CHAPTER 9 Shear and Torsion
The beam shown in Fig. 9.13, has four sections. If we consider the equilibrium conditions for systems of forces applied to the left cut-off part, we can write:
Section 1 |
a (Fig. 9.13, b). |
|||||||||||||||
Mx 0 : Mcr m x dx 0 ; Mkr |
dx. |
|||||||||||||||
Section 2 |
a x2 |
a b (Fig. 9.13, c). |
||||||||||||||
Mx 0 : Mcr m x dx M1 0 ; Mkr m x dx M1 . |
||||||||||||||||
Section 3 |
a b x2 |
a b c (Fig. 9.13, d). |
||||||||||||||
M0; |
x dx M . |
|||||||||||||||
Section 4 |
a b c x2 a b c d . |
|||||||||||||||
Mx 0 : Mcr m x dx M1 M2 0 ; |
||||||||||||||||
M cr |
m x dx M1 M2 . |
|||||||||||||||
Thus, the torque Mcr in the cross section of the beam is equal to the algebraic sum of the moments of all external forces acting on one side of the section.
9.2.2. Stresses and strains during torsion of a straight beam of circular cross-section
As already mentioned, the total tangential stresses could be determined from dependence (9.14) if the law of their distribution over the cross section of the beam was known. The impossibility of analytically determining this law forces one to turn to an experimental study of beam deformations.
V. A. Zhilkin
Let us consider a beam, the left end of which is rigidly clamped, and a torsional moment M cr is applied to the right end. Before loading the beam with a moment, an orthogonal mesh with cell dimensions a×b was applied to its surface (Fig. 9.14, a). After applying a twisting moment M cr, the right end of the beam will rotate relative to the left end of the beam by an angle, while the distances between the sections of the twisted beam will not change, and the radii drawn in the end section will remain straight, i.e. it can be assumed that the hypothesis of flat sections is satisfied (Fig. 9.14, b). Sections that are flat before the beam is deformed remain flat after deformation, turning like hard disks, one relative to the other at a certain angle. Since the distance between the sections of the beam does not change, the longitudinal relative deformation x 0 is equal to zero. The longitudinal lines of the grid take on a helical shape, but the distance between them remains constant (hence, y 0), the rectangular grid cells turn into parallelograms, the dimensions of the sides do not change, i.e. the selected elementary volume of any layer of timber is under conditions of pure shear.
Let's cut out a beam element with length dx in two cross sections (Fig. 9.15). As a result of loading the beam, the right section of the element will rotate relative to the left by an angle d. In this case, the generatrix of the cylinder will rotate at an angle
CHAPTER 9 Shear and Torsion
shift All generatrices of the internal cylinders of radius will rotate through the same angle.
According to Fig. 9.15 arc
ab dx d .
where d dx is called the relative twist angle. If the dimensions of the cross sections of a straight beam and the torques acting in them are constant in a certain area, then the value is also constant and equal to the ratio full angle twisting in this section to its length L, i.e. L.
Passing according to Hooke's law under shear (G) to stresses, we obtain
So, in cross sections When a beam is torsioned, tangential stresses arise, the direction of which at each point is perpendicular to the radius connecting this point with the center of the section, and the magnitude is directly proportional
V. A. Zhilkin
the distance of the point from the center. At the center (at 0 ) the tangential stresses are zero; at points located in close proximity to the outer surface of the beam, they are greatest.
Substituting the found stress distribution law (9.18) into equality (9.14), we obtain
Mkr G dF G 2 dF G J , |
||||||||||||||||
where J d 4 is the polar moment of inertia of the circular transverse |
||||||||||||||||
of a wide section of timber. |
||||||||||||||||
Product by G.J. |
called lateral stiffness |
|||||||||||||||
th section of the beam during torsion. |
||||||||||||||||
The units of measurement for hardness are |
||||||||||||||||
are N·m2, kN·m2, etc. |
||||||||||||||||
From (9.19) we find the relative angle of twist of the beam |
||||||||||||||||
M cr |
||||||||||||||||
and then, eliminating (9.18) from equality, we obtain the formula |
||||||||||||||||
for stresses during torsion of timber round section |
||||||||||||||||
M cr |
||||||||||||||||
The highest voltage values are reached at the end |
||||||||||||||||
tour points of the section at d 2: |
||||||||||||||||
M cr |
M cr |
M cr |
||||||||||||||
is called the moment of resistance to torsion of a shaft of circular cross-section.
The dimension of the moment of torsional resistance is cm3, m3, etc.
which allows you to determine the angle of twist of the entire beam
GJ cr. |
If the beam has several sections with different analytical expressions for M cr or different meanings cross-section stiffness GJ , then
Mkr dx |
|||||
For a beam of length L of constant cross-section, loaded at the ends by concentrated pairs of forces with a moment M cr,
Mkr L |
|||||||||||||||||||
| D and internal d. Only in this case J and W cr are necessary | |||||||||||||||||||
1 c 4 ; W cr |
1 c 4 ; c |
|||||
The diagram of tangential stresses in the section of a hollow beam is shown in Fig. 9.17.
