Solving problems in technical mechanics. Solving problems in theoretical mechanics Prz in technical mechanics solution

Theoretical mechanics is a section of mechanics that sets out the basic laws of mechanical motion and mechanical interaction of material bodies.

Theoretical mechanics is a science that studies the movement of bodies over time (mechanical movements). It serves as the basis for other branches of mechanics (theory of elasticity, strength of materials, theory of plasticity, theory of mechanisms and machines, hydroaerodynamics) and many technical disciplines.

Mechanical movement- this is a change over time in the relative position in space of material bodies.

Mechanical interaction- this is an interaction as a result of which the mechanical movement changes or the relative position of body parts changes.

Rigid body statics

Statics is a section of theoretical mechanics that deals with problems of equilibrium of solid bodies and the transformation of one system of forces into another, equivalent to it.

Basic concepts and laws of statics
• Absolutely rigid body(solid body, body) is a material body, the distance between any points in which does not change.
• Material point is a body whose dimensions, according to the conditions of the problem, can be neglected.
• Free body- this is a body on the movement of which no restrictions are imposed.
• Unfree (bound) body is a body whose movement is subject to restrictions.
• Connections– these are bodies that prevent the movement of the object in question (a body or a system of bodies).
• Communication reaction is a force that characterizes the action of a bond on a solid body. If we consider the force with which a solid body acts on a bond to be an action, then the reaction of the bond is a reaction. In this case, the force - action is applied to the connection, and the reaction of the connection is applied to the solid body.
• Mechanical system is a collection of interconnected bodies or material points.
• Solid can be considered as a mechanical system, the positions and distances between points of which do not change.
• Force is a vector quantity that characterizes the mechanical action of one material body on another.
  Force as a vector is characterized by the point of application, direction of action and absolute value. The unit of force modulus is Newton.
• Line of action of force is a straight line along which the force vector is directed.
• Focused Power– force applied at one point.
• Distributed forces (distributed load)- these are forces acting on all points of the volume, surface or length of a body.
  The distributed load is specified by the force acting per unit volume (surface, length).
  The dimension of the distributed load is N/m 3 (N/m 2, N/m).
• External force is a force acting from a body that does not belong to the mechanical system under consideration.
• Inner strength is a force acting on a material point of a mechanical system from another material point belonging to the system under consideration.
• Force system is a set of forces acting on a mechanical system.
• Flat force system is a system of forces whose lines of action lie in the same plane.
• Spatial system of forces is a system of forces whose lines of action do not lie in the same plane.
• System of converging forces is a system of forces whose lines of action intersect at one point.
• Arbitrary system of forces is a system of forces whose lines of action do not intersect at one point.
• Equivalent force systems- these are systems of forces, the replacement of which one with another does not change the mechanical state of the body.
  Accepted designation: .
• Equilibrium- this is a state in which a body, under the action of forces, remains motionless or moves uniformly in a straight line.
• Balanced system of forces- this is a system of forces that, when applied to a free solid body, does not change its mechanical state (does not throw it out of balance).
  .
• Resultant force is a force whose action on a body is equivalent to the action of a system of forces.
  .
• Moment of power is a quantity characterizing the rotating ability of a force.
• Couple of forces is a system of two parallel forces of equal magnitude and oppositely directed.
  Accepted designation: .
  Under the influence of a pair of forces, the body will perform a rotational movement.
• Projection of force on the axis- this is a segment enclosed between perpendiculars drawn from the beginning and end of the force vector to this axis.
  The projection is positive if the direction of the segment coincides with the positive direction of the axis.
• Projection of force onto a plane is a vector on a plane, enclosed between perpendiculars drawn from the beginning and end of the force vector to this plane.
• Law 1 (law of inertia). An isolated material point is at rest or moves uniformly and rectilinearly.
  The uniform and rectilinear motion of a material point is motion by inertia. The state of equilibrium of a material point and a rigid body is understood not only as a state of rest, but also as motion by inertia. For a solid body there are different kinds motion by inertia, for example, uniform rotation of a rigid body around a fixed axis.
• Law 2. A rigid body is in equilibrium under the action of two forces only if these forces are equal in magnitude and directed in opposite directions along a common line of action.
  These two forces are called balancing.
  In general, forces are called balanced if the solid body to which these forces are applied is at rest.
• Law 3. Without disturbing the state (the word “state” here means the state of motion or rest) of a rigid body, one can add and reject balancing forces.
  Consequence. Without disturbing the state of the solid body, the force can be transferred along its line of action to any point of the body.
  Two systems of forces are called equivalent if one of them can be replaced by the other without disturbing the state of the solid body.
• Law 4. The resultant of two forces applied at one point, applied at the same point, is equal in magnitude to the diagonal of a parallelogram constructed on these forces, and is directed along this
  diagonals.
  The absolute value of the resultant is:
  
