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TORSİON of SHAFTS

(tvid- 4.1.a , tvid- 4.1.b)

4.1

4.1.1 Shafts and Their Place in the Industry

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4.1 Torsion of Shafts

• The circular cross-sectional rod elements that we will examine within the scope of this topic are called shafts.
• Shafts are power transmission elements that are frequently used in industry.
• While the shafts can be connected to a motor, parts such as gears, pulleys, etc. are also placed on them.In this state, they transmit the power received from the engine to other parts and in the meantime are subjected to the loading we call torsion.

(Shafts may be subject to bending besides torsion at the same time.. This subject will be discussed later. Only shafts subject to torsion will be discussed here.)

shaft

motor

shaft

Figure 4.1.1

Figure 4.1.2

Figure 4.1.3

4.1.2 What is Torsion Loading?

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• Now, our first goal is to calculate the stresses occurring in circular cross-section rod elements (shafts) due to torsional moment…>>

• Forced to rotate a rod-like element from both ends in opposite directions around its own axis is called torsional loading.
• Torsion is caused by the torsional moments (also called torque or twisting moment) we apply from both ends.
• According to the right-hand rule, the torsional moment vector is always in the direction of the rod axis.
• Formulas for calculating stress or deformations in torsion vary depending on the cross-section geometry.

(Calculations for non-circular sections are shown in Strength 2 lesson.)

Spinning laundry technically means applying a torsional moment to the laundry from both ends.

4.1 Torsion of Shafts

Figure 4.1.4

Figure 4.1.5

Figure 4.1.6

Angle of twist (φ) varies in direct proportion to the applied torque T, that is, the value of the torsional moment, and the shaft length.

2- Before torsion, plane circular sections remain plane after torsion and their shape does not change. There is no distortion in the cross-sections. (Distortion occurs in non-circular cross-section elements)

3- A linear diameter line drawn (marked) on the section remains linear after torsion.

Based on these assumptions, the stress distributions and deformations that occur after torsion under elastic loading in a circular cross-section bar element can be calculated.

Attention: These assumptions cannot be made for non-circular cross-section elements. Calculation approaches and formulas for stress and deformation during torsion in this type of elements are completely different.

4.1.3 Acceptances made for torsion of shafts

• Based on observations, the following assumptions can be made within elastic limits for a shaft subject to torsion:

4.1 Torsion of Shafts

4

Cantilever Shaft

(Shaft with one end built-in or fix)

Figure 4.1.7

(a)

(b)

(c)

(d)

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4.1.4 Stress and Deformation Calculations in Torsion: :

• Let a shaft be subjected to torsional moments T within elastic limits at both ends.

Now, we will continue a little further and find the stress and deformation in terms of known ones..>>

4.1 Torsion of Shafts

• As a result of this moment T_int , an internal force of dF falls on each element dA. The sum of the moments of these dF forces relative to the center of the section is equal to the moment T_int.

(4.1.1)

(4.1.2)

• When we separate this shaft, which is in static equilibrium, imaginarily with the C plane, each part of it will be in balance under the influence of internal and external forces. (separation principle).

• If we look at the equlibrium of the left part, we see that the external torsional moment T at B is balanced by the internal moment T_int arising at the separation surface C. From static equilibrium, T_int = T

• The internal moment (T_int) forces the section to rotate (i.e. torsion) around the shaft axis. For this reason, dF forces must be parallel to the surface. In this case, the stress falling on the area dA becomes shear stress (𝜏).

Left part of the cut

internal moment

Figure 4.1.8

(a)

(b)

(c)

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• Examine the figure below carefully: We consider a circumference point C on the built-in section of a solid cantilever shaft with radius r, and a point B on the circle with radius ρ in the same section. A is another point on any cross-section and aligned with B.

• As a result of the deformation caused by the internal torsion moment, point 𝑏 comes to point 𝑏' and point 𝐴 comes to point 𝐴'. Points C and B remain in place due to the built-in connection. O𝐴′𝑏′ is also a straight line. We can also see from the figure:

• If we apply Hooke's Relation to the differential element at point Q, :

(4.1.3)

(4.1.4)

• If we substitute it into the equation we found on the previous page:

: is the polar moment of inertia of the section.

