Deformation during transverse bending of the beam is characterized. The concept of bending deformation
General concepts.
bending deformationconsists in the curvature of the axis of the straight rod or in changing the initial curvature of the straight rod(Fig. 6.1) . Let's get acquainted with the basic concepts that are used when considering bending deformation.
Bending rods are called beams.
clean called a bend, in which the bending moment is the only internal force factor that occurs in the cross section of the beam.
More often, in the cross section of the rod, along with the bending moment, a transverse force also occurs. Such a bend is called transverse.
flat (straight) called a bend when the plane of action of the bending moment in the cross section passes through one of the main central axes of the cross section.
With an oblique bend the plane of action of the bending moment intersects the cross section of the beam along a line that does not coincide with any of the main central axes of the cross section.
We begin the study of bending deformation with the case of pure plane bending.
Normal stresses and strains in pure bending.
As already mentioned, with pure flat bend in cross section of the six internal force factors are not zero only bending moment (Fig. 6.1, c):
; (6.1)
Experiments performed on elastic models show that if a grid of lines is applied to the surface of the model(Fig. 6.1, a) , then with pure bending it is deformed in the following way (Fig. 6.1, b):
a) longitudinal lines are curved along the circumference;
b) the contours of the cross sections remain flat;
c) the lines of the contours of the sections intersect everywhere with the longitudinal fibers at a right angle.
Based on this, it can be assumed that in pure bending, the cross sections of the beam remain flat and rotate so that they remain normal to the bent axis of the beam (flat section hypothesis in bending).
Rice. .
By measuring the length of the longitudinal lines (Fig. 6.1, b), it can be found that the upper fibers lengthen during the bending deformation of the beam, and the lower ones shorten. Obviously, it is possible to find such fibers, the length of which remains unchanged. The set of fibers that do not change their length when the beam is bent is calledneutral layer (n.s.). The neutral layer intersects the cross section of the beam in a straight line calledneutral line (n. l.) section.
To derive a formula that determines the magnitude of the normal stresses that arise in the cross section, consider the section of the beam in the deformed and non-deformed state (Fig. 6.2).
Rice. .
By two infinitesimal cross sections, we select an element of length. Before deformation, the sections bounding the element were parallel to each other (Fig. 6.2, a), and after deformation, they tilted somewhat, forming an angle. The length of the fibers lying in the neutral layer does not change during bending. Let us designate the radius of curvature of the trace of the neutral layer on the plane of the drawing by a letter. Let us determine the linear deformation of an arbitrary fiber spaced at a distance from the neutral layer.
The length of this fiber after deformation (arc length) is equal to. Considering that before deformation all fibers had the same length, we obtain that the absolute elongation of the considered fiber
Its relative deformation
Obviously, since the length of the fiber lying in the neutral layer has not changed. Then after substitution we get
(6.2)
Therefore, the relative longitudinal strain is proportional to the distance of the fiber from the neutral axis.
We introduce the assumption that the longitudinal fibers do not press each other during bending. Under this assumption, each fiber is deformed in isolation, experiencing a simple tension or compression, at which. Taking into account (6.2)
, (6.3)
i.e., normal stresses are directly proportional to the distances of the considered points of the section from the neutral axis.
We substitute dependence (6.3) into the expression for the bending moment in the cross section (6.1)
Recall that the integral is the moment of inertia of the section about the axis
Or
(6.4)
Dependence (6.4) is Hooke's law for bending, since it relates the deformation (curvature of the neutral layer) to the moment acting in the section. The product is called the bending stiffness of the section, N m 2.
Substitute (6.4) into (6.3)
(6.5)
This is the desired formula for determining the normal stresses in pure bending of the beam at any point in its section.
For In order to establish where the neutral line is in the cross section, we substitute the value of normal stresses in the expression for the longitudinal force and the bending moment
Insofar as,
then
(6.6)
(6.7)
Equality (6.6) indicates that the axis - the neutral axis of the section - passes through the center of gravity of the cross section.
Equality (6.7) shows that and are the main central axes of the section.
According to (6.5), the greatest stresses are reached in the fibers furthest from the neutral line
The ratio is the axial section modulus relative to its central axis, which means
The value for the simplest cross sections is as follows:
For rectangular cross section
, (6.8)
where is the section side perpendicular to the axis;
The side of the section is parallel to the axis;
For round cross section
, (6.9)
where is the diameter of the circular cross section.