• Law 5 (law of equality of action and reaction). The forces with which two bodies act on each other are equal in magnitude and directed in opposite directions along the same straight line.
  It should be kept in mind that action- force applied to the body B, And opposition- force applied to the body A, are not balanced, since they are applied to different bodies.
• Law 6 (law of solidification). The equilibrium of a non-solid body is not disturbed when it solidifies.
  It should not be forgotten that the equilibrium conditions, which are necessary and sufficient for a solid body, are necessary but insufficient for the corresponding non-solid body.
• Law 7 (law of emancipation from ties). A non-free solid body can be considered as free if it is mentally freed from bonds, replacing the action of the bonds with the corresponding reactions of the bonds.

Connections and their reactions
• Smooth surface limits movement normal to the support surface. The reaction is directed perpendicular to the surface.
• Articulated movable support limits the movement of the body normal to the reference plane. The reaction is directed normal to the support surface.
• Articulated fixed support counteracts any movement in a plane perpendicular to the axis of rotation.
• Articulated weightless rod counteracts the movement of the body along the line of the rod. The reaction will be directed along the line of the rod.
• Blind seal counteracts any movement and rotation in the plane. Its action can be replaced by a force represented in the form of two components and a pair of forces with a moment.

Kinematics

Kinematics- a section of theoretical mechanics that examines the general geometric properties of mechanical motion as a process occurring in space and time. Moving objects are considered as geometric points or geometric bodies.

Basic concepts of kinematics
• Law of motion of a point (body)– this is the dependence of the position of a point (body) in space on time.
• Point trajectory- This locus positions of a point in space as it moves.
• Speed ​​of a point (body)– this is a characteristic of the change in time of the position of a point (body) in space.
• Acceleration of a point (body)– this is a characteristic of the change in time of the speed of a point (body).

Determination of kinematic characteristics of a point
• Point trajectory
  In a vector reference system, the trajectory is described by the expression: .
  In the coordinate reference system, the trajectory is determined by the law of motion of the point and is described by the expressions z = f(x,y)- in space, or y = f(x)- in a plane.
  In a natural reference system, the trajectory is specified in advance.
• Determining the speed of a point in a vector coordinate system
  When specifying the movement of a point in a vector coordinate system, the ratio of movement to a time interval is called the average value of speed over this time interval: .
  Taking the time interval to be an infinitesimal value, we obtain the speed value at a given time (instantaneous speed value): .
  Vector average speed is directed along the vector in the direction of the point’s movement, the instantaneous velocity vector is directed tangentially to the trajectory in the direction of the point’s movement.
  Conclusion: the speed of a point is a vector quantity equal to the time derivative of the law of motion.
  Derivative property: the derivative of any quantity with respect to time determines the rate of change of this quantity.
• Determining the speed of a point in a coordinate reference system
  Rate of change of point coordinates:
  .
  The modulus of the total velocity of a point with a rectangular coordinate system will be equal to:
  .
  The direction of the velocity vector is determined by the cosines of the direction angles:
  ,
  where are the angles between the velocity vector and the coordinate axes.
• Determining the speed of a point in a natural reference system
  The speed of a point in the natural reference system is defined as the derivative of the law of motion of the point: .
  According to previous conclusions, the velocity vector is directed tangentially to the trajectory in the direction of the point’s movement and in the axes is determined by only one projection.

