From the strength conditions to determine the diameter of the shaft. Determine from the strength conditions the necessary dimensions of the diameters of the reducerning step shaft. We construct the fusion of bending moments

Task 4.

For steel shaft permanent cross section

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

2. Build a torus of torque;

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

P 1 \u003d 50 kW

P 3 \u003d 15 kW

P 4 \u003d 25 kW

w \u003d 18 Rad / s

w \u003d n \u003d 30 * 18 / 3.14 \u003d 172 rpm

[C 0] \u003d 0.02 Rad / M - spinning angle

G \u003d 8 * 10 4 MPa

Determine external points:

M 1 \u003d 9550 \u003d 9550 \u003d 2776 HM \u003d 2.8 kNm;

M 3 \u003d 9550 \u003d 9550 \u003d 832.8 HM \u003d 0.83 kNm;

M 4 \u003d 9550 \u003d 9550 \u003d 1388 HM \u003d 1.4 kNm;

We write the statics equation:

Mind \u003d m 1 + m 3 - m 2 + m 4 \u003d 0

And we will find the magnitude of the moment m 2:

M 2 \u003d m 3 + m 1 + m 4 \u003d 832.8 +2776 +1388 \u003d 4996.8 HM \u003d 5 kNm;

First of all, we build a plumb of torque. The values \u200b\u200bof torque in the sections are as follows:

T 1 \u003d -m 1 \u003d -2.8kn;

T 2 \u003d mm 1 - m 3 \u003d -2.8 - 0.83 \u003d - 3.63 kNm;

T 3 \u003d mm 1 - m 3 + m 2 \u003d -3.63 + 5 \u003d 1.37 kNm.

Build pliers:

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

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

W p \u003d \u003d 121 10 -6 m 3 \u003d 121 cm 3

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

W p 0.2d C 3 \u003d 121 cm 3,

d C 3 \u003d 8.46 cm 9 cm \u003d 90 mm.

Then the diameters are calculated along the shaft plots from the hardness condition, i.e. using formula

d gesture1 \u003d \u003d 0.1 m \u003d 100 mm

d gesture2 \u003d \u003d 0.1068 m \u003d 107 mm

d gesture1 \u003d \u003d 0.0837 m \u003d 84 mm

As the final one should choose the largest values \u200b\u200bof the diameters calculated from the hardness condition. Thus, the final size of the diameter of the shaft is: D 1 \u003d 107 mm.

From the standard row: D 1 \u003d 120 mm

Task 5.

The shaft is rigidly attached pulley and wheel,

Determine the forces F 2 .F 2R \u003d 0.4 F 1 If the value of the force F 1 is specified

Imagine the physical system:

The task is solved in the following sequence:

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

2. From the body's equilibrium condition having a fixed axis, we determine the values \u200b\u200bof the forces F 2, F R2;

3. We make six equilibrium equations;

4. solve equations and determine the reactions of supports;

5. Check the correctness of the problem.

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

Consider the system of forces acting in the system

Determine the components of the load from the pulley

P 1 \u003d (2F 1 + F 1) \u003d 3 F 1 \u003d 3 * 280 \u003d 840 H \u003d 0.84 kN

2. Determine F2 and FR2. From the condition of equilibrium body having a fixed axis:

F 2 \u003d \u003d 507.5 H

F R2 \u003d 0.4F 2 \u003d 0.4 * 507.5 \u003d 203 H

3. We make six equilibrium equations:

Oy \u003d -r 1 - f 2 + a y + b y \u003d 0 (1)

Ux \u003d -f 2r + a x + b x \u003d 0 (2)

Mind yc \u003d -r 1 * 32 + a y * 20 - in y * 10 \u003d 0 (3)

Mind yv \u003d - p 1 * 42 + a y * 30 - f 2 * 10 \u003d 0 (4)

Mind xc \u003d a x * 20 - in x * 10 \u003d 0 (5)

