5. Lines of influence and their application for calculations
statically definable beams
5.1. Loads and internal force factors
Strength of materials considers only single-span beams under action on them stationary loads. In the course of structural mechanics, these same beams are considered, but when acting on them and moving loads, and multi-span statically definable beams, trusses and arches under the action of moving and stationary loads on them.
Moving load is a load moving through a structure at a certain speed. For example, such a load is transport (Fig. 5.1, A), a train moving on a bridge; a crane moving along a crane beam, etc. It can be considered as a system of interconnected parallel forces moving along a structure (Fig. 5.1, b). In this case, the forces (as well as stresses and deformations) depend on the position of the moving load. To determine the calculated values of the forces, it is necessary to select from all possible load positions the one at which the calculated element will be in the most unfavorable conditions. This load position is called the most unprofitable , or dangerous.
Rice. 5.1
5.2. Methods for calculating structures for moving loads
A moving load causes variable internal forces in the elements of a structure. Calculation of a structure for a moving load, even without taking into account dynamic effects (for example, accelerations and inertial forces), is more difficult than calculation for a constant load. Because we have to solve several problems:
1) determine the most dangerous (design) load position;
2) determine the largest (calculated) value of this load;
3) calculate the structure for the design load.
Calculation of moving loads can be carried out using two methods.
General method . Essence of the method: the moving load is considered as a whole and is designated by one coordinate; the desired internal force is expressed as a function of this coordinate; this function is examined for its extremum and the calculated position of the load is determined; then the design value of the internal force is calculated.
This method is universal, but difficult to implement.
Influence line method . Essence of the method: the required quantity (internal force, reaction, etc.) is determined as a function of the moving unit force; a graph of this function is plotted, and then the calculated position and calculated value of this quantity are found.
The influence line method is simpler to implement; it allows one to quite simply determine the calculated position of the load and its magnitude. Therefore, further we will focus only on it.
Line of influence (LV) is a graph of the change in one force (support reaction, reaction in connection, bending moment, shearing and longitudinal forces) in a certain place (section) of the structure from a unit dimensionless force P=1, which moves along the structure without acceleration, while maintaining a constant direction.
The concepts of LP and diagram cannot be confused, because the diagram shows the value of internal force for all points (sections) from a constant load, and LP shows the value of internal force from a moving unit force P=1 for only one section.
Lines of influence, mainly p at the same time, are used in beam systems (as well as in arches, trusses and other rod systems), in which a concentrated force can move along the span, maintaining its direction. P p and Using influence lines, it is easy to calculate a beam for a moving load that occurs, for example, when a train or a stream of cars moves on a bridge span.
5.3. Constructing force influence lines for a simple beam
Example 5.1. Consider a cantilever beam subject to a moving load P=1 (Fig. 5.2, A).
Rice. 5.2
1) Lines of influence of support reactions
Sum of moments in the right support:
Σ M B =−R A ∙l + 1 ∙ (l – x)= 0.
From here
To plot a graph of this function, we find the position of two points:
If x=0, then R A=1;
If x=l, That R A=0.
We draw a straight line through these points and construct the LP reactions R A(Fig. 5.2, b).
To determine the right support reaction, we create the equation
Σ M A = R B ∙ l – 1 ∙ x = 0.
From here
If x=0, then R B=0;
If x=l, That R B=1.
We draw a straight line through these points and construct an l.v. reactions R B(Fig. 5.2, V).
2) Lines of influence of shear force and moment
They depend on the position of the section in which they are determined.
a) Unit force to the right of section K
In this case Q K = R A, M K = R A ∙a .
These functions define right branches LW shear force and moment in section TO (Fig. 5.2, g, d).
b) Unit force to the left of section K
In this case, the internal forces are determined through the right support reaction. Then Q K =– R B, M K =R B ∙b. These functions define left branches of the LV shear force and moment in section TO (Fig. 5.2, g, d).
If the section is located on the cantilever (left or right) parts of the beam (Fig. 5.3, A), the shear force and moment LPs will be completely different. We present the result of their construction for two sections K 1 And K 2(Fig. 5.3, b-d).
Rice. 5.3
In some design diagrams (for example, in floor diagrams of a split beam) there are consoles with terminations on the right or left. The LP of their efforts can be obtained without calculations, using the corresponding left and right parts of the previous lines of influence (Fig. 5.3, b-d), assuming that at points A And IN there are seals.
The obtained LPs of support reactions and internal forces are used as known solutions when calculating similar beams and as intermediate solutions when calculating multi-span beams.
Example 5.2. Consider a simple beam on two supports (Fig. 5.4, A).
Solution.
Load it with unit force R = 1. Since the force moves along the beam (say in the vertical direction), we fix its location with the coordinate X from the support A.
Fig.5.4
Solution.
Let's build l. V . for ground reactionR A.
Let's calculate the valueR A, having considered the statics equationΣ M B =0.
Σ M B =−R A ∙l + 1 ∙ (l – x)= 0.
From here
From the expressionR A we see that the magnitude of the support reaction changes according to a linear law. Therefore, you can specify two sections X and according to these valuesR A plot the reaction changeR A .
At x=0,R A=1.
At x= l(i.e. strength R = 1 will be on support B) R A=0.
Postponing these values R A on one graph and connecting their straight line (Fig. 5.4, b), we get l. V.R A within the length of the beam. When the power R= 1 will be at point C, the valueR A can be calculated from the similarity of triangles or analytically from the previously obtained formula:
The reader is invited to construct the l. V.Rb and compare with the graph shown in Fig. 5.4, V.
Let us analyze the construction of l. V . For M k. Section “K” at a distance of 4.0 m from support A (Fig. 5.5, A).
Because the R = 1 moves along the beam, it can end up either to the left of section “K” or to the right of it. It is necessary to consider both load positions relative to section “K”.
A) R = 1 to the left of section “K” (as shown in Fig. 5.5, A).
Fig.5.5
The bending moment in section “K” can be calculated from both left and right forces. It is more convenient to calculate the moment of starting forces - there are fewer terms (less forces):
From this expression it follows that
Therefore, it is necessary to construct a l.v.Rb and increase all its ordinates by 2 times (Fig. 5.5, b), but this graph will be valid only to the left of section “K”, i.e. where the load is located R = 1. This direct l.v. M K is called the left straight line. Let's consider the second position R = 1.
b) R= 1 to the right of section “K”.
or
i.e., l should be constructed. V.R A, the ordinates of which should be increased by 4 times, and this graph will be valid only to the right of the section “K” - the right straight line l.v. M K(Fig. 5.5, V).
To get the full schedule l. V. M K we combine both straight lines (left and right) on one axis. V. M K(Fig. 5.5, G).
