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Calculation of timber with a round cross-section for strength and torsional rigidity. Forces and stresses in the cross sections of the beam Determine the maximum stress in the cross section of the beam

Tension (compression)- this is a type of loading of a beam in which only one internal force factor appears in its cross sections - the longitudinal force N.

In tension and compression, external forces are applied along the longitudinal axis z (Figure 109).

Figure 109

Using the section method, it is possible to determine the value of the VSF - longitudinal force N under simple loading.

Internal forces (stresses) arising in an arbitrary cross section during tension (compression) are determined using Bernoulli's hypothesis of plane sections:

The section of the beam, flat and perpendicular to the axis before loading, remains the same during loading.

It follows that the fibers of the timber (Figure 110) elongate by the same amount. This means that the internal forces (i.e. stresses) acting on each fiber will be identical and distributed evenly over the cross section.

Figure 110

Since N is the resultant of internal forces, then N = σ A, which means the normal stresses σ in tension and compression are determined by the formula:

[N/mm 2 = MPa], (72)

where A is the cross-sectional area.

Example 24. Two rods: round section diameter d = 4 mm and square section with a side of 5 mm are stretched by the same force F = 1000 N. Which of the rods is loaded more?

Given: d = 4 mm; a = 5 mm; F = 1000 N.

Define: σ 1 and σ 2 – in rods 1 and 2.

Solution:

When stretching, the longitudinal force in the rods is N = F = 1000 N.

Cross-sectional areas of rods:

; .

Normal stresses in cross sections of rods:

, .

Since σ 1 > σ 2, the first round rod is loaded more.

Example 25. A cable twisted from 80 wires with a diameter of 2 mm stretches with a force of 5 kN. Determine the stress in the cross section.

Given: k = 80; d = 2 mm; F = 5 kN.

Define: σ.

Solution:

N = F = 5 kN, ,

Then .

Here A 1 is the cross-sectional area of one wire.

Note: the cable cross-section is not a circle!

2.2.2 Diagrams of longitudinal forces N and normal stresses σ along the length of the beam

To calculate the strength and rigidity of a complexly loaded beam under tension and compression, it is necessary to know the values of N and σ in various cross sections.

For this, diagrams are constructed: plot N and plot σ.

Diagram is a graph of changes in the longitudinal force N and normal stresses σ along the length of the beam.

Longitudinal force N in an arbitrary cross section of the beam is equal to the algebraic sum of all external forces applied to the remaining part, i.e. on one side of the section

External forces F, stretching the beam and directed away from the section, are considered positive.

The order of plotting N and σ

1 Using cross sections, we divide the timber into sections, the boundaries of which are:

a) sections at the ends of the beam;

b) where forces F are applied;

c) where the cross-sectional area A changes.

2 We number the sections starting from

free end.

3 For each site, using the method

sections we determine the longitudinal force N

and build a diagram N on a scale.

4 Determine normal stress σ

on each site and build in

scale diagram σ.

Example 26. Construct diagrams of N and σ along the length of the stepped beam (Figure 111).

Given: F 1 = 10 kN; F 2 = 35 kN; A 1 = 1 cm 2; A 2 = 2 cm 2.

Solution:

1) We divide the beam into sections, the boundaries of which are: sections at the ends of the beam, where external forces F are applied, where the cross-sectional area A changes - there are 4 sections in total.

2) We number the sections starting from the free end:

from I to IV. Figure 111

3) For each section, using the section method, we determine the longitudinal force N.

The longitudinal force N is equal to the algebraic sum of all external forces applied to the remaining part of the beam. Moreover, external forces F, tensile beams are considered positive.

Table 13

4) We construct a diagram N on a scale. We indicate the scale only with positive values N; on the diagram, the plus or minus sign (extension or compression) is indicated in a circle in the rectangle of the diagram. Positive values of N are plotted above the zero axis of the diagram, negative - below the axis.

5) Verification (oral): In sections where external forces F are applied, the diagram N will show vertical jumps equal in magnitude to these forces.

6) Determine the normal stresses in the sections of each section:

; ;

; .

We build a diagram σ on a scale.

7) Examination: The signs of N and σ are the same.

Think and answer the questions

1) it is impossible; 2) it is possible.

