and distributes the load evenly. What is distributed load. Uniformly distributed load. Graphical definition of support forces
In engineering calculations, one often encounters loads distributed along a given surface according to one law or another. Consider some of the simplest examples of distributed forces lying in the same plane.
A flat system of distributed forces is characterized by its intensity q, i.e., by the value of the force per unit length of the loaded segment. The intensity is measured in Newtons divided by meters.
1) Forces uniformly distributed along a straight line segment (Fig. 69, a). For such a system of forces, the intensity q has a constant value. In static calculations, this system of forces can be replaced by the resultant
Modulo
A force Q is applied in the middle of segment AB.
2) Forces distributed along a straight line segment according to a linear law (Fig. 69, b). An example of such a load can be the forces of water pressure on the dam, which have the greatest value at the bottom and drop to zero at the surface of the water. For these forces, the intensity q is a variable value growing from zero to a maximum value. The resultant Q of such forces is determined similarly to the resultant of the gravity forces acting on a uniform triangular plate ABC. Since the weight of a homogeneous plate is proportional to its area, then, modulo,
A force Q is applied at a distance from the side BC of triangle ABC (see § 35, item 2).
3) Forces distributed along a straight line segment according to an arbitrary law (Fig. 69, c). The resultant Q of such forces, by analogy with the force of gravity, is equal in absolute value to the area of \u200b\u200bthe figure ABDE, measured on an appropriate scale, and passes through the center of gravity of this area (the question of determining the centers of gravity of areas will be considered in § 33).
4) Forces uniformly distributed along the arc of a circle (Fig. 70). An example of such forces is the forces of hydrostatic pressure on the side walls of a cylindrical vessel.
Let the radius of the arc be , where is the axis of symmetry along which we direct the axis. The system of converging forces acting on the arc has the resultant Q, directed along the axis due to symmetry, while
To determine the value of Q, we select an element on the arc, the position of which is determined by the angle and length. The force acting on this element is numerically equal to and the projection of this force on the axis will be Then
But from Fig. 70 it can be seen that Therefore, since then
where is the length of the chord that subtends the arc AB; q - intensity.
Task 27. A uniformly distributed intensity load acts on the cantilever beam A B, the dimensions of which are indicated in the drawing (Fig. 71). If
Solution. We replace the distributed forces with their resultants Q, R and R, where according to formulas (35) and (36)
and compose the equilibrium conditions (33) for the parallel forces acting on the beam
Substituting here instead of Q, R and R their values and solving the resulting equations, we finally find
For example, if we get and if
Problem 28. A cylindrical cylinder, the height of which is H, and the inner diameter d, is filled with gas under pressure. The thickness of the cylindrical walls of the cylinder is a. Determine the tensile stresses experienced by these walls in the directions: 1) longitudinal and 2) transverse (the stress is equal to the ratio of the tensile force to the cross-sectional area), considering it small.
Solution. 1) Let us cut the cylinder by a plane perpendicular to its axis into two parts and consider the equilibrium of one of them (Fig.
72a). It is acted upon in the direction of the axis of the cylinder by the pressure force on the bottom and the forces distributed over the cross-sectional area (the action of the discarded half), the resultant of which is denoted by Q. At equilibrium
Assuming that the cross-sectional area is approximately equal, we obtain the value for the tensile stress
When solving practical problems, it is far from always possible to assume that the force acting on the body is applied at one point. Often, forces are applied to the whole area of the body (for example, snow load, wind load, etc.). Such a load is called distributed. A uniformly distributed load is characterized by intensity q (Fig. 1.29). Intensity is the total load per unit length of the structure.
Fx=Fcos(60), Fy=Fcos(30)
Solving this equation we get:
From equation (2) we find:
F х = 0; R ax = 0;
When solving practical problems, it is far from always possible to assume that the force acting on the body is applied at one point. Often, forces are applied to the whole area of the body (for example, snow load, wind load, etc.). Such a load is called distributed. A uniformly distributed load is characterized by intensity q (Fig. 1.29). Intensity is the total load per unit length of the structure.
Solution. We use the same plan that was used to solve problems for a convergent system of forces. The object of equilibrium is the entire beam, the load on which is shown in the drawing. We discard the connections - hinges A and B. We decompose the reaction of the fixed hinge A into two components -
, and the reaction of the movable hinge B is directed perpendicular to the reference plane. Thus, a plane arbitrary system of forces acts on the beam, for which three equilibrium equations can be drawn up. We choose the coordinate axes and compose these equations. Projection equations:
1. F kx = 0; R ax -Fcos(60) = 0;
2. Fky = 0; Ray + R B - Fcos(30) = 0;
(the pair is not included in the projection equation, since the sum of the projections of the forces of the pair on any axis is zero).
We compose the equation of moments relative to point A, since two unknown forces intersect at it. When finding the moment of a couple relative to point A, remember that the sum of the moments of forces of a couple relative to any point is equal to the moment of the couple, and the sign of the moment will be positive, since the couple tends to rotate the body counterclockwise. To find the moment of force It is convenient to decompose it into vertical and horizontal components:
Fx=Fcos(60), Fy=Fcos(30)
and use the Varignon theorem, and it should be taken into account that the moment from the force with respect to point A is zero, since its line of action passes through this point. Then the equation of moments will take the form:
; R in. 3-F B cos(30)2 + M = 0.
Solving this equation we get:
From equation (2) we find:
R ay \u003d Fcos (30) - R B \u003d 20.867 - 4 \u003d -2.67 kN,
and from equation (1) R ax = Fcos(60) = 20.5 = 1 kN.
Solution. Let us replace the uniformly distributed load with its resultant Q = 3q = 310 = 30 kN. It will be applied in the middle of the span, that is, at a distance AC = 1.5 m. We consider the equilibrium of the beam AB. We discard the connection - a rigid termination, and instead of it we apply two components of the reaction R ax and R ay and a reactive moment M a. A flat arbitrary system of forces will act on the beam, for which three equilibrium equations can be compiled, from which the required unknowns can be found.