L. are built according to the same principle. V. ForQ K(Fig. 5.5, d) and other efforts.
Example 5.3. Consider a cantilever beam (Fig. 5.6). Let's construct graphs of changes (l.v.) in support reactions and internal forces in section “K”.
Fig.5.6
Solution.
Lines of influence R A . .
The reaction of this support will be determined from the statics equation
Σ y=0;R A- 1=0or R A=1.
Please note that the coordinate is not included in the equation X. Therefore, the reaction of the support A is constant, wherever the force is located R = 1 (Fig. 5.6, b).
Line of influence H A . .
The equation Σ x=0 gives thatH A=0.
Line of influence M A
From Eq. Σ M A=0 we get thatM A+ 1 ∙ x=0, from whereM A= - x.
The minus sign indicates that we chose the direction of the reactive torque incorrectly, and the value itself M A depends on the coordinate X.
At x =0 M A=0.
At x = l M A= l(Wherel– console crash).
Line of influence M A shown in Fig. 5.6, V.
Line of influence Q K (cutting force in section K).
Consider the position of the load R = 1 to the left of the section (Fig. 5.6, G).
Shearing forceQ K it is more convenient to calculate from the right forces, then
Q K=0.
The left straight line is valid for finishing up to section K (Fig. 5.6, e).
When the cargo R= 1 will be to the right of section K (Fig. 5.6, d), we again calculate the cutting force from the right forces:
Q K=1.
Let us note again - the valueQ K does not depend on the position of the load in this area, i.e.Q K – constant (Fig. 5.6, e) and the right straight line is valid from section K to the end of the console. In section K on the graph l.v. there is a jump of R = 1.
Line of influence M K (bending moment in section K).
Here we will again consider two positions of the load R = 1.
A ) Cargo R = 1 to the left of the section (Fig. 5.6, G).
It is easier to calculate the bending moment in section “K” from the right forces (there are none), thenM K=0 . Therefore, on the graph (Fig. 5.6, and) to the left of the section we draw the zero line (left straight line).
b) Cargo R = 1 to the right of the section (Fig. 5.6, d).
Let's fix it from the section “K” with the coordinate X. Then the bending moment in section “K” is calculated:
M K = 1∙ x.
From here we have:
at x =0 M K=0.
atx = b M K = b .
Using these data we construct the right straight line (Fig. 5.6, and).
5.4. Constructing force influence lines in broken rods (frames)
Example 5.4. Let's consider the simplest frame (Fig. 5.7). We will assume that R = 1 moves along the horizontal rod 2-3 and is directed vertically.
Fig.5.7
Solution.
Because the R = 1 moves along line 2-3, then we construct all graphs according to the projection of this line (Fig. 5.7).
Line of influence H 1
Let's write down an expression to determine H 1:
Σ M 3 =0;
where do we find it from?
at x =0 H 1 = 1,5;
atx =6 H 1 = 0.
Change schedule H 1 shown in Fig. 5.7, b.
Line of influence N 3
Σ x =0; H 3 + H 1 =0, from whereH 3 =- H 1 .
The minus sign indicates that our chosen direction was unsuccessful. Let's change it to the opposite. In other words, the valueH 3 = H 1 .
Line of influence R 3
Σ y=0;R 3 - 1=0; R 3=1.
This means that the magnitude of the reactionR 3 does not depend on the position of the load (Fig. 5.7, V).
Line of influence M 21 (moment in section 2 of section 2-1)
We write the magnitude of the bending moment as the sum of the moments of the lower forces, i.e.
or the magnitude of the moment changes in the same way as l.v. H 1, the ordinates of which are multiplied by 4 (m) (Fig. 5.7, G).
Line of influence Q 21 (cutting force in section 2 of section 2-1)
The equation speaks for itself (Fig. 5.7, d).
Line of influence Q 23 (cutting force in section 2 of section 2-3)
Line of influence N 21 (longitudinal force in node 2 of section 2-1) (Fig. 5.7, and).
N 21 =0 (from the projection onto the axis of rod 2-1).
Line of influence N 21 (longitudinal force in node 2 of section 2-3) (Fig. 5.7, h).
(from the projection onto the axis of rod 2-3).
5.5. Constructing force influence lines in a double-disc structure
Example 5.5. Let's consider the construction using the example of a double-disc frame(Fig. 5.8).
Fig.5.8
Solution.
Lines of influence of support reactions
Line of influence R 1 .
Calculate the support reactionR 1:
Σ M 6 =0;
At R = 1 to the left of hinge 3:
At R = 1to the right of hinge 3:
Solution of a system of 2 equations with 2 unknowns at R = 1 to the left of hinge 3:
gives Giving the coordinate " X» extreme values in this area, we obtain the valueR 1:
at x =0 R 1 =1 ,
atx
=
4
At R =1 to the right of hinge 3 we obtain a system of two equations:
the solution of which gives:
at x =4 R 1 = 0,567;
at x =7 R 1 = 0;
at x =9 R 1 = -0,377.
Change scheduleR 1 look at Fig. 5.8, b.
Line of influence H 1
From the previously obtained equations with a known valueR 1 find the value H 1 :
At R = 1 to the left of hinge 3
at x =0 H 1 = 0;
at x =4
When loaded R = 1 to the right of hinge 3
at x =4 H 1 = 0,324;
at x =7 H 1 = -0,756+0,756=0;
at x =9 H 1 = -0,972+0,756=-0,216.
Based on the obtained values, the line of influence H 1 built in Fig. 5.8, V.
Line of influence N 6 .
From the general equilibrium equation of the structure:
Σ x =0;
Whence it follows that, and therefore,(Fig. 5.8, V).
Line of influence R6.
Let us use the equilibrium equation for the entire structure:
Σ y =0;
From here
Line of influence R 6 shown in Fig. 58, G.
Lines of influence of internal efforts
Let us outline the sections at node 4 on rod 4 - 6; in node 4 in section 4 - 3; at node 4 in section 4 – 5 (Fig. 5.9, A).
Line of influence Q 4-6 .
Magnitude of effort Q 4-6 calculated from the equilibrium condition of the lower part (rod 4-6):
Please note that the magnitude of the shearing force (Q 4-6) from the position of the force R = 1 does not depend, therefore,(Fig. 5.8, d).
Line of influence N 4-6 .
An effort N 4-6 is calculated as the sum of all forces on the axis of the rod located below section 4 of section 4 - 6.
and, since the magnitudeN 4-6 does not depend on the coordinate X, we can say:(Fig. 5.8, e).
Line of influence M 4-6 .
The bending moment in section 4 of sections 4 – 6 is calculated:
and again, it does not depend on the location R = 1. Thus,changes just like, but all ordinates l.v. N 6 increase by 4 (m), i.e.:(Fig. 5.8, and).
Fig.5.9
Section 4 in section 4 – 3 – 2.
