Statically indeterminate torsion problems. Statically indeterminate problems on torsion Sopromat statically indeterminate systems torsion

Design diagram and diagrams

Solution

Let's denote the longitudinal axis z, points A and B, section numbers 1, 2, 3. The ends of the rod are pinched, so reactive moments M A and M B arise, which need to be calculated. The number of unknown support reactions is two, and the statics equation for this system of forces is unique:

M A – M 1 + M 2 – M B = 0. (1)

Therefore, this system is once statically indeterminate. In addition to equation (1), it is necessary to create another equation containing the same unknowns M A and M B . To this end, we will proceed as follows. Let's discard the right pinching, but replace its influence with the moment M B , still unknown in magnitude and direction. Thus, we obtain design scheme 2), equivalent to the original scheme 1). Now three loads are applied to the rod: M 1, M 2, M B in the form of moments, including the desired one - M B. Since the right end of the rod is clamped, the angle of rotation of this section around the longitudinal axis of the rod should be equal to zero, i.e. . Such a rotation at point B is the result of the action of three force factors: M 1, M 2, M B.

According to the principle of independence of forces, the angle of rotation of section B can first be calculated from each moment and the results can then be summed up. By doing this, we obtain a second equation complementing (1):

When compiling this equation, it was taken into account that the moment M 1 twists only the first section of the rod, the moment M 2 twists sections 1 and 2, and the moment M B twists all three sections. Let us reduce the left side of equation (2) by and G and we get

Equations (1) and (3) form a system for determining M A and M B . To solve it, you first need to determine the moments of inertia J, J, J.

The first section of the rod is a hollow cylinder. For its section

The second section of the rod has a rectangular cross section. Its torsional moment of inertia

J (5)

Here is a tabulated coefficient depending on the aspect ratio of the rectangle. For a given ratio h/b = 2.0 the value taken from the table.

Formula (5) gives the result

J . (6)

The cross-section of the rod of the second section is solid round. That's why

(7)

The values ​​of torques and the found values ​​of the moments of inertia of the sections are substituted into (3)

We reduce b 4 in all terms, carry out simple arithmetic calculations and get

After transformations, the equation takes the form

14.89 M B = 17.78.

From here we have

M B = 1.194kNm.

From equation (1) we find the reactive moment in the pinching of the left end:

M A = M 1 – M 2 + M B = 6 – 7 + 1.194 = 0.194 kNm.

Now you can start constructing a torque diagram. In an arbitrary place of each section of the rod we will draw sections 1–1, 2–2, 3–3.

Let's take the left cut-off part and show the torque in the section M. Although its direction can be chosen arbitrarily, it is better to choose the positive direction, i.e. such that when looking at the end of the cut-off part it is visible directed counterclockwise.

The entire rod is in equilibrium. This means that any cut-off part must be in balance. Therefore, we can write the equilibrium equation:

From here we have

Section 2–2

Section 3–3

kNm.

Based on the results of calculations, we construct a diagram of torques. The dimensions of the cross section of the rod must be found from the strength condition

(8)

Here i is the site number. The left side of the inequality is the greatest absolute value of the shear stress for the entire rod. The right side is the permissible stress for the material based on tangential stresses. Let's install them. For each section, we find the maximum shear stress using the general formula

The torques have already been found. Let us determine the moments of resistance during torsion:

The second formula is a tabulated coefficient depending on the aspect ratio of the rectangle. For a given ratio h/b = 2.0 the value taken from the table.

For each section we determine local maxima of tangential stresses:

(9)

(10)

(11)

From a comparison of the results, we see that the sections of the second section are dangerous.

Allowable shear stress

.

Unlike the previously discussed round rods, the torsion of rods of non-circular transverse shape has its own peculiarities. The main one is deplanation. This is the phenomenon that sections cease to be flat and become deplaned. Formulas based on the hypothesis of plane sections lose their validity. Normal stresses arise.

There is free and constrained torsion. Free This is called torsion in which the deplanation is constant along the length of the rod and can be characterized by the amount of displacement in the axial direction. Torsion of a rod, in which the deplanation of the section along the length of the rod changes, is called constrained torsion. In this case, a special type of internal force arises - a bimoment, which affects the distribution of normal and tangential stresses over the section.

Rods with a non-circular cross-section can be varied (Fig. 11.1).

Rice. 11.1. Rods with a non-circular cross-section: a) thick-walled; b) thin-walled closed and open profile

Thick-walled are called rods having the dimensions of various section elements commensurate with the dimensions of the section itself. The deformation of thick-walled rods is complex; problems of torsion of such rods are solved analytically or numerically using elasticity theory methods.

