Guidelines for the implementation of settlement and graphic work for students of specialties 2903, 2906,2907, 2908, 2910
Kazan, 2006
Compiled by: R.A. Kayumov
UDC 539.3
Calculation of a statically indeterminate rod system containing an absolutely rigid element; Guidelines for the implementation of settlement and graphic work for students of specialties 2903, 2906, 2907, 2908, 2910 / KazGASU; comp. R.A. Kayumov. Kazan, 2005, 24 p.
These guidelines briefly outline the methodology for calculating the simplest truss structures with a rigid element and provide an example of calculation.
Fig.6.
Reviewer Candidate of Physics and Mathematics sciences, prof. Department of Theoretical Mechanics, KSUAE Shigabutdinov F.G.
ã Kazan State University of Architecture and Civil Engineering
TASK #3
CALCULATION OF A STATICALLY UNDETERMINATE HINGED-ROD SYSTEM
For a given hinge-rod system (see diagram), consisting of an absolutely rigid beam and elastic rods with given cross-sectional area ratios, it is required:
1. Set the degree of static indeterminacy.
2. Find the forces in the rods.
3. Write down the strength conditions for the rods from force effects and select the cross sections of the rods, taking into account the given area ratios. Material St-3, yield strength taken equal to 240 MPa = 24 kN/cm 2 , safety factor k = 1.5.
4. Find the stresses in the rods from the inaccuracy of the manufacturing of the rods d 1 = d 2 = d 3 = (see Table 3). If it has a plus sign, then the rod is made longer; if minus - shorter.
5.
Find the stresses in the rods from the temperature change in the rods by Dt° (see Table 3). Linear expansion coefficient for steel 
1/deg.
6. Check the strength of the system under various options for force and non-force effects: 1) the structure is assembled, not yet loaded, but a temperature difference has occurred; 2) the case when there is no temperature difference, and the structure is assembled and loaded. 3) the case when the structure is assembled, loaded and there is a temperature difference.
7. Determine the ultimate load capacity of the system and the true safety factor by assuming a constant ratio between and .
The task is carried out in full by students of the specialties PGS and AD. Students of other specialties perform the calculation of the system only for external loading according to the allowable stresses and allowable load, excluding rod 3.
The initial data for performing settlement and graphic work are selected according to the code issued by the teacher.
Schemes for task number 3
table 3
| BUT | B | IN | G | B | in | IN | |||||
| , kN | , kN/m | , m | , m | , m | , m | , m | , mm | ||||
| 0.3 | 3/2 | ||||||||||
| -30 | -0.4 | 1/2 | |||||||||
| 0.5 | 3/2 | ||||||||||
| -25 | -0.6 | 3/4 | 3/2 | ||||||||
| 0.7 | 5/4 | 1/2 | |||||||||
| -35 | -0.4 | 1/2 | 4/5 | ||||||||
| 0.5 | 2/3 | 1/2 | |||||||||
| -0.7 | 1/2 | 4/5 | |||||||||
| -20 | -0.3 | 3/2 | 2/3 | ||||||||
| 0.6 | 2/3 | 5/4 |
FORMULATION OF THE PROBLEM
A hinge-rod system (Fig. 1) is considered, consisting of a rigid beam and deformable rods made with a given ratio of cross-sectional areas, which is indicated in the task. Known Design Loads F , q ; construction dimensions h 1 , h 2 , L 1 , L 2 , L 3; design temperature fluctuations: D t 1 - in the first rod, D t 2 - in the second, D t 3 - in the third; inaccuracies in the manufacture of rods, namely d 1 - difference from the design length in the first bar, d 2 - in the second, d 3 - in the third. The mechanical characteristics of the material are known: modulus of elasticity E \u003d 2 × 10 4 kN / cm 2, yield strength s t\u003d 24 kN / cm 2, coefficient of thermal expansion a=125×10 -7 1/deg. safety factor k for this design is taken equal to 1.5.
 |
It is necessary to solve 3 tasks:
1. Select the sections of the rods for the manufacture of this system from the condition of the strength of these rods in terms of allowable stresses at design loads.
2. Make a conclusion about the admissibility of design temperature fluctuations and inaccuracies in the manufacture of rods.
3. Find the maximum load capacity of the structure, allowable loads and the true margin of safety.
Thus, the work consists of design calculation, verification calculation, calculation of limit loads for the system.
The RGR should contain 3 drawings (made to scale): the original diagram of the rod system, the power diagram and kinematic scheme structural deformation.
2. Method of sections.
3. Hooke's law.
4. Elongation from temperature change.
5. Tensile strength, allowable stress, strength condition.
6. Plastic flow, yield strength.
7. Static indefinability.
8. Condition of compatibility of deformations.
9. Calculation of allowable stresses.
10. Calculation according to the theory of limit equilibrium.
GENERAL DESIGN CALCULATION PLAN
First, the structure is freed from bonds, replacing them with reactions. The method of sections introduces into consideration the internal longitudinal forces (normal forces) arising in the rods. In this case, they need to be directed from the section, i.e. conditionally consider the rods to be stretched. It is not possible to determine the reactions and longitudinal forces from the equilibrium equations, because in a plane problem of statics, it is possible to compose 3 independent equilibrium equations, while the number of unknown force factors (reactions and longitudinal forces) is more than three. Therefore, it is necessary to compose additional equations arising from the assumption of the deformability of the rods (equations of the compatibility of deformations that relate the elongations of the rods to each other). They follow from geometric considerations. In this case, the assumption of smallness of deformations is used. In addition, the following rule of signs must be taken into account. The total difference between the design length of the rod l
and final true length l con denoted by D l
. Therefore, if the rod lengthens, then 
, if shortened, then 
.
As can be seen from Fig. 2, the change in the length of the rod D l made up of extension D l (N) , caused by the force of axial tension N , elongation D l(t) caused by temperature change, and manufacturing inaccuracies d.
 |
If the temperature drops, then D t < 0, то длина стержня уменьшается, т.е. ; если стержень сделан короче проектного, то d< 0. С учетом закона Гука это соотношение примет вид:
Since elongations are expressed in terms of longitudinal forces 
according to formulas (1), then from the compatibility equations follow the relationships that connect the desired efforts. Here and below, to simplify the notation, the following designations are used: longitudinal force and stress in the rod with the number i
.
