The external force acting on the discarded part of the beam and tending 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).

Compiling the sum of the moments, we use the rule of termekh signs: counterclockwise "+", clockwise "-". It's not a wording, but it's much easier to remember.

Many people have a problem: how to understand in which direction the force rotates the structure?

The question is not very complicated and if you know some tricks it is quite easy to understand.

Let's start simple, we have a scheme

And for example, we need the sum of moments about point A.

Let's go in order from left to right:

Ra and Ha will not give momentum, since they act at point A and they will not have a shoulder to this point.

This is an example: the green line is the Ra power line, the yellow one is Na. There are no shoulders to point A, because it lies on the lines of action of these forces.

Let's continue: the moment arising in the rigid termination of Ma. It is quite simple with the moments, in which direction it is directed, anyone will figure it out, in this case it is directed counterclockwise.

The force from the distributed load Q is directed downward with a shoulder of 2.5. Where does it rotate our design?

We discard all forces except Q. Remember that at point A we have a “nail” hammered in.

If we imagine that point A is the center of the clock face, then it can be seen that the force Q rotates our beam clockwise, which means the sign will be “-”.

Point A is the center of the dial and F rotates the beam counterclockwise, the sign will be "+"

Everything is clear with the moment, it is directed counterclockwise, which means it rotates the beam in the same direction.

There are other points:

Frame given. We need to sum the moments about point A.

We consider only the force F, we do not touch the reactions in the termination.

And so, in what direction does the force F rotate the structure relative to point A?

To do this, as before, we draw the axis from point A, and for F - the line of action of the force

Now everything is visible and clear - the structure rotates clockwise

Thus, there should be no problems with the direction.

The sign rule for bending moments is related to the nature of the deformation of the beam. So, the bending moment is considered positive if the beam is bent with a convexity down - the stretched fibers are located below. When bending with a bulge upwards, when the stretched fibers are on top, the moment is negative.

For the transverse force, the sign is also related to the nature of the deformation. When external forces tend to raise the left side of the beam or lower the right side, the shear force is positive. With the opposite direction of external forces, i.e. if they tend to lower the left side of the beam or raise the right side, the transverse force is negative.

To facilitate the construction of diagrams, you should remember a number of rules:

In the area where there is no uniformly distributed load, the diagram Q is depicted as a straight line parallel to the axis of the beam, and the diagram M from is an inclined straight line.

In the section where a concentrated force is applied, there should be a jump in the Q diagram by the magnitude of the force, and a break in the M out diagram.

In the area of ​​action of a uniformly distributed load, the diagram Q is an inclined straight line, and the diagram M from is a parabola, convexly facing the arrows depicting the intensity of the load q.

If the diagram Q on the inclined section crosses the line of zeros, then in this section on the diagram M from there will be an extremum point.

If there are no concentrated forces at the boundary of the action of a distributed load, then the inclined section of the diagram Q is connected to the horizontal one without a jump, and the parabolic section of the diagram M from is connected to the inclined one smoothly without a break.

In sections where concentrated pairs of forces are applied to the beam, on the diagram M from there will be jumps by the value of the acting external moments, and the diagram Q does not change.

EXAMPLE 5. For a given two-support beam, construct diagrams of transverse forces and bending moments and select the required size of two I-beams from the strength condition, assuming [σ]=230 MPa for steel, if q=20 kN/m, M=100 kNm.

SOLUTION:

Determining support reactions

From these equations we find:

Examination:

Therefore, the reactions of the supports are found correctly.

We divide the beam into three sections.

Plotting Q:

section 1-1: 0≤z 1 ≤2,
;

section 2-2: 0≤z 2 ≤10,
;

z 2 \u003d 0,
;

section 3-3: 0≤z 3 ≤2,
(from right to left);

z 3 \u003d 0,
;

z 3 \u003d 2,
.

We build a diagram of transverse forces.

Plot M from:

section 1-1: 0≤z 1 ≤2, ;

section 2-2: 0≤z 2 ≤10,
;

To determine the extremum:
,

,
;

section 3-3: 0≤z 3 ≤2;
.

We build a diagram of bending moments.

From the condition of bending strength, we select the size of the cross section - two I-beams:

,

Since there are two I-beams, then
.

In accordance with GOST, we select two I-beams No. 30, W x \u003d 472 cm 3 (see Appendix 4).

Tasks for performing control work Tasks 1-10

Select the section of the suspension rod or column supporting the beam AB according to the data of your option, shown in fig. 9. The material of the rod for shaped profiles is rolled steel C-245, for a round section - hot-rolled reinforcing steel of class A-I.

So, for the equilibrium of a body fixed on an axis, it is not the modulus of force itself that is essential, but the product of the modulus of force by the distance from the axis to the line along which the force acts (Fig. 115; it is assumed that the force lies in a plane perpendicular to the axis of rotation). This product is called the moment of force about the axis, or simply the moment of force. The distance is called the shoulder of the force. Denoting the moment of force by the letter , we get

Let us agree to consider the moment of force positive if this force, acting separately, would rotate the body clockwise, and negative otherwise (in this case, we must agree in advance which side we will look at the body from). For example, the forces and in Fig. 116 a positive moment must be attributed, and a negative moment must be attributed to the force.

Rice. 115. The moment of force is equal to the product of its module and the shoulder

Rice. 116. Moments of forces and are positive, moment of force is negative

Rice. 117. The moment of force is equal to the product of the module of the force component and the module of the radius vector

The moment of force can be given yet another definition. Let's draw a directed segment from a point lying on the axis in the same plane as the force to the point of application of the force (Fig. 117). This segment is called the radius vector of the force application point. The module of the vector is equal to the distance from the axis to the point of application of force. Now let's construct the force component perpendicular to the radius vector . Let's denote this component by . It can be seen from the figure that , a . Multiplying both expressions, we get that .

Thus, the moment of force can be represented as

where is the modulus of the force component perpendicular to the radius vector of the force application point, is the modulus of the radius vector. Note that the product is numerically equal to the area of ​​the parallelogram built on the vectors and (Fig. 117). On fig. 118 shows forces whose moments about the axis are the same. From fig. 119 shows that moving the point of application of force along its direction does not change its momentum. If the direction of the force passes through the axis of rotation, then the arm of the force is zero; therefore, the moment of force is also equal to zero. We have seen that in this case the force does not cause rotation of the body: a force whose moment about a given axis is equal to zero does not cause rotation about this axis.

Rice. 118. Forces and have the same moments about the axis

Rice. 119. Equal forces with the same shoulder have equal moments about the axis

Using the concept of the moment of force, we can formulate in a new way the conditions for the equilibrium of a body fixed on an axis and under the action of two forces. In the condition of equilibrium, expressed by formula (76.1), there is nothing but the shoulders of the corresponding forces. Therefore, this condition consists in the equality of the absolute values ​​of the moments of both forces. In addition, in order to avoid rotation, the directions of the moments must be opposite, i.e., the moments must differ in sign. Thus, for the equilibrium of a body fixed on an axis, the algebraic sum of the moments of the forces acting on it must be equal to zero.

Since the moment of force is determined by the product of the modulus of force and the arm, we will obtain the unit of the moment of force by taking a force equal to unity, the arm of which is also equal to one. Therefore, in SI, the unit of moment of force is the moment of force equal to one newton and acting on a shoulder of one meter. It is called the newton meter (Nm).

If many forces act on a body fixed on an axis, then, as experience shows, the equilibrium condition remains the same as for the case of two forces: for the balance of a body fixed on an axis, the algebraic sum of the moments of all forces acting on the body must be equal to zero. The resulting moment of several moments acting on the body (component moments) is called the algebraic sum of the constituent moments. Under the action of the resulting moment, the body will rotate around the axis in the same way as it would rotate under the simultaneous action of all component moments. In particular, if the resulting moment is zero, then the body fixed on the axis is either at rest or rotates uniformly.

Basic course of lectures on strength of materials, theory, practice, tasks.

3. Bend. Determination of stresses.

3.4. Sign rule for bending moments and shear forces.

The transverse force in the beam section mn (Fig. 3.7, a) is considered positive if the resultant of external forces to the left of the section is directed from bottom to top, and to the right - from top to bottom, and negative - in the opposite case (Fig. 3.7, b).

The bending moment in the beam section, for example, in the section mn (Fig. 3.8, a), is considered positive if the resultant moment of external forces is directed clockwise to the left of the section, and counterclockwise to the right, and negative in the opposite case (Fig. 3.8 , b). The moments depicted in fig. 3.8, a, bend the beam with a bulge down, and the moments shown in fig. 3.8, b, bend the beam with a bulge upwards. This can be easily checked by bending a thin ruler.

From this follows another, more convenient to remember, sign rule for the bending moment. The bending moment is considered positive if, in the considered section, the beam bends with a convexity downwards. Further, it will be shown that the fibers of the beam located in the concave part experience compression, and in the convex part they experience tension. Thus, agreeing to put the positive ordinates of the diagram M up from the axis, we get that the diagram is built from the side of the compressed fibers of the beam.