The external force acting on the discarded part of the beam and seeking to rotate it relative to the section in a clockwise direction is included in the algebraic sum for determining the shear force () with a plus sign (Fig. 7.5, a). Note that the positive transverse force () "tends to rotate" any of the parts of the beam also in a clockwise direction.

In simple terms: in the section of the beam arises, which must be determined and depicted on. In order for the rule of signs for transverse forces to be fulfilled, you need to remember:

If the transverse force occurs to the right of the section, it is directed downwards, and if the transverse force occurs to the left of the section, it is directed upwards (Fig. 7.5, a).

For the convenience of determining the sign of the bending moment, it is recommended to mentally represent the cross section of the beam in the form of a fixed one.

In other words: according to the rule of signs, the bending moment is positive if it “bends the beam” upwards, regardless of the part of the beam under study. If in the selected section the resulting moment of all external forces that generate the bending moment (it is an internal force) is directed opposite direction of the bending moment according to the sign rule, then the bending moment will be positive.

Let's say the left side of the beam is considered (Fig. 7.5, b). The moment of force P relative to the section is directed clockwise. According to the rule of signs for bending moments for the left side of the beam, the bending moment is positive if it is directed counterclockwise ("bends the beam" upwards). This means that the bending moment will be positive (the sum of the moments of external forces and the bending moment, according to the rule of signs, are oppositely directed).

Instruction

Let Q be the point relative to which the moment of force is considered. This point is called the pole. Draw the radius vector r from this point to the point of application of force F. Then the moment of force M is defined as the vector product of r and F: M=.

The result of a cross product is a vector. The length of a vector is expressed in modulus: |M|=|r|·|F|·sinφ, where φ is the angle between r and F. The vector M is orthogonal to both the vector r and the vector F: M⊥r, M⊥F.

The vector M is directed in such a way that the triple of vectors r, F, M is right. How to determine that the triple of vectors is right? Imagine that you (your eye) are at the end of the third vector and look at the other two vectors. If the shortest transition from the first vector to the second seems to be counter-clockwise, this is a right triple of vectors. Otherwise, you are dealing with a left three.

So, align the beginnings of the vectors r and F. This can be done by parallel transfer of the vector F to the point Q. Now draw an axis through the same point perpendicular to the plane of the vectors r and F. This axis will be perpendicular to the vectors at once. Here, in principle, only two options are possible to direct the moment of force: up or down.

Try to direct the moment of force F up, draw a vector arrow on the axis. From this arrow, as it were, look at the vectors r and F (you can use the symbolic eye). You can mark the shortest transition from r to F with a rounded arrow. Is the triple of vectors r, F, M right? Does the arrow point in a counter-clockwise direction? If yes, then you are in the right direction for the moment of force F. If not, then you need to change the direction to the opposite.

You can also determine the direction of the moment of force using the right hand rule. Align your index finger with the radius vector. Align the middle finger with the force vector. From the end of the thumb up, look at two vectors. If the transition from the index to the middle finger is counterclockwise, then the direction of the moment of force coincides with the direction that the thumb points. If the transition is clockwise, then the direction of the moment of force is opposite to it.

The gimlet rule is very similar to the hand rule. With four fingers of your right hand, as it were, rotate the screw from r to F. The vector product will have the direction in which the gimlet is twisted during such a mental rotation.

Now let the point Q be located on the same line that contains the force vector F. Then the radius vector and the force vector will be collinear. In this case, their vector product degenerates into a zero vector and is represented by a dot. The null vector has no specific direction, but is considered to be co-directional with any other vector.

To correctly calculate the action of a force rotating a body, determine the point of its application and the distance from this point to the axis of rotation. This is important for determining the technical characteristics of various mechanisms. The torque of an engine can be calculated if its power and speed are known.

You will need

• Ruler, dynamometer, tachometer, tester, teslameter.

Instruction

Determine the point or axis around which the body. Find the point of application of the force. Connect the point of application of the force and the point of rotation, or lower the perpendicular to the axis of rotation. Measure this distance, it is the "shoulder of power". Measure in meters. Measure the force in newtons using a dynamometer. Measure the angle between the shoulder and the force vector. To calculate the torque, find the product of the force and the sine of the angle between them M=F r sin(α). The result is in newtons per meter.

The moment of force relative to point O is a vector whose module is equal to the product of the module of force and the arm - the shortest distance from point O to the line of action of the force. The direction of the vector of the moment of force is perpendicular to the plane passing through the point and the line of action of the force, so that, looking in the direction of the moment vector, the rotation performed by the force around the point O occurs clockwise.