In the considered RGR, it is not required to search for reactions. Therefore, from the 3 equations of equilibrium, it is enough to leave one - the condition of equality to zero of the moments of all external and internal forces relative to the axis passing through the center of the hinge D (Fig. 1). The solution of the resulting system (equations of equilibrium and compatibility of deformations) makes it possible to find the forces in rods.
Further, design (task 1) and verification (task 2) calculations are carried out using the allowable stress method. The yield stress is taken as dangerous stress s t. According to the allowable stress method, the design considered out of order if the voltage has reached a dangerous value in at least one rod, i.e. turned out to be destroyed at least one from rods:
To ensure the safety of the structure, a safety margin is required, i.e. must be carried out strength condition kind
, (3)
where k - safety factor, [ s] - allowable voltage.
The destruction of one structural element does not always mean the loss of its operational properties (i.e. collapse). Other elements can take over the load, or part of it, that the destroyed element was supposed to carry. This consideration is used in Problem 3, which is solved limit equilibrium method, also called permissible load method.
In the formulation of the problem, it is assumed that the forces R And Q increase proportionally ( R / Q = const), the cross-sectional areas of the rods are known from the solution of problem 1, the material of the rods is elastic-ideal-plastic. With an increase in the load, one rod will first “flow”, the stress in it will not increase with further deformation and will remain equal in modulus to the yield strength s t(see fig. 3). The subsequent increase in loads will lead to the fact that, first, in the second, and then in the third rods, plastic flow will begin, i.e. stress has reached the yield point. Obviously, no matter what the installation or temperature stresses were at the beginning of the process, the moment finally comes when the stresses reach the yield strength in all rods (because they cannot take large values, according to the deformation diagram in Fig. 3). Achieved force values F = F etc And Q = Q etc are called limiting, because their increase is impossible, and the system will begin to deform indefinitely. Since efforts N i in the limiting state are known (because they are expressed in terms of stresses), then from the equilibrium equation is determined F etc. From the loading safety condition, the allowable loads are found
 |
As can be seen from the reasoning in solving problem 3, the presence of temperature changes or inaccuracies in the manufacture of rods does not reduce the load-bearing capacity of the structure if the rods are made of an elastic-ideal-plastic material.
NOTES
1. The teacher can specify the task of selecting rods by requiring the use of a rolled steel assortment, for example, to select a composite section from angles according to assortment tables (see calculation example).
2. When calculating, it is enough to leave 3 significant figures.
3. When selecting the dimensions of the rods, 5% overload is allowed.
Calculation example
Let a hinge-rod system be given (Fig. 4). It is known that
E \u003d 2 × 10 4 kN / cm 2, s t \u003d 24 kN / cm 2, a \u003d 125 × 10 -7 1 / deg. (5)
Rod systems in which support reactions and internal force factors cannot be found from equilibrium equations alone are called statically indeterminate.
The difference between the number of required unknown forces and independent equilibrium equations determines degree of static indeterminacy of the system. The degree of static indeterminacy is always equal to the number of redundant (superfluous) connections, the removal of which turns a statically indeterminate system into a statically determinable geometrically invariable system. Both external (reference) connections and internal ones, which impose certain restrictions on the movement of the system sections relative to each other, can be redundant.
Geometrically invariable such a system is called, the change of the form of which is possible only in connection with the deformations of its elements.
Geometrically variable such a system is called, the elements of which can move under the action of external forces without deformation (mechanism).
Shown in fig. 12.1 the frame has seven external (support) links. To determine the forces in these bonds (support reactions), it is possible to compose only three independent equilibrium equations. Therefore, this system has four redundant connections, which means that it is four times statically indeterminate. Thus, the degree of static uncertainty for flat frames is:
where R- number of support reactions.
A contour consisting of a number of elements (straight or curvilinear) rigidly (without hinges) connected to each other and forming a closed circuit is called a closed circuit. . The rectangular frame shown in Figure 12.2 is a closed loop. It is three times statically indeterminate, since to make it statically indeterminate it is necessary to cut one of its elements and eliminate three extra connections. The reactions of these bonds are: longitudinal force, transverse force and bending moment acting at the cut point; they cannot be determined using the equations of statics. In similar conditions, in the sense of static indeterminacy, there is any closed loop, which is always thrice statically indeterminate.
Inclusion of a hinge in a frame node in which two rods converge, or placing it in any place on the rod axis, removes one connection and reduces the overall degree of static indeterminacy by one. Such a hinge is called a single or simple (Fig. 12.3).
In general, each hinge included in a node connecting c rods, reduces the degree of static uncertainty by c-1 , since such a hinge replaces c-1 single hinges (Fig. 12.3). Thus, the degree of static indeterminacy of the system in the presence of closed loops is determined by the formula.
Statically indeterminate systems are called rod systems, for determining the reactions of supports in which only equilibrium equations are not enough. From a kinematic point of view, these are such rod systems, the number of degrees of freedom of which is less than the number of bonds. To reveal the static indeterminacy of such systems, it is necessary to compose additional deformation compatibility equations. The number of such equations is determined by the number of static indeterminacy of the rod system. Figure 8.14 shows examples of statically indeterminate beams and frames.
The beam shown in Fig. 8.14b is called continuous beam. This name comes from the fact that the intermediate support only supports the beam. At the point of support, the beam is not cut by a hinge, the hinge is not cut into the beam body. Therefore, the influence of stresses and deformations that the beam experiences on the left span also affects the right span. If a hinge is cut into the beam body at the place of the intermediate support, then as a result the system will become statically determinate - from one beam we will get two beams independent of each other, each of which will be statically determinate. It should be noted that continuous beams are less material-intensive compared to split beams, since they more rationally distribute bending moments along their length. In this regard, continuous beams are widely used in construction and engineering. However, continuous beams, being statically indeterminate, require a special calculation technique, which includes the use of system deformations.
Before proceeding to the calculation of statically indeterminate systems, it is necessary to learn how to determine the degree of their static indeterminacy. One of the simplest rules for determining the degree of static indeterminacy is the following:
,
(8.3)
where 
number of bonds imposed on the structure; 
the number of possible independent equilibrium equations that can be compiled for the system under consideration.