The mind is x \u003d a x * 30 + f 2r * 10 \u003d 0 (6)

Consider equations (3) and (4)

840 * 32 + and y * 20 - in * 10 \u003d 0

840 * 42 + and y * 30 - 507.5 * 10 \u003d 0

From the last equation:

And y \u003d 40355/30 \u003d 1345

From the first equation:

26880 + 26900 \u003d 10 * in y? In y \u003d 20/10 \u003d 2

Consider equations (5) and (6)

And x * 20 - in x * 10 \u003d 0

And x * 30 + 203 * 10 \u003d 0

From the last equation a x \u003d 2030/30 \u003d 67.7 n

From the first equation: 1353.3 \u003d 10 * in y? In y \u003d 1353/10 \u003d 135.3 n

Verification We will produce according to equations (1) and (2):

Oy \u003d -840 - 507,5 + 1345 + 2 \u003d 0

Ux \u003d -203 + 67,7 + 135.3 \u003d 0

Calculations are made true. Finally, the reaction of supports A and B:

A \u003d \u003d \u003d 1346.7 n

B \u003d \u003d \u003d 135.3 n

Example 1. From the calculations for strength and rigidity to determine the required diameter of the shaft for transmitting power of 63 kW at a speed of 30 rad / s. The material of the shaft is steel, allowable voltage when crashes 30 MPa; permissible relative spinning angle [φ О] \u003d 0.02rad / m; Elastic module with shift G. \u003d 0.8 * 10 5 MPa.

Decision

1. Determination of cross-sectional dimensions based on strength.

Circular strength condition:

We determine the torque from the power formula when rotating:

From the strength condition, we determine the moment of the resistance of the shaft when crashes

Values \u200b\u200bsubstitute in Newton and mm.

Determine the shaft diameter:

2. Determination of cross-sectional dimensions based on rigidity.

Stiffness condition when cutting:

From the condition of rigidity, we determine the moment of inertia of the section when cutting:

Determine the shaft diameter:

3. Selecting the required diameter of the shaft from the calculations for strength and rigidity.

To ensure the strength and stiffness at the same time from two found values, choose greater.

The resulting value should be rounded using a number of preferred numbers. We practically round the resulting value so that the number is ending with 5 or 0. Take the value of D shaft \u003d 75 mm.

To determine the diameter of the shaft, it is desirable to use the standard number of diameters shown in Appendix 2.

Example 2. In cross section of timber d. \u003d 80 mm largest tangent τ Tah \u003d 40 N / mm 2. Determine the tangent stress at the point removed from the center of the section by 20 mm.

Decision

b.. Obviously

Example 3. At the points of the inner contour of the cross section of the pipe (D 0 \u003d 60 mm; d \u003d 80 mm), tangent stresses occur equal to 40 N / mm 2. Determine the maximum tangent stresses arising in the pipe.

Decision

The escape of tangent stresses in cross section is presented in Fig. 2.37 in. Obviously

Example 4. In the annular cross section of the bar ( d 0. \u003d 30 mm; d \u003d70 mm) torque arises M Z.\u003d 3 kN. Calculate the tangent voltage at the point removed from the center of the section by 27 mm.

Decision

The tangent stress in an arbitrary cross section is calculated by the formula

In this example M Z.\u003d 3 kN \u003d 3-10 6 H mm,

Example 5. Steel pipe (D 0 \u003d l00 mm; d \u003d 120 mm) long l. \u003d 1.8 m twisted moments t.enabled in its end sections. Determine the quantity t.at which the angle of twisting φ \u003d 0.25 °. With found meaning t. Calculate maximum tangent stresses.

Decision

The spinning angle (in hail / m) for one site is calculated by the formula

In this case

Substituting numeric values

Calculate maximum tangent stresses:

Example 6. For a given timber (Fig. 2.38, but) Build plugs of torque, maximum tangent stresses, rotation angles of cross-sections.

