Home Convectors Find the diameter in torque. Calculation and graphic work

Find the diameter in torque. Calculation and graphic work

Task 4

For steel shaft of constant cross section

1. Determine the value of the moments M 1, M 2, M 3, M 4;

2. Construct a diagram of torques;

3. Determine the diameter of the shaft from calculations of strength and rigidity, taking the cross section of the shaft - a circle

P 1 = 50 kW

P 3 = 15 kW

P 4 = 25 kW

w = 18 rad/sec

w = n = = 30*18/3.14 = 172 rpm

[ts 0 ] =0.02 rad/m - angle of twist

G = 8*10 4 MPa

We determine the external moments:

M 1 = 9550 = 9550 = 2776 Nm = 2.8 kNm;

M 3 = 9550 = 9550 = 832.8 Nm = 0.83 kNm;

M 4 = 9550 = 9550 = 1388 Nm = 1.4 kNm;

Let's write the static equation:

UM = M 1 + M 3 - M 2 + M 4 = 0

And from it we find the value of the moment M 2:

M 2 = M 3 + M 1 + M 4 = 832.8 +2776 +1388 = 4996.8 Nm = 5 kNm;

First of all, we build a torque diagram. The torque values for the sections are as follows:

T 1 = -M 1 = -2.8 kNm;

T 2 = -M 1 - M 3 = -2.8 - 0.83 = - 3.63 kNm;

T 3 = -M 1 - M 3 + M 2 = -3.63 + 5 = 1.37 kNm.

We build diagrams:

The shaft is divided into three sections I, II, III.

We find the polar moment of resistance of the shaft required by the strength condition:

W p = = = 121 10 -6 m 3 = 121 cm 3

The diameter of the solid shaft is determined using the formula:

W p 0.2d c 3 = 121 cm 3,

d c 3 = = 8.46 cm 9 cm = 90 mm.

Then the diameters are calculated for the shaft sections based on the rigidity condition, i.e. using formula

d gesture1 = = 0.1 m = 100 mm

d gesture2 = = 0.1068 m = 107 mm

d gesture1 = = 0.0837 m = 84 mm

The largest diameter values calculated from the rigidity condition should be selected as the final ones. Thus, the final size of the shaft diameter is: d 1 = 107 mm.

From the standard range: d 1 = 120 mm

Task 5

A pulley and a wheel are rigidly mounted on the shaft,

Determine the forces F 2 .F 2r = 0.4 F 1 if the value of the force F 1 is given

Let's imagine a physical system:

We solve the problem in the following sequence:

1. We depict in the figure the body whose equilibrium is being considered, with active and reactive forces acting on it and select a system of coordinate axes;

2. From the condition of equilibrium of a body having a fixed axis, we determine the values of the forces F 2, F r2;

3. compose six equilibrium equations;

4. solve equations and determine support reactions;

5. check the correctness of the solution to the problem.

1. We depict the shaft with all the forces acting on it, as well as the coordinate axes

Let us consider a system of forces acting in the system

Determine the components of the load on the pulley side

P 1 = (2F 1 + F 1) = 3 F 1 = 3*280 = 840 N = 0.84 kN

2. Determine F2 and Fr2. From the condition of equilibrium of a body having a fixed axis:

F 2 = = = 507.5 H

F r2 = 0.4F 2 = 0.4*507.5 = 203 H

3. We compose six equilibrium equations:

YY = -P 1 - F 2 + A y + B y = 0 (1)

УX = -F 2r + A x + B x = 0 (2)

UM yC = -P 1 * 32 + A y * 20 - B y * 10 = 0 (3)

UM yB = - P 1 * 42 + A y * 30 - F 2 * 10 = 0 (4)

UM xC = A x * 20 - V x * 10 = 0 (5)

UM xB = A x * 30 + F 2r * 10 = 0 (6)

Consider equations (3) and (4)

840 * 32 + A y * 20 - B y * 10 = 0

840 * 42 + A y * 30 - 507.5 *10 = 0

From the last equation:

A y = 40355/30 = 1345 N

From the first equation:

26880 + 26900 = 10*V y? V y = 20/10 = 2 N

Consider equations (5) and (6)

A x * 20 - B x * 10 = 0

A x * 30 + 203 * 10 = 0

From the last equation A x = 2030/30 = 67.7 N

From the first equation: 1353.3 = 10*V y? V y = 1353/10 = 135.3 N

We will check using equations (1) and (2):

YY = -840 - 507.5 + 1345 + 2 = 0

УX = -203 + 67.7 + 135.3 = 0

The calculations were made correctly. The final reactions of supports A and B are:

A = = = 1346.7 N

B = = = 135.3 N

Torsional rigidity condition: .

From the conditions of strength and rigidity, the cross-sectional dimensions can be determined. The final diameter values should be rounded to the nearest standard according to GOST (30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100, 105, 110, 115, 120, 125, 130, 135, 140, 145, 150, 155, 160).