We use equation (8.3) to determine the degree of static indeterminacy of the systems shown in Figure 8.14.
The beam shown in Figure 8.14a is once statically indeterminate, as it has three ties on the left leg and one tie on the right leg. There are only three independent equilibrium equations for such a beam. Thus, the degree of static indeterminacy of the beam 
. The continuous beam shown in Fig. 8.14b is also once statically indeterminate, since it has two bonds on the left support and one connection each on the intermediate support and on the right support - a total of four connections. Thus, the degree of its static indeterminacy 
.
The frame shown in fig. 8.14c, is three times statically indeterminate, since it has six bonds in the supports. There are only three independent equilibrium equations for this frame. Thus, the degree of static uncertainty for this frame from equation (8.3) is: 
. The degree of static indeterminacy of the frame shown in Fig. 8.18, d is equal to four, since the frame has seven connections on the supports. Therefore, the degree of its static indeterminacy is equal to 
.
Rule (8.3) for determining the degree of static indeterminacy is used only for simple systems. In more complex cases, this rule does not work. Figure 8.15 shows a frame whose degree of static uncertainty cannot be determined using Equation (8.3).
Externally, the system shown in Figure 8.15 is five times statically indeterminate. This is easy to establish using equation (8.3): from six external bonds (three in section A, three in section B and two in section C), three possible equilibrium equations are subtracted. However, this system also has an internal static indeterminacy. It is impossible to take into account the internal static indeterminacy using equation (8.3). Before moving on to determining the degree of static indeterminacy of the frame shown in Figure 8.15, we introduce several definitions. The first of these definitions includes the notion of a simple hinge.
Simple called a hinge connecting two rods (Fig. 8.16).
Fig.8.16. simple hinge
A hinge connecting several rods is called difficult(Fig.8.17).
Fig.8.17. complex hinge
The number of simple hinges that can replace one complex hinge is determined from the formula:
,
(8.4)
where 
- the number of rods included in the node.
We recalculate the complex hinge shown in Fig. 8.17 into the number of simple hinges using formula (8.4): 
. Thus, the complex hinge shown in Figure 8.17 can be replaced by four simple hinges.
Let's introduce one more concept - closed loop.
Let's prove the theorem: any closed loop is three times statically indeterminate.
To prove the theorem, consider a closed circuit loaded with external forces (Fig.8.18).
Let us cut a closed contour with a vertical section and show the internal force factors that arise at the site of the section. Three internal factors arise in each of the sections: shear force 
, bending moment 
and longitudinal force 
. In total, each of the cut-off parts of the contour, in addition to external forces, is affected by six internal factors (Fig. 8.18, b, c). Considering the balance of one of the cut-off parts, for example, the left one (Fig. 8.18, b), we find out that the problem is three times statically indeterminate, since only three independent equilibrium equations can be compiled for the cut-off part, and there are six unknown forces acting on the cut-off part . Thus, the degree of static indeterminacy of a closed loop is equal to 
. The theorem has been proven.
Now, using the concept of a simple hinge and a closed loop, we can formulate another rule for determining the degree of static indeterminacy:
,
(8.5)
where 
number of closed loops; 
number of hinges in terms of simple ones (8.4).
Using equation (8.5), we determine the degree of static indeterminacy of the frame shown in Figure 8.15. The frame has five contours 
, including the contour formed by the support rods. The hinge at node D is simple, as it connects two rods. The hinge in section K is complex, as it connects four rods. The number of simple hinges that could replace the hinge in section K is equal to the formula (8.4): 
. Hinge C is also complex as it connects three rods. For this hinge 
. In addition, the system has two more simple hinges, with which it is attached to the base. Thus, the number of simple hinges in the system is 
. Substituting the number of closed loops 
and the number of simple hinges 
in formula (8.5) we determine the degree of static indeterminacy of the frame: 
. Thus, shown in Fig. 8.15 frame, seven times statically indeterminate. And this means that in order to calculate such a system, it is necessary to compose, in addition to the three equilibrium equations, seven equations of compatibility of deformations. By solving the system of 10 equations obtained in this way with respect to the unknowns included in these equations, it is possible to determine both the magnitude of the reactions in external bonds and the internal forces arising in the frame. The procedure for solving this problem can be somewhat simplified by excluding the equilibrium equations from the system of equations. However, this approach requires the use of special solution methods, one of which is the force method.
A task. Determine the stress in steel bars supporting an absolutely rigid beam. Material - steel St3, α=60°, [σ]=160MPa.
- We draw the scheme to scale. We number the rods.
In a hinged-fixed support BUT reactions occur R A And ON THE . In rods 1 And 2 efforts arise N 1 And N 2 . Applicable . Cut out with a closed cut middle part of the system. We will show a rigid beam schematically - by a line, efforts N 1 And N 2 send from section.
Compiling equilibrium equations
Number of unknowns exceeds number of equations of statics per 1 . Hence, the system , and for its solution it is required one additional equation. To compose additional equation to consider system deformation scheme. Hinged-fixed support BUT stays in place and rods deform under the action of force.
Scheme of deformations
According to the deformation scheme, we will compose deformation compatibility condition from consideration of the similarity of triangles ACC 1 And ABB 1 . From the similarity of triangles ABB 1 And ACC 1 write the ratio:
, where BB 1=∆ ℓ
1
(extension of the first rod)
Now we express SS 1 through deformation second rod. Let's enlarge a fragment of the scheme.
It can be seen from the figure that SS 2 = SS one · cos(90º- α )= SS one · sinα.
But SS 2 = ∆ ℓ 2 , then Δ ℓ 2 = SS one · sinα , where:
Let's turn deformation compatibility condition(4) in deformation compatibility equation via . In doing so, we must take into account character of deformations(shortening is written with a “-” sign, lengthening with a “+” sign).
Then it will be:
We shorten both parts by E , substitute numerical values and express N 1 across N 2
Substitute the ratio (6) into the equation (3) from where we find:
N 1 = 7.12kN (stretched),
N 2 = -20.35kN (compressed).
Let's define voltage in rods.