Thin-walled are called rods in which the length of the cross-sectional contour is much greater than the thickness of the section.

The calculation of thin-walled rods of open and closed profile for constrained torsion is studied in the theory of thin-walled rods developed by prof. V.Z. Vlasov.

The solution to the problem of free torsion of rods of non-circular cross-section was obtained by Saint-Venant.

Torsional rectangular cross section the greatest stress occurs in the middle of the long side of the circuit (Fig. 11.2). To calculate it, use formula (11.1).

Here W t =αhb 2- moment torsional resistance, α – Saint-Venant coefficient, h And b dimensions of a rectangular section (Fig. 11.2).

Twist angle of cargo section length l with constant internal force is found by formula (11.2)

Here I t =βhb 3- moment of inertia during torsion, β – Saint-Venant coefficient.

Ep. τ[MPa]

Rice. 11.2. Shear stress diagram

Saint-Venant coefficients α, β, γ are determined using Table 11.1 depending on the ratio h/b.

Table 11.1

h/b
α	0,208	0,246	0,267	0,282	0,299	0,307	0,313	0,333
β	0,140	0,229	0,263	0,281	0,299	0,307	0,312	0,333
γ	1,000	0,795	0,753	0,745	0,743	0,742	0,742	0,742

The calculation of various non-circular cross sections for strength and stiffness is carried out similarly to that described in the previous lecture. Using the conditions of strength and stiffness, problems are solved in order to select cross-sectional dimensions, determine the permissible load and check that the conditions are met. Depending on the profile of the cross section, the geometric characteristics of the cross section, which appear in the formulas for calculating stresses and displacements, are determined differently. (Look up these formulas yourself using the textbook).

Solving statically indeterminate problems in torsion. The problems of torsion of rods are statically indeterminate, if the torques arising in the cross sections of the rod cannot be determined using equilibrium equations alone. To solve such problems, it is necessary to consider the deformed state of the twisted rod. The solution algorithm is similar to that described in the topic axial tension–compression.

In the case of constant rigidity of the rod, it is convenient to use the initial parameters method to solve statically indeterminate problems (familiarize yourself with this method).

Problems may be statically indeterminate several times. Let us consider statically indeterminate problems once.

Rice. 11.3. Statically indeterminate rods in torsion

a) Disclosure of static indetermination

∑ m X = 0; M A - M + M V n st

It is impossible to move (angle of twist) point B (rigid embedding), then this movement can be represented as the sum of the twist angles of the load sections φ B =φ I+φ II = 0 (2).

M t =const can be represented as: (3). Let's substitute (3) into (2): . (4)

Let us write down the equations of torques on the load sections, while considering the equilibrium of the right side containing the support reaction M V: M t,I = M V- const M t,II = M V - M– const. If the stiffnesses on the load sections are equal, equation (4) will take the form:

M IN

b) Disclosure of static indetermination

1. Consider the static side of the problem

Let's create an equilibrium equation:

∑ m X = 0; M A + ml – M V = 0 (1), we find the degree of static indetermination as the difference between the unknown support reactions and the number of static equations n st = 2 – 1 = 1 – the problem is once statically indeterminate and one more equation is required to reveal the static indetermination.

2. Consider the geometric side of the problem

Moving (twist angle) of a point IN(rigid embedding) is impossible, then this movement can be represented as the sum of the angles of twist of the load sections φ B =φ I = 0 (2).

3. Let's consider the physical side of the problem

The angle of twist on the load section length, where M t described by a linear equation can be represented as:

(3). Let's substitute (3) into (2): . (4)

Let us write the equation of torques on the load section, considering the equilibrium of the right side containing the support reaction M V: M t, I = - M V + mx, we substitute the equation of internal force into (4):

Let's solve the resulting equation for one unknown M IN . Next, the problem is solved as statically determinable.

Calculation of rods in torsion based on the limit state. Let us consider the distribution of tangential stresses in the cross section of a round rod made of an elastoplastic material, subject to the idealized Prandtl diagram (Fig. 11.4).