Strength condition according to normal stresses in bending can be written in the form
(6.10)
All obtained formulas were obtained for the case pure bending straight rod. The action of the transverse force leads to the fact that the hypotheses underlying the conclusions lose their strength. However, calculation practice shows that transverse bend beams and frames, when in addition to the bending moment, a longitudinal force and a transverse force also act in the section, you can use the formulas given for pure bending. In this case, the error turns out to be insignificant.
Determination of transverse forces and bending moments.
As already mentioned, with a flat transverse bending in the cross section of the beam, two internal force factors u arise.
Before determining and determine the reactions of the beam supports (Fig. 6.3, a), compiling the equilibrium equations of statics.
To determine and apply the method of sections. In the place of interest to us, we will make a mental section of the beam, for example, at a distance from the left support. Let's discard one of the parts of the beam, for example, the right one, and consider the balance of the left side (Fig. 6.3, b). We will replace the interaction of the beam parts with internal forces and.
Let's install following rules signs for and:
- The transverse force in the section is positive if its vectors tend to rotate the considered section clockwise;
- The bending moment in the section is positive if it causes compression of the upper fibers.
Rice. .
To determine these forces, we use two equilibrium equations:
1. ; ; .
2. ;
Thus,
a) the transverse force in the cross section of the beam is numerically equal to the algebraic sum of the projections onto the transverse axis of the section of all external forces acting on one side of the section;
b) the bending moment in the cross section of the beam is numerically equal to the algebraic sum of the moments (calculated relative to the center of gravity of the section) of external forces acting on one side of the given section.
In practical calculations, they are usually guided by the following:
- If a external load tends to rotate the beam relative to the considered section clockwise, (Fig. 6.4, b) then in the expression for it gives a positive term.
- If an external load creates a moment relative to the considered section, causing compression of the upper fibers of the beam (Fig. 6.4, a), then in the expression for in this section it gives a positive term.
Rice. .
Construction of diagrams in beams.
Consider a double beam(Fig. 6.5, a) . A beam is acted upon at a point by a concentrated moment, at a point by a concentrated force, and at a section by a uniformly distributed load of intensity.
We define support reactions and(Fig. 6.5, b) . The resultant distributed load is equal, and its line of action passes through the center of the section. Let us compose the equations of the moments with respect to the points and.
Let's determine the transverse force and the bending moment in an arbitrary section located in a section at a distance from point A(Fig. 6.5, c) .
(Fig. 6.5, d). The distance can vary within ().
The value of the transverse force does not depend on the coordinate of the section, therefore, in all sections of the section, the transverse forces are the same and the diagram looks like a rectangle. Bending moment
The bending moment changes linearly. Let's determine the ordinates of the diagram for the boundaries of the plot.
Let us determine the transverse force and the bending moment in an arbitrary section located in a section at a distance from the point(Fig. 6.5, e). The distance can vary within ().
The transverse force changes linearly. Define for the boundaries of the site.
Bending moment
The diagram of bending moments in this section will be parabolic.
To determine the extreme value of the bending moment, we equate to zero the derivative of the bending moment along the abscissa of the section:
From here
For a section with a coordinate, the value of the bending moment will be
As a result, we get diagrams transverse forces (Fig. 6.5, e) and bending moments (Fig. 6.5, g).
Differential dependencies in bending.
(6.11)
(6.12)
(6.13)
These dependencies allow you to establish some features of the diagrams of bending moments and shear forces:
H in areas where there is no distributed load, the diagrams are limited to straight lines parallel to the zero line of the diagram, and the diagrams in the general case are inclined straight lines.
H in areas where a uniformly distributed load is applied to the beam, the diagram is limited to oblique straight lines, and the diagram is limited to quadratic parabolas with a bulge facing the side, opposite direction load action.
AT sections, where, the tangent to the diagram is parallel to the zero line of the diagram.
H and areas where, the moment increases; in areas where, the moment decreases.
AT sections where concentrated forces are applied to the beam, there will be jumps on the magnitude of the applied forces on the diagram, and fractures on the diagram.
In sections where concentrated moments are applied to the beam, there will be jumps in the diagram by the magnitude of these moments.
