter = 3:4

Power transmitted, P = 60kW

Number of revolutions, N = 200 rev/min

Length = 4m

For the shaft material, G = 80GN/m 2

Calculation and solution

Given the value for power, we can calculate the torque

P = therefore

T = = =


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Using the maximum shear stress approach:


But by change of subject

= Let us not forget that d: D = 3: 4 which can also be written as

So, going back to the question, we have to factorise which will become:

Substituting the ratios

Therefore,

= = Rearranging

D =

D = 0.06726m = 67.3mm

Recall from the ratios that d=0.75D

Therefore


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d =

Using the angle of twist approach:

From the torsion theory,

From the equation, But Therefore

Let us not forget that the angle of twist has been given in degrees and has to be converted to

radians before it is used in the calculation. The conversion is done as follows:

= radians

Substituting

= =

=

factorise

Substituting the ratios


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Therefore,

=

D = 0.07528m = 75.3mm

Recall from the ratios that d=0.75D

Therefore

d =

So we said from the preceding statement that the dimensions with the largest values should bechosen, the dimensions of the shaft should be internal diameter, d = 56.5mm and externaldiameter, D =75.3mm.

In order to prove, the statement let us substitute the values from the two limiting conditions andsee which values exceed them.

a. d = 56.5mm, D = 75.3mm

= /m 2


= 0.066 radians

Converting it to degrees = 0.066 = 3.78


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b. d= 50.5mm, D = 67.3mm

= /m 2


= 0.104 radians

Converting it to degrees = 0.104 = 5.958

From the calculations, it is obvious that the values derived from the limiting condition of theshear stress exceed the angle of twist condition given and must be rejected. Hence, the proof.


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Example 6

a. A steel transmission shaft of a Mercedes Benz articulator truck is 570mm long and 50mmexternal diameter. For part of its length, it is bored to a diameter of 25mm and for the restto a diameter of 38mm. Find the maximum power that can be transmitted at the speed of

210 rev / min if the shear stress is not to exceed 70MN/m2.

Data given

It is important to note and understand that the case of this shaft presented is one of a shaftconnected in series because the torque generated would be shared.

Length of transmission shaft = 570mm = 0.57m

External diameter of shaft, D = 50mm = 0.05m

Internal diameter 1, d 1 = 25mm = 0.025m

Internal diameter 2, d 2 = 38mm = 0.038m

From the torsion theory,

Note that maximum shear stress and the radius at which the torque would occur will be the samefor both shafts. Therefore the allowable torque for a know value of shear stress will depend on J.What we need is the value of J which would be least to gain the maximum power.

T = 1.3kNm

Therefore the maximum power

P = therefore


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P = =28.5kW

b. If the angle of twist in the length of 25mm bore is equal to that in the length of the 38mmbore, find the length bored to the latter diameter.

Let L 1 and J 1 refer respectively to the length and polar second moment of area of the 25mm bore

Let L 2 and J 2 refer respectively to the length and polar second moment of area of the 25mm bore

Mathematically, if torque is shared

T = T 1 + T 2


If angle of twist is equal in each portion

Conditions

1. Where the lengths are the same, it would mean that




But we choose not to cancel out the Ls, the immediate preceding equation can be rewrittenas

therefore substituting values


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= ( )

But the total length of shaft = 510mm which implies that

Substituting for

Therefore


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Example 7

a. A circular bar ABC, 3m long is rigidly fixed at its ends A and C. the portion AB is 1.8mlong and of 50mm diameter whilst BC is 1.2m long and of 25mm diameter. If a twistingmoment of 680Nm is applied at B, determine the values of the resisting moments at A

and C and the maximum stresses in each section of the shaft.b. What will be the angle of twist in each portion?

Take G = 80GN/m 2

Data given

Required

a. Values of resisting moments at A and Cb. Maximum stress in each portionc. Angle of twist in each portion


Notes:

1. Shaft is connected in parallel2. The torque is shared3. Angle of twist is the same in each portion

AB

C

Shaft 1: 50 mm diameter

Length: 1.8m

Shaft 2: 25 mm diameter

Length: 1.2mTorque, T =680Nm


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4. Same materials is used for both shafts, therefore G is the same

From the torsion theory:

So for the composite shaft, since angle of twist is the same

Since the material used to make both shafts are the same, G 1 = G 2

Therefore Substituting values

( )( ) ( )

If the torque is shared it means T =

Therefore 680 =

Substituting for , 680 =

Factorizing , 680 =

= = 58.3Nm

Therefore, T

680 - 58.3 = 621.7Nm

Therefore torque in shaft 1 = 58.3Nm and torque in shaft 2 = 621.7Nm


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For portion AB, the maximum shear stress is given as