Line of influence Q 4-3 (Fig. 5.9, b).
The magnitude of the shearing force in section 4 of section 4 – 3 – 2 (Q 4-3 ) will depend on the position of the force R = 1.
Force R = 1 to the left of section 4.
Got it like this called left straight.
Force R = 1 to the right of section 4 – 3.
Line of influence N 4-3 (Fig. 5.9, V).
Regardless of load position R = 1, valueN 4-3 will be equal to either H 1, or N 6, i.e.
Line of influence M 4-3 (Fig. 5.9, G).
Force R = 1 to the left of the section: (left straight).
Force R = 1 to the right of the section.
There are two possible calculation options here:
A) , i.e.
b) Force R = 1, located to the right of section 4 of rods 4 – 3, fix by ordinate X from node 4 (Fig. 4.9, A). Then
Line of influence already built. Remains at X= 2add to the value –0.864 value 2 , i.e.:
atx =2
atx =0
For the forces of section 4 of sections 4 - 5, influence lines are drawn as for the console (Fig. 5.9, d ,e,and). We suggest building them yourself.
H a few more difficult construction lines of influence effort in elements statically definable farms, arch, and statically indefinable systems.
Note also that the lines of influence yc ily V statically definable systems at movement load By direct are depicted in segments direct lines, then time as lines of influence effort V statically indefinable systems, How rule, curved.
5.6. Calculation of forces along influence lines from a stationary load
Let's turn to l.v. effortsR A simple beam (Fig. 5.10). Note that when finding the force R= 1 on support A the magnitude of the reaction is equal to 1, and when the force is found R= 1 at a distance X from the support A magnitudeR A will be equal to the valueR A (X) , taken from the graph (Fig. 5.10). If force R = 1 increase in "n " times, then the graph (its values) will increase by "n " once.
Fig.5.10Fig.5.11
Then at loading with one concentrated force, say, R = 5 kN (Fig. 5.11), valueR A will be equal to the product of force 5 (kN) by the ordinate of L.V.R A , taken by force, i.e.
or, calculating analytically, we get the same valueR A .
If a beam or other structure is loaded with concentrated forces (Fig. 5.12) and, using the principle of independence of the action of forces, we calculate the values of the force from each force and add the results, i.e.
where: P i– meaning of concentratedi-th strength;
y i – ordinate L.V. effortsS taken by force R i , i.e.:
From p as allocated loads q(x) the force through the influence lines is determined by:
Where a And b - coo p dinats start and end points actions distributed loads.
For p uniformly distributed loads(Fig. 5.13) q= const:
Where - square, og p limited line of influence axis abscissa And direct x = a And x = b.
Rice. 5.12Fig. 5.13
So for the circuit in Fig. 5.14 with a uniformly distributed load, the forceS will be calculated as the product of load intensity and area (-Ω ) l.v. efforts (in Fig. 5.14 l.v. efforts M k ), i.e.S = Ω ∙ q or for M k :
Fig.5.14
It is necessary to establish a sign rule when calculating internal forces along influence lines.
If concentrated forces and distributed load are directed from top to bottom, then the sign of the ordinates of the line of influence and area determines the sign of the force.
If the positive branch of the line of influence is laid below the axis of the rod and the concentrated moment falls on it, then when the beam axis rotates at the shortest angle to l. V. matches With direction of the concentrated moment, we have a positive internal force.
C follows underline difference between concepts of line of influence and diagrams, which By definition Also is graphic image law of change efforts or movements.
ABOUT p dinats y i and lines of influence, and diagrams moments are here functions from coordinates x. However, in c better lines of influence this coordinate defines position load P= 1, and in case diagrams- position sections, V which is located moment.
Example 5.6. Let's turn for example (Fig. 5.15).
Fig.5.15
Solution.
Let's calculate the magnitude of the support reaction C. Multiply the value of the force of 15 kN to the value of the line of influence under the force (0.5) and get:
R With= 15 ∙ 0,5 =7,5 kN.
For comparison, it is easy to calculate the reaction from the equation: bending moment at the hinge IN right forces is zero:
M B = R With ∙ 3 - 15 ∙1 ,5 =0, where we findR With= 7,5 kN.
Similarly we find:
M B = 8 ∙ 3 +15 ∙ 2 +2 ∙ (4 ∙ 4/2) = 70 kNm.
Example 5.7. Design (Fig. 5.16, V) is loaded by a system of forces (option a and option b). Let's calculate the effort values along the lines of influence N 3 (Fig. 5.16, G), M To(Fig. 5.16, d), M F(Fig. 5.16, e).
Fig.5.16
Solution.
Loadingaccording to option "a".
Loadingaccording to option "b"
5.7. Construction of influence lines for nodal load transfer
Cha c That load transmitted on design Not directly, A through system statically definable beams ( pic. 5.17, A). Then, e c whether unit cargo is located at first flight beams, i.e. at the point A, then he is entirely transmitted on basically design and calls force, For which built line of influence, numerically equal y a - ordinate lines of influence, correspondingI main designs (pic. 5.17, b).
Rice. 5. 17
E c whether cargo is located at the end flight beams (point b), then he also transmitted on basically design, calling force, numerically equal y b - ordinate lines of influence at a point b main structure.
H finally, if cargo is located V overflight beams on distancet from point a(pic. 5.17, V), then left reaction beams there will be equal , and the right , (l 1 - flight beams). Meaning yc or me V main designs:
those. line of influence on y part movement load along the beam there will be rectilinear. E c whether main line of influence on this area broken line or curved, That at transmission loads through statically definable beam at transition from ordinates y a To ordinate y b this line of influence straightens up.
Opie c annay way transfers loads on basically design called nodal transmission loads. He O c especially often occurs V farms, Where supports beams flooring are located above nodes farms, and beams serve themselves panels upper or lower belts(Fig. 5.18).
Rice. 5. 18
P p avilo construction lines of influence efforts S at nodal transmission loads is as follows:
1. By c triple in advance line of influence what you're looking for efforts at movement load By main parts designs;
2. Fix the ordinates of the constructed line of influence under the load transfer nodes;
3. Connect P p I my line ordinates lines of influence under nodes transfers loads.
This line is called transmission line lines of influence. An example of applying this rule to draw a line influence bending moment for section K beams are shown in Fig. 5.19.
Rice. 5. 19
5.8. Unfavorable or dangerous load position
In the process of designing rod structures, the question often arises about this loading external load, when internal forces in the section under consideration (or support reaction) take on maximum (minimum) values. This problem is studied primarily using influence lines.
Let's assume that l. V. consists of from individual linear sections, consider various cases loading.
P .
