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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/Torque8/4/2019 Torsion Repaired)
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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_constant8/4/2019 Torsion Repaired)
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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
Calculation and Solution
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
The resultant equation by this condition would be
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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
The resultant equation by this condition would be
2. Where the material used is said to be the same, it means
The resultant equation by this condition would be
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
External diameter, D = 25cm = 0.25m
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,
Calculation and Solution
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
External diameter, D = 100cm = 0.1m
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
Calculation and Solution
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:
From the torsion theory
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
Therefore maximum shear stress in the shaft is MN/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
Therefore maximum shear stress in the shaft is MN/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
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
The resultant equation by this condition would be
2. Where the material used is said to be the same, it means
The resultant equation by this condition would be
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
Calculation and solution
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
Calculation and solution
From the torsion theory
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
Calculation and solution
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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