To ensure strength and rigidity at the same time, we select the larger of the two found diameters.

Example 1. For a steel transmission shaft with a constant cross-section along its length and rotating at a constant angular velocity. Construct a diagram of torques, determine the required shaft diameter from calculations of strength and rigidity, assuming that the cross-section of the shaft is a circle and the cross-section of the shaft is a ring with a diameter ratio of . Compare how many times lighter an annular shaft will be than a solid one. Accept: [τ To ] = 30 MPa R 2 = 0.5 R 1, R 3 = 0.3 R 1 R 4 = 0.2 R 1

G= 8·10 4 MPa [φ 0 ] = 0.02 rad/m

Given: R 2 = 52 kW

R 3 = 50 kW

R 4 = 20 kW

R 1 = 132 kW

ω = 20 rad/s

T 3 T 1 T 2 T 4

3.6·10 3 10 3

ep Mk, Nּ m

– 2.510 3

Solution:

Determine torques.

We divide the shaft into sections and determine the torque value in each section.

We build a diagram of torques.

We determine the shaft diameter from the conditions of strength and rigidity.

The dangerous section is the section IIM To max = 3.6· 10 3 N· m

Shaft cross section - circle

We accept d= 85 mm

We accept d 1 = 70 mm.

The required diameter turned out to be larger based on strength, so we accept d 1 = 85 mm.

Shaft section - ring

We determine the shaft diameter from the strength condition:

We accept D=105 mm.

We determine the shaft diameter from the stiffness:

We accept D= 80 mm.

The required diameters are finally taken based on strength

Example 2. For a steel shaft (Figure 11, A) determine from the strength conditions the required diameters of each section and the angles of twist of these sections. Take the angular velocity of the shaft  = 100 rad/s, permissible voltage [  ] = 30 MPa, shear modulus of elasticity G= 0.8  10 5 MPa.

Select the dimensions of the cross section of the shaft (Fig. 1) according to the strength condition. In sections from section 1 to section 3 and from section 5 to section 6, the outer diameter of the shaft is constructive considerations must be the same size.

In the section from section 1 to section 2, the shaft has an annular cross-section with n=d B /d=0.4. In sections from section 3 to section 5, the shaft is selected only according to strength conditions.

M = 1 kN∙m, [τ ] = 80 MPa.

Solution

We divide the shaft into power sections and build a torque diagram (Fig. 1, b).

Determine the shaft diameters. In sections I, II and V, the outer diameter of the shaft is the same. For them, it is not possible to indicate in advance the section with the highest tangential stress value, since different sections have various types cross section: section I – circular, section II and V – solid round.

It is necessary to determine separately, according to the strength condition, the diameters for each type of cross section for the most loaded power section (that is, the one on which the maximum absolute value of the torque acts). We will finally accept the largest diameter obtained.

For a section with a ring cross-section:

For a shaft of solid cross-section

We finally accept highest value the resulting diameter, rounded up to the nearest whole value:

d 1 = d 2 = d 5 = 61 mm;

d B1 = n∙d 1 = 0.4∙61 = 24.4 mm.

The highest voltage acting in these areas is:

Shaft diameter in section III (M K3 = 5M = 5 kNm).

3. Determine the shaft diameter from the strength condition.

= ≤ → ≥ ;

= → d = ≈73mm.

4. Determine the shaft diameter from the rigidity condition

= ≤ → Jp ≥ = =1458125

Jp = → d = = = 62mm

5. We finally accept the shaft diameter d=75 mm.

4. Tasks for independent decision

Task No. 1

For given beams, construct a torque diagram and determine the dangerous section.

Answer: Mz max a) 2m; b) 4m; c) 4m; e) 18kNM; e) 45kNM

Task No. 2

Determine the ratio of the diameters and masses of two shafts of the same strength and length, transmitting the same power, if one shaft rotates n 1 = 800 min -1, the other with n 2 = 1200 min -1.

Answer: d 1:d 2 =1.15; m 1:m 2 =1.31

Task No. 3

The steel shaft rotates with a rotation speed n=980min -1 and transmits power P=40kW. Determine the required shaft diameter if the permissible tangential stress [τ to ]=25MPa

Answer: d=43mm.

Task No. 4

A steel beam of annular cross-section (d=100mm and d0=80mm) 3M long is twisted at an angle of 30. Calculate the highest shear stresses occurring in the beam.

Answer: τ max =70MPa

Problem #5

The steel shaft d=60mm has a rotation speed n=900min -1. Determine the permissible value of transmitted power if [φ 0 ]=0.5

Answer: [P]=83.4 kW

Problem #6

Check the strength and rigidity of steel beams, if [τ k]=40MPa; [φ 0 ]=0.6

Answer: a) τ max =68.4MPa; φ 0 max =1.63;

b) τ max =27.6 MPa; φ 0 max =0.4.