In this case, the reasoning O disadvantageous loading protozoa:
– the maximum force will be when the concentrated force is located above the maximum positive (y max) ordinate of the line of influence:
S max = P ∙ y max;
– the minimum force will be when the concentrated force is located above the maximum negative (y min) ordinate of the line of influence:
S min = P ∙ y min.
2. The case of action of a system of rigidly coupled concentrated forces.
This load models the load from a car, train, etc.
In general, the force influence line can represent broken line.
Let's consider the case when two associated concentrated forces act (Fig. 5.20). LetP 2 > P 1 .
Rice. 5.20
To determine a dangerous situation their cargo are installed over unambiguous sections of the line of influence so that the largest load is located above the largest ordinate. From Fig. 5.20 everything becomes clear.
With a larger number of loads, the desired dangerous position is established by searching through several options for their position, in which one of the loads must necessarily located above one of the vertices of the line of influence (Fig. 5.21).
Rice. 5.21
The following considerations will help reduce the number of provisions considered. Let us establish a moving system of associated forces in the assumption of the occurrence of a dangerous loading(Fig. 5.21). Let's move the weight system right on∆ x . The force increment will be equal to
∆ S = Σ P i∙ ∆ h i = Σ P i ∙ ∆ x ∙ tgαi=∆ x ∙ Σ P i ∙ tgαi,
Where∆ h i– the amount of change in coordinates underP i ;
α i– angle of inclination of the body under forceP i .
Let's assume that the increment∆ S >0. Mentally revenge weight system to the left of the original position on∆ x . If the force increment∆ N will be negative, then the initial position of the loads corresponds dangerous loading.
Indeed, if dangerous loading uniquely for a given section, then the desired function of changing the internal force depending on the position of the load system must have a single extremum. The condition of changing the sign of the force increment when passing through the extremum allows us to reduce the number of searches.
3. The case of a moving uniformly distributed load acting on a structure q .
An effortN from a uniformly distributed load, as shown earlier, is calculated by the formula
Maximum force valueS will be determined by the area , since the valueq is constant. Consequently, a moving constant distributed load must be placed above that section of the force influence line where the area under it will be maximum (minimum).
5.9. Matrix form of effort calculation
P p And carrying out calculations With using computing technology wide apply matrices influence, those. matrix, the elements of which are the ordinates of the lines of influence. Task p accounts designs being formed next thus.
Let it be required to produce calculation Which- or a statically determinable system under the action of a given load (Fig. 5.22, A).
Let us replace the given system with its discrete circuit, for which we outline the sections i = 1, 2, 3,..., n, in which it is necessary to calculate the efforts S i (i = 1, 2, 3,..., n).
By replacing the distributed load with concentrated forces, and the moment in the form of a pair of forces, the system of external forces is represented as a system of concentrated forces (Fig. 5.22, b) P T = ( P 1 ,P 2 ,P 3 ,..., Pn ), Where R i - the value of the external force applied in i - ohm section.
Rice. 5.22
Further c triplets influence lines of the desired force for sections i = 1, 2, 3,..., n given beam. C publicly principle independence actions strength for each i - Wow sections Can compile expression what you're looking for efforts V next form:
Where y ik - meaning And c whom efforts V i - ohm section from single strength Pk = 1, attached V k - oh point ( pic. 5.22, b).
Enter vecto p s S t = ( S 1 ,S 2 ,S 3 ,..., S n );P t = (P 1 ,P 2 ,P 3 , ..., Pn ) And matrix L s , elements which are ordinates of influence lines:
This mat p itza called matrix influence effortsS. P p And using the introduced notations ratios(1) possible write down as:
In practice, a matrix of the influence of bending moments is constructed L M . Next, using this matrix, you can use the formula , and make the transition from the matrix of influence of bending moments to the matrix of influence of shearing forces. To determine the shear force acting on an arbitrary i - ohm section of the beam limited by sections i And i - 1, using the discrete analogue of the last formula in the form
it is numerically equal to the tangent of the angle of inclination of the moment diagram.
The transformed moment matrix can be obtained by multiplying two matrices:
Where - coefficient matrix for transforming the moment influence matrix into the matrix of the influence of shearing forces. It has a bidiagonal structure: there are ones on the diagonal, and under the diagonal Theory of machines and mechanisms
Lines of influence of forces in a given section of a structure are constructed using two methods: static and kinematic.
2.1.1. Static method for constructing influence lines
Cargo F=1 is installed in an arbitrary section, the position of which is fixed by a variable X(Fig. 10). From the equilibrium condition of the system, the analytical expression for the determined force is written J=f(x). Substituting into it the value of the coordinates fixing the position of the load F=1, calculate the ordinates BGN, located under load, and build a graph.
Rice. 1.10. Lines of influence of efforts
When constructing force influence lines M To ,Q To for fixed section “K” located between the supports, should be considered two positions of the load F=1 – to the left and to the right of section “K”, while considering equilibrium of the right and left cut-off parts, respectively. In this case, writing the equations M To ,Q To easier. In the case when the section is located on the console, when the load moves F=1 to the left and right of the section expedient consider the balance of the cantilever part, considering that the load moves away from the section.
Outside the structure, the influence lines are zero.
Lines of influence of efforts R A , R B , M K ,Q K , M n ,Q nshown in Fig. 2.1.
Lines of influence R A
From the statics equation we determine the reaction R A.
equation of a straight line, for which two points are enough to construct.
At x = 0,
lv R A (0) = 1(L-0) / L = 1,
at x = L;
lv R A (L) = 1(L-L) / L = 0.
Cargo F=1 is on the console, x = -d,
lv R A (-d) = 1(L + d) / L.
Based on the obtained values, we build a line of influence of the support reaction R A .
RB influence line
,
lv R B (x) = x/L.
Line of influence R b (x) changes according to a linear law. Substituting coordinates X into the equation BGN Rb:
x = 0,LWR B (0) = 0/L; x = L,LWR B (L) = L / L = 1;
x = –d, lv R B (–d) = –d/L.
Characteristics reaction influence lines R:
consists of one branch;
above the support for which the force R is determined, cuts off the ordinate equal to plus 1;
on the opposite support the ordinate is zero.
Bending moment influence line M TO .
Section “K” is located between the supports: . Cargo F=1 to the left of section K, the equilibrium of the right side of the beam is considered.
M K = R B (x) X b or BGN M K = lev R B X b- equation of a straight line.
At x = a, lv M K = a X b/L, at x = -d, lv M K = -d X b/L.
Line of influence constructed under the assumption that the load F=1 moves to the left of the section TO, is called the left branch of the line of influence. Left branch lv M TO represents lv R B, increased by b once.
Cargo F=1 to the right of section K, the equilibrium of the left side, .
M K = R A (x) X a = a X (L - x) / L or lv M K = lev R A X a.
At x = a,BGNM K = (L - a) X a/L=a X b/L,
at x = L, lv M K = (L - L) X a/L =0.