Decision

The specified bar has sections I, II, III, IV, V (Fig. 2. 38, but). Recall that the boundaries of the plots are sections in which external (twisting) moments and locations of the cross-sectional dimensions are applied.

Taking advantage of the ratio

we build torque incurns.

Building Epura. M Z. We start from the free end of the bar:

for plots III and IV

for the site V.

The torque mats are presented in Fig, 2.38, b.. We build the maximum tangent stresses in the length of the bar. Conditionally attributed τ Shah the same signs as the corresponding torque. Location on I.

location on II.

location on III

location on IV

location on V.

The maximum tangent stress is shown in Fig. 2.38, in.

The angle of rotation of the cross section of a bar with constant (within each section) of the diameter of the section and torque is determined by the formula

We build the corners of the rotation of the cross sections. The angle of rotation of the section A Φ. L \u003d 0, since in this section the timber is fixed.

The rotation of the cross sections rotation angles is shown in Fig. 2.38, g..

Example 7. On pulley IN stepped shaft (Fig. 2.39, but)transmitted from the engine power N. B \u003d 36 kW, pulleys BUT and FROM accordingly transmit to power machines N A. \u003d 15 kW and N C. \u003d 21 kW. Rotation frequency of Vala p \u003d 300 rpm. Check the strength and rigidity of the shaft if [ τ K j \u003d 30 n / mm 2, [θ] \u003d 0.3 hail / m, g \u003d 8.0-10 4 N / mm 2, d 1. \u003d 45 mm, d 2. \u003d 50 mm.

Decision

Calculate external (twisting) moments attached to the shaft:

We build torque incurns. At the same time, moving from the left end of the shaft, conditionally consider the moment corresponding to N. And positive N C. - Negative. EPUR M Z is shown in Fig. 2.39, b.. Maximum voltages in cross sections

which is less [t to] on

Relative Corner of Spinning Site AB

what is significantly larger [θ] \u003d\u003d 0.3 hail / m.

Maximum stresses in the cross sections of the site Sun

which is less [t to] on

Relative area spinning angle Sun

what is significantly larger [θ] \u003d 0.3 hail / m.

Consequently, the shaft strength is ensured, and stiffness is not.

Example 8. From the electric motor with a belt to the shaft 1 Power transmitted N. \u003d 20 kW, with shaft 1 Enters Val. 2 power N 1 \u003d 15 kW and working machines - power N 2. \u003d 2 kW and N 3. \u003d 3 kW. From Vala 2 Power comes to work machines N 4. \u003d 7 kW, N 5. \u003d 4 kW, N 6. \u003d 4 kW (Fig. 2.40, but). Determine the diameters of the shafts D 1 and D 2 from the strength and hardness, if [ τ K j \u003d 25 N / mm 2, [θ] \u003d 0.25 hail / m, g \u003d 8.0-10 4 N / mm 2. Sections of shafts 1 and 2 count along the length of constant. Motor Rotation Rotation Frequency n \u003d970 rpm, pulleys diameters D 1 \u003d 200 mm, d 2 \u003d 400 mm, d 3 \u003d 200 mm, D 4 \u003d 600 mm. Slide in belt transmission neglected.

Decision

Naris. 2.40, b. depicted Val. I.. Power comes on it N. and power is removed from it N L., N 2, N 3.

We define the angular speed of rotation of the shaft 1 and external twisting moments

Rod Rod Tour - Task Condition

To the steel shaft of a constant cross section (Fig. 3.8), four external twisting torches are applied: kN · m; kn · m; kn · m; kn · m. The length of the rod plots: m; m, m, m. It is required: to construct a torque of torque, to determine the diameter of the shaft with a BC / cm2 and construct the aid of the angles of spinning cross sections of the rod.

Round Rod Till - Calculation Scheme

Fig. 3.8.