Problem No. 7

Determine the required cross-sectional dimensions of the beam if the yield strength τ m = 140 MPa, and the required safety factor [n] = 2.5

Answer: d=65mm

Problem No. 8

The shaft transmits torque M=10kNM

Select the dimensions of the cross-section of the shaft for 2 x cases: a) solid circular section; b) rings with d 1 = D.

Compare sections in terms of material savings.

Allowable tangential stress [τ to ]=60MPa.

Answer: d=94mm; D=127mm; d 1 =111mm; ≈ 2.35.

References

1. Itskovich G.M. “Strength of Materials” M.: Higher School, 2005.

2. Arkusha A.I. “Technical mechanics”, “Theoretical mechanics and strength of materials”. M.: Higher School., 2002

3. Vereina L.M., Krasnov M.M. “Technical Mechanics” M.: Academy., 2008

Solid lines correspond to positive values of w, and dotted lines correspond to negative values, according to the sign rule. §1.3 Membrane analogy From the example discussed in the previous paragraph, it becomes obvious that the problem of torsion of a rod of a more complex cross-sectional shape can be very difficult. For an approximate solution of problems of torsion of rods of various sections, often encountered in...

The diameter of the bolts and the permissible stress of the bolt material for shear (shear) will be indicated accordingly. GEOMETRICAL CHARACTERISTICS OF FLAT SECTIONS When considering tensile, compressive, and shear deformations, it was found that the strength and rigidity of structural elements depends only on the size of the cross section and the properties of the material of the elements. For torsional and bending deformations,...

5.1 (Option 08)

Instructions: take the power on the gears P 2 =0.5P 1, P 3 =0.3P 1 and P 4 =0.2P1. Round the resulting calculated diameter value (in mm) to the nearest larger number ending in 0, 2, 5, 8 or ST SEV 208-75 [τ]=30 MPa.

Table 20 - Initial data

Task No. and diagrams in Fig. 35	R, kW	ω, rad/s	Distance between pulleys, m
			l 1	l 2	l 3
100.X	28	26	0,2	0,1	0,3

Answer: d 1 =45.2 mm, d 2 =53.0 mm, d 3 =57.0 mm, φ I =0.283º, φ II =0.080º, φ III =0.149º.

5.2

d) determine the diameter of the shaft, taking [σ]=60 N/mm² (in problem 117) and assuming F r =0.4F t. In problem 117, the calculation is carried out according to the hypothesis of the greatest tangential stresses.

Table 22 - Initial data

Task No. and diagrams in Fig. 37	Option	R, kW	ω 1 , rad/s
117.VII	08	8	35

Answer: R By =7145 H, R Ay =3481 H, d=51 mm.

5.3 For a steel shaft of constant cross-section (Fig. 7.17), transmitting power P (kW) at an angular velocity ω (rad/s) (take the numerical values of these quantities for your version from Table 7.4):

a) determine the vertical and horizontal components of the bearing reaction;

b) construct a diagram of torques;

c) construct diagrams of bending moments in the vertical and horizontal planes;

d) determine the diameter of the shaft, taking [σ]=70 MPa (in problems 41, 43, 45, 47, 49) or [σ]=60 MPa (in problems 42, 44, 46, 48, 50). For the forces acting on the gear, take F r =0.36F t, for belt tension S 1 =2S 2. In problems 42, 44, 46, 48, 50, the calculation is carried out according to the hypothesis of potential energy of shape change, and in problems 41, 43, 45, 47, 49 according to the hypothesis of the highest tangential stresses.

Table 22 - Initial data

Task number and diagrams in Fig. 7.17	Option	R, kW	ω, rad/s
Problem 45, scheme V	47	30	24

Answer: R By =4000 H, R Ay =14000 H, d=64 mm.

5.4 For one of the schemes (Fig. 35, Table 20), construct a torque diagram; determine the diameter of the shaft at each section and the full angle of twist.

Instructions: take the power on the gears P 2 =0.5P 1 , P 3 =0.3P 1 and P 4 =0.2P 1 . Round the resulting calculated diameter value (in mm) to the nearest larger number ending in 0, 2, 5, 8 or ST SEV 208-75 [τ]=30 MPa.

Table 20

No. of the task and diagram in Fig. 35	Option	R, kW	ω, rad/s	Distance between pulleys, m
				l 1	l 2	l 3
91.I	29	20	30	0,2	0,9	0,4

Answer: d 1 =28.5 mm, d 2 =43.2 mm, d 3 =48.5 mm, φ I =0.894º, φ II =0.783º, φ III =0.176º.

5.5 For a steel shaft of constant cross-section with one gear (Fig. 37), transmitting power P (kW) at angular velocity ω 1 (rad/s) (take the numerical values of these quantities for your version from Table 22):

a) determine the vertical and horizontal components of the bearing reaction;

b) construct a diagram of torques;

c) construct diagrams of bending moments in the vertical and horizontal planes;

d) determine the diameter of the shaft, taking [σ]=70 N/mm² (in problem 112) and assuming F r =0.4F t. In problem 112, the calculation is carried out according to the hypothesis of potential energy of shape change.