Rigid body kinematics
• In the kinematics of rigid bodies, two main problems are solved:
  1) setting the movement and determining the kinematic characteristics of the body as a whole;
  2) determination of kinematic characteristics of body points.
• Translational motion of a rigid body
  Translational motion is a motion in which a straight line drawn through two points of a body remains parallel to its original position.
  Theorem: during translational motion, all points of the body move along identical trajectories and at each moment of time have the same magnitude and direction of speed and acceleration.
  Conclusion: the translational motion of a rigid body is determined by the movement of any of its points, and therefore, the task and study of its motion is reduced to the kinematics of the point.
• Rotational motion of a rigid body around a fixed axis
  Rotational motion of a rigid body around a fixed axis is the motion of a rigid body in which two points belonging to the body remain motionless during the entire time of movement.
  The position of the body is determined by the angle of rotation. The unit of measurement for angle is radian. (A radian is the central angle of a circle, the arc length of which is equal to the radius; the total angle of the circle contains 2π radian.)
  The law of rotational motion of a body around a fixed axis.
  We determine the angular velocity and angular acceleration of the body using the differentiation method:
  — angular velocity, rad/s;
  — angular acceleration, rad/s².
  If you dissect the body with a plane perpendicular to the axis, select a point on the axis of rotation WITH and an arbitrary point M, then point M will describe around a point WITH circle radius R. During dt there is an elementary rotation through an angle , and the point M will move along the trajectory a distance .
  Linear speed module:
  .
  Point acceleration M with a known trajectory, it is determined by its components:
  ,
  Where .
  As a result, we get the formulas
  tangential acceleration: ;
  normal acceleration: .

Dynamics

Dynamics is a section of theoretical mechanics in which the mechanical movements of material bodies are studied depending on the causes that cause them.

Basic concepts of dynamics
• Inertia- this is the property of material bodies to maintain a state of rest or uniform rectilinear motion until external forces change this state.
• Weight is a quantitative measure of the inertia of a body. The unit of mass is kilogram (kg).
• Material point- this is a body with mass, the dimensions of which are neglected when solving this problem.
• Center of mass of a mechanical system- a geometric point whose coordinates are determined by the formulas:
  
  Where m k , x k , y k , z k— mass and coordinates k-that point of the mechanical system, m— mass of the system.
  In a uniform field of gravity, the position of the center of mass coincides with the position of the center of gravity.
• Moment of inertia of a material body relative to an axis is a quantitative measure of inertia during rotational motion.
  The moment of inertia of a material point relative to the axis is equal to the product of the mass of the point by the square of the distance of the point from the axis:
  .
  The moment of inertia of the system (body) relative to the axis is equal to the arithmetic sum of the moments of inertia of all points:
  
• Inertia force of a material point is a vector quantity equal in modulus to the product of the mass of a point and the acceleration modulus and directed opposite to the acceleration vector:
• The force of inertia of a material body is a vector quantity equal in modulus to the product of the body mass and the modulus of acceleration of the center of mass of the body and directed opposite to the acceleration vector of the center of mass: ,
  where is the acceleration of the center of mass of the body.
• Elementary impulse of force is a vector quantity equal to the product of the force vector and an infinitesimal period of time dt:
  .
  The total force impulse for Δt is equal to the integral of the elementary impulses:
  .
• Elementary work of force is a scalar quantity dA, equal to the scalar proi

Many university students encounter certain difficulties when their courses begin to teach basic engineering subjects such as strength of materials and theoretical mechanics. This article will discuss one of these subjects - the so-called technical mechanics.

Technical mechanics is a science that studies various mechanisms, their synthesis and analysis. In practice, this means combining three disciplines - strength of materials, theoretical mechanics and machine parts. It is convenient because each educational institution chooses in what proportion to teach these courses.

Accordingly, in most tests the tasks are divided into three blocks, which must be solved separately or together. Let's look at the most common tasks.