Calculation of a beam with a gap. For a statically indeterminate steel stepped beam, construct diagrams of longitudinal forces, normal stresses, and displacements. Check the strength of the beam. Before loading, there was a gap Δ=0.1 mm between the upper end and the support. Material - steel St 3, modulus of longitudinal elasticity E=2·10 5 MPa, allowable stress [σ]=160 MPa.
- After loading the gap will close And reactions arise and at the bottom, and in top support. Let's show them arbitrarily, these are reactions R A And R B . Let's compose equation of statics.
∑at=0 R A- F 1 + F 2 - R B=0
In the equation 2 unknowns, and the equation one, so the task 1 once statically indeterminate, and its solution requires 1 additional equation.
This deformation compatibility equation. In this case, the compatibility of deformations of the beam sections is that the change in the length of the beam (elongation) cannot exceed the gap, i.e. Δ ℓ =Δ , this deformation compatibility condition.
- Now we will divide the beam into sections and draw sections on them - their 4 in count characteristic plots. Each section is considered separately, moving in one direction- from the bottom support up. In each section we express the force N across unknown reaction. Directing N from section.
We write out separately the values longitudinal forces in sections:
N 1 = -R A
N 2 = 120 -R A
N 3 = 120 -R A
N 4 = 30-R A
3. Back to compiling deformation compatibility conditions. We have 4 area, which means
Δ ℓ 1 + ∆ ℓ 2+∆ ℓ 3+∆ ℓ 4 = Δ (gap size).
Using the formula for definition of absolute strain 
compose deformation compatibility equation, is exactly that additional equation that is needed to solve the problem.
Let's try simplify the equation. Remember that the size of the gap Δ=0.1 mm = 0.1 10 -3 m
E- elastic modulus, E\u003d 2 10 5 MPa \u003d 2 10 8 kPa.
We substitute instead N their values, written through the support reaction R A .
4. Calculate N and build longitudinal force diagram.
N 1 =-R A =-47.5kN
N 2 =120 -R A = 72.5kN
N 3 =120 -R A = 72.5kN
N 4 =30-R A =-17.5kN.
5. Define normal stresses σ according to the formula and build their diagrams
We are building diagram normal stresses.
Checking strength.
σ max= 90.63 MPa< [σ]=160МПа.
Strength guaranteed.
- Calculate displacement, using the formula for deformations.
Let's go from the wall BUT to the gap.
Got the value ω 4 equal to the gap, this is a check of the correctness of the definition of displacements.
We are building displacement diagram.
A longitudinal force P and its own weight (γ = 78 kN / m 3) act on the steel rod. Find the displacement of the section 1 –1.
Given: E \u003d 2 10 5 MPa, A \u003d 11 cm 2, a \u003d 3.0 m, b \u003d 3.0 m, c \u003d 1.3 m, P \u003d 2 kN.
Section displacement 1–1 will be made up of displacement from the action of the force R, from the action of its own weight above section and from the action of its own weight below section. moving from the action of the force R will be equal to the elongation of the section of the rod length b+a located above section 1–1. Load P causes elongation only area a, since it only has longitudinal force from this load. According to Hooke's law the elongation from the action of the force P will be equal to: Define elongation from the own weight of the rod below the section 1–1.
Let's denote it as . It will be called own weight of the plot with And the weight of the rod in the section a + b
Let's define elongation from the own weight of the rod above the section 1–1.
Let's denote it as It will be called own weight of section a+b
Then full displacement of section 1-1:
Those, section 1-1 will drop by 0.022 mm.
An absolutely rigid beam rests on a pivotally fixed support and is attached to two rods with the help of hinges. It is required: 1) to find the forces and stresses in the rods, expressing them in terms of the force Q; 2) Find the allowable load Q add by equating the greater of the stresses in the two rods to the allowable stress 
; 3) find the ultimate load capacity of the system if the yield strength 
4) compare both values obtained in the calculation of allowable stresses and ultimate loads. Dimensions: a=2.1 m, b=3.0 m, c=1.8 m, cross-sectional area A=20 cm 2
This system once statically indeterminate. For disclosure of static indeterminacy it is necessary to solve jointly the equilibrium equation and the equation of compatibility of rod deformations.
(1) -equilibrium equation
Let's compose deformation scheme- see fig. Then from the schema: 
(2)
By Hooke's law we have:
Rod lengths:
Then we get:
Substitute the resulting relation into the equation (1):
We define voltage in rods:
In the limit state: We substitute the obtained relations into the equation (1):
When compared, we see an increase in the load: 
A column consisting of a steel rod and a copper pipe is compressed by a force P. The length of the column is ℓ. Express the forces and stresses that occur in a steel rod and a copper pipe. 
Let us draw a section 1 - 1 and consider the equilibrium of the cut-off part 
Let's compose static equation: N C + N M - P= 0 , N C + N M = P (1)
The problem is statically indeterminate. Deformation compatibility equation write from the condition that elongations of steel rod and copper pipe are the same:
(2) or
Let us cancel both parts by the length of the rod and express force in a copper pipe through force in a steel rod:
(3)
Substitute the found value into the equation (1), we get:
Always working together the element made of a material with a high modulus of elasticity is stressed more strongly. At E C \u003d 2 10 5 MPa, E M \u003d 1 10 5 MPa:
For the column, determine the stresses in all sections. After applying the force P, the gap closes, P = 200 kN, E = 2. 10 5 MPa, A \u003d 25 cm 2 
After applying the force P, there will be pinching efforts. Let's call them C and B.
Let's compose static equation: ∑y = 0; C + B - P \u003d 0; (one)
Additional deformation compatibility equation: ∆ℓ 1 +∆ℓ 2 =0.3 mm (2);
To find absolute deformation, you need to know longitudinal force Location on. On the first section, the longitudinal force is equal to FROM, on the second differences (S-R). Let us substitute these values into the expressions for absolute deformations: 
(3)
We substitute the expression (3 ) into expression ( 2) and find: C = 150 kN, and from (1) B = 50 kN .
