Stress on Torsion Bar Section

Solution:

Calculate the number of rivets in shear, apply the formula

C F

n

 d 2

 C 



 



720 . 10  3

20 2 . 10  6 4

. 100

Draw:

n n 

24

Calculate the number of rivets to be crushed, apply formula 10-4

P d F d

n d 





720 . 10  3

Draw:

n 

db d

n 15

 3

20 . 10  3 . 10 . 10 . 100

Thus, if calculated by cutting, 24 rivets are required, and if calculated by stamping, 15 rivets are required. To satisfy both the cutting and stamping durability conditions, we must choose 24 rivets.

4. PURE TWIST

4.1 Concept of pure torsion

4.1.1 Definition

A balanced bar under the action of external forces is the torques located in the cross-sections of the bar, the bar will be subjected to torsion.

For example:

Figure 2.29

section method

A circular cross-section bar with one fixed end and one free end is subjected to the action of a torque m = Pa (Figure 2.29).

4.1.2 Internal force

To determine internal force we use the method

Imagine cutting the torsion bar AB into 2 parts A, B remove end A and keep end B for consideration.

To balance end B, we need to apply an internal force M x to the cross section with a moment value equal to and opposite to the couple (P,P),

(Figure 2.30).

M x = m = Pa

A B

M x

Figure 2.30

4.1.3 Deformation in torsion bar

Consider a bar with a circular cross-section, draw the generating lines representing the longitudinal fibers, the lines perpendicular to the bar axis representing the cross-sections of the bar, those lines form rectangular boxes (Figure 2.31). At the end of the bar, draw a radius r.

Acting on the coupler bar, we find:

- When subjected to torsion, the cross-sections of the bar rotate around the axis at a certain angle but remain round with the old radius, still flat and perpendicular to the axis of the bar.

- The distance between the two cross-sections before and after torsion remains constant.

- Before and when subjected to torsion, the radius of the cross-section remains straight and has a constant length.

The angle of rotation of the face radius is called the absolute torsion angle, denoted (Figure 2.31).



Figure 2.31

Score

   is called the relative torsion angle, where l is the length of the bar.

Under the effect of torque, the material elements on the cross-sections move at a relative angle ê.

We have the relationship between and on the outside of the bar:

 .l  .r

good

 . r

good

 .r

A point at a distance from the axis will make a sliding angle ê = ì .

Thus, the shear strain in the torsion bar changes continuously and increases gradually from the inside of the bar to the outside of the bar.

At the axis, the slip is the smallest: ê = 0. At a point a distance from the axis: ê = ì.

At the outer surface, ê reaches its maximum value: ê max = ỉ.r

4.1.4 Stress on torsion bar cross section

 pi

M x

 pi



Observing the deformation of the torsion bar, it can be concluded that on the cross-section of the bar there is no normal stress but only shear stress, the direction of the stress is perpendicular to the radius passing through.

point under consideration.

According to the butt law of shear deformation: = ê .G

Because the value of e in the cross section of the bar changes from 0 to the maximum value corresponding to the position of the points from the center to the outside. Therefore, the value of shear stress also changes from 0 ÷ max .

max = ê max .G = ỉ.rG

We can represent the change of stress by a diagram (Figure 11.11).

According to the chart we have:

Figure 11.11

Figure 2.32

= max ./r

 max

To solve the strength problem, we must find the relationship between internal moment Mx and the maximum stress max generated on the outside of the bar. To get that relationship, we do the following:

Divide the surface F into n parts F 1 , F 2 , F 3 , ... (Fi ) so that the elements are small enough to have

The internal force can be considered to be uniformly distributed by the corresponding stress as:

  1 ,   2 ,   3 ,...(   i ) .

F = F 1 +F 2 +...+F n the distances from the divided elements to the center are 1 , 2 , 3 , ... n .

after:

Thus, the internal torsional force on each area element F i is Mi determined as

Mi = i .F i . i

But M x =

M i   i .F i .  i (i = 1, 2, ..., n)

From that we have:

M x    i .F i .  i

i  1

According to the formula: = max ./r

We c:

M  

 max .  i .F . 

 max  n

F .  2

i  1

r i  1

Let J 0 = F i .  i

i  1

(J 0 is called the single pole moment of inertia, unit is m 4 )

Substitute into the above expression and we have:

M   max .J

x r 0

Put

0  W 0

Hence:

M x  

max

.W 0

good

M x





max

W 0 characterizes the torsional resistance of the bar, called the torsional moment, in m 3 . For a bar with a circular cross-section:

J 0 

 .d 4

 0.1.d 4

and W 0 

 .d 3

 0.2.d 3

4.2 Torsion calculation

4.2.1) Strength conditions:

For a bar to withstand torsion, the maximum stress generated in the bar must be less than or at most equal to the allowable stress of the material used to make the bar.





M x

max

  

4.2.2) Select cross section:

From the intensity condition we deduce:

W  M x 0   

Note: For shafts transmitting rotational motion with power P, external torque is calculated according to the formula:

m = 9.55. P (Nm)

In which: Power P is calculated in watts, angular velocity n is calculated in rpm.