Section one. Theoretical mechanics

Of the variety of problems in theoretical mechanics, you can most often find problems from the section of kinematics and statics. These are problems on the balance of a flat frame, determination of the laws of motion of bodies and kinematic analysis of a lever mechanism.

To solve problems on the equilibrium of a flat frame, it is necessary to use the equation of equilibrium of a plane system of forces:

The sum of the projections of all forces on the coordinate axes is equal to zero and the sum of the moments of all forces relative to any point is equal to zero. Solving these equations together, we determine the magnitude of the reactions of all supports of the flat frame.

In tasks to determine the basic kinematic parameters of the motion of bodies, it is necessary, based on a given trajectory or the law of motion of a material point, to determine its speed, acceleration (total, tangential and normal) and radius of curvature of the trajectory. The laws of motion of a point are given by the trajectory equations:

Projections of the point's velocity onto the coordinate axes are found by differentiating the corresponding equations:

By differentiating the velocity equations, we find the projections of the acceleration of the point. The tangential and normal accelerations, the radius of curvature of the trajectory are found graphically or analytically:

Kinematic analysis of the lever mechanism is carried out according to the following scheme:

1. Dividing the mechanism into Assur groups
2. Construction of speed and acceleration plans for each group
3. Determination of speeds and accelerations of all links and points of the mechanism.

Section two. Strength of materials

Strength of materials is a rather difficult section to understand, with many different problems, most of which are solved using their own methods. In order to simplify their solution for students, most often in the course of applied mechanics they give elementary problems on the simple resistance of structures - and the type and material of the structure, as a rule, depends on the profile of the university.

The most common tasks are tension-compression, bending and torsion.

In tension-compression problems, it is necessary to construct diagrams of longitudinal forces and normal stresses, and sometimes also displacements of sections of the structure.

To do this, it is necessary to divide the structure into sections, the boundaries of which will be the places where the load is applied or the cross-sectional area changes. Next, using the formulas for equilibrium of a rigid body, we determine the magnitude of internal forces at the boundaries of sections, and, taking into account the cross-sectional area, internal stresses.

Based on the data obtained, we construct graphs - diagrams, taking the axis of symmetry of the structure as the axis of the graph.

Torsion problems are similar to bending problems, except that instead of tensile forces, torques are applied to the body. Taking this into account, it is necessary to repeat the calculation stages - dividing into sections, determining torques and angles of twist, and constructing diagrams.

In bending problems, it is necessary to calculate and determine shear forces and bending moments for a loaded beam.
First, the reactions of the supports in which the beam is fixed are determined. To do this, you need to write down the equilibrium equations of the structure, taking into account all the acting forces.

After this, the beam is divided into sections, the boundaries of which will be the points of application of external forces. By considering the equilibrium of each section separately, shear forces and bending moments at the boundaries of the sections are determined. Diagrams are constructed based on the data obtained.

The cross-section is checked for strength as follows:

1. The location of the dangerous section is determined - the section where the greatest bending moments will act.
2. From the condition of bending strength, the moment of resistance of the cross section of the beam is determined.
3. The characteristic size of the section is determined - diameter, side length or profile number.

Section three. Machine parts

The “Machine Parts” section combines all the tasks for calculating mechanisms operating in real conditions - this could be a conveyor drive or a gear drive. The task is greatly simplified by the fact that all formulas and calculation methods are given in reference books, and the student only needs to select those that are suitable for a given mechanism.

Literature

1. Theoretical mechanics: Guidelines and test assignments for part-time students of mechanical engineering, construction, transport, instrument-making specialties of higher education educational institutions/ Ed. prof. S.M. Targa, - M.: Higher School, 1989. Fourth edition;
2. A. V. Darkov, G. S. Shpiro. "Strength of materials";
3. Chernavsky S.A. Course design of machine parts: Proc. manual for students of mechanical engineering specialties of technical schools / S. A. Chernavsky, K. N. Bokov, I. M. Chernin and others - 2nd ed., revised. and additional - M. Mechanical Engineering, 1988. - 416 p.: ill.