The ordinates of the diagram are proportional to the tangent of the slope of the tangent to the diagram.
Building a diagram Q.
Let's build a plot M method characteristic points. We arrange points on the beam - these are the points of the beginning and end of the beam ( D,A ), concentrated moment ( B ), and also note as a characteristic point the middle of a uniformly distributed load ( K ) is an additional point for constructing a parabolic curve.
Determine bending moments at points. Rule of signs cm. - .
The moment in AT will be defined as follows. First let's define:
Point To let's take in middle area with a uniformly distributed load.
Building a diagram M . Plot AB – parabolic curve(rule of "umbrella"), plot BD – straight oblique line.
For a beam, determine the support reactions and plot bending moment diagrams ( M) and shear forces ( Q).
- We designate supports letters BUT and AT and direct the support reactions R A and R B .
Compiling equilibrium equations.
Examination
Write down the values R A and R B on the calculation scheme.
2. Plotting transverse forces method sections. We place the sections on characteristic areas(between changes). According to the dimensional thread - 4 sections, 4 sections.
sec. 1-1 move left.
The section passes through the section with uniformly distributed load, note the size z 1 to the left of the section before the beginning of the section. Plot length 2 m. Rule of signs for Q - cm.
We build on the found value diagramQ.
sec. 2-2 move right.
The section again passes through the area with a uniformly distributed load, note the size z 2 to the right of the section to the beginning of the section. Plot length 6 m.
Building a diagram Q.
sec. 3-3 move right.
sec. 4-4 move to the right.
We are building diagramQ.
3. Construction diagrams M method characteristic points.
characteristic point- a point, any noticeable on the beam. These are the dots BUT, AT, With, D , as well as the point To , wherein Q=0 and bending moment has an extremum. also in middle console put an additional point E, since in this area under a uniformly distributed load the diagram M described crooked line, and it is built, at least, according to 3 points.
So, the points are placed, we proceed to determine the values in them bending moments. Rule of signs - see..
Plots NA, AD – parabolic curve(the “umbrella” rule for mechanical specialties or the “sail rule” for construction), sections DC, SW – straight slanted lines.
Moment at a point D should be determined both left and right from the point D . The very moment in these expressions Excluded. At the point D we get two values from difference by the amount m – jump to its size.
Now we need to determine the moment at the point To (Q=0). However, first we define point position To , denoting the distance from it to the beginning of the section by the unknown X .
T. To belongs second characteristic area, shear force equation(see above)
But the transverse force in t. To is equal to 0 , a z 2 equals unknown X .
We get the equation:
Now knowing X, determine the moment at a point To on the right side.
Building a diagram M . The construction is feasible for mechanical specialties, postponing positive values up from the zero line and using the "umbrella" rule.
For a given scheme of a cantilever beam, it is required to plot the diagrams of the transverse force Q and the bending moment M, perform a design calculation by selecting a circular section.
Material - wood, design resistance material R=10MPa, M=14kN m,q=8kN/m
There are two ways to build diagrams in a cantilevered beam with rigid termination - the usual one, having previously determined the support reactions, and without determining the support reactions, if we consider the sections, going from the free end of the beam and discarding the left side with the termination. Let's build diagrams ordinary way.
1. Define support reactions.
Uniformly distributed load q replace the conditional force Q= q 0.84=6.72 kN
In a rigid embedment, there are three support reactions - vertical, horizontal and moment, in our case, the horizontal reaction is 0.
Let's find vertical support reaction R A and reference moment M A from the equilibrium equations.
In the first two sections on the right, there is no transverse force. At the beginning of a section with a uniformly distributed load (right) Q=0, in the back - the magnitude of the reaction R.A.
3. To build, we will compose expressions for their definition on sections. We plot the moment diagram on the fibers, i.e. down. 
(the plot of single moments has already been built earlier)
We solve equation (1), reduce by EI
Static Indeterminacy Revealed, the value of the "extra" reaction is found. You can start plotting Q and M diagrams for a statically indeterminate beam... We sketch the given beam scheme and indicate the reaction value Rb. In this beam, the reactions in the termination can not be determined if you go to the right.
Building plots Q for a statically indeterminate beam
Plot Q.