53 Do the tensile (compressive) stresses of rods depend on the shape of their cross-section (square, rectangle, circle, etc.)?

1) depend; 2) do not depend.

54 Does the magnitude of the stress in the cross section depend on the material from which the rod is made?

1) depends; 2) does not depend.

55 Which points of the cross section of a round rod are loaded more under tension?

1) on the axis of the beam; 2) on the surface of the circle;

3) at all points of the cross-section the stresses are the same.

56 Steel and wood rods of equal cross-sectional area are stretched by equal forces. Will the stresses arising in the rods be equal?

1) in steel the stress is greater;

2) in wood the tension is greater;

3) equal stresses will arise in the rods.

57 For timber (Figure 112), construct diagrams N and σ, if F 1 = 2 kN; F 2 = 5 kN; A 1 = 1.2 cm 2; A 2 = 1.4 cm 2.

Obliquely called this type of bending in which all external loads causing bending act in one force plane that does not coincide with any of the main planes.

Consider a beam clamped at one end and loaded at the free end with a force F(Fig. 11.3).

Rice. 11.3. Design diagram for oblique bending

External force F applied at an angle to the axis y. Let's break down the power F into components lying in the main planes of the beam, then:

Bending moments in an arbitrary section taken at a distance z from the free end will be equal:

Thus, in each section of the beam, two bending moments simultaneously act, which create bending in the main planes. Therefore, oblique bending can be considered as a special case of spatial bending.

Normal stresses in the cross section of a beam during oblique bending are determined by the formula

To find the highest tensile and compressive normal stresses during oblique bending, it is necessary to select a dangerous section of the beam.

If bending moments | M x| and | M y| reach highest values in a certain section, then this is a dangerous section. Thus,

Dangerous sections also include sections where bending moments | M x| and | M y| simultaneously reach quite large values. Therefore, with oblique bending there may be several dangerous sections.

In general, when – asymmetrical section, i.e. the neutral axis is not perpendicular to the force plane. For symmetrical sections, oblique bending is not possible.

11.3. Position of the neutral axis and dangerous points

in cross section. Strength condition for oblique bending.

Determination of cross-section dimensions.

Movements during oblique bending

The position of the neutral axis during oblique bending is determined by the formula

where is the angle of inclination of the neutral axis to the axis X;

Angle of inclination of the force plane to the axis at(Fig. 11.3).

In the dangerous section of the beam (in the embedment, Fig. 11.3), the stresses at the corner points are determined by the formulas:

With oblique bending, as with spatial bending, the neutral axis divides the section of the beam into two zones - a tension zone and a compression zone. For a rectangular section, these zones are shown in Fig. 11.4.

Rice. 11.4. Diagram of the cross-section of a clamped beam during oblique bending

To determine extreme tensile and compressive stresses, it is necessary to draw tangents to the section in the tension and compression zones, parallel to the neutral axis (Fig. 11.4).

The most distant points of contact from the neutral axis A And WITH– dangerous points in the compression and tension zones, respectively.

For plastic materials, when the calculated resistances of the timber material under tension and compression are equal to each other, i.e. [ σ р] = = [σc] = [σ ], in the dangerous section is determined and the strength condition can be represented in the form

For symmetrical sections (rectangle, I-section), the strength condition has the following form:

Three types of calculations follow from the strength condition:

Check;

Design - definition geometric dimensions sections;

Definition bearing capacity timber (permissible load).

If the relationship between the sides of the cross section is known, for example, for a rectangle h = 2b, then from the condition of the strength of the pinched beam it is possible to determine the parameters b And h as follows:

finally .

The parameters of any section are determined in a similar way. The total displacement of a beam section during oblique bending, taking into account the principle of independence of the action of forces, is determined as the geometric sum of displacements in the main planes.

Let us determine the displacement of the free end of the beam. Let's use Vereshchagin's method. We find the vertical displacement by multiplying the diagrams (Fig. 11.5) according to the formula

Similarly, we define horizontal displacement:

Then we determine the total displacement using the formula

Rice. 11.5. Diagram for determining total displacement

with oblique bending

The direction of complete movement is determined by the angle β (Fig. 11.6):

The resulting formula is identical to the formula for determining the position of the neutral axis of the beam section. This allows us to conclude that , i.e., the direction of deflection is perpendicular to the neutral axis. Consequently, the deflection plane does not coincide with the loading plane.