Then voltage in areas:
A rigid beam is suspended on three steel rods; rod 2 is made shorter than the design one. Determine the stresses in the rods after the assembly of the system. Given:
After completion of the assembly in this system, the rigid beam will turn and take new position.
points C, D And TO move to positions С 1 , D 1 And K 1
According to the deformation pattern SS 1 =Δℓ 1, DD 1 =Δ−D 1 D 2 = Δ−Δℓ 2, KK 1 \u003d ℓ 3, while the rods 1 and 3 experiencing compression, and the rod 2 – stretching.
According to the deformation scheme equilibrium equation will take the form: 
Additional equations can be obtained based on analysis of the deformation scheme; from similar triangles VSS 1 And BDD 1, triangles VSS 1 And BKK 1 follows:
According to Hooke's law absolute deformations: 
Then the additional equations will be written as follows: 
Solving together this system of obtained additional equations and the equilibrium equation, we obtain:
N 1 \u003d 14.3 kN (the rod is compressed), N 2 \u003d 71.5 kN (the rod is stretched), N 3 \u003d 42.9 kN (the rod is compressed).
Thus, the desired stresses in the rods have meanings: 
Problem solved.
The stepped copper rod is heated from temperature t H =20ºС to t К =50ºС. Check the strength of the rod. Given:
Let's compose rod equilibrium equation assuming the replacement of external links by reactive forces: 
As you can see, the system is statically indeterminate, and an additional equation is required to solve it.
The strain compatibility equation follows from the condition that the displacements of external links are equal to 0 - W B =0 or W K =0. In this way: 
Where:
As a result R B \u003d 20723N.
Normal Forces and Stresses in areas: 
According to the results of calculations σ max =│69.1│MPa, wherein σmax< σ adm , (69,1<80). Consequently, the strength condition of the rod is satisfied.
Calculation of a bar with a gap. For a steel stepped bar with a gap between the lower end and the support, it is required: to construct diagrams of normal forces and stresses, displacements; check the strength. Given:
Let's compose equilibrium equation rod:
In him two unknown, system once statically indeterminate,required the additional equation is the strain equation.
An additional equation can be written from the condition of closing the gap in the process of deformation of the rod:
For the areas under consideration absolute deformations:
Let's define normal (longitudinal) forces, go from the wall to the gap: 
Substitute all found values into additional equation:
After substituting the initial data and abbreviations: 
From equilibrium equations we get:
In this way, R B \u003d 40.74 kN, R K \u003d 9.26 kN.
Payment normal forces: 
We are building plot N
Payment normal stresses: 
We are building normal stress diagram
Payment movements characteristic sections.
The rule of signs for displacements is adopted: down - positive, up - negative. 
We are building movement diagram.
A statically indeterminate rod system is given (part BCD is rigid). It is required to select the cross-sectional areas of bars 1 and 2. 
Denote efforts in rods 1 and 2, respectively N 1 and N 2.
Let us show the scheme of the system with efforts N 1 and N 2 
Compose for this system balance Equation, excluding from consideration the reactive forces in support C This equation contains two unknowns: N 1 and N 2. Therefore, the system once statically indeterminate, and for its solution it is required additional equation. This strain equation. Let's show the system in deformable state under load :
From analysis of the system in a deformable state follows that:
Since , and given that we can write: 
The last entry is the necessary additional deformation equation.
Let us write down the values of the absolute deformations of the rods: 
Then, taking into account the initial data additional equation will take the form: 
Pay attention to equilibrium equation, we get the system: 
From the solution of this system of equations follows:
N 1 \u003d 48kN (rod stretched), N 2 \u003d -36.31kN (rod compressed).
According to strength condition of the rod 1:
then, taking into account the condition A 1 \u003d 1.5A 2 given the assignment, we get 
According to strength condition of the rod 2:Then 
We finally accept:
In order for the rod systems (beams, frames, etc.) to serve as structures and withstand external loads, it is necessary to impose certain bonds on them, which divide them into external and internal bonds. A connection is usually understood as bodies (obstacles) that restrict the movement of other bodies, points or sections of a structure. In practice, such bodies are called supporting devices, foundations, etc. In engineering calculations, the concept of ideal connections is introduced. If, for example, a condition is imposed on the left end of the beam (Fig. 1.1, a), which prohibits vertical movement, then they say that there is one external connection at this point. Conventionally, it is depicted as a rod with two hinges. If vertical and horizontal displacements are prohibited, then two external links are imposed on the system (Fig. 1.1, b). Embedding in a flat system gives three external connections (Fig. 1.1, c), which prevent vertical, horizontal displacements and rotation of the embedding section. ld Fig. 1.1 In order to fix the body (rod) on the plane and ensure its geometric invariability, it is necessary and sufficient to impose three bonds on it (Fig. 1.2), and all three bonds should not be mutually parallel and should not intersect at one point. In what follows, the connections that ensure the geometric immutability of the system and its static definability will be understood as necessary connections. A geometrically invariable system is a system that can change its shape only due to the deformation of its elements (Fig. 1.2), while a geometrically variable system can allow movement even in the absence of deformation (Fig. 1.3). Such a system is a mechanism (Fig. 1.3, a). 5 Fig. 1.2 Along with the noted, instantaneous systems are also distinguished, which are understood as systems that allow infinitesimal displacements without deformation of its elements (Fig. 1.4). Rice. 1.3 So, for example, under the action of a force P applied in the hinge D (Fig. 1.4, a), the rods DV and DS without deformation will rotate relative to the hinges B and C through an infinitely small angle d. Then, from the equilibrium condition cut out at a small value of the force P, the forces in the rods of the DW and DS will tend to infinity, causing axial deformation of the rods and changing the position of the system. 6 Fig. 1.4 For the frame in fig. 1.4, b, when considering the equation of statics, the moment of force P is not balanced (the reaction R1, cannot cause a moment relative to the point under consideration, since the line of its action passes through this point). A similar feature is also manifested for the system shown in Fig. 1.4, c. The moment of force P relative to the point k is not balanced. Thus, these systems also allow infinitesimal displacements (relative to the moment point) without deformation of their elements. In buildings and structures, such systems are unacceptable. If a geometrically invariable system has, in addition to the necessary additional connections, then the independent equations of statics are not enough to determine the unknown forces (reactions of the connections) and such a system is called statically indeterminate. The difference between the number of unknown forces to be determined and the number of independent equations of statics characterizes the degree of static indeterminacy, which is usually denoted by the symbol n. Thus, the beam and frame shown in Fig. 1.5 are two times (twice) statically indeterminate. In these schemes, the number of unknown reactions is five, and the number of independent static equations that can be written for each of them is three. Any closed circuit is a system three times statically indeterminate (Fig. 1.6). Rice. 1.6 Setting a single hinge reduces the degree of static indeterminacy of the system by one (Fig. 1.7, a), since there is no bending moment in the hinge. A single hinge is understood to mean a hinge connecting the ends of two rods. Rice. 1.7 A hinge included in a node where the ends of several rods meet reduces the degree of static indeterminacy of the system by the number of single hinges, determined by the formula O=C–1. Here, C is understood as the number of rods converging at a node. For example, in a frame (Fig. 1.7, b) the number of single hinges is O=C–1=3-1=2, so the degree of static uncertainty is reduced by two units and becomes equal to n4.Calculation of statically determinate frames