As a result;

Angle of Twist:

(The amount of rotation of two sections relative to each other)

(4.1.5)

and

Shear Stress:

at a point r from the center of the section

(4.1.6)

4.1 Torsion of Shafts

• OAb is a straight line, b is the outer circumference point and is aligned with C.

Figure 4.1.9

Solid Shaft

Hollow Shaft

• Equations 4.1.5 and 4.1.6 are valid only for circular sections, isotropic materials and in the elastic region.
• Equation 4.1.5 is valid if there is no other moment, section or material change between the sections.
• T_int. in the formulas is the internal torsion moment in the examined section.

(4.1.7)

(4.1.8)

4.1 Torsion of Shafts

7

Polar moment of inertia for Circular Sections

Figure 4.1.10

Figure 4.1.11

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4.1.5 Shear stress distribution in circular section:

4. At the center, the stress is zero because ρ = 0.

5. Shear stresses are in the tangent direction of the circle. Their direction is in the direction of rotation of the T_int moment.

«Sometimes we can use c instead of r».

The shear stress at a point ρ away from the center of the section is found from equation 4.1.6:

Now we will interpret this equation: :

(4.1.9)

4.1 Torsion of Shafts

1. The stresses at points at the same distance from the center, that is, on the same circle (e.g. points a, b, c), are equal to each other.

3. Maximum stress occurs in the outermost circle. Because ρ_max = r in the outer circle.

2. Since ρ increases as you move outwards, the stresses also increase (linearly).

(4.1.6)

The internal moment (𝑇_𝑖ç ) and Polar moment of inertia ( J ) values in a cross section are constant. The only thing that changes is the distance (ρ) value from the center. According to this:

Figure 4.1.12

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Example 4.1.1

radius

Solution:

We have shown that maximum shear stresses occur at the outermost points of the circular section. Accordingly, if we write Equation 4.1.9;

From Equation 4.1.8, the polar moment of inertia of the hollow circular section is:

Minimum stress at safety limit:

(Torsional moment at any instant):

4.1 Torsion of Shafts

Figure 4.1.13

Figure 4.1.14

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Example 4.1.2 ⁽ⁿ⁾:The BC part of the stepped shaft in the figure is hollow and its inner diameter is 90 mm and its outer diameter is 120 mm. The other parts are solid and their diameter is "d". According to this;

a-) If the allowable (safe) shear stress value for parts AB and CD is 65 MPa, calculate the allowable diameter value of these parts (d_all. = ?)

b-) Find the maximum and minimum shear stresses in section BC.

Solution: We will divide the shaft into regions, starting from end A to D. When the moment (T) and shaft cross-section (J) change, the region changes. Accordingly, there are 3 regions: AB, BC and CD. We will now apply the separation principle to each region, determine the internal moments from static equilibrium, and substitute this internal moment into the stress formula.

BC region (left part)

CD region (right part)

4.1 Torsion of Shafts

AB region (left part)

Figure 4.1.15

Figure 4.1.16

(a)

(b)

(c)

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a)

b)

* For the CD part, the same result is found since the internal moment value is the same.

For the solid section AB of the shaft, the stresses at the outermost point are maximum at any instant:

At the safety limit:

For the part BC:

T_int = 20kNm was found

The greatest (maximum)stress of all at that instant.

4.1 Torsion of Shafts

and

allowable

Figure 4.1.17

Figure 4.1.18

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Example 4.1.3

The moments acting on the pulleys on the steel shaft, which is connected to a fixed wall at the D end, are shown in the figure. Shear yield strength of steel τ_ak =200MPa,

Since the safety factor is n=2, d2 = 18mm, d3 = 15mm,

a-) Determine the allowable diameter values of the DC part (d_1-em = ?).

b-) Check whether the AB part remains within the safety limits.

c-) To prevent material waste, what is the maximum diameter of a hole that can be drilled in the BC section, provided that it does not exceed the safety limits?

Answers: a-) 12.67mm, b-) It remains within safety limits, c-) 17.15mm

4.1 Torsion of Shafts

Figure 4.1.19

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Angle of twist (𝜙) shows the amount of rotation of shaft sections relative to each other and is especially useful in solving hyperstatic problems in torsion.