F х = 0; R ax = 0;
F ku = 0; R ay - Q = 0; R ay = Q = 30 kN;
M a (F k) = 0; M a - 1.5Q = 0; M a = 1.5Q = 1.530 = 45 kNm.
Stress distribution in the case of a plane problem
This case corresponds to the stress state under wall foundations, retaining walls, embankments and other structures, the length of which significantly exceeds their transverse dimensions:
Where l- the length of the foundation; b- foundation width. In this case, the distribution of stresses under any part of the structure, separated by two parallel sections perpendicular to the axis of the structure, characterizes the stress state under the entire structure and does not depend on the coordinates perpendicular to the direction of the loaded plane.
Consider the action of a linear load in the form of a continuous series of concentrated forces R, each of which is per unit length. In this case, the stress components at any point M with coordinates R and b can be found by analogy with the spatial problem:
If the ratios of the geometric characteristics of the considered points z, y, b represent in the form of coefficients of influence K, then the formulas for stresses can be written as follows:
Influence coefficient values Kz,K y,Kyz tabulated according to relative coordinates z/b, y/b(Table II.3 of Appendix II).
An important property of the plane problem is that the stress components t and s y in the considered plane z 0y do not depend on the coefficient of transverse expansion n 0 , as in the case of a spatial problem.
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Fig.3.15. Arbitrary distribution
bandwidth loads b
If the load is propagated from a point A(b=b 2) to the point B(b \u003d b 1), then, summing up the stresses from its individual elements, we obtain expressions for stresses at any point in the array from the action of a continuous strip-like load.
For a uniformly distributed load, integrate the above expressions with P y = P= const. In this case, the main directions, i.e. the directions in which the largest and smallest normal stresses act will be the directions located along the bisector of the "visibility angles" and perpendicular to them (Fig. 3.16). The angle of visibility a is the angle formed by the straight lines connecting the point under consideration M with band load edges.
We obtain the values of the principal stresses from expressions (3.27), assuming b=0 in them:
These formulas are often used in assessing the stress state (especially the limit state) in the foundations of structures.
On the values of the main stresses as semi-axes, it is possible to construct stress ellipses that clearly characterize the stress state of the soil under a uniformly distributed load applied along the strip. The distribution (location) of stress ellipses under the action of a local uniformly distributed load in a plane problem is shown in Fig. 3.17.
Fig.3.17. Stress ellipses under the action of a uniformly distributed load in a plane problem
By formulas (3.28) one can determine sz, s y And t yz at all points of the section perpendicular to the longitudinal axis of the load. If we connect points with the same values of each of these quantities, we get lines of equal voltages. Figure 3.18 shows lines of identical vertical stresses sz, called isobars, of horizontal stresses s y, called spacers, and tangential stresses t zx called shifts.
These curves were constructed by D.E. Pol'shin using the theory of elasticity for a load uniformly distributed over a strip of width b, extending indefinitely in a direction perpendicular to the drawing. The curves show that the effect of compressive stresses sz intensity 0.1 external load R affects the depth of about 6 b, while the horizontal stresses s y and tangents t propagate at the same intensity 0.1 R to a much shallower depth (1.5 - 2.0) b. Curvilinear surfaces of equal stresses will have similar outlines for the case of a spatial problem.
Fig.3.18. Lines of equal stresses in a linearly deformable array:
and for sz(isobars); b - for s y(spread); in - for t(shift)
The influence of the width of the loaded strip affects the depth of stress propagation. For example, for a foundation 1 m wide, transferring to the base a load with intensity R, voltage 0.1 R will be at a depth of 6 m from the sole, and for a foundation 2 m wide, with the same load intensity, at a depth of 12 m (Fig. 3.19). If there are weaker soils in the underlying layers, this can significantly affect the deformation of the structure.
where a and b / are, respectively, the angles of visibility and inclination of the line to the vertical (Fig. 3.21).
Fig.3.21. Diagrams of the distribution of compressive stresses over vertical sections of the soil mass under the action of a triangular load
Table II.4 of Annex II shows the dependences of the coefficient TO| z depending on z/b And y/b(Fig.3.21) to calculate s z by the formula.
Stress distribution in the case of a plane problem
This case corresponds to the stress state under wall foundations, retaining walls, embankments and other structures, the length of which significantly exceeds their transverse dimensions:
Where l- the length of the foundation; b- foundation width. In this case, the distribution of stresses under any part of the structure, separated by two parallel sections perpendicular to the axis of the structure, characterizes the stress state under the entire structure and does not depend on the coordinates perpendicular to the direction of the loaded plane.
Consider the action of a linear load in the form of a continuous series of concentrated forces R, each of which is per unit length. In this case, the stress components at any point M with coordinates R and b can be found by analogy with the spatial problem:
(3.27)
If the ratios of the geometric characteristics of the considered points z, y, b represent in the form of coefficients of influence K, then the formulas for stresses can be written as follows:
(3.28)
Influence coefficient values Kz,K y,Kyz tabulated according to relative coordinates z/b, y/b(Table II.3 of Appendix II).
An important property of the plane problem is that the stress components t and s y in the considered plane z 0y do not depend on the coefficient of transverse expansion n 0 , as in the case of a spatial problem.
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Fig.3.15. Arbitrary distribution
bandwidth loads b
If the load is propagated from a point A(b=b 2) to the point B(b \u003d b 1), then, summing up the stresses from its individual elements, we obtain expressions for stresses at any point in the array from the action of a continuous strip-like load.
(3.29)
For a uniformly distributed load, integrate the above expressions with P y = P= const. In this case, the main directions, i.e. the directions in which the largest and smallest normal stresses act will be the directions located along the bisector of the "visibility angles" and perpendicular to them (Fig. 3.16). The angle of visibility a is the angle formed by the straight lines connecting the point under consideration M with band load edges.