Basic concepts A frame is a rod system in which all or some of the nodal connections are rigid (Fig. 1.8 a). A rigid knot is characterized by the fact that the angle between the axes of the rods that form it does not change under the action of a load (Fig. 1.8 a). The angle between the tangents to the elastic lines of the crossbar and the inclined post at node B remains unchanged α, and the angle between the tangents to the elastic lines of the same crossbar and the right post at node D retains the same value β. Frames can be flat, when all the axes of the rods lie in the same plane (Fig. 1.8 a, b, c) and spatial (Fig. 1.8 d). The horizontal rod of the frame is called the crossbar, and the rods that support it are called the rack. The left stance is slanted and the right stance is vertical. Frames can be simple, consisting of three rods (Figure 1.8), complex, multi-span (Figure 1.8 b) and multi-tiered (Figure 1.8 c). They are also divided into statically determinate (Figure 1.8 b), when the number of unknown reactions, efforts is less than or equal to the number of independent static equations that can be compiled for a given frame, and statically indeterminate if this condition is not met (Figure 1.8 a, c, d) This will be discussed later. Unlike beams, in the cross sections of frames, along with bending moments, transverse force, there is also a longitudinal force. Rice. 1.8 Determination of forces (M, Q, N) are performed in the same way as in beams using the section method (ROSE). In this case, the sign rule for the bending moment M and the transverse force Q is the same as for beams, and for the longitudinal force N, as in 9 rods in tension - compression. The determination of normal n and shear stresses is carried out according to the same dependencies as in beams, if the rod is bent. In the case of complex resistance, when, along with the bending moment, a longitudinal force also arises in the rod, then the calculation is carried out as in case of bending with tension - compression, described in the "Complex resistance" section. Example 1.1 For a given frame (Fig. 1.9), plot diagrams internal forces and find the magnitude and direction of the total displacement of the section K, if P = 5 kN; q = 10 kN/m; EIz = const; sections of posts and crossbars are the same I = 8000 cm4: 1. Find the support reactions: a) vertical reactions V1, V2: b) horizontal reactions H1 and H2: 2. We build diagrams of internal forces M, Q, N. a. Construction of a diagram of bending moments M.Calculation of statically indeterminate bar systems by the force method
We select the observation point, assuming that it is inside the contour. In this case, the fields are located above sections 1-3, 3-4, 4-K, 4-2, are considered as external, and inside the contour - internal. When determining bending moments, we follow the same rules as in beams. We calculate the moments in the characteristic sections of each of the sections of the frame. Plot 1-3. The moment at the end from the side of the support is 1, M13 = 0. The moment at the node is 3, The sign is minus because in section 1-3 the lower cut-off part is bent upward with a convexity towards the observer. Plot 3-4 (crossbar). Moment at the beginning of the section (in the section of node 3) M34, the same as on the rack 1 - Moment In the hinge, the moment is zero. Section 2-4 (inclined post) Section 4-K At the beginning of the section, the moment MK4 = 0. At the end of the section, the curve of bending moments is shown in (Fig. 1.10, a) 1.10 We check the correctness of the construction of the diagram M. If the diagram M is built correctly, then any off-support node or any part of the frame under the action of external and internal forces must be in balance. Let us cut out from the frame sections infinitely close to the node, for example, node (4) and consider its equilibrium. We take the values of the moments in the corresponding sections from the diagram M (Fig. 1.10, b). The knot moment equations (4) have the formFeatures of the calculation by the method of forces of multi-span continuous beams
The condition is satisfied, which means that in the sections adjacent to the node (4) the moments are determined correctly. Similarly, a check is performed in the node (3), etc. Note If concentrated external forces (moment or forces) are applied in the node, then they must be taken into account when checking. The distributed load is not shown because dx is a small value. b. Construction of a diagram of transverse forces Q. We adhere to the same sign rule as for beams: if the resultant of external forces to the left of the section is directed upwards, and to the right downwards, the transverse force Q > 0, if vice versa - m Section 1–3. When considering the left cut-off part 10 kN. (minus because the left cut-off part is under the influence of force H1 12 directed downwards, if you look at the cut-off part from the observer's point). The transverse force is constant along the length of this section (Fig. 1.11, a) 1.11 Section 3-4 The shear force in any section, taken at a distance x from the node (3), when considering the forces acting from the section to the left, is equal to 103 01QV xqx. At x = 0, we get the transverse force in the section to the left of the node (3), i.e. Q34 30kN; at x = 3 m, we obtain the transverse force Q, i.e., in the section to the left of the node (4). The transverse force in section 3-4 changes according to a linear law (Fig. 1.11, a). Plot 4-K. In a section at a distance x from the right end of the section (Fig. 1.11, a), the transverse force is equal to (linear law). At x = 0, we get, and at x = 3 m, we get Section 2–4. We obtain the transverse force in the section of this section by projecting the external forces H2, V2 applied at point 2 (Fig. 1.11, a) on the Y axis, perpendicular to the longitudinal axis of the rod. Along the length of section 3–4, the transverse force is constant. The diagram of the transverse forces is shown in (Fig. 1.11, a).The use of symmetry properties in the disclosure of the static indeterminacy of rod systems