SOLUTION OF THE PROBLEM TURING ROING ROME

Determine the jet moment that occurs in a tight dressing

Denote the moment in the sealing and send it, for example, against the course of the clockwise (when viewed to meet the Z axis).

We write the equilibrium shaft equation. In this case, we will use next rule Signs: External twisting moments (active points, as well as a jet in the sealing), rotating the shaft against the time of the clockwise (when looking at it to meet the z axis), we consider positive.

The plus sign in the expression we received says that we guessed the direction of the reactive moment that occurs in the seal.

Build torque

Recall that the internal torque arising in some cross section of the rod is equal to the algebraic amount of external twisting moments attached to any of the parts under consideration (that is, the left or right of the cross section of the cross section). In this case, the external twisting moment, rotating the considered part of the rod against the course of the clockwise (when looking at the cross section), enters this algebraic amount with the "plus" sign, and in the course - with the "minus" sign.

Accordingly, the positive inner torque, opposing the external twisting moments, is directed along the clockwise arrow (when looking at the cross section), and negative - against its stroke.

We divide the length of the rod into four sections (Fig. 3.8, a). The boundaries of the plots are those sections in which external moments are applied.

We make one cross section in an arbitrary place of each of the four pieces of the rod.

CONEMENT 1 - 1. Mentally throwing (or close a piece of paper) left part of the rod. To balance the twisting moment of the KN · M, in the cross section of the rod should occur equal to it and the oppositely directional torque. Taking into account the rule of signs mentioned above

kn · m.

Sections 2 - 2 and 3 - 3:

Section 4 - 4. To determine the torque, in section 4 - 4, we throw the right side of the rod. Then

kn · m.

It is easy to make sure that the resulting result will not change if we score now not right, but the left side of the rod. Receive

To construct an incision of torque, we carry out a thin line of the axis parallel to the axis of the rod z (Fig. 3.8, b). The calculated values \u200b\u200bof the torque in the selected scale and, taking into account their sign, lay off this axis. Within each of the portions of the rod, the torque is constant, so we are "hatching" the corresponding plot "hatch" vertical lines. Recall that each segment of the "hatching" (ordinate of the group) gives the value of the torque in the appropriate cross section of the rod. Obtained by the fat line.

Note that in the places of application of external twisting moments on the stage, we obtained a jump-like change in the inner torque by the value of the corresponding external moment.

Determine the diameter of the shaft from the strength condition

Crucity strength condition has the form

where - Polar moment of resistance (moment of resistance when cutting).

The largest torque of the torque occurs in the second section of the shaft: kn · see

Then the required diameter of the shaft is determined by the formula

cm.

Rounding the resulting value to standard, we accept the shaft diameter of Equal MM.

Determine the angles of twisting cross-sections A, B, C, D and E, and build the aid of the spinning angles

First, calculate the rod rigid rigidity, where G is the shift module, and - Polar moment of inertia. Receive

Spinning angles in separate rod sections are equal:

glad;

glad.

The angle of twisting in the seal is zero, that is. Then

Epura spinning angles are shown in Fig. 3.8, in. Note that within the length of each of the sections of the shaft, the spinning angle varies according to the linear law.

Example Task on the "Round" rod for self-solving

The condition for the task of the "round" rod

The steel rod is rigidly pinched in one end (KN / CM2 shift module) of the round cross section twisted with four moments (Fig. 3.7).

Requires:

· Build an aple of torque moments;

· With a given permissible tangent voltage of the KN / cm2 from the strength of the strength to determine the diameter of the shaft, rounded it to the nearest of the following values \u200b\u200bof 30, 35, 40, 45, 50, 60, 70, 80, 90, 100, 200 mm;

· To construct the aid of the angles of spinning cross sections of the rod.