Rice. 11.6. Scheme for determining the deflection plane

with oblique bending

Deviation angle of the deflection plane from the main axis y will be greater, the greater the displacement. Therefore, for a beam with an elastic cross-section, in which the ratio J x/Jy is large, oblique bending is dangerous, as it causes large deflections and stresses in the plane of least rigidity. For timber with J x= Jy, the total deflection lies in the force plane and oblique bending is impossible.

11.4. Eccentric tension and compression of a beam. Normal

stresses in beam cross sections

Eccentric stretch (compression) is a type of deformation in which the tensile (compressive) force is parallel to the longitudinal axis of the beam, but the point of its application does not coincide with the center of gravity of the cross section.

This type of problem is often used in construction when calculating building columns. Let us consider the eccentric compression of the beam. Let us denote the coordinates of the force application point F through x F And y F, and the main cross-sectional axes are through x and y. Axis z let's direct it in such a way that the coordinates x F And y F were positive (Fig. 11.7, a)

If you transfer the force F parallel to itself from a point WITH to the center of gravity of the section, then eccentric compression can be represented as the sum of three simple deformations: compression and bending in two planes (Fig. 11.7, b). In this case we have:

Stresses at an arbitrary cross-section point under eccentric compression lying in the first quadrant, with coordinates x and y can be found based on the principle of independence of the action of forces:

squares of the radii of inertia of the section, then

Where x And y– coordinates of the cross-section point at which the stress is determined.

When determining stresses, it is necessary to take into account the signs of the coordinates of both the point of application of the external force and the point where the stress is determined.

Rice. 11.7. Diagram of a beam under eccentric compression

In the case of eccentric tension of the beam, the “minus” sign in the resulting formula should be replaced with a “plus” sign.

If, during direct or oblique bending, only a bending moment acts in the cross section of the beam, then, accordingly, there is a pure straight or pure oblique bend. If a transverse force also acts in the cross section, then there is a transverse straight or transverse oblique bend. If the bending moment is the only internal force factor, then such bending is called clean(Fig. 6.2). When there is a shear force, bending is called transverse. Strictly speaking, to simple types resistance relates only to pure bending; transverse bending is conventionally classified as a simple type of resistance, since in most cases (for sufficiently long beams) the effect of transverse force can be neglected when calculating strength. See plane bending strength condition. When calculating a beam for bending, one of the most important tasks is to determine its strength. Plane bending is called transverse if two internal force factors arise in the cross sections of the beam: M - bending moment and Q - transverse force, and pure if only M occurs. In transverse bending, the force plane passes through the axis of symmetry of the beam, which is one of the main axes of inertia of the section.

When a beam bends, some of its layers are stretched, others are compressed. Between them there is a neutral layer, which only bends without changing its length. The line of intersection of the neutral layer with the cross-sectional plane coincides with the second main axis of inertia and is called the neutral line (neutral axis).

Due to the action of the bending moment, normal stresses arise in the cross sections of the beam, determined by the formula

where M is the bending moment in the section under consideration;

I – moment of inertia of the cross section of the beam relative to the neutral axis;

y is the distance from the neutral axis to the point at which the stresses are determined.

As can be seen from formula (8.1), the normal stresses in the section of the beam along its height are linear, reaching a maximum value at the most distant points from the neutral layer.

where W is the moment of resistance of the cross section of the beam relative to the neutral axis.

27.Tangential stresses in the cross section of a beam. Zhuravsky's formula.

Zhuravsky's formula allows you to determine the shear stresses during bending that arise at points in the cross section of the beam located at a distance from the neutral axis x.

DERIVATION OF THE ZHURAVSKI FORMULA

Let's cut an element with a length and an additional longitudinal section into two parts from a beam of rectangular cross-section (Fig. 7.10, a) (Fig. 7.10, b).

Let us consider the equilibrium of the upper part: due to the difference in bending moments, different compressive stresses arise. In order for this part of the beam to be in equilibrium (), a tangential force must arise in its longitudinal section. Equilibrium equation for part of the beam:

where integration is carried out only over the cut-off part of the cross-sectional area of the beam (shaded in Fig. 7.10), – static moment of inertia of the cut-off (shaded) part of the cross-sectional area relative to the neutral x-axis.