Right branch BGN M TO- This lv R A, increased in A once.
Characteristics of the influence line M TO , section “K” is located between the supports:
consists of two branches: the left branch is valid from the left support to the section, the right branch is valid from the right support to the section;
the branches cut off the distance from this support to the section above the support.
Line of influence of shear force Q K
Section TO located between the supports: . Cargo F=1 to the left of the section, the equilibrium of the right side.
; lv mQ = -LV R B ;
x =a,BGNQ K = -a/L; x = -d,BGNQ K =d/L.
Cargo F=1 to the right of the section TO, balance of the left side .
Q K = R A (x) = 1(L-x)/L; BGN Q K = LV R A ;
x = a,BGNQ K (a) = (L-a) / L = b / L.
Characteristics of lv Q K , cross-section between supports:
– consists of two parallel branches;
– the right branch cuts off an ordinate equal to plus 1 above the left support, and the left branch under the right support cuts off an ordinate equal to minus 1;
– in the cross section there is a jump equal to 1.
Line of influence M n
Section n located on the console, .
Cargo F=1 to the left of the section n
M n = -Fx 1 ; lv M n = -x 1 ;
x 1 =0 , LV M n = 0 ;
x 1 =-C, LV M n = -C
Cargo F=1 to the right of the section n, equilibrium of the cantilever (left) part of the beam.
M n = 0 - right branch.
The right branch from the side of the supports is zero, since the equilibrium of that part of the beam on which there is no load is considered. Consequently, the branch from the side of the supports coincides with the axis of the line of influence.
Characteristics of LV M n , section on the console:
consists of two branches;
branches always intersect under the section;
branch from the side of the supports, the termination is always zero;
the branch from the console side cuts off an ordinate at the end of the console equal to the distance from the section to the end of the console.
Line of influence Q n
Load F =1 to the left of the section n
Q n (x 1) = - F=- 1- left branch.
Load F =1 to the right of the section n, balance of the cantilever part.
Q n (x 1) = 0 - right, zero branch.
Characteristics of drug Q n , section on the console:
consists of two parallel branches;
the branch from the side of the supports is always zero;
the branch on the cantilever part is parallel to the axis of the influence line and cuts off an ordinate in the section equal to minus 1 if the console is located to the left of the supports, and plus 1 if the console is to the right of the supports;
in the cross section – a jump equal to one.
2.1.2. Kinematic method for constructing influence lines
The kinematic method is based on the principle of possible displacements: if the system is in equilibrium, then the sum of the work of all forces acting on the system at any possible infinitesimal displacements is zero.
The essence of the kinematic method for constructing influence lines is as follows:
the connection in which the force is determined is discarded, and a mechanism with one degree of freedom is obtained;
instead of the discarded connection, the required force is applied;
in the direction of the required force, the system is given a single displacement and a displacement diagram of the resulting mechanism is constructed. The constructed displacement diagram gives the appearance of a line of influence;
to obtain the ordinates of the line of influence, the equation of work for a certain position of the load F = 1 is written;
The characteristic ordinates of the line of influence are determined from geometric constructions.
Type of displacement diagrams in accordance with Fig. 2.2 is obtained to construct lines of influence:
ground reaction R– by throwing away the support rod, the action of which is replaced by force R;
bending moment M– in any section, by cutting a hinge into a given section, the effect of the broken connection is compensated by the application of two equal and oppositely directed moments;
shear force Q– in any section by introducing a slider into a given section, while the rods of the system always remain parallel. Replacing the broken connection is carried out by applying two equal and oppositely directed concentrated forces to the ends of the resulting parts of the beam.
Preface.... 3
Introduction.... 7
Chapter 1. Kinematic analysis of structures.... 14
§ 1.1. Supports.... 14
§ 1.2. Conditions for geometric immutability of rod systems.... 16
§ 1.3. Conditions for static definability of geometrically invariable rod systems.... 23
Chapter 2. Beams.... 27
§ 2.1. General information.... 27
§ 2.2. Lines of influence of support reactions for single-span and cantilever beams.... 31
§ 2.3. Lines of influence of bending moments and shear forces for single-span and cantilever beams.... 34
§ 2.4. Lines of influence during nodal load transfer.... 38
§ 2.5. Defining forces using influence lines...... 41
§ 2.6. Determination of the unfavorable position of the load on the structure. Equivalent load.... 45
§ 2.7. Multi-span statically determinate beams.... 51
§ 2.8. Determination of forces in multi-span statically determinate beams from a stationary load.... 55
§ 2.9. Force influence lines for multi-span statically determinate beams.... 59
§ 2.10. Determination of forces in statically determinate beams with broken axes from a stationary load.... 62
§ 2.11. Construction of influence lines in beams using the kinematic method.... 64
Chapter 3. Three-hinged arches and frames.... 70
§ 3.1. The concept of an arch and its comparison with a beam.... 70
§ 3.2. Analytical calculation of a three-hinged arch.... 73
§ 3.3. Graphic calculation of a three-hinged arch. Pressure polygon.... 82
§ 3.4. Equation of the rational axis of a three-hinged arch.... 87
§ 3.5. Calculation of three-hinged arches for a moving load.... 88
§ 3.6. Sound moments and normal stresses.... 95
Chapter 4. Flat trusses.... 98
§ 4.1. Farm concept. Classification of farms.... 98
§ 4.2. Determination of forces in the rods of the simplest trusses.... 101
§ 4.3. Determination of forces in the rods of complex trusses.... 118
§ 4.4. Distribution of forces in truss elements of various shapes.... 121
§ 4.5. Investigation of the invariability of trusses.... 125
§ 4.6. Lines of influence of forces in the rods of the simplest trusses.... 133
§ 4.7. Lines of influence of forces in the rods of complex trusses.... 142
§ 4.8. Trunnion systems.... 146
§ 4.9. Three-hinged arch trusses and combined systems.... 152
Chapter 5. Determination of displacements in elastic systems.... 159
§ 5.1. The work of the vernal forces. Potential energy.... 159
§ 5.2. Theorem on reciprocity of work.... 163
§ 5.3. Theorem on the reciprocity of displacements.... 166
§ 5.4. Determination of movements. Mohr's integral.... 168
§ 5.5. Vereshchagin's rule.... 173
§ 5.6. Calculation examples.... 179
§ 5.7. Temperature movements.... 185
§ 5.8. Energy method for determining displacements.... 188
§ 5.9. Movements of statically determined systems caused by movements of supports.... 189
Chapter 6. Calculation of statically indeterminate systems by the force method.... 193
§ 6.1. Static indetermination.... 193
§ 6.2. Canonical equations of the method of forces.... 199