Example 1:

A circular cross-section bar is subjected to two torques acting on the cross-section at the two free ends M x = 2kNm. The cross-section has a diameter d = 6.5cm. Check the strength of the bar knowing that     40MN / m 2 .

Prize:

With circular cross section we have:

W 0 ≈ 0.2 d 3 = 0.2 . 0.065 3 = 54.10 -6 m 2

Applying the formula we have:

 max

 M x

W 0

 2000

54.10  6

 36.10 6  40.10 6 N / m 2    

So the bar is guaranteed in strength.

Example 2:

A steel shaft with a capacity of 295 kW rotates at n = 300 rpm. Calculate the shaft diameter according to the strength condition, knowing that     80MN / m 2 .

Prize:

From the formula m = 9.55. P we have:

m = 9.55. P=

To calculate the shaft diameter we have:

9.55.

295.10 3 

300

939Nm

W  M x 0   

=> 0.2d3 M x





M x

3 0.2.   

0.2.80

939.10  6

 2

=> d  

Choose d = 4 cm

 3.9.10 m

Questions, exercises

1. What is a true tension and compression bar? Name, symbol and how to calculate stress in tension and compression bars.

2. Write and explain the formula for calculating the deformation of tensile and compressive bars. State Hook's law on tensile and compressive forces.

3. What is a shear bar? Name, symbol and how to calculate stress in a shear bar. How to calculate shear and stamping?

4. What is a torsion bar? What stress arises in a torsion bar?

? What is the distribution rule? Write and explain the formula for calculating the maximum stress in the cross section of the bent bar.

5. What is a bending bar? What are the properties of the internal forces in the bending bar? Clearly state the stress components that arise. What is the distribution rule?

6. State and write the expression for the strength conditions of bars subjected to tension, compression, shear, crushing, and torsion.

7. With a steel bar of length l = 4m, diameter d = 12m, what force P must be applied so that it stretches 0.5cm? The stress generated when that force is applied. Knowing E = 2.10 - 5 MN/m 2 .

Answer: P = 2.8.10 3 kN; = 200MN/m 2

8. Draw the shear force and bending moment diagram of the beam under uniformly distributed load q = 10kN/m (Figure 11.21).

Prize:Y A qY B

a. Determine the reactions Y A , Y B :1

Let the resultant force be R: A 1 B

R = ql = 10.2 = 20kN

 m A Y B .2  1.R  0

 m B Y A.2  1.R  0

Y A

Maybe you are interested!

=> Y B = R 20  10KN

2 2

YA = R 20  10KN 2 2

b. Set up the expression Q(x), M(x): Use 1-1 section with:

10kN

Qx (+)

(+)

(-)

10kN

0  x  2

Q(x) = Y A - qx = 10 - 10x

5kN

Figure 2.42

M(x) = Y A .x -

q .x  10  5x 2

Figure 11.21

c. Draw the graph Q(x), M(x):

Shear force Q(x) = 10 - 10x so the graph is a diagonal line determined by 2 points (Figure 11.21).

x = 0 then Q(x) = 10 x = 2 then Q(x) = -10

Bending moment M(x) = 10 - 5x2 so the graph is a quadratic parabola (figure 11.21). Determined by the following points:

x = 0 then M(x) = 0

x = 1

then M(x) = 15 4

x= 1 then M(x) = 5

x = 3

then M(x) = 15 4

x = 2 then M(x) = 0

 Qmax = 10 and Mmax = 5

9. Draw the diagram of shear force Q and bending moment M of the beam subjected to concentrated moment m (Figure 2.42).

Prize:

Y A

Y B

m = 20 kNm

10kN 10kN

(+)

10kN

(-)

(+)

Figure 11.22

10kN

Figure 2.42

CHAPTER 4: MACHINE DETAILS

Target:

- Present the concepts, working principles and application scope of gear transmission, chain transmission, worm gear - screw transmission, rack - pinion transmission, transmission mechanisms of planers, hammers, and slotters.

- Determine the transmission ratio of each transmission.

- Analyze the structure and working principles of some motion conversion mechanisms such as: cam mechanism, rocker arm, and pulley mechanism.

- Explain the application of motion conversion mechanisms.

- Practice discipline, perseverance, carefulness, seriousness, initiative and creativity in learning.

1 Belt mechanism

1.1 Concept

1.1.1. Definition

Figure 3.9 Conventional belt drive

Figure 3.10 Cross and semi-cross belt drive

Belt drive works based on the principle of friction force between the belt and the pulleys that moves and mechanical energy from the leading pulley to the driven pulley. Because the belt is a soft link, after a period of operation it will be driven, so it is necessary to have a belt tensioning measure to fix it.

Belt drive is used to transmit rotary motion between shafts by the friction force between the belt and the pulleys. When the drive pulley rotates, the drive belt makes the driven wheel rotate.

1.1.2 Types of belt drives

* Based on the transmission position divided

- Transmission between two parallel shafts

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