Plotting M
We define M at the point of extremum - at the point To. First, let's define its position. We denote the distance to it as unknown " X". Then
We plot M.
Determination of shear stresses in an I-section. Consider the section I-beam. S x \u003d 96.9 cm 3; Yx=2030 cm 4; Q=200 kN
To determine the shear stress, it is used formula
, where Q is the transverse force in the section, S x 0 is the static moment of the part of the cross section located on one side of the layer in which shear stresses are determined, I x is the moment of inertia of the entire cross section, b is the width of the section in the place where shear stress is determined
Compute maximum shear stress:
Let us calculate the static moment for top shelf: 
Now let's calculate shear stresses: 
We are building shear stress diagram:
Design and verification calculations. For a beam with built diagrams of internal forces, select a section in the form of two channels from the condition of strength in terms of normal stresses. Check the beam strength using the shear strength condition and energy strength criterion. Given:
Let's show a beam with constructed plots Q and M 
According to the diagram of bending moments, the dangerous is section C, wherein M C \u003d M max \u003d 48.3 kNm.
Strength condition for normal stresses for this beam has the form σ max \u003d M C / W X ≤σ adm . It is necessary to select a section from two channels.
Determine the required calculated value axial section modulus:
For a section in the form of two channels, according to accept two channels №20a, the moment of inertia of each channel I x =1670cm 4, then axial moment of resistance of the entire section:
Overvoltage (undervoltage) at dangerous points, we calculate according to the formula: Then we get undervoltage:
Now let's check the strength of the beam, based on strength conditions for shear stresses. According to diagram of shear forces dangerous are sections in section BC and section D. As can be seen from the diagram, Q max \u003d 48.9 kN.
Strength condition for shear stresses looks like:
For channel No. 20 a: static moment of area S x 1 \u003d 95.9 cm 3, moment of inertia of section I x 1 \u003d 1670 cm 4, wall thickness d 1 \u003d 5.2 mm, average shelf thickness t 1 \u003d 9.7 mm , channel height h 1 \u003d 20 cm, shelf width b 1 \u003d 8 cm.
For transverse sections of two channels:
S x \u003d 2S x 1 \u003d 2 95.9 \u003d 191.8 cm 3,
I x \u003d 2I x 1 \u003d 2 1670 \u003d 3340 cm 4,
b \u003d 2d 1 \u003d 2 0.52 \u003d 1.04 cm.
Determining the value maximum shear stress:
τ max \u003d 48.9 10 3 191.8 10 -6 / 3340 10 -8 1.04 10 -2 \u003d 27 MPa.
As seen, τmax<τ adm (27MPa<75МПа).
Hence, strength condition is met.
We check the strength of the beam according to the energy criterion.
Out of consideration diagrams Q and M follows that section C is dangerous, in which M C =M max =48.3 kNm and Q C =Q max =48.9 kN.
Let's spend analysis of the stress state at the points of section С
Let's define normal and shear stresses at several levels (marked on the section diagram)
Level 1-1: y 1-1 =h 1 /2=20/2=10cm.
Normal and tangent voltage:
Main voltage:
Level 2-2: y 2-2 \u003d h 1 / 2-t 1 \u003d 20 / 2-0.97 \u003d 9.03 cm.
Main stresses:
Level 3-3: y 3-3 \u003d h 1 / 2-t 1 \u003d 20 / 2-0.97 \u003d 9.03 cm.
Normal and shear stresses: 
Main stresses:
Extreme shear stresses:
Level 4-4: y 4-4 =0.
(in the middle, the normal stresses are equal to zero, the tangential stresses are maximum, they were found in the strength test for tangential stresses)
Main stresses: 
Extreme shear stresses:
Level 5-5:
Normal and shear stresses: 
Main stresses:
Extreme shear stresses: 
Level 6-6:
Normal and shear stresses: 
Main stresses:
Extreme shear stresses: 
Level 7-7:
Normal and shear stresses:
Main stresses:
Extreme shear stresses:
According to the performed calculations stress diagrams σ, τ, σ 1 , σ 3 , τ max and τ min are presented in fig.
Analysis these diagram shows, which is in the cross section of the beam dangerous points are at level 3-3 (or 5-5), in which:
Using energy criterion of strength, we get
From a comparison of the equivalent and allowable stresses, it follows that the strength condition is also satisfied 
(135.3 MPa<150 МПа).