= 25.33 MN/m 2

= 19.0 MN/m

2

The shafts would have the same angle of twist so we can use any formula = 0.0228 rad

This is an angle so it would be better to be converting it to degrees

0.0228

Proof that yields the same results

= = 0.0228 rad This is an angle so it would be better to be converting it to degrees

0.0228


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Example 8

A hollow low carbon steel shaft is subjected to a torque of 0.25MNm. If the ratio of internal to

external diameter is 1:3 and the shear stress due to torque has to be limited to 19.0 MN/m 2.

Determine the required diameters and the angle of twist per meter length of shaft. Take G =80GN/m 2.

Given data

Shear stress,

Shear modulus,

d = 0.333D

Requireda. Internal diameter, d = ?

External diameter, D = ?

b. Angle of twist per meter length of shaft



Torque, T =0.25MNm


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But

Substituting given values

D = D = 0.264m = 264mm

But d = 0.333D

Therefore

So

a. External diameter, D = 264mmInternal diameter, d = 88mm

b. Angle of twist per meter length of shaft,


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Remember that now we know d and D and therefore we should be able to calculate for J andhence the angle of twist.

= =

This is an angle so it would be better to be converting it to degrees


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Example

A solid steel shaft of diameter, d = 60mm and length, L = 4m is to be designed using an

allowable shear stress, and an allowable angle of twist per length, , = per

meter.

a. Determine the maximum permissible torque, T max that may be applied to the shaft,

assuming G = 80GPa

b. Determine the angle of twist

Data given

Diameter of shaft, d = 60mm = 0.06m

Length, L = 4m

Allowable angle of twist per length, = per meter

Allowable shear stress, = Shear modulus, G = 80GPa =

Required

Maximum permissible torque, T max that may be applied to the shaft Angle of twist


In this case our usual =


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Substituting values

= =

If for the design the allowable angle of twist per length is 1 o per meter, it implies that it is equal

to

. This design consideration will give us a different torque other than what the

diameter just gave us. Apply the torsion theory therefore we will get

= = 1776Nm

So for the design the maximum permissible torque, T max , should be the smaller of the two values

calculated which is equal to

.

Therefore the angle of twist corresponding to this permissible torque can be determined using the

torsion theory given as

The alternative equation given by the torsion theory which involves shear stress and radius will

also yield the same results.


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Example

A hollow shaft and a solid shaft constructed of the same material have the same length and the

same outside radius, r. The inside radius of the hollow shaft is 0.6r. Assuming that both shafts

are subjected to the same torque, compare the maximum shear stresses, the angles of rotation,and the weights of the two shafts.

Solution

It can be inferred from the torsion theory, that the maximum shear stresses areinversely proportional to J, the polar second moment of area. Better still, it can be reasoned outthat the shear stresses are proportional to . Therefore, if the torque, T and radii, r are the same For the solid shaft,

For the hollow shaft

We can therefore write the relationship properly and mathematically as

= = 1.15

This means that the ratio of the shear stress in the hollow shaft is approximately 1.15 times that

of the solid shaft.

% of more shear stress carried by hollow shaft =

Also since the lengths are the same and they are made of the same material, from the torsion

theory, it can be inferred that the angle of twist for hollow and solid shafts would be inthe same proportions.


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Further, the weights of the shafts are proportional to the cross-sectional areas; hence

the weight of the solid shaft is proportional to

the weight of the hollow shaft is proportional to

% ratio of weight of hollow shaft to solid shaft = = 64%

This means that the weight of the hollow shaft is only 64% that of the solid shaft. The advantage

of using a hollow shaft is obvious. In this example, the hollow shaft carries 15% more shear

stress and can take on 15% times more of the angle of twist at 36% less weight.

In addition, the relative efficiency of a structure is measured by the strength to weight ratio. In

this example we would define it as the allowable load divided by the weight.

The allowable torque for the hollow shaft is

=

=

The allowable torque for the solid shaft is

=

=

Weight of the hollow shaft

Remember that Let us represent specific gravity with

Weight of the solid shaft


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Same material so densities and specific gravities are the same

Denoting the strength to weight ratio of the hollow and solid shafts to be s 1 and s 2 respectively,

% increase in strength to weight ratio =

It is obvious from the above that the hollow shaft has a greater strength to weight ratio. It canalso be seen that a hollow shaft of the same length and torque has of 36% more strength than asolid shaft.

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