in. Construction of a diagram of longitudinal forces N. We calculate the longitudinal force in the section of each section. Plot 1–3. We consider the lower part (Fig. 1.12) The minus is taken because the longitudinal force balancing the reaction V1 is directed towards the section, i.e. towards the reaction V1, which means that the cut-off section is under compression. If the longitudinal force were directed away from the section, then the sign of N is positive. Plot 3-4 (on the crossbar). Longitudinal force N30 kN, negative, as compressive. In section x (Fig. 1.12, b) in section 4-K: perpendicular to the longitudinal axis of the section. Plot 2–4. Rice. 1.12 On an inclined post in section x, we find the longitudinal force by projecting external forces V2 and H2 onto the X axis, coinciding with the axis of the rod (Fig. 1.12): 34 5 4 (compression), Therefore, we assign a minus sign N24 kN. 14 The diagram of longitudinal forces is shown in (Fig. 1.11, b). 3. We determine the displacements of the section K. For this, we use the Mohr integral, the formulas of A.K. Vereshchagin, Simpson, (see section "Direct bending"). We determine the vertical displacement of section K. To do this, we release the frame from all external loads (q, P) and apply a single dimensionless force in this section (Fig. 1.13, a). Direction we accept the forces ourselves, for example, to the bottom.Calculation by the method of forces of statically indeterminate systems operating in tension or compression
Rice. 1.13 In fig. 1.13, a plot of the bending moments M1 from this force is presented. We multiply the diagrams M and M1 according to the Vereshchagin method, we find the vertical displacement of the section K. In the 4-K section, the Simpson formula was used, and in the 2-4 section, the Vereshchagin formula. We determine the horizontal displacement of the section K. To do this, we release the frame from external loads, load it with a single dimensionless force applied horizontally (Fig. 1.13, b). The plot of this force is shown in Fig. 1.13b. We calculate the horizontal displacement using the formulas of Vereshchagin and Simpson. The minus sign indicates that the actual horizontal displacement is directed in the opposite direction of the application of a unit force, i.e., to the left. 15 We find the total displacement of the section K as the geometric sum of the displacements found. The direction of full movement is determined by the angle (Figure 1.14, b). We determine the angle of rotation of the section K. We apply a single dimensionless moment in the section K (Fig. 1.14, a) and build a diagram of bending moments from it.Calculation of statically indeterminate bar systems by force method in matrix form
Rice. 1.14 We multiply the diagrams M and M3, using the Vereshchagin formula, we find the angle of rotation of the section K: 16 1.3. Calculation of statically indeterminate rod systems by the method of forces The most widely used method for revealing the static indeterminacy of rod systems is the method of forces. It lies in the fact that a given statically indeterminate system is freed from additional (extra) connections, both external and internal, and their action is replaced by forces and moments. Their value is further determined so that the displacements correspond to the restrictions that are imposed on the system by the discarded links. Thus, with the indicated method of solution, the forces or moments acting in the places of discarded or cut bonds are unknown. Hence the name "method of forces". Let us consider the essence of the force method using the example of calculating a statically indeterminate frame shown in Fig. 1.15. We assume that the external load, dimensions and stiffness of the rods are known. Calculation procedure 2.1. We set the degree of static indeterminacy, for which we use the expression, where X is the number of unknowns (there are 5 external links); Y is the number of independent static equations that can be compiled for the system under consideration. For a given frame, the number of unknown reactions is five, and the number of independent equations is three, since the system of forces is flat and arbitrarily located, therefore the System is twice statically indeterminate. 2.2. We transform the given system into a statically determinate, geometrically invariable and equivalent to the given system, i.e. we form the main system. To do this, we remove unnecessary connections by discarding or cutting them. On fig. 1.15 shows the main system obtained by discarding unnecessary support links, and in fig. 1.16 the main systems are formed by discarding and cutting links. For example, (Fig. 1.16, a) in support A, a horizontal connection is discarded and in support C, a connection is cut that prevents the rotation of the section. Thus, for each statically indeterminate rod system, one can 1.15 17 select several options for the main systems (Fig. 1.15, 1.16). It is necessary to pay special attention to the fact that in the formation of the main system of the method of forces, the introduction of new connections is unacceptable. It is desirable that the main system be rational, i.e., one for which it is easier to build diagrams of internal force factors and the amount of calculations is the smallest. Such a system is shown in Fig. 1.15 (option I). There is no need to determine the support reactions here if you build diagrams from the free (loose) end of the frame. Rice. 1.16 2.3. We form an equivalent system by loading the main system with external forces and the forces of discarded (cut) bonds (Fig. 1.17). Unknown force factors will be denoted by the symbol Xi, where i is the number of the unknown. If the rejected constraints prohibit linear displacements, then the unknowns are the forces, if the angular displacements are forbidden, the moments. If the main system was obtained by cutting the extra connections, then equal and opposite forces and moments are applied to both the right and left parts of the dissected system at the places of cutting. In the example under consideration, X1 and X2 represent the vertical and horizontal components of the reaction of the pivot support A. 2.4. We compose the canonical equations of the force method, which express in mathematical form the conditions for the equivalence of the main and given systems. Otherwise, they express conditions denoting that the relative displacements in the direction of remote superfluous links from the joint action of an external load and unknown forces must be equal to zero. For the equivalent system of the considered example, based on the principle of independence of the action of forces and fig. 1.18 the canonical equations will be written in the formwhere 11 is the relative displacement in the main system in the direction of the extra unknown X1, caused by the same force; 12 - relative movement in the direction of the extra unknown X1, caused by the force X2; 1P - relative displacement in the direction of action of the unknown X1, caused by a given load. Rice. 1.18 Physical meaning of these equations. The first equation denies the possibility of vertical movement of the support section A in the direction of the excess unknown X1 from the combined action of the given load P and the full values of the unknowns X1 and X2. The second equation has a similar meaning. In this form (1.1), the use of equations in engineering calculations is difficult, so we will transform them to a new form. Taking into account the fact that for linear systems the expression can be written: where 11 is the relative displacement in the main system in the direction of the force X1 from the action of the force X1 1 (Fig. 1.19); 21 - relative displacement in the main system in the direction of the force X2 from the force X1 1. Here X1 and X2 are the actual values of the reactions of the discarded bonds. Then the canonical equations of the force method (1.1) can be written in the form By analogy, for n times statically indeterminate systems, the canonical equations have the form The leading coefficients are always positive. Side factors can be positive, negative or zero. 1P - are called free or load coefficients. 