Custom technical mechanics solution

Our company also offers services for solving problems and tests in mechanics. If you have difficulty understanding this subject, you can always order detailed solution we have. We take on difficult tasks!
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Tasks for calculation-analytical and calculation-graphical work in all sections of the technical mechanics course are given. Each task includes a description of problem solving with brief methodological instructions, examples of solutions are given. The appendices contain the necessary reference material. For students of construction specialties of secondary vocational educational institutions.

Determination of reactions of ideal bonds by an analytical method.
1. Indicate the point whose equilibrium is being considered. In tasks for independent work such a point is the center of gravity of the body or the point of intersection of all rods and threads.

2. Active forces are applied to the point under consideration. In tasks for independent work, the active forces are the body’s own weight or the weight of the load, which are directed downwards (more correctly, towards the center of gravity of the earth). If there is a block, the weight of the load acts on the point in question along the thread. The direction of action of this force is determined from the drawing. Body weight is usually denoted by the letter G.

3. Mentally discard connections, replacing their action with reactions of connections. In the proposed problems, three types of connections are used - an ideally smooth plane, ideally rigid rectilinear rods and ideally flexible threads - hereinafter referred to as a plane, a rod and a thread, respectively.

TABLE OF CONTENTS
Preface
Section I. Independent and test work
Chapter 1. Theoretical mechanics. Statics
1.1. Analytical determination of ideal bond reactions
1.2. Determination of support reactions of a beam on two supports under the action of vertical loads
1.3. Determining the position of the center of gravity of the section
Chapter 2. Strength of materials
2.1. Selection of cross-sections of rods based on strength
2.2. Determination of the main central moments of inertia of a section
2.3. Constructing diagrams of shear forces and bending moments for a simple beam
2.4. Determination of the permissible value of the central compressive force
Chapter 3. Statics of structures
3.1. Constructing diagrams of internal forces for the simplest single-circuit frame
3.2. Graphic determination of forces in truss rods by constructing a Maxwell-Cremona diagram
3.3. Determination of linear movements in the simplest cantilever frames
3.4. Calculation of a statically indeterminate (continuous) beam using the three-moment equation
Section II. Calculation and graphic works
Chapter 4. Theoretical mechanics. Statics
4.1. Determination of forces in the rods of the simplest cantilever truss
4.2. Determination of support reactions of a beam on two supports
4.3. Determining the position of the center of gravity of the section
Chapter 5. Strength of materials
5.1. Determination of forces in rods of a statically indeterminate system
5.2. Determination of the main moments of inertia of a section
5.3. Selection of beam section from a rolled I-beam
5.4. Selection of the cross-section of a centrally compressed composite rack
Chapter 6. Statics of structures
6.1. Determination of forces in sections of a three-hinged arch
6.2. Graphic determination of forces in the rods of a flat truss by constructing a Maxwell-Cremona diagram
6.3. Calculation of a statically indeterminate frame
6.4. Calculation of a continuous beam using the three-moment equation
Applications
Bibliography.

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Content

Kinematics

Kinematics of a material point

Determining the speed and acceleration of a point using the given equations of its motion

Given: Equations of motion of a point: x = 12 sin(πt/6), cm; y = 6 cos 2 (πt/6), cm.

Set the type of its trajectory for the moment of time t = 1 s find the position of the point on the trajectory, its speed, total, tangential and normal acceleration, as well as the radius of curvature of the trajectory.

Translational and rotational motion of a rigid body

Given:
t = 2 s; r 1 = 2 cm, R 1 = 4 cm; r 2 = 6 cm, R 2 = 8 cm; r 3 = 12 cm, R 3 = 16 cm; s 5 = t 3 - 6t (cm).

Determine at time t = 2 the velocities of points A, C; angular acceleration of wheel 3; acceleration of point B and acceleration of rack 4.

Kinematic analysis of a flat mechanism

Given:
R 1, R 2, L, AB, ω 1.
Find: ω 2.