Let's assume: the tangential stresses () arising in the longitudinal section of the beam are uniformly distributed across its width () at the cross section:

We obtain an expression for tangential stresses:

, and , then the formula for tangential stresses () arising at points of the cross section of the beam located at a distance y from the neutral axis x:

Zhuravsky's formula

Zhuravsky's formula was obtained in 1855 by D.I. Zhuravsky, therefore bears his name.

Calculation of timber with a round cross-section for strength and torsional rigidity

The purpose of calculations for strength and torsional rigidity is to determine the cross-sectional dimensions of the beam at which stresses and displacements will not exceed specified values allowed by operating conditions. The strength condition for permissible tangential stresses is generally written in the form This condition means that the highest tangential stresses arising in a twisted beam should not exceed the corresponding permissible stresses for the material. The permissible stress during torsion depends on 0 ─ the stress corresponding to the dangerous state of the material, and the accepted safety factor n: ─ yield strength, nt - safety factor for a plastic material; ─ tensile strength, nв - safety factor for brittle material. Due to the fact that it is more difficult to obtain values in torsion experiments than in tension (compression), then, most often, the permissible torsional stresses are taken depending on the permissible tensile stresses for the same material. So for steel [for cast iron. When calculating the strength of twisted beams, three types of problems are possible, differing in the form of using strength conditions: 1) checking stresses (test calculation); 2) selection of section (design calculation); 3) determination of the permissible load. 1. When checking stresses for given loads and dimensions of the beam, the largest tangential stresses occurring in it are determined and compared with those specified according to formula (2.16). If the strength condition is not met, then it is necessary to either increase the cross-sectional dimensions, or reduce the load acting on the beam, or use a material of higher strength. 2. When selecting a section for a given load and a given value of permissible stress, from the strength condition (2.16), the value of the polar moment of resistance of the cross section of the beam is determined. The diameters of the solid round or annular section of the beam are determined by the value of the polar moment of resistance. 3. When determining the permissible load from a given permissible stress and polar moment of resistance WP, based on (3.16), the value of the permissible torque MK is first determined and then, using a torque diagram, a connection is established between K M and external twisting moments. Calculation of timber for strength does not exclude the possibility of deformations that are unacceptable during its operation. Large angles of twist of the beam are very dangerous, as they can lead to a violation of the precision of processing parts if this beam is a structural element of a processing machine, or torsional vibrations may occur if the beam transmits torsional moments that vary in time, so the beam must also be calculated on its rigidity. The stiffness condition is written in the following form: where ─ the largest relative angle of twist of the beam, determined from expression (2.10) or (2.11). Then the rigidity condition for the shaft will take the form The value of the permissible relative angle of twist is determined by the standards for various structural elements and different types loads vary from 0.15° to 2° per 1 m of beam length. Both in the strength condition and in the rigidity condition, when determining max or max  we will use geometric characteristics: WP ─ polar moment of resistance and IP ─ polar moment of inertia. Obviously, these characteristics will be different for round solid and annular cross sections with the same area of these sections. Through specific calculations, one can be convinced that the polar moments of inertia and the moment of resistance for the annular section are significantly greater than for the irregular circular section, since the annular section does not have areas close to the center. Therefore, a beam with an annular cross-section during torsion is more economical than a beam with a solid circular cross-section, i.e., it requires less material consumption. However, the production of such beams is more difficult, and therefore more expensive, and this circumstance must also be taken into account when designing beams operating in torsion. We will illustrate the methodology for calculating timber for strength and torsional rigidity, as well as discussions about efficiency, with an example. Example 2.2 Compare the weights of two shafts, the transverse dimensions of which are selected for the same torque MK 600 Nm at the same permissible stresses 10 R and 13 Tension along the fibers p] 7 Rp 10 Compression and crushing along the fibers [cm] 10 Rc, Rcm 13 Collapse across the fibers (at a length of at least 10 cm) [cm]90 2.5 Rcm 90 3 Chipping along the fibers during bending [and] 2 Rck 2.4 Chipping along the fibers when cutting 1 Rck 1.2 – 2.4 Chipping across the cuts fibers

From the formula for determining stresses and the diagram of the distribution of tangential stresses during torsion, it is clear that the maximum stresses occur on the surface.