§ 6.3. Calculation of statically indeterminate systems under the action of a given load.... 202
§ 6.4. Calculation of statically indeterminate systems under the influence of temperature.... 213
§ 6.5. Comparison of canonical equations when calculating systems for support movements.... 215
§ 6.6. Determination of displacements in statically indeterminate systems.... 219
§ 6.7. Construction of diagrams of transverse and longitudinal forces. Checking diagrams.... 222
§ 6.8. Elastic center method.... 228
§ 6.9. Lines of influence of the simplest statically indeterminate systems.... 231
§ 6.10. Using symmetry.... 238
§ 6.11. Group of unknowns.... 241
§ 6.12. Symmetrical and inversely symmetrical loads.... 243
§ 6.13. Load conversion method.... 245
§ 6.14. Checking the coefficients and free terms of the system of canonical equations.... 247
§ 6.15. Examples of frame calculations.... 249
§ 6.16. “Models” of force influence lines for continuous beams.... 263
Chapter 7. Calculation of statically indeterminate systems by displacement and mixed methods.... 265
§ 7.1. Selecting unknowns in the displacement method.... 265
§ 7.2. Determining the number of unknowns.... 266
§ 7.3. Main system.... 269
§ 7.4. Canonical equations.... 276
§ 7.5. A static method for determining the coefficients and free terms of a system of canonical equations.... 280
§ 7.6. Determination of coefficients and free terms of a system of canonical equations by multiplying diagrams.... 283
§ 7.7. Checking the coefficients and free terms of the system of canonical equations of the displacement method.... 286
§ 7.8. Construction of diagrams M, Q and N in a given system.... 287
§ 7.9. Calculation by the displacement method for the effect of temperature.... 288
§ 7.10. Using symmetry when calculating frames using the displacement method.... 292
§ 7.11. An example of calculating a frame using the displacement method.... 295
§ 7.12. Mixed calculation method.... 302
§ 7.13. Combined solution of problems using methods of forces and displacements.... 307
§ 7.14. Constructing influence lines using the displacement method.... 309
Chapter 8. Complete system of equations of structural mechanics of rod systems and methods for its solution.... 313
§ 8.1. General notes.... 313
§ 8.2. Drawing up equilibrium equations, static equations. Study of systems education.... 313
§ 8.3. Drawing up compatibility equations, geometric equations. The principle of duality.... 321
§ 8.4. Hooke's law. Physical equations.... 326
§ 8.5. System of equations of structural mechanics. Mixed method.... 328
§ 8.6. Movement method.... 333
§ 8.7. Method of forces.... 341
§ 8.8. Equations of the theory of elasticity and their connection with the equations of structural mechanics.... 345
Chapter 9. Calculation of rod systems using a computer.... 352
§ 9.1. Introductory remarks.... 352
§ 9.2. Semi-automated calculation of statically indeterminate systems using calculators.... 353
§ 9.3. Automation of calculations of rod systems. A complete system of structural mechanics equations for a rod.... 363
§ 9.4. Reaction (stiffness) matrices for plane and spatial rods and their use.... 372
§ 9.5. Description of the educational complex for the calculation of rod systems. Internal and external representation of source data. Block diagram of the complex for calculating rod systems.... 389
Chapter 10. Taking into account geometric and physical nonlinearity when calculating rod systems.... 397
§ 10.1. 0general remarks.... 397
§ 10.2. Calculation of rod systems taking into account geometric nonlinearity.... 398
§ 10.3. Stability of rod systems.... 411
§ 10.4. Calculation of rod systems taking into account physical nonlinearity. Ultimate condition.... 419
Chapter 11. Finite element method (FEM) .... 435
§ 11.1. General notes.... 435
§ 11.2. Connection of FEM with the equations of structural mechanics.... 435
§ 11.3. Construction of a stiffness magnet for solving a plane problem in the theory of elasticity.... 456
§ 11.4. Passage to the limit for a plane problem.... 464
§ 11.5. Construction of stiffness matrices for solving a volumetric problem in the theory of elasticity.... 467
§ 11.6. Complex elements, construction of stiffness matrices for elements with curved boundaries.... 471
§ 11.7. Construction of reaction matrices for calculating plates and shells.... 485
§ 11.8. Features of complexes for calculating structures using FEM. Superelement approach.... 493
Chapter 12. Fundamentals of the dynamics of structures.... 501
§ 12.1. Types of dynamic influences. The concept of degrees of freedom.... 501
§ 12.2. Free vibrations of systems with one degree of freedom....
§ 12.3. Calculation of systems with one degree of freedom under the action of a periodic load.... 518
§ 12.4. Calculation of systems with one degree of freedom under the action of an arbitrary load. Duhamel integral.... 524
§ 12.5. Movement of a system with two degrees of freedom. Reduction from systems with two degrees of freedom to two systems with one degree of freedom.... 529
§ 12.6. Kinetic energy. Lagrange equation.... 536
§ 12.7. Bringing kinematic action to force.... 544
§ 12.8. Reducing a system of differential equations of dynamics to separable equations by solving the problem of eigenvalues.... 546
§ 12.9. The method of constant acceleration and its use for solving dynamic problems.... 550
Chapter 13. Information from computational mathematics used in structural mechanics.... 554
§ 13.1. General notes.... 554
§ 13.2. Matrices, their types, simple operations on matrices.... 555
§ 13.3. Matrix multiplication. Inverse matrix.... 557
§ 13.4. Gauss method for solving systems of linear equations. Decomposition of a matrix into a product of three matrices.... 562
§ 13.5. Study of systems of linear equations. Homogeneous equations. Solving n equations in m unknowns using the Gaussian method.... 574
§ 13.6. Square shape. Matrix of quadratic form. Derivative of a quadratic form.... 578
§ 13.7. Eigenvalues and eigenvectors of a positive definite matrix.... 581
§ 13.8. Homogeneous coordinates and integration over a triangular region.... 594
§ 13.9. Relationships between trigonometric, hyperbolic functions and exponential functions.... 599
Conclusion.... 600
Literature.... 601
Subject index.... 602
The study of the method of analytical calculation of multi-span statically determinate beams for a fixed load showed that the main task of the calculation is to determine the design forces Mmax And Qmax. This problem is solved by constructing diagrams M And Q from a given stationary load.
At the same time, a large number of engineering structures, the load-bearing parts of which are welded metal structures, including beams, operate under the influence of moving loads. These are railway and road bridges, crane beams and crane bridges, etc. In this case, determine the design forces using diagrams M And Q almost impossible. Therefore, calculations for moving loads are made in a different way.
The calculation of a structure for a moving load is greatly facilitated by the possibility of applying the principle of independence of forces, the essence of which is that the internal forces, stresses and deformations caused by the impact of various loads on the structure can be summed up.