Options for calculated schemes for the task of the round of the rod rod for self-decisions

Example Task on the circling of a round rod - baseline conditions for self-solving

Scheme number



			Before solving the problem on the routine, it is necessary to rewrite its condition with numerical data, make a sketch on scale and indicate in numbers in numbers all the values \u200b\u200brequired for further calculation, Solving tasks on the concomitant complement the brief explanations and drawings, on which the inbox included in the calculation of the value are visualized Before using the formula to determine the stress-strain state, it is necessary to study the appropriate lectures on the construct, to understand the physical meaning of all the values \u200b\u200bincluded in it, When substituting in the formula used, the values \u200b\u200bof force, the moment or length must be translated into one units system, When solving problems on the rusticity, the accuracy of calculations should not exceed three significant digits (the result of the solution of the problem cannot be more accurate in the calculated formulas of the prerequisites), Finishing the calculations need to be analyzing the results - they taught on the routine in this way check your work. Analysis of the results of the solution will help avoid ridiculous errors and quickly eliminate them.

The task

For the steel shaft of the circular cross section, determine the values \u200b\u200bof the external moments corresponding to the transmitted capacities, and balanced moment (Table 7.1 and Table 7.2).

Build a plot of torque over the length of the shaft.

Determine the diameters of the shaft in cross sections from the calculations for strength and rigidity. The resulting greater result is rounded to the nearest even or ending on the 5th number.

When calculating, use the following data: the shaft rotates with an angular velocity 25 rad / s; The material of the shaft is steel, allowable twisting voltage 30 MPa, modulus of elasticity during a shift of 8 10 4 MPa; Allowed twisting angle \u003d 0.02 rad / m.

Conduct the calculation for the ring section shaft by accepting from\u003d 0.9. To draw conclusions about the feasibility of performing the shaft of the round or ring section, comparing the cross-sections area.

purpose of work - learn to perform design and verification calculations of a round timber for statically definable systems, check the hardness check.

Theoretical Justification

Circling is called loading, in which only one internal power factor appears in the cross section of the bar - torque. External loads are also two opposite directed pairs of forces.

Distribution of tangent stresses in cross section when cutting (Fig. 7.1)

Tangent voltage at point BUT:

Fig.7.1

(7.1)

where is the distance from the point BUTbefore

centers.

Crucity strength condition

; (circle), (7.2)

(Ring), (7.3)

where m k is a torque in section, n-m, n-mm;

W P.- moment of resistance when cutting, m 3, mm 3;

[T to] - allowable voltage when cutting, N / m 2, H / mm 2.

Design calculation, definition of cross-sectional size

(7.4)

where d.- the outer diameter of the circular cross section;

D B N.- the inner diameter of the annular section; c \u003d D bk / d.

Determination of rational wheel location shaft

The rational arrangement of the wheels is the location at which the maximum torque value on the shaft is the smallest of possible.

Conditioning hardness

; G ≈ 0.4e(7.5)

where G.- module of elasticity during shift, n / m 2, n / mm 2;

E.- modulus of elasticity when tensile, N / m 2, H / mm 2.

[φ] - Allowable spinning angle, [φO] \u003d 0, 54-1 deg / m;

J P.- Polar moment of inertia in section, m 4, mm 4.

(7.6)

Design calculation, definition of outer cross section diameter

Procedure for performing work

1. To construct the torque of torque over the length of the shaft for the scheme proposed in the task.

2. Select the rational arrangement of the wheels on the shaft and further calculations for the shaft with rationally positioned pulleys.

3. Determine the required diameters of the circular shaft at the rate of strength and rigidity and select the largest of the obtained values, rounded the size of the diameter.

4. Compare metal costs for the case of round and annular sections. Comparison to carry out the cross-sections of the shafts.

Control questions

1. What deformations occur when cutting?

2. What hypothesis are performed during twisting deformation?

3. Will the length and diameter of the shaft shaft change after twisting?

4. What domestic power factors occur when you take a crash?

5. What is the rational location of the spikes on the shaft?

6. What is the polar moment of inertia? What physical meaning is this value?

7. Which units is measured?

Example of execution

For a given timber (Fig. 7.1), build pieces of torque, the rational arrangement of pulleys on the shaft to reduce the value of the maximum torque. Build an aid of torque at the rational position of the pulleys. From the strength of the strength to determine the diameters of the shafts for solid and annular sections, accepting C \u003d. Compare the results obtained on the obtained cross-sections. [τ] \u003d 35 MPa.