Rice. 11.4. Prandtl diagram

τ max < τs τ max = τ s. τ sτ s

M t = τ sWρ Elastic core Plastic hinge

(M t , lim)

Rice. 11.5. Distribution of tangential stresses in cross section

At shear angles γ ≤ γ s the material obeys Hooke's law, i.e. τ = G γ, with γ = γ s shear stress reaches the yield point τ s, for γ > γ s the material “flows” at a constant voltage τ = τ s. This ends the purely elastic stage of work (Fig. 11.5 b) and the moment reaches a dangerous value. With a further increase in torque, the stress diagram takes on the form shown in Fig. 11. 5th century As the torque increases, the elastic core decreases, and the fluidity of the material occurs throughout the entire section; a state of limiting equilibrium occurs, corresponding to the maximum load-bearing capacity of the rod. For a solid circular section in the case shown in Fig. 11. 5 g load-bearing capacity of the rod increases by 33% compared to the load-bearing capacity calculated for the situation shown in Fig. 11.5

4.4. Statically indeterminate torsion problems

Such problems usually arise if the movement of the shaft is limited in some sections, for example, (Fig. 4.9), when its ends are pinched. IN

one equilibrium equation: :

there are two unknown moments in the supports, so the problem is statically indeterminate. To solve it, we create an additional displacement equation. Let's consider the displacements (rotation angles) of the sections that are the boundaries of the shaft sections..gif" width="99" height="27 src=">.

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Since the shaft section is pinched, then from: https://pandia.ru/text/78/579/images/image011_42.gif" align="left" width="258" height="186">

The potential deformation of the shaft section with length dz will be:
Since during torsion τ = (MK / IP) r, then

Reducing by IP, we obtain an expression for the potential energy of deformation during torsion

4.6 . Torsion of rods of non-circular cross-section

https://pandia.ru/text/78/579/images/image018_20.gif" align="left" width="324" height="237 src="> When torsion of rods (shafts) not round and not annular cross sections, the assumptions accepted for torsion of round and annular shafts are not met: flat cross sections of the rod do not remain flat during torsion, but deplane (curve); straight radii drawn in flat sections are bent; the distance between sections changes (Fig. 4If a rod of constant cross-section along its entire length is not pinched anywhere and the twisting moments are located at its ends, then all sections deplane equally and normal stresses do not arise. Such torsion is called free. However, with sufficient accuracy for practical purposes, for non-circular rods you can use formulas derived for a round rod, replacing both https://pandia.ru/text/78/579/images/image021_17.gif" width="23" height="27 src=">- moment of inertia during torsion, and - moment of resistance during torsion.

https://pandia.ru/text/78/579/images/image024_18.gif" width="90" height="49">, ,

For a rectangular cross section (Fig. 4.12)

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Here and - depend on the relationship.

Coefficients.	The ratio of the larger side of the section to the smaller side.

Differential" href="/text/category/differentcial/" rel="bookmark">differential equation, the same as the problem of the equilibrium of a thin film stretched over a contour of the same outline as the contour of the cross section of the rod and loaded with uniformly distributed pressure. Analogue voltage is the angle made by the tangent to the surface of the film with the plane of the contour, and the analogue of torque is the volume enclosed between the plane of the contour and the surface of the film. Fig. 4.13, a shows the behavior of the film under pressure, Fig. 4.13, b shows the qualitative distribution stresses during torsion of a rod of a complex profile. Using a special device and calibrated film, quantitative results can also be obtained. To do this, in order to take into account the rigidity of the film, the same experiment is carried out with a round hole, from where the required film rigidity is obtained, since the solution in this case can be get exactly.

4.7. Free torsion of thin-walled rods

Thin-walled rods are those that have one cross-sectional dimension - profile thickness , and less than another - cross-sectional contour length s. Rods come in open (Fig. 4.14) and closed (Fig. 4.15) profiles. Let's use the membrane analogy. The nature of the behavior of the film and, accordingly, the tangential stresses in thin-walled rods of open and closed profiles is fundamentally different (Fig. 4.16 and Fig. 4. If the rod of an open profile is straightened into a long rectangle, then the shape of the film will not change.

Then for a rectangular section at , we have: ,..gif" width="22" height="25"> rectangles, then

..gif" width="42" height="26"> .

Systems in which the number of superimposed connections is greater, the number of independent equilibrium equations, are called stat undefined.Compared with statistically definable systems, in one hundred indefinable. systems have additional extra connections. The term “extra connections” is conditional. These connections are redundant from the point of view of the calculation premises. In fact, these connections create additional reserves for structures, both in terms of ensuring its rigidity and strength. In Fig. 2.5, and shows a bracket consisting of 2 rods hingedly connected to each other. Due to the fact that only vertical force acts on the structure R, and the system is flat, it turns out that the forces in the rods are easily determined. from the equilibrium conditions of the node A, i.e. x= 0, y= 0. Expanding these equations, we obtain a closed system of linear equations for unknown forces N 1 and N 2 in which the number of equations is equal to the number of unknowns: N 1  N 2 sin  = 0; N 2 cos   R = 0.