Equation 4.1.5 can be applied to such a part of the shaft as shown in the figure;

in that part, there is an equal and opposite torsional moment T at both end sections.

There is no other moment in between.

T in the formula is actually the internal moment (T_int.) in the section. Material and section geometry are the same along L.

One of the sections A or B does not have to be built-in and can be two sections in the middle region of the shaft.

a and b are the points in the same alignment before torsion, and after torsion, these points move to 𝑎′ and b′ positions. Point 𝑎′′ is the projection of a on section B.

4.1.6 Deformation and Hyperstatic Problems in Torsion

Now we will interpret equation 4.1.5.

We have shown that rotation of two sections relative to each other in a shaft subject to torsion, that is, angle of twist (also known as the twist angle or angle of rotation), is calculated from equation 4.1.5.

Angle of twisit - φ_B/A: It is the angle between b′ and 𝑎′′, and represents the amount of rotation of section B relative to A.

Examine the figure on the side and try to understand exactly where the φ_B/A angle is.

4.1.6.1 Angle of Twist (φ) :

4.1 Torsion of Shafts

(G: shear modulus)

Figure 4.1.20

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4.1.6.2 Sign of Angle of Twist:

When we look at the section we are examining perpendicularly, if the internal moment (T_int.) in the section is clockwise, the φ angle is taken as positive (+), otherwise it is taken as negative (-). Whether we take the right or left part of the cut does not change this rule. Try to understand this rule thoroughly by examining the figures below.

Cuts from plane K

Clockwise

Clockwise

4.1 Torsion of Shafts

Figure 4.1.21

Figure 4.1.22

Figure 4.1.23

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• If at least one of the shaft diameter, material or T torsional moments changes along the shaft axis, these types of shafts are called stepped shafts.
• Stepped miles are first divided into regions.
• The φ angle is calculated separately for each region and these are summed.
• Thus, the rotation amount of one end section relative to the other, that is, the angle of twist φ, is found.

4.1.6.3 Angle of Twist in Stepped Shafts:

- Additionally, static equilibrium must be ensured between torsional moments.

Total twist angle:

• Attention should be paid to the sign of the angle f in each region. (sign rule was explained on the previous page)

- The internal moment (T_int-i) in a region i is found by the separation principle and substituted in equation 4.1.10.

This subject will be better understood as you solve sample problems. >>

(4.1.10)

4.1 Torsion of Shafts

Left side of cut I-I

For the example above:

Figure 4.1.24

Figure 4.1.25

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Example 4.1.4

If the shear modulus of the stepped steel shaft in the figure is G = 77 GPa, find the total rotation (twist) angle of end A.

Solution: From static equilibrium, the reaction moment at the fixed end is T_D = 2250 kNm. However, if we take the right part in the sections, we do not need to find TD.

We can divide the shaft into areas AB no. 1, CB no. 2 and CD no. 3.

Actually, what is asked is the rotation (twist) angle of plane A relative to plane D (∅=∅_𝐴𝐷=?).

This is equal to the sum of the rotations of the regions from D to A relative to each other. Well;

.

If you noticed this, you understand this topic very well:

In fact, the B plane shifts by the amount φ_CD and changes shape by the amount φ_BC. In this case, the total rotation (φ_BD) of plane B is the sum of these two angles. Try to make the same comparison for plane A.

4.1 Torsion of Shafts

Examine the figure on the side carefully and try to understand the rotation (twist) angles of the planes relative to each other.

Rotation angle of C relative to A

Figure 4.1.26

Figure 4.1.27

Figure 4.1.28

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L1 =400mm (total length of the 1st region )

1st Region: AB

Static equlibrium requirement :

L2 =200mm (total length of the 2nd region )

2nd Region: CB

4.1 Torsion of Shafts

If the II-II section is viewed vertically and from the front, it is understood that φ is positive since T_int-2 is clockwise.

If the I-I section is viewed vertically and from the front, it is understood that f is positive since T_int-1 is clockwise.

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3rd Region: CD

L₃ =600mm ((total length of the 3rd region )

looking direction

If the III-III section is viewed vertically and from the front, it is understood that φ is positive since T_int-3 is clockwise.

Total twist angle:

4.1 Torsion of Shafts

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Those who study the subject carefully may have a question like this:

4.1.7 Are torsion formulas valid for rotating shafts?

Shafts used in industry are mobile and rotate around their own axes at a constant angular speed.