Decision

Section 2 (Fig. 7.2b):

Section 3 (Fig.7.3B):

Fig.7.2.

A B C

Fig.7.3.

We build torque incurns. Values \u200b\u200bof torque sets down from the axis, because Moments are negative. The maximum torque value on the shaft in this case is 1000 N · m (Fig.7.1).
Choose the rational location of the pulleys on the shaft. This placement of pulleys is most appropriate, in which the greatest positive and negative values \u200b\u200bof torque in the plots will be the same as possible. Of these considerations, the leading pulley, transmitting a moment 1000 N · m, is placed closer to the center of the shaft, the slave pullees 1 and 2 are placed on the left of the lead with the moment 1000 N · m, the pulley 3 remains in the same place. We build torque incurries with the selected position of the pulleys (Fig. 7.3).

The maximum torque value on the shaft with the selected position of the pulleys - 600 N * m.

Fig.7.4.

Torque torque torque:

Determine the diameters of the shaft in sections:

Round the resulting values \u200b\u200b:,

Determine the diameters of the shaft in cross sections, provided that the cross section is ring

The moments of resistance remain the same. By condition

The polar moment of rings resistance:

The formula for determining the outer diameter of the ring segment shaft:

The calculation can be carried out according to the formula:

Shaft diameters for sections:

The outer diameters of the ring section shaft have practically not changed.

For an annular section: ,,

For the conclusion about the savings of the metal, during the transition to the annular section, comparable of the cross sections (Fig. 7.4)

Provided that the cross section is a circle (Fig. 7.4a)

Solid round section:

Provided that the cross section is ring, (Fig. 7.4b)

Ring section:

Comparative rating:

Consequently, when moving from a circular cross section, the savings of metal in weight will be 1.3 times.

fig.7.4.

Table 7.1.

Table 7.2.

Option	Parameters
a \u003d b \u003d s, m	P1, kw	P2, kW	P3, kW
	1,1	2,1	2,6	3,1
	1,2	2,2	2,7	3,2
	1,3	2,3	2,8	3,3
	1,4	2,4	2,9	3,4
	1,5	2,5	3,0	3,5
	1,6	2,6	3,1	3,6
	1,7	2,7	3,2	3,7
	1,8	2,8	3,3	3,8
	1,9	2,9	3,4	3,9
	2,0	3,0	3,5	4,0
	1,1	3,1	3,4	4,1
	1,2	3,2	3,3	4,2
	1,3	3,3	3,2	4,3
	1,4	3,4	3,1	4,5
	1,5	3,5	2,8	2,9
	1,3	2,1	2,6	3,1
	1,4	2,2	2,7	3,2
	1,5	2,3	2,8	3,3
	1,6	2,4	2,9	3,4
	1,7	2,5	3,0	3,5
	1,8	2,6	3,1	3,6
	1,9	2,7	3,2	3,7
	2,0	2,8	3,3	3,8
	1,1	2,9	3,4	3,9
	1,2	3,0	3,5	4,0
	1,3	3,1	3,4	4,1
	1,4	3,2	3,3	4,2
	1,5	3,3	3,2	4,3
	1,4	3,4	3,1	4,5
	1,9	3,5	2,8	2,9

Appendix A.

5.1 (option 08)

NOTES: Power on gear wheels take p 2 \u003d 0.5P 1, p 3 \u003d 0.3р 1 and p 4 \u003d 0.2Р1. The resulting calculated value of diameter (in mm) is rounded to the nearest larger number, ending with 0, 2, 5, 8 or ST SEV 208-75 [τ] \u003d 30 MPa.