The continuous beam is loaded in all spans. Build diagrams Q and M for a continuous beam.
1. Define degree of static uncertainty beams according to the formula:
n= Sop -3= 5-3 =2, where Sop - the number of unknown reactions, 3 - the number of equations of statics. To solve this beam, it is required two additional equations.
2. Denote numbers supports with zero in order ( 0,1,2,3 )
3. Denote span numbers from the first in order ( v 1, v 2, v 3)
4. Each span is considered as simple beam and build diagrams for each simple beam Q and M. What pertains to simple beam, we will denote with index "0", which refers to continuous beam, we will denote without this index. Thus, is the transverse force and the bending moment for a simple beam.
Consider beam of the 1st span
Let's define fictitious reactions for the beam of the first span according to tabular formulas (see table "Fictitious support reactions....»)
Beam 2nd span
Beam 3rd span 
5. Compose 3 x moment equation for two points– intermediate supports – support 1 and support 2. This will be two missing equations to solve the problem.
The equation of 3 moments in general form:
For point (support) 1 (n=1):
For point (support) 2 (n=2):
We substitute all the known values, taking into account that the moment on the zero support and on the third support are equal to zero, M 0 =0; M3=0
Then we get: 
Divide the first equation by the factor 4 for M 2
We divide the second equation by the factor 20 for M 2
Let's solve this system of equations: 
Subtract the second equation from the first equation, we get: 
We substitute this value in any of the equations and find M2
Chapter 1
1.1. Basic dependencies of the theory of beam bending
Beams It is customary to call rods working in bending under the action of a transverse (normal to the axis of the rod) load. Beams are the most common elements of ship structures. The axis of the beam is the locus of the centers of gravity of its cross sections in the undeformed state. A beam is called straight if the axis is a straight line. The geometric location of the centers of gravity of the cross sections of the beam in a bent state is called the elastic line of the beam. The following direction of the coordinate axes is accepted: axis OX aligned with the axis of the beam, and the axis OY and oz- with the main central axes of inertia of the cross section (Fig. 1.1).
The theory of beam bending is based on the following assumptions.
1. The hypothesis of flat sections is accepted, according to which the cross sections of the beam, initially flat and normal to the axis of the beam, after its bending remain flat and normal to the elastic line of the beam. Due to this, the beam bending deformation can be considered regardless of the shear deformation, which causes distortion of the beam cross-sectional planes and their rotation relative to the elastic line (Fig. 1.2, a).
2. Normal stresses in areas parallel to the axis of the beam are neglected due to their smallness (Fig. 1.2, b).
3. Beams are considered sufficiently rigid, i.e. their deflections are small compared to the height of the beams, and the angles of rotation of the sections are small compared to unity (Fig. 1.2, in).
4. Stresses and strains are connected by a linear relationship, i.e. Hooke's law is valid (Fig. 1.2, G).
Rice. 1.2. Beam bending theory assumptions
We will consider the bending moments and shearing forces that appear during the bending of the beam in its section as a result of the action of the part of the beam mentally discarded along the section on the remaining part of it.
The moment of all forces acting in the section relative to one of the main axes is called the bending moment. The bending moment is equal to the sum of the moments of all forces (including support reactions and moments) acting on the rejected part of the beam, relative to the specified axis of the considered section.
The projection onto the plane of the section of the main vector of forces acting in the section is called the shear force. It is equal to the sum of the projections onto the sectional plane of all forces (including support reactions) acting on the discarded part of the beam.
We confine ourselves to considering the beam bending occurring in the plane XOZ. Such bending will take place in the case when the transverse load acts in a plane parallel to the plane XOZ, and its resultant in each section passes through a point called the center of the bend of the section. Note that for sections of beams with two axes of symmetry, the center of bending coincides with the center of gravity, and for sections with one axis of symmetry, it lies on the axis of symmetry, but does not coincide with the center of gravity.
The load of the beams included in the ship's hull can be either distributed (most often evenly distributed along the axis of the beam, or changing according to a linear law), or applied in the form of concentrated forces and moments.