If, for example, two groups of forces simultaneously act on a structure, then the resulting force in any element of the structure will be equal to the sum of the forces arising in it under the action of each group of forces separately. We begin our study of the effect of a moving load on a structure by considering the simplest case, when only one vertical load moves through the structure. R, equal to one (Fig. 3.14). We study how one or another factor changes (for example, the support reaction, the bending moment in a certain section of the beam, the deflection of the beam at a given point, etc.) when the load moves P = 1 by construction. The established law of change of the studied factor depending on the position of the moving load P = 1 We will depict it graphically.
A graph depicting the law of change of any force factor (for example, a bending moment in a section) when a force moves along a structure P = 1, is called the line of influence of this factor.
The concept of lines of influence. It is obvious that the magnitude of any force in the elements of supporting structures depends on the position of the external moving load. For example, in a single-span beam on two supports (Fig. 3.14), the magnitude of the support reaction R A will be greater, the closer to the support the moving load is located R, and vice versa, R A the less, the farther from the support A there is a moving load R.
A graph expressing the law of changes in forces (support reactions, bending moments, transverse forces in a given section of a beam) depending on the position of a moving unit load on the beam P = 1, is called the line of influence.
Let us consider the procedure for constructing lines of influence of support reactions of single-span beams.
Single-span statically determinate beam AB(Fig. 3.14 A). Beam load - moving unit load P = 1. Let's determine the magnitude of the support reaction R A depending on position P = 1(in current coordinates).
∑М В = 0; R A · L - P (L - X) = 0; R A = (L - X)/L. (3.12)
Equation (3.12) is the equation of a straight line. Let's determine its position in coordinates X–Y.
At X = 0.75L R A = 0.25P, at X = 0.5L R A = 0.5P., At Х =0.25L R A = 0.75Р, which is presented on the left side of Fig. 3.14.
Rice. 3.14. Analysis of changes in support reactions R A And R B depending on the position of a single load P = 1 with the construction of graphs of lines of influence of support reactions R A ( b) And R B (V) depending on the position of a unit load at R = 1
Place it on the left support ( X = 0) ordinate equal to + 1, on an arbitrary scale, on the right support ( X = L) - ordinate equal to zero. The two points found determine the position of the straight line, which is the line of influence of the support reaction R A(Fig. 3.14 b). Using the resulting graph, you can determine the magnitude of the support reaction at any position of the load P = 1. To do this, it is enough to measure the ordinate under the load. This ordinate (on the accepted scale) will be equal to the support reaction R A in this situation P = 1. The line of influence is shown in Figure 3.14 V.
Let's look at an example of using the line of influence for practical purposes. Single span beam AB(Fig. 3.15) is loaded with three stationary concentrated forces.
Rice. 3.15. Using the line of influence to determine the ground reaction force R A
Using the line of influence we determine the value R A from the action of this load. To do this, we will use one of the consequences of the principle of independence of the action of forces: the results of the influence of various loads on a structure can be summed up. Based on this
R A = P 1 y 1 + P 2 y 2 + P 3 y 3 = 8 0.75 + 6 0.5 + 8 0.125 = 10 t(3.13) Let us consider the procedure for constructing the line of influence of the bending moment in an arbitrarily chosen section of the beam.
Statically determined beam on two supports AB(Fig. 3.16 A). Let's find the bending moment in the section I - I, which is at a distance A from the left support. If a moving unit load P = 1 located to the right of the section (Fig. 3.16 A), then the bending moment in the section is equal to
M 1 = R A · a = a · (L - X)/ L.(3.14)
The graph of equation (3.14) is also a straight line, which is the line of influence of the bending moment in the section I - I (Fig. 3.16 V). But this is not the entire line of influence, but only its right branch. It is valid from support B to the section, since equation (3.14) is compiled under the condition that the load P=1 is on this (right) part of the beam. Let's move the cargo P = 1 to the part of the beam to the left of the section I - I. Then the moment in section I - I equals
M 1 = R B · b. (3.15)
Figure 3.16. Construction of the line of influence of the bending moment in section I - I
We plot the graph of equation (3.15). On the right support we lay off an ordinate equal to the segment V. Straight line connecting points to ordinate V on the right support and with the ordinate equal to zero, the left support, is the line of influence of the moment in the section I - I. But, as is now clear, this is also not the entire line of influence, but its left branch (Fig. 3.16 V). Combining both branches, we obtain the full line of influence of the bending moment in the section I - I(Fig. 3.16 G). The ordinate dimension of the line of influence of the bending moment is meters (centimeters).
It is necessary to pay attention to the following circumstance. Line of influence M 1 its outline is similar to a diagram of bending moments due to the action of a concentrated force. But this similarity is only external. There is a fundamental difference between the bending moment diagram and the bending moment influence line. If a moment diagram is a graph of the distribution of moments in all sections of a beam from a fixed fixed load, then the moment influence line is a graph of the moment values in one specific section of a beam depending on the position of a moving unit load P = 1.
Let's consider constructing the line of influence of the shear force.
Rice. 3.17. Constructing the line of influence of the shear force Q
Statically determined beam on two supports AB(Fig. 3.17). Let's build a line of influence of shear force Q I for section I - I located at a distance a left support. If a moving unit load P = 1 is located to the right of the section I - I, then the magnitude of the transverse force in the section is equal to
Q I = + R A . (3.16)
Let us recall that the rule for determining the signs of transverse forces in a section was discussed above (section 3.3.3, Fig. 3.13).
From equation (3.16) it follows that the transverse force Q I and ground reaction R A depending on the position of the moving unit load P = 1 change according to the same law. Therefore, the line of influence R A will also be the right branch of the line of influence Q I(Fig. 3.17 A).
Let's move the cargo P =1 to the part of the beam to the left of the section I - I. Then
Q I = - R V. (3.17)
From equation (2.17) it follows that the line of influence R B(with the opposite sign) will also be the left branch of the line of influence Q I(Fig. 3.17 b). Combining both branches, we obtain the complete line of influence of the shear force in the section I - I(l.l. Q I) ( Figure 3.17 V).
Let's consider the construction of influence lines for single-span beams with consoles (Fig. 3.18).
Figure 3.18. Beam AB with influence lines R A,, R B, M and Q in section I – I between supports
Construction of lines of influence of support reactions, bending moment and shear force for sections located within the main span AB, is carried out according to the same rules as for beams without consoles.
The magnitude of the ground reaction R A in current coordinates is determined by formula (3.12) given above.
R A = (L - X)/L,
Formula (3.12) is valid for all positions of the load P = 1, including consoles (Fig. 3.18 A). Constructing a line of influences of the ground reaction R A: we connect two points with a straight line - the first with an ordinate equal to + 1 , on the left support, and the second with an ordinate equal to zero, on the right support. Then we continue straight to the ends of the consoles. Within the right console, the ordinates are negative. It means that R A pointing down , when the cargo P = 1 is located within this console.