A comparison of diagrams of tangential stresses in solid and hollow beams indicates the advantages of hollow shafts, since in such shafts the material is used more rationally (material in the area of low stress is removed). As a result, the distribution of stresses across the cross-section becomes more uniform, and the beam itself becomes lighter,
than a solid beam of equal strength - Fig. 9.17 cross-section, despite some
swarm increase in outer diameter.
But when designing beams that work in torsion, it should be taken into account that in the case of an annular section, their production is more difficult, and therefore more expensive.
The longitudinal force N arising in the cross section of the beam is the resultant of the internal normal forces distributed over the cross-sectional area, and is related to the normal stresses arising in this section by dependence (4.1):
here is the normal stress at an arbitrary cross-sectional point belonging to an elementary area - the cross-sectional area of the beam.
The product represents the elementary internal force per area dF.
The magnitude of the longitudinal force N in each particular case can be easily determined using the section method, as shown in the previous paragraph. To find the values of stresses a at each point of the cross section of the beam, you need to know the law of their distribution over this section.
The law of distribution of normal stresses in the cross section of a beam is usually depicted by a graph showing their change along the height or width of the cross section. Such a graph is called a normal stress diagram (diagram a).
Expression (1.2) can be satisfied for infinitely large number types of stress diagrams a (for example, with diagrams a shown in Fig. 4.2). Therefore, to clarify the law of distribution of normal stresses in the cross sections of a beam, it is necessary to conduct an experiment.
Let us draw lines on the side surface of the beam, before loading it, perpendicular to the axis of the beam (Fig. 5.2). Each such line can be considered as a trace of the cross-sectional plane of the beam. When the beam is loaded with an axial force P, these lines, as experience shows, remain straight and parallel to each other (their positions after loading the beam are shown in Fig. 5.2 with dashed lines). This allows us to assume that the cross sections of the beam, flat before it is loaded, remain flat under the action of the load. This experience confirms the hypothesis of plane sections (Bernoulli's hypothesis), formulated at the end of § 6.1.
Let's imagine a beam consisting of countless fibers parallel to its axis.
When a beam is stretched, any two cross sections remain flat and parallel to each other, but move away from each other by a certain amount; Each fiber lengthens by the same amount. And since the same elongations correspond to the same stresses, the stresses in the cross sections of all fibers (and, consequently, at all points of the cross section of the beam) are equal to each other.
This allows us to take the value a out of the integral sign in expression (1.2). Thus,
So, in the cross sections of the beam, during central tension or compression, uniformly distributed normal stresses arise, equal to the ratio of the longitudinal force to the cross-sectional area.
If there are weakening of some sections of the beam (for example, by holes for rivets), when determining the stresses in these sections, one should take into account the actual area of the weakened section equal to the total area reduced by the value of the weakening area
To visually depict changes in normal stresses in the cross sections of the rod (along its length), a diagram of normal stresses is constructed. The axis of this diagram is a straight line segment, equal to length rod and parallel to its axis. For a rod of constant cross-section, the normal stress diagram has the same form as the diagram longitudinal forces(it differs from it only in the accepted scale). With a rod of variable cross-section, the appearance of these two diagrams is different; in particular, for a rod with a stepwise law of change in cross sections, the normal stress diagram has jumps not only in sections in which concentrated axial loads are applied (where the longitudinal force diagram has jumps), but also in places where the dimensions of the cross sections change. The construction of a diagram of the distribution of normal stresses along the length of the rod is considered in example 1.2.
Let us now consider the stresses in the inclined sections of the beam.
Let us denote a the angle between the inclined section and the cross section (Fig. 6.2, a). We agree to consider angle a to be positive when the cross section must be rotated counterclockwise by this angle to align with the inclined section.
As is already known, the elongations of all fibers parallel to the axis of the beam when it is stretched or compressed are the same. This allows us to assume that the stresses p at all points of the inclined (as well as the cross) section are the same.
Let's consider the lower part of the beam, cut off by a section (Fig. 6.2, b). From the conditions of its equilibrium it follows that the stresses are parallel to the axis of the beam and are directed in the direction opposite to the force P, and inner strength acting in the section is equal to P. Here, the area of the inclined section is equal to (where is the cross-sectional area of the beam).
Hence,
where are the normal stresses in the cross sections of the beam.
Let us decompose the stress into two stress components: normal, perpendicular to the section plane, and tangent, parallel to this plane (Fig. 6.2, c).
We obtain the values of and from the expressions
Normal stress is usually considered positive in tension and negative in compression. The tangential stress is positive if the vector representing it tends to rotate the body about any point C lying on the internal normal to the section, clockwise. In Fig. 6.2, c shows the positive shear stress ta, and in Fig. 6.2, g - negative.
From formula (6.2) it follows that normal stresses have values from (at to zero (at a). Thus, the greatest (at absolute value) normal stresses arise in the cross sections of the beam. Therefore, the strength of a tensile or compressed beam is calculated according to normal voltages in its cross sections.
Calculation of timber with a round cross-section for strength and torsional rigidity
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 shear 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 elements structures 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 considerations about cost-effectiveness, 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
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 only applies pure bend; 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 arises. 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).
From 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:
, a , 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.