The flat mechanism consists of rods 1, 2, 3, 4 and a slider E. The rods are connected using cylindrical hinges. Point D is located in the middle of rod AB.
Given: ω 1, ε 1.
Find: velocities V A, V B, V D and V E; angular velocities ω 2, ω 3 and ω 4; acceleration a B ; angular acceleration ε AB of link AB; positions of instantaneous speed centers P 2 and P 3 of links 2 and 3 of the mechanism.

Determination of absolute speed and absolute acceleration of a point

A rectangular plate rotates around a fixed axis according to the law φ = 6 t 2 - 3 t 3. The positive direction of the angle φ is shown in the figures by an arc arrow. Rotation axis OO 1 lies in the plane of the plate (the plate rotates in space).

Point M moves along the plate along straight line BD. The law of its relative motion is given, i.e. the dependence s = AM = 40(t - 2 t 3) - 40(s - in centimeters, t - in seconds). Distance b = 20 cm. In the figure, point M is shown in a position where s = AM > 0 (at s< 0 point M is on the other side of point A).

Find the absolute speed and absolute acceleration of point M at time t 1 = 1 s.

Dynamics

Integration of differential equations of motion of a material point under the influence of variable forces

A load D of mass m, having received an initial speed V 0 at point A, moves in a curved pipe ABC located in vertical plane. In section AB, the length of which is l, the load is acted upon by constant force T (its direction is shown in the figure) and the force R of the resistance of the medium (the modulus of this force R = μV 2, the vector R is directed opposite to the speed V of the load).

The load, having finished moving in section AB, at point B of the pipe, without changing the value of its speed module, moves to section BC. In section BC, the load is acted upon by a variable force F, the projection F x of which on the x axis is given.

Considering the load to be a material point, find the law of its motion in section BC, i.e. x = f(t), where x = BD. Neglect the friction of the load on the pipe.

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Theorem on the change in kinetic energy of a mechanical system

The mechanical system consists of weights 1 and 2, a cylindrical roller 3, two-stage pulleys 4 and 5. The bodies of the system are connected by threads wound on the pulleys; sections of threads are parallel to the corresponding planes. The roller (a solid homogeneous cylinder) rolls along the supporting plane without sliding. The radii of the stages of pulleys 4 and 5 are respectively equal to R 4 = 0.3 m, r 4 = 0.1 m, R 5 = 0.2 m, r 5 = 0.1 m. The mass of each pulley is considered to be uniformly distributed along its outer rim . The supporting planes of loads 1 and 2 are rough, the sliding friction coefficient for each load is f = 0.1.

Under the action of a force F, the modulus of which changes according to the law F = F(s), where s is the displacement of the point of its application, the system begins to move from a state of rest. When the system moves, pulley 5 is acted upon by resistance forces, the moment of which relative to the axis of rotation is constant and equal to M 5 .

Determine the value of the angular velocity of pulley 4 at the moment in time when the displacement s of the point of application of force F becomes equal to s 1 = 1.2 m.

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Application of the general equation of dynamics to the study of the motion of a mechanical system

For a mechanical system, determine the linear acceleration a 1 . Assume that the masses of blocks and rollers are distributed along the outer radius. Cables and belts should be considered weightless and inextensible; there is no slippage. Neglect rolling and sliding friction.

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Application of d'Alembert's principle to determining the reactions of the supports of a rotating body

The vertical shaft AK, rotating uniformly with an angular velocity ω = 10 s -1, is fixed by a thrust bearing at point A and a cylindrical bearing at point D.

Rigidly attached to the shaft are a weightless rod 1 with a length of l 1 = 0.3 m, at the free end of which there is a load with a mass of m 1 = 4 kg, and a homogeneous rod 2 with a length of l 2 = 0.6 m, having a mass of m 2 = 8 kg. Both rods lie in the same vertical plane. The points of attachment of the rods to the shaft, as well as the angles α and β are indicated in the table. Dimensions AB=BD=DE=EK=b, where b = 0.4 m. Take the load as a material point.

Neglecting the mass of the shaft, determine the reactions of the thrust bearing and the bearing.