Moment influence line in section I-I Let's build it as for a regular beam, but we will continue the left and right branches to the ends of the consoles (Fig. 3.18 V). Within the consoles, the ordinates of the influence line are negative. This means that the moment of annihilation I - I negative when the load P = 1 is on consoles.
When constructing the line of influence of the shear force in the section I - I the right and left branches must be continued to the end of the consoles (Fig. 3.18, G).
The construction of lines of influence of bending moment and shear force for sections located on consoles is carried out according to different rules (Fig. 3.19).
Rice. 3.19. Lines of influence of bending moments M 1 And M 1 I and shear forces Q I And QII for sections I–I And II–II beams on consoles
Line of influence of the bending moment in the section I - I will only be within the limits of the section I - I to the end of the console. It seems obvious that when the load P = 1 located to the left of the section I - I, the section does not work, there is no bending moment (and shear force) in it.
Therefore, the ordinates of the line of influence M 1 to the left of the section I - I are equal to zero. The magnitude of the bending moment in the section I - I in current coordinates (Fig. 3.19 A), is equal to
M 1 = -P X = -X
When the cargo P = 1 is located above the section ( X = 0), M 1 = 0 when the load is on the edge of the console ( X = d), M 1 = -d. Line of influence M 1 And M 1 I are shown in Fig. 2.19 b; lines of influence Q I And Q II - in Fig. 3.19 V. (Ordinate signs of the lines of influence of bending moments M 1 And M 1 I and shear forces Q I And QII determined in accordance with the diagrams shown in Fig. 3.13).
Let us consider the construction of influence lines for multi-span statically determinate beams.
The construction of influence lines for multi-span statically determinate beams is based on the same principles that are used in the study of single-span beams.
Consider a beam A-N(Fig. 3.20 A). The beam is statically determinate and geometrically invariable. Let's draw up an interaction diagram (Fig. 3.20 b), which helps to identify the main and auxiliary elements.
When constructing lines of influence, you should be guided by the following rules:
a) the influence lines for a secondary element do not differ in construction rules from the influence lines for a regular single-span beam and do not extend beyond the element;
b) when constructing lines of influence for the main element, we first build it without paying attention to the secondary elements, as for a regular single-span beam, and then take into account their influence (the secondary elements).
Let's look at the construction of influence lines using an example for a beam A-N(Fig. 3.20 A).
Lines of influence of support reactions R A And R B(Fig. 3.20 c, d), we first build within the main element ABC, as for a regular beam with consoles. When the cargo P = 1 will move to a secondary element SD, its impact on the magnitude of support reactions R A And R B will begin to decrease and become equal to zero when the load is positioned at the point D. Accordingly, equal to zero at this position of the load P = 1 the magnitudes of the support reactions will also become R A And R V. To the right of the hinge D ordinates of influence lines R A And R B are equal to zero, since at the position of the load P = 1 to the right of the hinge D it has no effect on these support reactions.
Lines of influence M 1 II And Q 1 II for section III - III located on the secondary beam SD, do not differ from the influence lines for a conventional single-span beam (Fig. 3.20 d).
Lines of influence M 1 And Q 1 for section I - I, located within the main span of the main element ABC, we build, adhering to the rules applied when constructing lines of influence R A And R B(Fig. 3.20 e).
Lines of influence M 1 I And Q 1 I for section II - II located on the console part of the main element ABC, we build first as for a regular beam, then we take into account the influence of the secondary element SD. When the cargo P = 1 reaches the hinge D, its impact through the element SD by the amount M 1 I And Q 1 I will stop (Fig. 3.20 and).
Lines of influence R E, M 1 V And Q 1 V are similar in construction to influence lines, respectively R A , M 1 And Q 1, since the element DEFG is also basic. Only by the amount R E, M 1 V And Q 1 V in addition to the secondary element SD the second minor element influences G.H.(Fig. 3.20 h, i, k).
Line of influence M V similar in construction of the line of influence M 1 I, and the line of influence M 1 V - respectively, the line of influence M 1 II(Fig. 3.20 l, m).
The correctness of the construction of influence lines can be checked statically. To do this, placing the load P = 1 in arbitrarily selected sections on the beam, it is necessary to compile and solve the corresponding static equations (according to the method discussed in section 3.3.3).
Rice. 3.20. Construction of lines of influence of support reactions, bending moments and shear forces for a multi-span beam in sections I, II, III, IV, V and VI
Let's consider the construction of influence lines in a multi-span beam using a specific example (Fig. 11 A).
It is more convenient to construct the line of influence of reactions of supports, bending moments and shear forces in any section in a multi-span statically determinate beam using its floor diagram, which gives a visual representation of the interaction of spans (Fig. 11 b).
Rice. 11. Lines of influence in a multi-span beam
Suspended beams B.C. (beam insert) and KLT relative to the main two beams AB And CDEK are transferable and experience load only when it acts directly on these beams.
When a single load moves along a suspended beam KLT , the resulting support reaction Rk will exert pressure on the beam CDEK , changing in particular the support reactions R B And R E . As soon as the unit load reaches
supports L , support reaction R L = 1, and the support reaction R K = 0, and, therefore, the pressure on the beam CDEK will be missing ( R B = 0, R E = 0).
When a single load moves along the main beam CDEK the latter no pressure on the hanging beams KLT And B.C. does not provide.
Using similar reasoning, we can formulate the basic principles for constructing influence lines in a multi-span beam:
1. For a multi-span beam, we build a floor diagram.
2. For an elementary beam in which a section is specified, we construct influence lines using Fig. 10.
3. Lines of influence are added only to the beams located above them according to the following rules:
Under the connecting hinges, the lines of influence always have a fracture;
Under the next support of the above beams, all influence lines have zero ordinates;
Within each overlying beam, the influence lines are straight.
The ordinates of the line of influence on the supports of the secondary beams (hinges) are determined from the relationships of similar sides of similar triangles.
For the beam shown in Fig. 11, we will construct lines of influence of the support reaction R E and lines of influence of bending moments and shear forces in sections 1 And 2 .
Line of influence of the support reaction R E
Support R E belongs to the beam CDEK - This is a two-support beam with hanging consoles. According to Fig. 8 V put the unit under the support E , connect to zero on the support D and extend left and right by the amount of cantilever overhangs. Ordinates of the line of influence in sections C And K beams CDEK determine from the ratios of the sides of similar triangles. We complete the line of influence on the above beams B.C. And KLT . We connect the ordinate of the line of influence in the section C with zero in the hinge B , and the ordinate of the line of influence in the section K with zero on support L and extend to the right by the amount of cantilever overhang LT . The ordinate of the line of influence in the section T determine from the ratios of the sides of similar triangles.