Let us denote the intensity of the distributed load (the load per unit length of the beam axis) through q(x), an external concentrated force - as R, and the external bending moment as M. A distributed load and a concentrated force are positive if their directions of action coincide with the positive direction of the axis oz(Fig. 1.3, a,b). The external bending moment is positive if it is directed clockwise (Fig. 1.3, in).
Rice. 1.3. Sign rule for external loads
Let us denote the deflection of a straight beam when it is bent in the plane XOZ through w, and the angle of rotation of the section through θ. We accept the rule of signs for bending elements (Fig. 1.4):
1) the deflection is positive if it coincides with the positive direction of the axis oz(Fig. 1.4, a):
2) the angle of rotation of the section is positive if, as a result of bending, the section rotates clockwise (Fig. 1.4, b);
3) bending moments are positive if the beam under their influence bends with a convexity upwards (Fig. 1.4, in);
4) shear forces are positive if they rotate the selected beam element counterclockwise (Fig. 1.4, G).
Rice. 1.4. Sign rule for bend elements
Based on the hypothesis of flat sections, it can be seen (Fig. 1.5) that the relative elongation of the fiber ε x, located at z from the neutral axis, will be equal to
ε x= −z/ρ ,(1.1)
where ρ is the beam curvature radius in the considered section.
Rice. 1.5. Beam bending scheme
The neutral axis of the cross section is the locus of points for which the linear deformation during bending is equal to zero. Between curvature and derivatives of w(x) there is a dependence
By virtue of the accepted assumption about the smallness of the angles of rotation for sufficiently rigid beams, the valuesmall compared to unity, so we can assume that
Substituting 1/ ρ from (1.2) to (1.1), we obtain
Normal bending stresses σ x according to Hooke's law will be equal
Since it follows from the definition of beams that there is no longitudinal force directed along the axis of the beam, the main vector of normal stresses must vanish, i.e.
where F is the cross-sectional area of the beam.
From (1.5) we obtain that the static moment of the cross-sectional area of the beam is equal to zero. This means that the neutral axis of the section passes through its center of gravity.
The moment of internal forces acting in the cross section relative to the neutral axis, M y will
If we take into account that the moment of inertia of the cross-sectional area relative to the neutral axis OY is equal to , and substitute this value in (1.6), then we obtain a dependence that expresses the basic differential equation for the beam bending
Moment of internal forces in the section relative to the axis oz will
Since the axes OY and oz by condition are the main central axes of the section, then 
.
It follows that under the action of a load in a plane parallel to the main bending plane, the elastic line of the beam will be a flat curve. This bend is called flat. Based on dependences (1.4) and (1.7), we obtain
Formula (1.8) shows that the normal bending stresses of beams are proportional to the distance from the neutral axis of the beam. Naturally, this follows from the hypothesis of flat sections. In practical calculations, to determine the highest normal stresses, the section modulus of the beam is often used
where | z| max is the absolute value of the distance of the most distant fiber from the neutral axis.
Further subscripts y omitted for simplicity.
There is a connection between the bending moment, the shearing force and the intensity of the transverse load, which follows from the equilibrium condition of the element mentally isolated from the beam.
Consider a beam element with a length dx (Fig. 1.6). Here it is assumed that the deformations of the element are negligible.
If a moment acts in the left section of the element M and cutting force N, then in its right section the corresponding forces will have increments. Consider only linear increments 
.
Fig.1.6. Forces acting on the beam element
Equating to zero the projection on the axis oz of all efforts acting on the element, and the moment of all efforts relative to the neutral axis of the right section, we get:
From these equations, up to values of a higher order of smallness, we obtain
From (1.11) and (1.12) it follows that
Relationships (1.11)–(1.13) are known as the Zhuravsky–Schwedler theorem. It follows from these relationships that the shear force and bending moment can be determined by integrating the load q:
where N 0 and M 0 - shear force and bending moment in the section corresponding tox=x 0 , which is taken as the origin; ξ,ξ 1 – integration variables.
Permanent N 0 and M 0 for statically determinate beams can be determined from the conditions of their static equilibrium.
If the beam is statically determinate, the bending moment in any section can be found from (1.14), and the elastic line is determined by integrating the differential equation (1.7) twice. However, statically determinate beams are extremely rare in ship hull structures. Most of the beams that are part of ship structures form repeatedly statically indeterminate systems. In these cases, to determine the elastic line, equation (1.7) is inconvenient, and it is advisable to go over to a fourth-order equation.