2.5. We determine the coefficients of the canonical equations. These coefficients represent the displacements of the points of the system in the direction of the dropped links, therefore, they can be found using the Mohr integral: The procedure for determining the coefficients: Fig. 1.19 20 a) we plot bending moment diagrams for the main system from a given external load P and from unit forces of dropped bonds X11 (Fig. 1.20); Rice. 1.20 b) we calculate the coefficients of the canonical equations. Since the system under consideration consists only of rectilinear rods and the rigidity of the rods within their lengths are constant, then the calculation of the Mohr integral is carried out according to the method of A.K. Vereshchagin by multiplying the corresponding diagrams using Simpson's formulas and trapezoids: 2.6. We write down the system of canonical equations. After substituting the found coefficients into equation (1.3), we obtain: We solve the system of equations and find the unknown forces, kN: Note. If the force sign turned out to be negative, then this means that the actual force (reaction) is directed in the opposite direction than the force Xi adopted in the equivalent system. Thus, the static indefinability of the system is revealed. 2.7. We build the final (real) diagrams of internal force factors for a given system. Plotting can be done in two ways. The first way We load the main system with a given load and the found forces X1 and X2 (Fig. 1.17), after which we build diagrams M, Q, and N in the same way as for a conventional statically determinate system. The diagrams constructed in this way are shown in Fig. 1.21, where the ordinates of the bending moment diagram are plotted from the side of the stretched fibers. This method is most convenient for simple systems. The second way We calculate the values of bending moments in any (usually characteristic) section based on the principle of independence of the action of forces according to the formula 22 where k is the number of the section for which the value of the bending moment is determined; n is the degree of static indeterminacy of the system. Rice. 1.21 In this case, if the found force Xi has a negative sign, then the corresponding diagram Mi must be mirrored with respect to the axes of the rods. When determining the actual values of bending moments, the ordinates of the moments in the calculated sections are taken from the diagrams M1, M2 and MP, taking into account their signs. The signs of the moments in the section under consideration are determined depending on which side of the base line the ordinates of the moments are located and on the position of the observer's point. In our case, we assume that the observer's point is located inside the contour, therefore, the positive values of the moments are taken to be the moments that cause tension in the calculated section of the internal fibers, and the negative values of the external fibers of the contour. For example, for section D of the frame, we obtain Similarly for other sections. The final diagram of bending moments for a given system is shown in fig. 1.21 a. 23 2.8. We carry out a deformation check of the correctness of constructing a real diagram of bending moments. The meaning of the deformation test is to confirm the absence of displacements in the main system in the direction of the discarded (cut) bonds at the found values of the unknown forces. So, if the unknown forces are found correctly, then for the example under consideration the equalities must be satisfied: If we build a diagram of single moments 2, then the check is called a check for group displacement (Fig. 1.22): The absence of displacement confirms the correctness of the solution of the problem. If the performed calculations do not confirm the absence of displacements of the points of the main system in the direction of the discarded links, then in order to identify the calculation error, it is necessary to check the correctness of determining the coefficients of the canonical equations according to the formula If there is no equality in this equation, a line-by-line check of the coefficients of the canonical equations is performed. First line: . If there is no calculation error in this line, then the condition must be met: Similarly, you can check the 2nd and other lines. When performing these checks, you should check the correctness of the calculation of the load coefficients: 2.9. We build a diagram of transverse forces Q according to the diagram of bending moments M by sequentially cutting out the rods from a given system and considering them as hinged statically determinate beams. We apply moments at the ends of the rods, the values and directions of which are selected from the diagram M in the corresponding sections. In the presence of external forces, we apply them in the appropriate areas. We determine the support reactions from the condition of static equilibrium and plot Q as usual for statically determinate beams. For a given frame (Fig. 1.15), when constructing a diagram of transverse forces for a rack, we cut out section AB and in section B we apply a moment B 3, 56 M P taken from the diagram of real moments M (Fig. 1.21, b). We determine the support reactions from the consideration of the equilibrium 3 P and build a diagram of the transverse forces Q (Fig. 1.23). Rice. 1.22 25 In a similar way, we cut out the horizontal rod (crossbar) BC, consider its balance and plot Q for this section of the frame (Fig. 1.24). We transfer Q diagrams for individual rods to a given system. The final diagram of the transverse forces for a given frame is shown in Figure 7.14, b. The construction of a diagram of transverse forces according to the diagram of bending moments is also possible on the basis of a differential dependence: where α is the angle of inclination of the straight line outlining the diagram of bending moments to the base line (beam axis). The transverse force is considered positive if the bending moment increases in the direction of the axis. For the considered example: 2.10. We construct a diagram of longitudinal forces N.Truss with reservations include trussed beams, which are a combination of a two- or three-span continuous beam and spring traction; they are typical for steel and wooden structures, with an upper chord of a continuous rolled profile (sawn timber or glued board packages). There may also be reinforced concrete trusses of small spans.
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Rice. 7.16 Fig. 1.24 26 To do this, we use the method of cutting nodes (we cut out only off-support nodes with sections infinitely close to the node) and consider their equilibrium under the action of an external load (if any is applied to the nodes) and forces in discarded (cut) links. We cut out node B. We apply to it transverse forces taken in the corresponding sections from diagram Q (Fig. 1.23, b). The node must be in equilibrium (Fig. 1.25) under the action of transverse and longitudinal forces (unknown). We determine the unknown longitudinal forces from the condition of static equilibrium. The diagram of longitudinal forces is shown in fig. 1.23, c. 2.11. We carry out a final check of the correctness of the solution of the problem. The system (frame), an off-support unit or some part of the system must be in balance under the action of an external load and the forces of discarded (cut) links. For a given example, we consider the balance of the frame using the equations of statics (Fig. 1.26):
The equilibrium condition is satisfied. Notes. 1. If the frame has several off-support nodes, then all nodes are covered by the check.