As if we only derived the formulas for the shafts at rest.

Are the formulas we have derived for rotating shafts also valid?

Because we know from Newton's 2nd law that:

Total force and total moment are zero not only for stationary objects but also for moving objects whose acceleration of the center of gravity is zero. Therefore, static equilibrium equations are again satisfied for objects performing this type of motion. Shafts rotating with constant angular velocity also fall into this class. For such shafts the equations of static equilibrium apply, so the principle of separation can be applied to these shafts as well, the internal torsional moments can be found from static equilibrium and all the formulas we have derived for torsion can be deduced in the same way. As a result, all the torsion calculations and equations we have made in this section are also valid for shafts rotating around their own axis at constant angular speed. This situation also shows us that static and strength calculations have a very wide application area.

Answer: Static equilibrium equations can also be written for rotating shafts, and the formulas we have derived for the stationary state are also valid for them.

4.1 Torsion of Shafts

Figure 4.1.29

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• An important trick here is to correctly determine the boundary conditions of the rotating shaft. The shaft transmits the power it receives from one end to the other parts of the mechanism from the other end. If we apply the force coming from other components to one end, we must build-in (fix) the end connected to the engine, so that static balance is achieved and the operating conditions are reflected in the solution. (For shafts that are not directly connected to the engine, different boundary conditions appropriate to the operating conditions can be considered, but static balance must be ensured in all cases.)

• In this case, the moving shaft will be transformed into a static state and resolved in this state, but the stresses or deformations found will correspond to their actual values on the rotating shaft. When we separate the shaft connected to the motor and the gear at the end from the system in the figure above, we fix the left end of the shaft connected to the motor from which it receives power. We apply the force F coming from the other gear to the gear connected to the other end of the shaft through which power is transmitted. In this way, we define the correct loading and boundary conditions to the system.

4.1 Torsion of Shafts

Figure 4.1.30

Figure 4.1.31

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The torque (torsional moment) taken from the electric motor in the figure is transmitted to the other parts of the mechanism as 200Nm and 300Nm by gears B and C. The shear modulus of the aluminum shaft is G=27GPa. Bearing A allows rotation. In this situation,

a-) In which direction and how much torque does the engine produce,

b-) Calculate the amount of rotation (angle of twist) of end A relative to D.

Solution:

Although this mechanism is in motion, the total acceleration of the center of gravity is zero because the shaft rotates at a constant angular velocity around its own axis. In this case the total torsional moment is zero and hence the static equilibrium equations are satisfied for this system as well. In addition, the engine part is built-in (fix) to ensure static equilibrium and working conditions are reflected in the system...>>

Example 4.1.5 :

4.1 Torsion of Shafts

Figure 4.1.32

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a-) From static equilibrium: The reaction torque at cantilever end D is equal to the torque produced by the motor.

b-) angle of twist (amount of rotation of end A relative to D) .

Since bearing A allows rotation, no reaction moment occurs at point A. From the static equlibriumof the Shaft:

I

II

II

III

III

D

Why were φ angles taken negative?

Try to understand the sign rule by examining it (see 4.1.6.2).

clockwise

4.1 Torsion of Shafts

4.1.8 Hyperstatic shaft problems in torsion: There are also hyperstatic torsion problems in which the number of static equations is not sufficient to find the unknown forces. Unknown forces can be found by obtaining additional equations with the help of a twist angle whose value can be seen beforehand. We will try to understand this issue with examples.

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The shaft, which is hollow up to its halfway point, has an outer diameter of 22 mm and an inner diameter of the hollow part of 12 mm. The shaft is fixed from its ends A and B. A torque of 120 Nm is applied to the middle of this shaft. Accordingly, calculate the reaction moments occurring at points A and B?

Example 4.1.6 ^:

4.1 Torsion of Shafts

Solution..>>

Figure 4.1.33

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Solution:

From static equlibrium:

Since equation 1 is not enough, this is a hyperstatic problem and 1 more independent equation is needed. Since B and A are built-in, there is no rotation in these parts. Therefore, the total rotation angle (twist angle of B relative to A) is zero.