Let us determine the maximum voltage, taking into account that ρ ta X =d/ 2, where d- diameter of round beam.

For a circular cross-section, the polar moment of inertia is calculated using the formula (see lecture 25).

The maximum stress occurs on the surface, so we have

Usually JP/pmax denote Wp and call moment of resistance in torsion, or polar moment of resistance sections

Thus, to calculate the maximum stress on the surface of a round beam, we obtain the formula

For round section

For annular section

Torsional strength condition

Fracture of a beam during torsion occurs from the surface; when calculating the strength, the strength condition is used

Where [ τ k ] - permissible torsional stress.

Types of strength calculations

There are two types of strength calculations.

1. Design calculation - the diameter of the beam (shaft) in the dangerous section is determined:

2. Verification calculation - the fulfillment of the strength condition is checked

3. Determination of load capacity (maximum torque)

Stiffness calculation

When calculating rigidity, the deformation is determined and compared with the permissible one. Let us consider the deformation of a round beam under the action of an external pair of forces with a moment T(Fig. 27.4).

In torsion, the deformation is estimated by the angle of twist (see lecture 26):

Here φ - twist angle; γ - shear angle; l- beam length; R- radius; R =d/2. Where

Hooke's law has the form τ k = G γ. Let's substitute the expression for γ , we get

Work GJP called section stiffness.

The elastic modulus can be defined as G = 0,4E. For steel G= 0.8 10 5 MPa.

Usually the angle of twist per one meter of beam (shaft) length is calculated. φ o.

The torsional stiffness condition can be written as

Where φ o - relative twist angle, φ o = φ/l; [φ o ]≈ 1 deg/m = 0.02 rad/m - permissible relative angle of twist.

Examples of problem solving

Example 1. From calculations of strength and rigidity, determine the required shaft diameter to transmit power of 63 kW at a speed of 30 rad/s. Shaft material - steel, permissible torsional stress 30 MPa; permissible relative twist angle [φ o ]= 0.02 rad/m; shear modulus G= 0.8 * 10 5 MPa.

Solution

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

Torsional strength condition:

We determine the torque from the rotational power formula:

From the strength condition we determine the moment of resistance of the shaft during torsion

We substitute the values in newtons and mm.

Determine the shaft diameter:

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

Torsional rigidity condition:

From the rigidity condition we determine the moment of inertia of the section during torsion:

Determine the shaft diameter:

3. Selecting the required shaft diameter based on strength and rigidity calculations.

To ensure strength and rigidity simultaneously, we select the larger of the two values found.

The resulting value should be rounded using a range of preferred numbers. In practice, we round the resulting value so that the number ends in 5 or 0. We take the value d of the shaft = 75 mm.

To determine the shaft diameter, it is advisable to use the standard range of diameters given in Appendix 2.

Example 2. In the cross section of the beam d= 80 mm highest shear stress τ max= 40 N/mm 2. Determine the shear stress at a point 20 mm away from the center of the section.

Solution

b. Obviously,

Example 3. At points of the internal contour of the cross section of the pipe (d 0 = 60 mm; d = 80 mm), tangential stresses equal to 40 N/mm 2 arise. Determine the maximum shear stresses occurring in the pipe.

Solution

The diagram of tangential stresses in the cross section is shown in Fig. 2.37, V. Obviously,

Example 4. In the annular cross section of the beam ( d 0= 30 mm; d = 70 mm) torque occurs M z= 3 kN-m. Calculate the shear stress at a point 27 mm away from the center of the section.

Solution

The tangential stress at an arbitrary point of the cross section is calculated by the formula

In the example under consideration M z= 3 kN-m = 3-10 6 N mm,

Example 5. Steel pipe(d 0 = l00 mm; d = 120 mm) length l= 1.8 m twists moments T, applied in its end sections. Determine the value T, at which the twist angle φ = 0.25°. When the value is found T calculate the maximum shear stress.

Solution

The twist angle (in degrees/m) for one section is calculated using the formula

IN in this case

Substituting numeric values, we get

We calculate the maximum shear stress:

Example 6. For a given beam (Fig. 2.38, A) construct diagrams of torques, maximum shear stresses, and rotation angles of cross sections.

Solution

The given beam has sections I, II, III, IV, V(Fig. 2. 38, A). Let us recall that the boundaries of the sections are the sections in which external (torsional) moments are applied and the places where the cross-sectional dimensions change.