If the design of the bracket is complicated by adding another rod (Fig. 2.5, b), then the forces in the rods N 1 ,N 2 and N 3 can no longer be determined using the previous method, because with the same two equilibrium equations (2.16), there are 3 unknown forces in the rods. A semi-system is once a hundred indeterminate. The difference between the number of unknown forces and the number of independent (meaningful) equilibrium equations connecting these forces is called the degree c of the indeterminate system. In the general case, under na statically indeterminate system is understood as a system in which the number of unknown external support reactions and internal forces exceeds the number of independent and meaningful equilibrium equations by n units. The solution of statically indeterminate problems by the method of forces is carried out in the following sequence.1 Set the degree st of the indeterminate system as the difference between the number of unknown forces sought and the number of independent equilibrium equations. It is taken into account that a simple hinge connecting 2 rods of the system reduces the degree of st by 1, since it removes one connection that prevents the rotation of one part of the system relative to the other. A simple hinge allows you to add to Eq. equal of the entire system, the equilibrium equation of the part of the system connected by this hinge.2. From the given st. undef. system, the main system is isolated by removing unnecessary connections and external load.3. The equivalent system corresponding to the selected main one is depicted, in which forces are applied instead of the removed extra bonds and in their direction X i, if the connections prevented linear movement, and pairs Xk, if they excluded section rotations.4. The canonical equations of the force method are compiled.5. The coefficients of canonical equations are calculated analytically

IN TORSION (Task No. 11)

The task

A steel shaft of circular cross-section consists of three sections with different polar moments of inertia (Fig. 3.6, A). The ends of the shaft are rigidly secured against rotation relative to the longitudinal axis of the shaft. Loads are specified: pairs of forces and , acting in the plane of the cross-section of the shaft; the relationship between the polar moments of inertia of the shaft sections and ; lengths of sections , , .

Required:

1) build a diagram of torques;

2) select the dimensions of the cross sections based on the strength conditions;

3) construct a diagram of the twist angles.

Solution

Due to the presence of two rigid support fastenings, under the influence of load, reactive pairs and arise in each of them. Having created the equilibrium condition for the shaft

We are convinced that the written equation cannot be solved uniquely, since it contains two unknown quantities: and . The remaining equilibrium equations for a given load are carried out identically. Consequently, the problem is once statically indeterminate.

To reveal static indetermination, we create a condition for the compatibility of deformations. Due to the rigidity of the supporting fastenings, the end sections of the shaft do not rotate. This is equivalent to the fact that the total angle of rotation of the shaft in the area A–B equal to zero: , or .

The last equation is the condition for compatibility of deformations. To connect it with the equilibrium equation, we write down the physical equations relating torques and angles of twist (3.3) (Hooke’s law for torsion) for each section of the rod:

, , .

Substituting the physical relations into the condition of compatibility of deformations, we find the reactive moment , and then from the equilibrium equation we determine . The torque diagram is shown in Fig. 3.6, b.

To solve the problem of selecting a section, we write down formulas for determining the maximum tangential stresses (3.5) on each section of the shaft:

; ; .

The coefficients and , representing the ratio of the polar moments of resistance of the sections of the second and third sections of the shaft to the polar moment of resistance of the section of the first section, will be determined through the known parameters and .

The polar moment of inertia can be written in two ways:

; ,

where , are the radii of the first and second sections of the rod. From here we express the radius through:

Then the polar moment of resistance of the second section

,

that is . Likewise.

Now we can compare the maximum tangential stresses in individual sections and write down the strength condition (3.13) for the largest of them. From this condition we find the required polar moment of resistance, and then, using formula (3.8), the radii of the shaft in each section.

; ; .

To construct a diagram of twist angles, we calculate the twist angles at each section of the rod using formula (3.3). The ordinates of the diagram are obtained by sequentially summing the results for individual sections, starting from one of the ends of the shaft. The correctness of the solution is checked by the equality of the twist angle to zero at the other end of the shaft. The diagram of the twist angles is shown in Fig. 3.6, V.

For a structure with a rigid rod, the rational equilibrium equation, which includes one unknown force, is the equation where A- a hinge around which a rigid rod rotates.

As the name implies, this method is applicable to structures whose rods are made of plastic material.

Obviously, the relationship between the deformations of the rods will be the same as in the first part of the problem, therefore the equation for the compatibility of deformations in the third part of the problem can be written using the previously obtained equation, replacing it with .

When solving this problem, correspondence students perform only calculations based on the plastic limit state. The remaining students solve problem No. 6 in accordance with the teacher’s requirement. Point 2, marked *, is optional and is carried out at the request of the student.

Modern building design standards provide for a more complex approach (introduction of separate safety factors for loads, material properties, operating conditions of the structure). The student will become familiar with this when studying courses on metal, reinforced concrete and other structures.