If the radius vector is known point of application of force relative to the point O, then the moment of this force relative to O is expressed as follows:

Indeed, the modulus of this vector product is:

. (1.9)

According to the figure, therefore:

The vector , as well as the result of the cross product, is perpendicular to the vectors that belong to the plane Π. The direction of the vector is such that looking in the direction of this vector, the shortest rotation from k is clockwise. In other words, the vector completes the system of vectors () to the right triple.

Knowing the coordinates of the point of application of the force in the coordinate system, the origin of which coincides with the point O, and the projection of the force on these coordinate axes, the moment of force can be determined as follows:

. (1.11)

Moment of force about the axis

The projection of the moment of force about a point on some axis passing through this point is called the moment of force about the axis.

The moment of force about the axis is calculated as the moment of the projection of the force onto the plane Π, perpendicular to the axis, relative to the point of intersection of the axis with the plane Π:

The sign of the moment is determined by the direction of rotation, which the force F⃗ Π tends to impart to the body. If, looking in the direction of the Oz axis, the force rotates the body clockwise, then the moment is taken with the sign ``plus"", otherwise - ``minus"".

1.2 Statement of the problem.

Determination of reactions of supports and hinge C.

1.3 Algorithm for solving the problem.

We divide the structure into parts and consider the balance of each of the structures.

Consider the balance of the entire structure as a whole. (fig.1.1)

We will compose 3 equilibrium equations for the entire structure as a whole:

Consider the equilibrium of the right side of the structure. (Figure 1.2)

Let us compose 3 equilibrium equations for the right side of the structure.

The moment of force relative to a point is determined by the product of the modulus of force and the length of the perpendicular dropped from the point to the line of action of the force (Figure 4).

Figure 4 - Moment of force F relative to point O

When a body is fixed at point O, the force F tends to rotate it around this point. The point O, relative to which the moment is taken, is called the center of the moment, and the length of the perpendicular a is called the shoulder of the force relative to the center of the moment.

The moment of force F relative to O is determined by the product of the force and the arm.

M O (F) = F a.

The moment is considered to be positive if the force tends to rotate the body clockwise, and negative - counterclockwise. When the line of action of the force passes through a given point, the moment of force relative to this point is equal to zero, since in the case under consideration, the arm a \u003d 0 (Figure 5).

Figure 5 - Determining the sign of the moment of force relative to a point

There is one significant difference between the moment of a couple and the moment of force. The numerical value and direction of the moment of a pair of forces do not depend on the position of this pair in the plane. The value and direction (sign) of the moment of force depend on the position of the point relative to which the moment is determined.

Equilibrium equations for a plane system of forces

Conditions for the equilibrium of forces on a plane: for the equilibrium of a system of forces arbitrarily located in a plane, it is necessary and sufficient that the main vector and the main moment of these forces relative to any center each individually equal to zero.

F GL = 0; M GL = Σ M O (F i) = 0.

We obtain the basic form of the equilibrium equation:

Theoretically, an infinite set of equations of moments can be written, but in practice, three equilibrium equations are sufficient to solve problems on a plane. In each specific case, equations with one unknown are used.

For different cases, three groups of equilibrium equations are used:

1. The first form of equilibrium equations

2. The second form of equilibrium equations

3. The third form of equilibrium equations

For a system of parallel forces (Figure 43), only two equilibrium equations can be drawn up:

Example.

Given: F = 24 kN; q = 6 kN/m; M = 12 kN m α = 60°; a = 1.8 m; b = 5.2 m; c = 3.0 m. Determine the reactions V A , H A and V B (Figure 6).

Figure 6 - Given two-support beam

We discard the connections (supports A and B), replace their action with reactions: the fixed support has reactions V A (vertical) and H A (horizontal). Movable support - reaction V B (vertical). We choose the XY coordinate system with the origin in the left support, determine the resultant of the distributed load:

Q \u003d q a 2 \u003d 6 5.2 \u003d 31.2 kN.

We draw the calculation scheme of the beam (Figure 7).

Figure 7 - Calculation diagram of the beam

For the obtained arbitrary flat system of forces, we compose the equilibrium equations:

∑ F ix = 0; H A – F cos60° = 0;

∑ F i y = 0; V A – F cos30° – Q + V B = 0;

∑M A (F i) = 0; Q (1.8 + 2.6) + F cos30 ° (1.8 + 5.2) - M - V B (1.8 + 5.2 + 3) = 0.

We solve a system of equations.

H A \u003d F cos60 ° \u003d 24 0.5 \u003d 12 kN;

V A \u003d F cos30 ° + Q - V B \u003d 24 0.866 + 31.2 - 27.08 \u003d 24.9 kN.

To check the correctness of the solution, we compose the sum of the moments relative to the point of application of the inclined force F:

∑M A (F i) \u003d V A (1.8 + 5.2) - Q 2.6 - M - V B 3 \u003d 24.9 7 - 31.2 2.6 - 12 - 27, 08 3 = - 0.06.

Answer: the support reactions of the beam are V A = 24.9 kN; V B \u003d 27.08 kN; N A = 12 kN.

Test questions:

1. What determines the effect of a pair of forces?

2. Does the effect of a pair of forces depend on its position in the plane?

3. Do the values ​​and direction of the moment of force relative to a point depend on the relative position of this point and the line of action of the force?

4. When is the moment of force relative to a point equal to zero?

5. How many independent equilibrium equations can be composed for a flat system of parallel forces?

In mechanics, there is a concept of the moment of force about a point.

The moment of force about a point is the product of the modulus of force taken with a sign (plus or minus) and the shortest distance from the point to the line of action of the force(Fig. 12), i.e.

M 0()= ± Ph.h.

Dot O, relative to which the moment of force is taken is called center moment; RH = h The shortest distance from the center of the moment to the line of action of the force is called shoulder of strength relative to this point; a plus sign is placed if the force tends to rotate the shoulder h counterclockwise and the minus sign in the opposite direction. Moment of force about a point O in fig. 12 positive.

It follows from the last equality that h=0, i.e. when O- the center of moments is located on the line of action of the force, M 0() =0. As you know, the force is a sliding vector, therefore, when transferring the force along the lines of action from the point BUT to any other point A 1 , A 2 etc. (Fig. 12) the length of the arm will not change, which means that the value of the moment of force relative to the point will not change either. The moment of force, like the moment of a couple, is measured in Newtonometers.

Fig.12. Moment of force about a point O.

1.12. Equilibrium equations for a plane system of parallel forces

Let a system of parallel forces be applied to a given body , , , , (Fig. 13). Through an arbitrary point O, taken in the plane of action of forces, we draw the axis Oh, perpendicular to the forces, and the axis OU, parallel to these forces. Let us write the equilibrium equations for this system of forces

Fig.13. Parallel force system.

Each force is perpendicular to the Ox axis, and its projection on this axis is zero. Therefore, the first equation turns into the identity 0 = 0 and is satisfied regardless of whether the forces are balanced or not. Thus, for a flat system of parallel forces, only two equilibrium equations remain, and for the axis OU forces are projected in full size, since this axis is parallel to the given forces.

The system of equilibrium equations for a plane system of parallel forces takes the form

The equilibrium equations for a plane system of parallel forces can be written as

Points A and B are arbitrary points, it is preferable to take them on the axis X, the equation =0 serves to check the correctness of the calculations.

So, for an arbitrary flat system of forces we have three equilibrium equations, and for a flat system of parallel forces we have only two equilibrium equations. Accordingly, when solving problems for the equilibrium of an arbitrary plane system of forces, one can find three unknowns, and when considering the equilibrium of a plane system of parallel forces, no more than two.

If the number of unknowns exceeds the number of static equations, the problem becomes statically indeterminate.

1.13. Types of beam supports

In machines and structures, very often there are elongated bodies called beams. They are mainly designed to absorb transverse loads. Beams have special support devices for mating with other elements and transferring forces to them. The supports of beams, considered as flat systems, are of three main types.

· Movable hinged support (Fig. 14, a). Such a support does not prevent the rotation of the end of the beam and its movement along the rolling plane. Only one reaction can occur in it, which is perpendicular to the rolling plane and passes through the center of the roller.

A schematic representation of a movable hinged support is given in fig. fourteen, b.

Rice. 14. Types of beam supports.

Movable supports allow the beam to freely change its length with temperature changes and thereby eliminate the possibility of thermal stresses.

· Fixed hinged support (Fig. 14, c). Such a support allows rotation of the end of the beam, but eliminates its translational movement in any direction. The reaction arising in it can be decomposed into two components - horizontal and vertical

· Rigid termination, or pinching (Fig. 14, G). Such fastening does not allow either linear or angular displacements of the reference section. In this support, a reaction can generally occur, which is usually decomposed into two components (vertical and horizontal) and a pinching moment (reactive moment).

A beam with one closed end is called cantilever beam or simply console.

If the support reactions can be found from the equations of statics alone, then the beams are called statically determined. If the number of unknown support reactions is greater than the number of static equations possible for a given problem, then the beams are called statically indeterminate.

Example.

Determine the unknown parameters of the reactions of supports A and B for a given (Fig. 15) beam design, loaded with parallel forces and.