Using the ratio

We build a diagram of torques.

Constructing a diagram M z we start from the free end of the beam:

for plots III And IV

for the site V

The diagram of torques is shown in Fig. 2.38, b. We construct a diagram of the maximum tangential stresses along the length of the beam. We conditionally attribute τ check the same signs as the corresponding torques. On the site I

on site II

on site III

on site IV

on site V

The diagram of maximum tangential stresses is shown in Fig. 2.38, V.

The angle of rotation of the cross section of the beam at constant (within each section) cross-section diameter and torque is determined by the formula

We construct a diagram of the angles of rotation of the cross sections. Section rotation angle A φ l = 0, since the beam is fixed in this section.

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

Example 7. On the pulley IN stepped shaft (Fig. 2.39, A) power is transmitted from the engine N B = 36 kW, pulleys A And WITH accordingly transfer power to the machines N A= 15 kW and N C= 21 kW. Shaft speed n= 300 rpm. Check the strength and rigidity of the shaft if [ τ K J = 30 N/mm 2, [Θ] = 0.3 deg/m, G = 8.0-10 4 N/mm 2, d 1= 45 mm, d 2= 50 mm.

Solution

Let's calculate the external (torsional) moments applied to the shaft:

We build a diagram of torques. In this case, moving from the left end of the shaft, we conditionally calculate the moment corresponding to N Ah, positive N c- negative. The M z diagram is shown in Fig. 2.39, b. Maximum stresses in cross sections of section AB

which is less [tk] by

Relative angle of twist of section AB

which is significantly greater than [Θ] ==0.3 deg/m.

Maximum stresses in cross sections of the section Sun

which is less [tk] by

Relative angle of twist of the section Sun

which is significantly greater than [Θ] = 0.3 deg/m.

Consequently, the strength of the shaft is ensured, but the rigidity is not.

Example 8. From the electric motor using a belt to the shaft 1 power is transmitted N= 20 kW, From shaft 1 enters the shaft 2 power N 1= 15 kW and to working machines - power N 2= 2 kW and N 3= 3 kW. From the shaft 2 power is supplied to working machines N 4= 7 kW, N 5= 4 kW, N 6= 4 kW (Fig. 2.40, A). Determine the diameters of the shafts d 1 and d 2 from the conditions of strength and rigidity, if [ τ K J = 25 N/mm 2, [Θ] = 0.25 deg/m, G = 8.0-10 4 N/mm 2. Shaft sections 1 And 2 be considered constant along the entire length. Motor shaft rotation speed n = 970 rpm, pulley diameters D 1 = 200 mm, D 2 = 400 mm, D 3 = 200 mm, D 4 = 600 mm. Neglect slippage in the belt drive.

Solution

Fig. 2.40, b depicts a shaft I. It receives power N and power is removed from it Nl, N 2 , N 3.

Let us determine the angular velocity of the shaft rotation 1 and external torsional moments m, m 1, t 2, t 3:

We build a diagram of torques for shaft 1 (Fig. 2.40, V). At the same time, moving from the left end of the shaft, we conditionally calculate the moments corresponding to N 3 And N 1, positive, and N- negative. Rated (maximum) torque N x 1 max = 354.5 H * m.

Shaft diameter 1 from strength conditions

Shaft diameter 1 from stiffness condition ([Θ], rad/mm)

We finally accept rounding to the standard value d 1 = 58 mm.

Shaft speed 2

In Fig. 2.40, G depicts a shaft 2; power is supplied to the shaft N 1, and the power is removed from it N 4, N 5, N 6.

Let's calculate the external twisting moments:

Torque diagram for the shaft 2 shown in Fig. 2.40, d. Estimated (maximum) torque M i max " = 470 N-m.

Shaft diameter 2 from the strength condition

Shaft diameter 2 from the rigidity condition

We finally accept d 2 = 62 mm.

Example 9. Determine the power from the conditions of strength and stiffness N(Fig. 2.41, A), which can be transmitted by a steel shaft with a diameter d = 50 mm, if [t k] = 35 N/mm 2, [ΘJ = 0.9 deg/m; G = 8.0* I0 4 N/mm 2, n= 600 rpm.

Solution