We obtain the values of the principal stresses from expressions (3.27), assuming b=0 in them:
. (3.30)
These formulas are often used in assessing the stress state (especially the limit state) in the foundations of structures.
On the values of the main stresses as semi-axes, it is possible to construct stress ellipses that clearly characterize the stress state of the soil under a uniformly distributed load applied along the strip. The distribution (location) of stress ellipses under the action of a local uniformly distributed load in a plane problem is shown in Fig. 3.17.
Fig.3.17. Stress ellipses under the action of a uniformly distributed load in a plane problem
By formulas (3.28) one can determine sz, s y And t yz at all points of the section perpendicular to the longitudinal axis of the load. If we connect points with the same values of each of these quantities, we get lines of equal voltages. Figure 3.18 shows lines of identical vertical stresses sz, called isobars, of horizontal stresses s y, called spacers, and tangential stresses t zx called shifts.
These curves were constructed by D.E. Pol'shin using the theory of elasticity for a load uniformly distributed over a strip of width b, extending indefinitely in a direction perpendicular to the drawing. The curves show that the effect of compressive stresses sz intensity 0.1 external load R affects the depth of about 6 b, while the horizontal stresses s y and tangents t propagate at the same intensity 0.1 R to a much shallower depth (1.5 - 2.0) b. Curvilinear surfaces of equal stresses will have similar outlines for the case of a spatial problem.
Fig.3.18. Lines of equal stresses in a linearly deformable array:
and for sz(isobars); b - for s y(spread); in - for t(shift)
The influence of the width of the loaded strip affects the depth of stress propagation. For example, for a foundation 1 m wide, transferring to the base a load with intensity R, voltage 0.1 R will be at a depth of 6 m from the sole, and for a foundation 2 m wide, with the same load intensity, at a depth of 12 m (Fig. 3.19). If there are weaker soils in the underlying layers, this can significantly affect the deformation of the structure.
where a and b / are, respectively, the angles of visibility and inclination of the line to the vertical (Fig. 3.21).
Fig.3.21. Diagrams of the distribution of compressive stresses over vertical sections of the soil mass under the action of a triangular load
Table II.4 of Annex II shows the dependences of the coefficient TO| z depending on z/b And y/b(Fig.3.21) to calculate s z using the formula:
sz = TO| z × R.
Surface and volume forces represent a load distributed over a certain surface or volume. Such a load is given by intensity, which is the force per unit of some volume, or some area, or some length.
A special place in solving a number of practically interesting problems is occupied by the case of a flat distributed load applied along the normal to a certain beam. If you direct the axis along the beam 
, then the intensity will be a function of the coordinate 
and is measured in N/m. Intensity is the force per unit length.
A flat figure bounded by a beam and a load intensity graph is called a distributed load diagram (Fig. 1.28). If, by the nature of the problem being solved, deformations can be ignored, i.e. Since the body can be considered absolutely rigid, then the distributed load can (and should) be replaced by the resultant.
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Let's split the beam into 
length segments 
, on each of which we assume that the intensity is constant and equal to 
, Where 
- segment coordinate 
. In this case, the intensity curve is replaced by a broken line, and the load per segment 
, is replaced by a concentrated force 
, applied at the point 
(Figure 1.29). The resulting system of parallel forces has a resultant equal to the sum of the forces acting on each of the segments, applied at the center of the parallel forces.
It is clear that such a representation describes the real situation the more accurately, the smaller the segment 
, i.e. the more segments 
. We obtain the exact result by passing to the limit at the length of the segment 
tending to zero. The limit resulting from the described procedure is an integral. Thus, for the module of the resultant we get:
To determine the coordinates of a point 
application of the resultant we use the Varignon theorem:
if the system of forces has a resultant, then the moment of the resultant about any center (any axis) is equal to the sum of the moments of all forces of the system about this center (this axis)
Writing this theorem for a system of forces 
in projections on the axis 
and passing to the limit with the length of the segments tending to zero, we obtain:
Obviously, the modulus of the resultant is numerically equal to the area of the distributed load diagram, and the point of its application coincides with the center of gravity of a homogeneous plate having the form of a distributed load diagram.
We note two frequently encountered cases.
,
(Fig. 1.30). The resultant module and the coordinate of its application point are determined by the formulas:
In engineering practice, such a load is quite common. In most cases, the weight and wind load can be considered uniformly distributed.
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(Figure 1.31). In this case:
In particular, the pressure of water on a vertical wall is directly proportional to the depth 
.
Example 1.5
Determine the reactions of the supports 
And 
beam under the action of two concentrated forces and a uniformly distributed load. Given:
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Find the resultant of the distributed load. The resultant modulus is equal to
shoulder of strength 
relative to the point 
equals 
Consider the balance of the beam. The power circuit is shown in Fig. 1.33.
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Example 1.6
Determine the response of the cantilever beam termination under the action of a concentrated force, a pair of forces and a distributed load (Fig. 1.34).
Let us replace the distributed load with three concentrated forces. To do this, we divide the distributed load diagram into two triangles and a rectangle. We find
The power circuit is shown in Fig. 1.35.
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Calculate the shoulders of the resultant relative to the axis 
The equilibrium conditions in the case under consideration have the form:
QUESTIONS FOR SELF-CHECKING:
1. What is called the intensity of the distributed load?
2. How to calculate the module of the resultant distributed load?
3. How to calculate the coordinate of the point of application of the resultant distributed
load?
4. What is the module and what is the coordinate of the point of application of a uniformly distributed load?
5. What is the module and what is the coordinate of the point of application of a linearly distributed load?
From the collection of problems by I.V. Meshchersky: 4.28; 4.29; 4.30; 4.33; 4.34.
From the textbook "THEORETICAL MECHANICS - theory and practice": sets СР-2; SR-3.
PRACTICAL STUDIES #4-5
Along with the concentrated forces discussed above, building structures and structures can be exposed to distributed loads- by volume, by surface or along a certain line - and determined by it intensity.
Load example, distributed over the area, is the snow load, wind, liquid or ground pressure. The intensity of such a surface load has the dimension of pressure and is measured in kN / m 2 or kilopascals (kPa \u003d kN / m 2).
When solving problems, there is often a load, distributed along the length of the beam. Intensity q such load is measured in kN/m.
Consider a beam loaded on the section [ a, b] distributed load, the intensity of which varies according to the law q= q(x). To determine the support reactions of such a beam, it is necessary to replace the distributed load with an equivalent concentrated one. This can be done according to the following rule:
Consider special cases of a distributed load.
A) general case of distributed load(fig.24)
Fig.24
q(x) - distributed force intensity [N/m],
Elemental power.
l- segment length
The intensity force q(x) distributed along the line segment is equivalent to the concentrated force
A concentrated force is applied at a point WITH(center of parallel forces) with coordinate
b) constant intensity of distributed load(fig.25)
Fig.25
V) the intensity of the distributed load, changing according to a linear law(fig.26)
Fig.26
Calculation of composite systems.
Under composite systems we will understand structures consisting of several bodies connected to each other.
Before proceeding to the consideration of the features of the calculation of such systems, we introduce the following definition.
statically determinatesuch problems and systems of statics are called for which the number of unknown reactions of constraints does not exceed the maximum allowable number of equations.
If the number of unknowns is greater than the number of equations, relevant tasks and systems are called statically indeterminate. The difference between the number of unknowns and the number of equations is called degree of static uncertainty systems.
For any plane system of forces acting on a rigid body, there are three independent equilibrium conditions. Consequently, for any flat system of forces, from the equilibrium conditions, one can find no more than three unknown coupling reactions.
In the case of a spatial system of forces acting on a rigid body, there are six independent equilibrium conditions. Consequently, for any spatial system of forces, it is possible to find no more than six unknown coupling reactions from the equilibrium conditions.
Let's explain this with the following examples.
1. Let the center of a weightless ideal block (example 4) be held by not two, but three rods: AB, sun And BD and it is necessary to determine the reactions of the rods, neglecting the dimensions of the block.
Taking into account the conditions of the problem, we get a system of converging forces, where to determine the three unknowns: S A, S C And S D it is still possible to compose a system of only two equations: Σ X = 0, Σ Y=0. Obviously, the problem posed and the system corresponding to it will be statically indeterminate.
2. The beam, rigidly fixed at the left end and having a hinged-fixed support at the right end, is loaded with an arbitrary planar system of forces (Fig. 27).
To determine the support reactions, only three equilibrium equations can be compiled, which will include 5 unknown support reactions: X A , Y A,M A,XB And Y B. The stated problem will be twice statically indeterminate.
Such a problem cannot be solved within the framework of theoretical mechanics, assuming the body under consideration to be absolutely rigid.
Fig.27
Let us return to the study of composite systems, a typical representative of which is a three-hinged frame (Fig. 28, A). It consists of two bodies: AC And BC connected key hinge C. For this frame, consider two ways to determine the support reactions of composite systems.
1 way. Consider a body AC, loaded with a given force R, discarding, in accordance with Axiom 7, all bonds and replacing them, respectively, with the reactions of external ( X A, Y A) and internal ( X C, Y C) connections (Fig. 28, b).
Similarly, we can consider the equilibrium of the body BC under the influence of support reactions IN - (XB, Y B) and reactions in the connecting hinge C - (X C', Y C’) , where, according to Axiom 5: X C= X C', Y C= Y C’.
For each of these bodies, three equilibrium equations can be compiled, so the total number of unknowns is: X A, Y A , X C=X C', Y C =Y C’, XB, Y B equals the total number of equations, and the problem is statically determinate.
Recall that, according to the condition of the problem, it was required to determine only 4 support reactions, but we had to do additional work, determining the reactions in the connecting hinge. This is the disadvantage of this method for determining support reactions.
2 way. Consider the balance of the entire frame ABC, discarding only external constraints and replacing them with unknown support reactions X A, Y A,XB, Y B .
The resulting system consists of two bodies and is not an absolutely rigid body, since the distance between the points A And IN can change due to the mutual rotation of both parts relative to the hinge WITH. Nevertheless, we can assume that the totality of forces applied to the frame ABC forms a system if we use the hardening axiom (Fig. 28, V).
Fig.28
So for the body ABC there are three equilibrium equations. For example:
Σ M A = 0;
Σ X = 0;
These three equations will include 4 unknown support reactions X A, Y A,XB And Y B. Note that an attempt to use as a missing equation, for example: Σ M B= 0 will not lead to success, since this equation will be linearly dependent with the previous ones. To obtain a linearly independent fourth equation, it is necessary to consider the equilibrium of another body. As it, you can take one of the parts of the frame, for example - sun. In this case, it is necessary to compose an equation that would contain the "old" unknowns X A, Y A,XB, Y B and did not contain new ones. For example, the equation: Σ X (sun) = 0 or more:- X C' + XB= 0 is not suitable for these purposes, because it contains a "new" unknown X C’, but the equation Σ M C (sun) = 0 meets all the necessary conditions. Thus, the desired support reactions can be found in the following sequence:
Σ M A = 0; → Y B= R/4;
Σ M B = 0; → Y A= -R/4;
Σ M C (sun) = 0; → XB= -R/4;
Σ X = 0; →X A= -3R/4.
To check, you can use the equation: Σ M C (AC) = 0 or, in more detail: - Y A∙2 + X A∙2 + R∙1 = R/4∙2 -3R/4∙2 +R∙1 = R/2 - 3R/2 +R = 0.
Note that this equation includes all 4 found support reactions: X A And Y A- explicitly, and XB And Y B- implicitly, since they were used in determining the first two reactions.
Graphical definition of support reactions.
In many cases, the solution of problems can be simplified if, instead of the equilibrium equations or in addition to them, the equilibrium conditions, axioms, and theorems of statics are directly used. The corresponding approach is called the graphical definition of support reactions.
Before proceeding to the consideration of the graphical method, we note that, as for a system of converging forces, only those problems that allow an analytical solution can be graphically solved. At the same time, the graphical method for determining support reactions is convenient for a small number of loads.
So, the graphical method for determining support reactions is based mainly on the use of:
Axioms about the balance of a system of two forces;
Axioms about action and reaction;
Theorems about three forces;
Conditions for the equilibrium of a plane system of forces.
In the graphical determination of the reactions of composite systems, the following is recommended. sequence of consideration:
Choose a body with a minimum number of algebraic unknown reactions of bonds;
If there are two or more such bodies, then start the solution by considering the body to which a smaller number of forces are applied;
If there are two or more such bodies, then choose the body for which the larger number of forces is known from the direction.
Problem solving.
When solving the problems of this section, you should keep in mind all those general instructions that were made earlier.
When starting to solve, it is necessary, first of all, to establish the equilibrium of which particular body should be considered in this problem. Then, having singled out this body and considering it as free, one should depict all the given forces acting on the body and the reactions of the discarded bonds.
Next, you should compose the equilibrium conditions, applying the form of these conditions that leads to a simpler system of equations (the simplest will be a system of equations, each of which includes one unknown).
To obtain simpler equations, it follows (unless this complicates the calculation):
1) compiling projection equations, draw a coordinate axis perpendicular to some unknown force;
2) when compiling the moment equation, it is advisable to choose as the moment equation the point where the lines of action of two of the three unknown support reactions intersect - in this case they will not be included in the equation, and it will contain only one unknown;
3) if two of the three unknown support reactions are parallel, then when drawing up the equation in projections onto the axis, the latter should be directed so that it is perpendicular to the first two reactions - in this case, the equation will contain only the last unknown;
4) when solving the problem, the coordinate system must be chosen so that its axes are oriented in the same way as the majority of the forces of the system applied to the body.
When calculating moments, it is sometimes convenient to decompose a given force into two components and, using the Varignon theorem, find the moment of force as the sum of the moments of these components.
The solution of many problems of statics comes down to determining the reactions of supports, with the help of which beams, bridge trusses, etc. are fixed.
Example 7 To the bracket shown in Fig. 29, A, in knot IN suspended load weighing 36 kN. The connections of the elements of the bracket are hinged. Determine the forces arising in the rods AB And sun, considering them weightless.
Solution. Consider the equilibrium of the node IN, in which the rods converge AB And sun. Knot IN represents a point in the drawing. Since the load is suspended from the node IN, then at the point IN We apply a force F equal to the weight of the suspended load. rods VA And sun, pivotally connected in a node IN, limit the possibility of any linear movement in the vertical plane, i.e. are links with respect to the node IN.
Rice. 29. Calculation scheme of the bracket for example 7:
A - design scheme; b - system of forces in a node B
We mentally discard connections and replace their actions with forces - reactions of connections R A And R C. Since the rods are weightless, the reactions of these rods (the forces in the rods) are directed along the axis of the rods. Let us assume that both rods are stretched, i.e. their reactions are directed from the hinge to the inside of the rods. Then, if after the calculation the reaction turns out with a minus sign, then this will mean that in fact the reaction is directed in the direction opposite to that indicated in the drawing, i.e. the rod will be compressed.
On fig. 29, b it is shown that at the point IN applied active force F and bond reactions R A And R C. It can be seen that the depicted system of forces is a flat system of forces converging at one point. We choose arbitrarily the coordinate axes OX And OY and compose equilibrium equations of the form:
Σ F x = 0;-R a - R c cos𝛼 = 0;
Σ F y = 0; -F - R c cos(90 - α) = 0.
Given that cos (90 -α ) = sinα, from the second equation we find
Rc = -F/sinα = - 36/0,5 = -72 kN.
Substituting the value Rc into the first equation, we get
R a \u003d -R c cosα \u003d - (-72) ∙ 0.866 \u003d 62.35 kN.
So the rod AB- stretched, and the rod sun- compressed.
To check the correctness of the found forces in the rods, we project all the forces on any axis that does not coincide with the axes X And Y e.g. axis U:
Σ F u = 0; -R c - R a cosα -Fcos(90-α) = 0.
After substituting the values of the found forces in the rods (dimension in kilonewtons), we obtain
- (-72) – 62,35∙0,866 - 36∙0,5 = 0; 0 = 0.
The equilibrium condition is satisfied, thus, the found forces in the rods are correct.
Example 8 A scaffold beam whose weight is negligible is held in a horizontal position by a flexible rod CD and pivotally rests on the wall at the point A. Find the pull force CD if a worker weighing 80 kg ≈0.8 kN stands on the edge of the scaffold (Fig. 30, A).
Rice. thirty. Calculation scheme of scaffolding for example 8:
A– calculation scheme; b– system of forces acting on the scaffold
Solution. We select the object of balance. In this example, the object of balance is the scaffold beam. At the point IN active force acting on the beam F equal to the weight of a person. The links in this case are the fixed support hinge A and thrust CD. Let's mentally discard the bonds, replacing their action on the beam with the reactions of the bonds (Fig. 30, b). It is not necessary to determine the reaction of a fixed hinged support according to the condition of the problem. Thrust response CD directed along the line. Let's assume the rod CD stretched, i.e. reaction R D directed away from the hinge WITH inside the rod. Let's decompose the reaction R D, according to the parallelogram rule, into horizontal and vertical components:
R Dx mountains \u003d R D cosα ;
R Dy vert = R D cos(90-α) = R D sinα .
As a result, we have obtained an arbitrary flat system of forces, the necessary condition for the equilibrium of which is the equality to zero of three independent equilibrium conditions.
In our case, it is convenient to first write the equilibrium condition as the sum of moments about the moment point A, since the moment of the support reaction R A relative to this point is equal to zero:
Σ m A = 0; F∙3a - R dy ∙ a = 0
F∙3a - R D sinα = 0.
The value of trigonometric functions is determined from the triangle ACD:
cosα = AC/CD = 0,89,
sinα = AD/CD = 0,446.
Solving the equilibrium equation, we obtain R D = 5.38 kN. (Tyazh CD- stretched).
To check the correctness of the calculation of the effort in the strand CD it is necessary to calculate at least one of the components of the support reaction R A. We use the equilibrium equation in the form
Σ Fy = 0; VA + R Dy- F= 0
VA = F- Rdy.
From here VA= -1.6 kN.
The minus sign means that the vertical component of the reaction R A pointing down on the support.
Let's check the correctness of the calculation of the effort in the strand. We use one more equilibrium condition in the form of equations of moments about the point IN.
Σ m B = 0; VA∙3a + R Dy ∙ 2a = 0;
1,6∙3A + 5,38∙0,446∙2A = 0; 0 = 0.
The equilibrium conditions are met, thus, the force in the strand is found correctly.
Example 9 The vertical concrete pillar is concreted with its lower end into a horizontal base. From above, the load from the wall of the building weighing 143 kN is transferred to the pole. The column is made of concrete with density γ= 25 kN/m 3 . The dimensions of the column are shown in fig. 31, A. Determine the reactions in a rigid embedment.
Rice. 31. Calculation scheme of the column for example 9:
A- loading scheme and dimensions of the column; b- calculation scheme
Solution. In this example, the object of balance is the column. The column is loaded with the following types of active loads: at the point A concentrated force F, equal to the weight of the wall of the building, and the own weight of the column in the form of a load uniformly distributed along the length of the beam with intensity q for each meter of column length: q = 𝛾A, Where A is the cross-sectional area of the column.
q= 25∙0.51∙0.51 = 6.5 kN/m.
The ties in this example are a rigid anchor at the base of the post. Mentally discard the termination and replace its action with the reactions of the bonds (Fig. 31, b).
In our example, we consider a special case of the action of a system of forces perpendicular to the embedment and passing along one axis through the point of application of support reactions. Then two support reactions: the horizontal component and the reactive moment will be equal to zero. To determine the vertical component of the support reaction, we project all forces onto the axis of the element. Align this axis with the axis Z, then the equilibrium condition can be written in the following form:
Σ F Z = 0; V B - F - ql = 0,
Where ql- resultant distributed load.
V B = F+ql= 143 + 6.5∙4 = 169 kN.
The plus sign indicates that the reaction V B directed upward.
To check the correctness of the calculation of the support reaction, one more equilibrium condition remains - in the form of an algebraic sum of the moments of all forces relative to any point that does not pass through the axis of the element. We suggest you do this test yourself.
Example 10 For the beam shown in Fig. 32, A, it is required to determine the support reactions. Given: F= 60 kN, q= 24 kN/m, M= 28 kN∙m.
Rice. 32. Calculation scheme and beam dimensions for example 10:
Solution. Consider the balance of the beam. The beam is loaded with an active load in the form of a flat system of parallel vertical forces, consisting of a concentrated force F, uniformly distributed load intensity q with the resultant Q, applied in the center of gravity of the cargo area (Fig. 32, b), and concentrated moment M, which can be represented as a pair of forces.
The connections in this beam are hinged-fixed support A and articulated support IN. We single out the object of equilibrium, for this we discard the support links and replace their actions with reactions in these links (Fig. 32, b). Bearing reaction R B is directed vertically, and the reaction of the hinged-fixed support R A will be parallel to the active system of acting forces and also directed vertically. Let's assume that they are directed upwards. Resultant distributed load Q= 4.8∙q is applied at the center of symmetry of the cargo area.
When determining the support reactions in beams, it is necessary to strive to compose the equilibrium equations in such a way that each of them includes only one unknown. This can be achieved by making two equations of moments about the reference points. Support reactions are usually checked by composing an equation in the form of the sum of the projections of all forces on an axis perpendicular to the axis of the element.
Let us conditionally accept the direction of rotation of the moment of support reactions around moment points as positive, then the opposite direction of rotation of forces will be considered negative.
A necessary and sufficient condition for equilibrium in this case is the equality to zero of independent equilibrium conditions in the form:
Σ m A = 0; V B ∙6 - q∙4,8∙4,8 + M+F∙2,4 = 0;
Σ m B = 0; VA∙6 - q∙4,8∙1,2 - M - F∙8,4 = 0.
Substituting the numerical values of the quantities, we find
V B= 14.4 kN, VA= 15.6 kN.
To check the correctness of the found reactions, we use the equilibrium condition in the form:
Σ Fy = 0; V A + V B - F -q∙4,8 =0.
After substituting the numerical values into this equation, we obtain an identity of the type 0=0. From this we conclude that the calculation was performed correctly and the reactions on both supports are directed upwards.
Example 11. Determine the support reactions for the beam shown in Fig. 33, A. Given: F= 2.4 kN, M= 12 kN∙m, q= 0.6 kN/m, a = 60°.
Rice. 33. Calculation scheme and dimensions of the beam for example 11:
a - design scheme; b - object of balance
Solution. Consider the balance of the beam. We mentally release the beam from the bonds on the supports and select the equilibrium object (Fig. 33, b). The beam is loaded with an active load in the form of an arbitrary planar system of forces. Resultant distributed load Q = q∙3 is attached at the center of symmetry of the cargo area. Strength F decompose according to the parallelogram rule into components - horizontal and vertical
F z = F cosα= 2.4 cos 60°= 1.2 kN;
Fy=F cos(90-α) = F sin 60°= 2.08 kN.
We apply to the object of equilibrium instead of the discarded bonds of the reaction. Suppose the vertical reaction VA articulated movable support A directed upwards, vertical reaction V B articulated fixed support B is also directed upwards, and the horizontal reaction H B- to the right.
Thus, in fig. 33, b an arbitrary flat system of forces is depicted, the necessary condition for the equilibrium of which is the equality to zero of three independent equilibrium conditions for a flat system of forces. Recall that, according to the Varignon theorem, the moment of force F with respect to any point is equal to the sum of the moments of the components Fz and Fy about the same point. Let us conditionally accept the direction of rotation of the moment of support reactions around moment points as positive, then the opposite direction of rotation of forces will be considered negative.
Then it is convenient to write the equilibrium conditions in the following form:
Σ fz = 0; - Fz + HB= 0; from here H B= 1.2 kN;
Σ m A = 0; V B∙6 + M - Fy∙2 + 3q∙0.5 = 0; from here V B= - 1.456 kN;
Σ m B = 0; VA ∙6 - 3q∙6,5 - Fy ∙4 - M= 0; from here VA= 5.336 kN.
To check the correctness of the calculated reactions, we use one more equilibrium condition, which was not used, for example:
Σ Fy = 0; V A + V B - 3q - Fy = 0.
Vertical support reaction V B turned out with a minus sign, this shows that in this beam it is directed not up, but down.
Example 12. Determine the support reactions for a beam rigidly embedded on one side and shown in fig. 34, A. Given: q=20 kN/m.
Rice. 34. Calculation scheme and beam dimensions for example 12:
a - design scheme; b - object of balance
Solution. Let's single out the object of equilibrium. The beam is loaded with an active load in the form of a flat system of parallel forces located vertically. We mentally release the beam from the connections in the termination and replace them with reactions in the form of a concentrated force V B and pairs of forces with the desired reactive moment M B(see fig. 34, b). Since active forces act only in the vertical direction, the horizontal reaction H B equals zero. Let us conditionally accept the direction of rotation of the moment of support reactions around the moment points clockwise as positive, then the opposite direction of rotation of forces will be considered negative.
We compose the equilibrium conditions in the form
Σ Fy = 0; V B- q∙1,6 = 0;
Σ m B = 0; M B - q∙1,6∙1,2 = 0.
Here q∙1.6 - the resultant of the distributed load.
Substituting the numerical values of the distributed load q, we find
V V= 32 kN, M B= 38.4 kN∙m.
To check the correctness of the found reactions, we compose one more equilibrium condition. Now let's take some other point as the moment point, for example, the right end of the beam, then:
Σ m A = 0; M B – V B∙2 + q∙1,6∙0,8 = 0 .
After substituting the numerical values, we obtain the identity 0=0.
Finally, we conclude that the support reactions are found correctly. Vertical reaction V B directed upwards, and the reactive moment M V- clockwise.
Example 13 Determine the support reactions of the beam (Fig. 35, A).
Solution. The resultant distributed load acts as an active load Q=(1/2)∙aq\u003d (1/2) ∙ 3 2 \u003d 3 kN, the line of action of which passes at a distance of 1 m from the left support, the thread tension force T = R= 2 kN applied at the right end of the beam and concentrated moment.
Since the latter can be replaced by a pair of vertical forces, the load acting on the beam, together with the reaction of the movable support IN forms a system of parallel forces, so the reaction R A will also be directed vertically (Fig. 35, b).
To determine these reactions, we use the equilibrium equations.
Σ M A = 0; -Q∙1 + R B∙3 - M + T∙5 = 0,
R B = (1/3) (Q + M-R∙5) = (1/3) (3 + 4 - 2∙5) = -1 kN.
Σ M B = 0; - R A∙3 +Q∙2 - M+ T∙2 = 0,
R A= (1/3) (Q∙2 - M+R∙2) = (1/3) (3∙2 - 4 + 2∙2) = 2 kN.
Fig.35
To check the correctness of the obtained solution, we use an additional equilibrium equation:
Σ Y i = R A - Q + R B+T = 2 - 3 - 1 + 2 = 0,
that is, the problem is solved correctly.
Example 14 Find the support reactions of a cantilever beam loaded with a distributed load (Fig. 36, A).
Solution. The resultant distributed load is applied at the center of gravity of the load diagram. In order not to look for the position of the center of gravity of the trapezoid, we represent it as the sum of two triangles. Then the given load will be equivalent to two forces: Q 1 = (1/2)∙3∙2 = 3 kN and Q 2 = (1/2)∙3∙4 = 6 kN, which are applied in the center of gravity of each of the triangles (Fig. 36, b).
Fig.36
The support reactions of rigid pinching are represented by the force R A and moment M A, for which it is more convenient to use the equations of equilibrium of a system of parallel forces, that is:
Σ M A = 0; M A= 15 kN∙m;
Σ Y= 0, R A= 9 kN.
To check, we use the additional equation Σ M B= 0, where the point IN located at the right end of the beam:
Σ M B = M A - R A∙3 + Q 1 ∙2 + Q 2 ∙1 = 15 - 27 + 6 +6 = 0.
Example 15 Homogeneous beam weighing Q= 600 N and length l= 4 m rests with one end on a smooth floor, and an intermediate point IN on a pole high h= 3 m, forming an angle of 30 ° with the vertical. In this position, the beam is held by a rope stretched across the floor. Determine rope tension T and post reactions - R B and sex - R A(fig.37, A).
Solution. A beam or rod in theoretical mechanics is understood as a body whose transverse dimensions in comparison with its length can be neglected. So the weight Q homogeneous beam attached at a point WITH, Where AC= 2 m.
Fig.37
1) Since two of the three unknown reactions are applied at the point A, the first thing to do is to formulate the equation Σ M A= 0, since only the reaction R B:
- R B∙AB+Q∙(l/2)∙sin30° = 0,
Where AB = h/cos30°= 2 m.
Substituting into the equation, we get:
R B∙2 = 600∙2∙(1/2) = 600,
R B\u003d 600 / (2) \u003d 100 ≅ 173 N.
Similarly, from the moment equation one could also find the reaction R A, choosing as the moment the point where the lines of action intersect R B And T. However, this will require additional constructions, so it is easier to use other equilibrium equations:
2) Σ X = 0; R B∙cos30° - T = 0; → T = R B∙cos30°= 100 ∙( /2) = 150 N;
3) Σ Y= 0, R B∙sin30°- Q +R A= 0; → R A = Q- R B∙sin30°= 600 - 50 ≅ 513 N.
Thus we have found T And R A through R B, so you can check the correctness of the solution obtained using the equation: Σ M B= 0, which explicitly or implicitly includes all found reactions:
R A∙AB sin30°- T∙AB cos30° - Q∙(AB - l/2)∙sin30°= 513∙2 ∙(1/2) - 150∙2 ∙( /2) - 600∙ (2 - 2)∙(1/2) = 513∙ - 150∙3 - 600∙( -1) ≅ 513∙1.73 - 450 - 600∙0.73 = 887.5 - 888 = -0.5.
Received as a result of rounding discrepancy∆= -0.5 is called absolute error calculations.
In order to answer the question of how accurate the result is, calculate relative error, which is determined by the formula:
ε=[|∆| / min(|Σ + |, |Σ - |)]∙100% =[|-0.5| / min(|887.5|, |-888|)]∙100% = (0.5/887.5)∙100% = 0.06%.
Example 16 Determine the support reactions of the frame (Fig. 38). Here and below, unless otherwise stated, all dimensions in the figures will be considered indicated in meters, and forces - in kilonewtons.
Fig.38
Solution. Let us consider the equilibrium of the frame, to which the thread tension force is applied as the active force T equal to the weight of the load Q.
1) The reaction of the movable support R B find from the equation Σ M A= 0. In order not to calculate the shoulder of the force T, we use the Varignon theorem, decomposing this force into horizontal and vertical components:
R B∙2 + T sin30°∙3 - T cos30°∙4 = 0; → R B = (1/2)∙ Q(cos30°∙4 - sin30°∙3) = (5/4) ∙ (4 - 3) kN.
2) To calculate Y A write the equation Σ M C= 0, where the point WITH lies at the intersection of the lines of action of the reactions R B And X A:
- Y A∙2 + T sin30°∙3 - T cos30°∙2 = 0; → Y A= (1/2)∙ Q(sin30°∙3 -cos30°∙2) = (5/4) ∙ (3 -2) kN.
3) Finally, we find the reaction X A:
Σ X = 0; X A - T sin30° = 0; → X A =Q sin30° = 5/2 kN.
Since all three reactions were found independently of each other, for verification, you need to take the equation that includes each of them:
Σ M D = X A∙3 - Y A∙4 - R B∙2 = 15/2 - 5∙(3 -2 ) - (5/2)∙ (4 - 3) = 15/2 - 15 + 10 -10 +15/2 = 0.
Example 17. Determine the support reactions of the rod with a broken outline (Fig. 39, A).
Solution. We replace the distributed load on each section of the rod with concentrated forces Q 1 = 5 kN and Q 2 \u003d 3 kN, and the action of the discarded hard pinch - by reactions X A,Y A And M A(fig.39, b).
Fig.39
1) Σ M A = 0; M A -Q 1 ∙2,5 - Q 2 ∙5,5 = 0; → M A= 5∙2.5 + 3∙5.5 = 12.5 + 16.5 = 29 kNm.
2) Σ X = 0; X A + Q 1 sina = 0; → X A= -5∙(3/5) = -3 kN.
3) Σ Y= 0; Y A - Q 1 cosa- Q 2 = 0; →Y A= 5∙(4/5) + 3 = 4 + 3 = 7 kN, since sinα = 3/5, cosα = 4/5.
Check: Σ M B = 0; M A + X A∙3 - Y A∙7 +Q 1 cosα∙4.5 + Q 1 sinα∙1.5 + Q 2 ∙1,5 = 29 -3∙3 - 7∙7 + 5∙(4/5)∙5 + 5∙(3/5)∙1,5 + 3∙1,5 = 29 - 9 - 49 + 20 + 4,5 + 4,5 = 58 - 58 = 0.
Example 18. For the frame shown in Fig. 40, A, support reactions need to be determined. Given: F= 50 kN, M= 60 kN∙m, q= 20 kN/m.
Solution. Consider the balance of the frame. Mentally release the frame from the bonds on the supports (Fig. 40, b) and select the equilibrium object. The frame is loaded with an active load in the form of an arbitrary planar system of forces. Instead of discarded bonds, we apply reactions to the object of equilibrium: on a hinged-fixed support A- vertical VA and horizontal H A, and on a hinged movable support IN- vertical reaction V B The expected direction of the reactions is shown in Fig. 40, b.
Fig.40. Calculation scheme of the frame and the balance object for example 18:
A– calculation scheme; b- the object of balance
We formulate the following equilibrium conditions:
Σ Fx = 0; -H A + F = 0; H A= 50 kN.
Σ m A = 0; V B∙6 + M - q∙6∙3 - F∙6 = 0; V B= 100 kN.
Σ Fy = 0; VA + V B - q∙6 = 0; VA= 20 kN.
Here, the direction of rotation around the moment points counterclockwise is conventionally taken as positive.
To check the correctness of the calculation of the reactions, we use the equilibrium condition, which would include all the support reactions, for example:
Σ m C = 0; V B∙3 + M – H A∙6 – VA∙3 = 0.
After substituting the numerical values, we obtain the identity 0=0.
Thus, the direction and magnitude of the support reactions are determined correctly.
Example 19. Determine the support reactions of the frame (Fig. 41, A).
Fig.41
Solution. As in the previous example, the frame consists of two parts connected by a key hinge WITH. The distributed load applied to the left side of the frame is replaced by the resultant Q 1 , and to the right - the resultant Q 2 , where Q 1 = Q 2 = 2kN.
1) Finding a reaction R B from the equation Σ M C (sun) = 0; → R B= 1kN;