( 1 )

From equa. (1) and (2) ..>>

( 2 )

φ < 0

φ > 0

Internal moment: T_int-1 counterclockwise

T_int-2 clockwise

(For a vertical view of the section)

4.1 Torsion of Shafts

We divide the shaft into two regions, AC and CB.

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Example 4.1.7

An aluminum shaft is placed inside a steel pipe, and the system is connected to a rigid cover at its left end and to a fixed wall at its right end. Calculate the torsional moment T that can be applied to the system from the fixed cover within the safety limits.

G_alum.=27 GPa, G_steel= 81 GPa , τ_steel-all. =100 MPa, τ_alum-all. =50 MPa

Solution:

In what proportions is the total torsional moment T distributed over steel and aluminum? We need to calculate this first.

Static Equilibrium of the cover

(1)

The important tip for the solution: End B of steel and aluminum rotates the same amount compared to end A due to the rigid cover:

I

I

Cut I-I

Cut II-II

(2)

4.1 Torsion of Shafts

Figure 4.1.34.a

Figure 4.1.34.b

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First, if we make the calculation according to steel safety:

Maximum stress at any time:

Maximum stress at safety limit for steel:

In this case, from equation 2, amount of moment carried by aluminum:

Now let's make the calculation according to aluminum safety:

However, we do not yet know whether aluminum can withstand this moment value.

4.1 Torsion of Shafts

Allowable moment that steel can carry within its safety limits:

From the last equation above:

Figure 4.1.35

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We equalize the maximum stress in aluminum to the safety stress and calculate the allowable moment value:

In this case, from equation 2, amount of moment carried by steel :

According to the safety of steel, aluminum is also safe because

The safe moment value that can be applied to the cover from equation (1):

4.1 Torsion of Shafts

Allowable moment that aluminum can carry within its safety limits:

According to aluminum safety:

Final Comment :

According to the safety of aluminum, steel is unsafe because

Therefore, steel safety must be taken into consideration for the safety of the entire system.

Figure 4.1.36

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The stepped shaft, made of copper and steel parts, is placed between two fixed walls and a torsional moment of T = 2.5kNm is applied from point B. Accordingly, check whether yielding occurs in the shaft.

Example 4.1.8

Answer:No yield occurs.

4.1 Torsion of Shafts / Cevaplı Sorular

Figure 4.1.37

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A steel shaft is placed inside an aluminum tube with a gap, both of them were connected to a fixed wall from the right side and a rigid cover from the left end. Since the allowable shear stresses for the steel shaft and aluminum tube are 120 MPa and 70 MPa, respectively, find the maximum torque (torsional moment) value that can be applied to the system through the cover. (G_Alum.=27 GPa, G_steel=77 GPa) Answer: 6319 Nm

Example 4.1.9*

4.1 Torsion of Shafts

Figure 4.1.38

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Calculate the maximum stresses occurring in sections AB and BC in the stepped shaft in the figure.

(Answers: 56.6 MPa, 36.6 MPa )

Example 4.1.10

4.1 Torsion of Shafts / Cevaplı Sorular

Figure 4.1.39

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Example 4.1.11

(Industrial Eng. 2006 2nd visa )

4.1 Torsion of Shafts / Cevaplı Sorular

In the power transmission system shown in the figure, pulleys A, B, C and D are connected to a solid shaft of constant diameter. The torsion moment on each pulley is shown in the figure. The system rotates with a constant angular velocity. Taking the shear yield stress of the shaft material as τ_akma=150MPa and the safety factor as n=3,

1. Calculate the moment T_B acting on gear B ?

b-) Determine the safe diameter of the shaft?

Answers: a-) 5kNm, b-) 93.42mm*

*You can find the solution to this question among the exam questions on mehmetzor.com.

​

Figure 4.1.40

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G_bronze = 35 GPa , G_aluminum = 28 GPa, G_steel = 83 GPa

The stepped shaft, made of three different materials, is connected to fixed walls at both ends as shown in the figure. Calculate the maximum stress that will occur in each material with the effect of the torsional moments applied from points C and D.

4.1 Torsion of Shafts / Cevaplı Sorular

Cevaplar: τ_bronze=111.79MPa, : τ_Alum.=1.75MPa, τ_steel 214.16MPa

Example 4.1.12

Figure 4.1.41