Table 20 - Initial data

№ Problem I. Schemes in Fig.35	R, kW	Ω, Rad / C	Distance between pulleys, m
			l 1.	l 2.	l 3.
100, H.	28	26	0,2	0,1	0,3

Answer: d 1 \u003d 45.2 mm, d 2 \u003d 53.0 mm, d 3 \u003d 57.0 mm, φ i \u003d 0.283º, φ ii \u003d 0.080º, φ iii \u003d 0.149º.

5.2

d) determine the diameter of the shaft, adopting [Σ] \u003d 60 H / mm² (in problem 117) and believing f r \u003d 0.4f t. In the task 117, the calculation is calculated on the hypothesis of the greatest tangent stresses.

Table 22 - source data

№ Problem I. Schemes in Fig.37.	Option	R, kW	ω 1, rad / c
117, VII.	08	8	35

Answer: R BY \u003d 7145 H, R AY \u003d 3481 H, D \u003d 51 mm.

5.3 For the steel shaft of a constant cross section (Fig.7.17), transmitting power P (kW) at an angular velocity ω (rad / c) (numerical values \u200b\u200bof these values \u200b\u200bfor their option, take from Table 7.4):

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

b) to build an aple of torque;

c) construct an epures of bending moments in vertical and horizontal planes;

d) Determine the diameter of the shaft, adopting [σ] \u003d 70 MPa (in problems 41, 43, 45, 47, 49) or [σ] \u003d 60 MPa (in tasks 42, 44, 46, 48, 50). For efforts acting on the gear wheel, take f r \u003d 0.36f t, for tensioning belts S 1 \u003d 2S 2. In tasks 42, 44, 46, 48, 50, calculat on the hypothesis of the potential energy of formation, and in problems 41, 43, 45, 47, 49 on the hypothesis of the greatest tangent stresses.

Table 22 - source data

Task number and schemes in Fig. 7.17	Option	R, kW	Ω, Rad / C
Task 45, scheme V	47	30	24

Answer: R BY \u003d 4000 H, R AY \u003d 14000 H, D \u003d 64 mm.

5.4 For one of the schemes (Fig. 35, Table.20) to build an aid of torque moments; Determine the shaft diameter on each site and the full twisting angle.

Specifications: Power on gear wheels take p 2 \u003d 0.5P 1, p 3 \u003d 0.3р 1 and p 4 \u003d 0.2Р 1. The resulting calculated value of diameter (in mm) is rounded to the nearest larger number, ending with 0, 2, 5, 8 or ST SEV 208-75 [τ] \u003d 30 MPa.

Table 20.

Number of tasks and schemes in Fig.35	Option	R, kW	Ω, Rad / C	Distance between pulleys, m
				l 1.	l 2.	l 3.
91, I.	29	20	30	0,2	0,9	0,4

Answer: d 1 \u003d 28.5 mm, d 2 \u003d 43.2 mm, d 3 \u003d 48.5 mm, φ i \u003d 0.894º, φ ii \u003d 0.783º, φ iii \u003d 0.176º.

5.5 For a steel shaft of a constant cross section with one gear wheel (Fig.37), transmitting power P (kW) at an angular velocity ω 1 (rad / c) (numerical values \u200b\u200bof these values \u200b\u200bfor their option to take from Table.22):

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

b) to build an aple of torque;

c) construct an epures of bending moments in vertical and horizontal planes;

d) determine the diameter of the shaft, adopting [Σ] \u003d 70 H / mm² (in problem 112) and believing F r \u003d 0.4f t. In problem 112, calculat on the hypothesis of the potential energy of formation.

Table 22.

Problems and schemes in Fig.37	Option	R, kW	ω 1, rad / c
112, II.	29	20	50

Answer: R BY \u003d 1143 H, R AY \u003d 457 H, D \u003d 40.5 mm.

Perhaps it will be useful to read: