It can be seen from the formula for determining the stresses and the diagram of the distribution of shear stresses during torsion that the maximum stresses occur on the surface.

Let us determine the maximum voltage, taking into account that ρ and X = d/ 2, where d- bar diameter round section.

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

The maximum stress occurs on the surface, so we have

Usually JP /pmax designate Wp and call moment of resistance when twisting, 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 an annular section

Torsional strength condition

The destruction of the beam during torsion occurs from the surface, when calculating the strength, the strength condition is used

Where [ τ k ] - allowable 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. Check calculation - the fulfillment of the strength condition is checked

3. Determination of load capacity (maximum torque)

Stiffness calculation

When calculating the stiffness, the deformation is determined and compared with the allowable one. Consider the deformation of a round beam under the action outer pair forces with moment T(Fig. 27.4).

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

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

Hooke's law has the form τ k = Gγ. Substitute the expression for γ , we get

Work GJP called the stiffness of the section.

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

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

The torsional rigidity condition can be written as

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

Examples of problem solving

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

Solution

1. Determining the dimensions of the cross section based on strength.

Torsional strength condition:

We determine the torque from the power formula during rotation:

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. Determining the dimensions of the cross section based on stiffness.

Torsional stiffness condition:

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

Determine the shaft diameter:

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

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

The resulting value should be rounded off using a range of preferred numbers. We practically round off the obtained value so that the number ends with 5 or 0. We take the value d of the shaft = 75 mm.

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

Example 2 In the cross section of the beam d= 80 mm maximum shear stress τ max\u003d 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 the points of the inner contour of the pipe cross section (d 0 = 60 mm; d = 80 mm), shear stresses equal to 40 N/mm 2 arise. Determine the maximum shear stresses that occur 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 ( d0= 30 mm; d= 70 mm) torque occurs Mz= 3 kN-m. Calculate the shear stress at a point 27 mm away from the center of the section.

Solution

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

In this example Mz= 3 kN-m = 3-10 6 N mm,

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

Solution

The angle of twist (in deg/m) for one section is calculated by the formula

In this case

Substituting numerical values, we get

We calculate the maximum shear stresses:

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

Solution

A given beam has sections I, II, III, IV, V(Fig. 2. 38, A). Recall that the boundaries of the sections are sections in which external (twisting) moments and places of change in the dimensions of the cross section are applied.

Using the ratio

we build a diagram of torques.

Plotting Mz 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 build 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. Location on I

Location on II

Location on III

Location on IV

Location on V

The plot of maximum shear stresses is shown in fig. 2.38 V.

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

We build a diagram of the angles of rotation of the cross sections. Section rotation angle A φ l \u003d 0, since the beam is fixed in this section.

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

Example 7 per pulley IN stepped shaft (Fig. 2.39, A) power transferred from the engine N B = 36 kW, pulleys A And WITH respectively transferred to the power machines N A= 15 kW and N C= 21 kW. Shaft speed P= 300 rpm. Check the strength and rigidity of the shaft, if [ τ K J \u003d 30 N / mm 2, [Θ] \u003d 0.3 deg / m, G \u003d 8.0-10 4 N / mm 2, d1= 45 mm, d2= 50 mm.

Solution

Let us calculate the external (twisting) moments applied to the shaft:

We build a diagram of torques. At the same time, moving from the left end of the shaft, we conditionally consider the moment corresponding to N A, positive Nc- negative. The diagram M z is shown in fig. 2.39 b. Maximum stresses in the cross sections of section AB

which is less [t k ] by

Relative angle of twist of section AB

which is much more than [Θ] ==0.3 deg/m.

Maximum stresses in the cross sections of the section sun

which is less [t k ] by

Relative twist angle of the section sun

which is much more than [Θ] = 0.3 deg/m.

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

Example 8 From the motor with a belt to the shaft 1 transmitted power N= 20 kW, From the 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 the working machines N 4= 7 kW, N 5= 4 kW, No. 6= 4 kW (Fig. 2.40, A). Determine the diameters of the shafts d 1 and d 2 from the condition of strength and stiffness, if [ τ K J \u003d 25 N / mm 2, [Θ] \u003d 0.25 deg / m, G \u003d 8.0-10 4 N / mm 2. Shaft sections 1 And 2 be considered constant over the entire length. Motor shaft speed n = 970 rpm, pulley diameters D 1 = 200 mm, D 2 = 400 mm, D 3 = 200 mm, D 4 = 600 mm. Ignore slip in the belt drive.

Solution

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

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

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

Shaft diameter 1 from strength condition

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

Finally, we accept with rounding up to the standard value d 1 \u003d 58 mm.

Shaft speed 2

On fig. 2.40 G the shaft is shown 2; power is applied to the shaft N 1, and power is removed from it N 4 , N 5 , N 6 .

Calculate the external torsional moments:

Shaft torque diagram 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 stiffness condition

We finally accept d2= 62 mm.

Example 9 Determine from the conditions of strength and rigidity the power N(Fig. 2.41, A), which can be transmitted by a steel shaft with a diameter d=50 mm, if [t to] \u003d 35 N / mm 2, [ΘJ \u003d 0.9 deg / m; G \u003d 8.0 * I0 4 N / mm 2, n= 600 rpm.

Solution

Let us calculate the external moments applied to the shaft:

Design scheme shaft is shown in fig. 2.41, b.

On fig. 2.41, V the diagram of torques is presented. Estimated (maximum) torque Mz = 9,54N. Strength condition

Rigidity condition

The limiting condition is rigidity. Therefore, the allowed value of the transmitted power [N] = 82.3 kW.

When stretching (squeezing) the timber in its cross sections arise only normal stresses. The resultant of the corresponding elementary forces o, dA - longitudinal force N- can be found using the section method. In order to be able to determine the normal stresses for a known value of the longitudinal force, it is necessary to establish the law of distribution over the cross section of the beam.

This problem is solved on the basis flat section prostheses(hypotheses of J. Bernoulli), which reads:

the beam sections, which are flat and normal to its axis before deformation, remain flat and normal to the axis even during deformation.

When a beam is stretched (made, for example, For greater visibility of the rubber experience), on the surface whom a system of longitudinal and transverse scratches has been applied (Fig. 2.7, a), you can make sure that the risks remain straight and mutually perpendicular, change only

where A is the cross-sectional area of ​​\u200b\u200bthe beam. Omitting the index z, we finally obtain

For normal stresses, the same sign rule is adopted as for longitudinal forces, i.e. when stretched, the stresses are considered positive.

In fact, the distribution of stresses in the beam sections adjacent to the place of application of external forces depends on the method of application of the load and may be uneven. Experimental and theoretical studies show that this violation of the uniformity of stress distribution is local character. In the sections of the beam, spaced from the place of loading at a distance approximately equal to the largest of the transverse dimensions of the beam, the stress distribution can be considered almost uniform (Fig. 2.9).

The considered situation is a special case principle of Saint Venant, which can be formulated as follows:

the distribution of stresses essentially depends on the method of application of external forces only near the place of loading.

In parts sufficiently remote from the place of application of forces, the distribution of stresses practically depends only on the static equivalent of these forces, and not on the method of their application.

Thus, applying Saint Venant principle and digressing from the question of local stresses, we have the opportunity (both in this and in subsequent chapters of the course) not to be interested in specific ways of applying external forces.

In places of a sharp change in the shape and dimensions of the cross section of the beam, local stresses also arise. This phenomenon is called stress concentration, which we will not consider in this chapter.

In cases where the normal stresses in different cross sections of the beam are not the same, it is advisable to show the law of their change along the length of the beam in the form of a graph - diagrams of normal stresses.

EXAMPLE 2.3. For a beam with a step-variable cross section (Fig. 2.10, a), plot longitudinal forces And normal stresses.

Solution. We break the beam into sections, starting from the free messenger. The boundaries of the sections are the places where external forces are applied and the dimensions of the cross section change, i.e., the beam has five sections. When plotting only diagrams N it would be necessary to divide the beam into only three sections.

Using the method of sections, we determine the longitudinal forces in the cross sections of the beam and build the corresponding diagram (Fig. 2.10.6). The construction of the diagram And is fundamentally no different from that considered in Example 2.1, so we omit the details of this construction.

We calculate normal stresses using formula (2.1), substituting the values ​​of forces in newtons, and areas - in square meters.

Within each section, the stresses are constant, i.e. e. the plot in this area is a straight line, parallel to the abscissa axis (Fig. 2.10, c). For strength calculations, first of all, those sections in which the greatest stresses occur are of interest. It is significant that in the considered case they do not coincide with those sections where the longitudinal forces are maximum.

In cases where the cross section of the beam along the entire length is constant, the diagram A similar to a diagram N and differs from it only in scale, therefore, naturally, it makes sense to build only one of the indicated diagrams.

Stretch (compression)- this is the type of loading of the beam, in which only one internal force factor arises 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 method of sections, it is possible to determine the value of the VSF - the longitudinal force N under simple loading.

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

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

It follows that the fibers of the beam (Figure 110) are elongated by the same amount. This means that the internal forces (i.e., stresses) acting on each fiber will be the same and distributed uniformly over the cross section.

Figure 110

Since N is the resultant internal forces, then N \u003d σ A, 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: a circular section with a diameter of d = 4 mm and a square section with a side of 5 mm are stretched with the same force F = 1000 N. Which of the rods is more loaded?

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

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

Solution:

In tension, the longitudinal force in the rods is N = F = 1000 N.

Cross-sectional areas of rods:

; .

Normal stresses in the cross sections of the 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 is stretched 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: cable section is not a circle!

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

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

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

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

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. one side of the cut

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

The order of plotting N and σ

1 Cross-sections divide the beam into sections, the boundaries of which are:

a) sections at the ends of the beam;

b) where the forces F are applied;

c) where the cross-sectional area A changes.

2 We number the sections, starting with

free end.

3 For each plot, using the method

sections, we determine the longitudinal force N

and plot the plot N on a scale.

4 Determine the normal stress σ

on each site and build in

plot scale σ.

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

Given: F 1 \u003d 10 kN; F 2 = 35 kN; A 1 \u003d 1 cm 2; A 2 \u003d 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 method of sections, we determine the longitudinal force N.

The longitudinal force N is equal to the algebraic sum of all external forces applied to the rest of the beam. Moreover, the external forces F, stretching the beam are considered positive.

Table 13

4) We build the diagram N on a scale. The scale is indicated only by positive values ​​of 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, on the diagram N there will be vertical jumps equal in magnitude to these forces.

6) We determine the normal stresses in the sections of each section:

; ;

; .

We build the diagram σ on a scale.

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

Think and answer questions

1) it is impossible; 2) is possible.

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

1) depend; 2) do not depend.

54 Does the amount of 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 in 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 with equal cross-sectional area are stretched by the same 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 appear in the rods.

57 For a bar (Figure 112), plot N and σ diagrams if F 1 = 2 kN; F 2 \u003d 5 kN; A 1 \u003d 1.2 cm 2; A 2 \u003d 1.4 cm 2.

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

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

Rice. 11.3. Design scheme for an oblique bend

External force F applied at an angle to the axis y. Let's decompose the force 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 to:

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

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

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

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

Dangerous sections also include sections where bending moments | M x| and | M y| reach sufficiently large values ​​at the same time. Therefore, with oblique bending, there may be several dangerous sections.

In general, when - asymmetric 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.

Determining the dimensions of the cross section.

Movements in oblique bending

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

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

The 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:

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

Rice. 11.4. Scheme of a section of a pinched beam at an oblique bend

To determine the 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).

Points of contact furthest from the neutral axis A And WITH are dangerous points in the compression and tension zones, respectively.

For plastic materials, when the design resistance of the beam material in tension and compression are equal to each other, i.e. [ σ p] = = [s c] = [σ ], in the dangerous section is determined and the strength condition can be represented as

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

Three types of calculations follow from the strength condition:

Checking;

Design - determination of the geometric dimensions of the section;

Determination of the bearing capacity of the beam (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 in the following way:

or

definitively .

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

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

Similarly, we define horizontal displacement:

Then the total displacement is determined by the formula

Rice. 11.5. Scheme for determining the full displacement

at an oblique bend

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 deflection direction 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

at an oblique bend

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

11.4. Eccentric tension and compression of the beam. Normal

stresses in the cross sections of the beam

Eccentric tension (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. Consider the eccentric compression of a beam. We denote the coordinates of the point of application of force F through x F And at F , and the main axes of the cross section - through x and y. Axis z direct in such a way that the coordinates x F And at F were positive (Fig. 11.7, a)

If you transfer the power 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 doing so, we have:

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

squared radii of inertia of the section, then

Where x And y are the coordinates of the 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. Scheme of a beam with eccentric compression

In the case of eccentric tension of the beam in the resulting formula, the "minus" sign should be replaced by the "plus" sign.

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

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

The purpose of calculations for strength and torsional rigidity is to determine such dimensions of the cross-section of the beam, at which stresses and displacements will not exceed the specified values ​​\u200b\u200ballowed by the operating conditions. The strength condition for allowable shear stresses is generally written as This condition means that the highest shear stresses that occur in a twisted beam should not exceed the corresponding allowable stresses for the material. The allowable torsional stress depends on 0 ─ the stress corresponding to the dangerous state of the material, and the accepted safety factor n: ─ yield strength, nt is the safety factor for 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 allowable torsional stresses are taken depending on the allowable tensile stresses for the same material. So for steel [for cast iron. When calculating the strength of twisted beams, three types of tasks are possible, differing in the form of using the strength conditions: 1) checking stresses (testing calculation); 2) section selection (design calculation); 3) determination of the permissible load. 1. When checking stresses for given loads and dimensions of a beam, the largest shear stresses arising in it are determined and compared with those given by 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 allowable 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 circular or annular section of the beam are found by the magnitude of the polar moment of resistance. 3. When determining the allowable load for a given allowable voltage and polar moment of resistance WP, the allowable torque MK is first determined on the basis of (3.16) and then, using the torque diagram, a connection is established between K M and external torsional moments. The calculation of the beam for strength does not exclude the possibility of deformations that are unacceptable during its operation. Large angles of twisting of the beam are very dangerous, as they can lead to a violation of the accuracy of machining parts if this beam is a structural element of the processing machine, or torsional vibrations can occur if the beam transmits time-varying torsional moments, so the beam must also be calculated for rigidity. The stiffness condition is written in the following form: where ─ the largest relative angle of beam twisting, determined from expression (2.10) or (2.11). Then the stiffness condition for the shaft will take the form different types loads vary from 0.15° to 2° per 1 m of beam length. Both in the strength condition and in the stiffness 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. By specific calculations, it can be seen that the polar moments of inertia and the moment of resistance for an annular section are much larger than for a round circular section, since the annular section does not have areas close to the center. Therefore, a bar of annular section in torsion is more economical than a bar of a solid round section, i.e., it requires less material consumption. However, the manufacture of such a bar is more complicated, and therefore more expensive, and this circumstance must also be taken into account when designing bars operating in torsion. We will illustrate the methodology for calculating the beam for strength and torsional rigidity, as well as reasoning 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 allowable stresses across the fibers (over a length of at least 10 cm) [cm] 90 2.5 Rcm 90 3 Splitting along the fibers when bending [u] 2 Rck 2.4 Splitting along the fibers when cutting 1 Rck 1.2 - 2.4 fibers