1.2. Differential equation for beam bending
Differentiating equation (1.7) for the general case, when the moment of inertia of the section is a function of x, taking into account (1.11) and (1.12), we obtain:
where the dashes denote differentiation with respect to x.
For prismatic beams, i.e. beams of constant section, we obtain the following differential equations of bending:
An ordinary inhomogeneous fourth-order linear differential equation (1.18) can be represented as a set of four first-order differential equations:
We further use equation (1.18) or the system of equations (1.19) to determine the beam deflection (its elastic line) and all unknown bending elements: w(x), θ (x), M(x), N(x).
Integrating (1.18) successively 4 times (assuming that the left end of the beam corresponds to the sectionx= x a ), we get:
It is easy to see that the integration constants N a ,M a ,θ a , w a have a certain physical meaning, namely:
N a- cutting force at the origin, i.e. at x=x a ;
M a- bending moment at the origin;
θ a – angle of rotation at the origin;
w a - deflection in the same section.
To determine these constants, it is always possible to make four boundary conditions - two for each end of a single-span beam. Naturally, the boundary conditions depend on the arrangement of the ends of the beam. The simplest conditions correspond to hinged support on rigid supports or a rigid attachment.
When the end of the beam is hinged on a rigid support (Fig. 1.7, a) beam deflection and bending moment are equal to zero:
With rigid termination on a rigid support (Fig. 1.7, b) deflection and angle of rotation of the section are equal to zero:
If the end of the beam (console) is free (Fig. 1.7, in), then in this section the bending moment and the shearing force are equal to zero:
A situation associated with a sliding or symmetry termination is possible (Fig. 1.7, G). This leads to the following boundary conditions:
Note that the boundary conditions (1.26) concerning deflections and angles of rotation are called kinematic, and conditions (1.27) power.
Rice. 1.7. Types of boundary conditions
In ship structures, one often has to deal with more complex boundary conditions, which correspond to the support of the beam on elastic supports or elastic termination of the ends.
Elastic support (Fig. 1.8, a) is called a support having a drawdown proportional to the reaction acting on the support. We will consider the reaction of the elastic support R positive if it acts on the support in the direction of the positive direction of the axis oz. Then you can write:
w =AR,(1.29)
where A- coefficient of proportionality, called the coefficient of compliance of the elastic support.
This coefficient is equal to the drawdown of the elastic support under the action of the reaction R= 1, i.e. A=w R = 1 .
Elastic supports in ship structures can be beams that reinforce the beam under consideration, or pillars and other structures that work in compression.
To determine the compliance coefficient of an elastic support A it is necessary to load the corresponding structure with a unit force and find the absolute value of the subsidence (deflection) at the place of application of the force. A rigid support is a special case of an elastic support with A= 0.
Elastic seal (Fig. 1.8, b) is such a support structure that prevents the free rotation of the section and in which the angle of rotation θ in this section is proportional to the moment, i.e. there is dependence
θ = Â M.(1.30)
Proportionality multiplier  is called the coefficient of compliance of the elastic seal and can be defined as the angle of rotation of the elastic seal at M= 1, i.e.  = θ M= 1 .
A special case of elastic embedding at  = 0 is a hard termination. In ship structures, elastic embeddings are usually beams normal to the one under consideration and lying in the same plane. For example, beams, etc., can be considered elastically embedded on the frames.
Rice. 1.8. Elastic support ( a) and elastic embedding ( b)
If the ends of the beam are long L supported on elastic supports (Fig. 1.9), then the reactions of the supports in the end sections are equal to the shear forces, and the boundary conditions can be written:
The minus sign in the first condition (1.31) is accepted because the positive shear force in the left reference section corresponds to the reaction acting on the beam from top to bottom, and on the support from bottom to top.
If the ends of the beam are long Lresiliently embedded(Fig. 1.9), then for the reference sections, taking into account the sign rule for the angles of rotation and bending moments, we can write:
The minus sign in the second condition (1.32) is adopted because, with a positive moment in the right reference section of the beam, the moment acting on the elastic attachment is directed counterclockwise, and the positive angle of rotation in this section is directed clockwise, i.e. the directions of the moment and the angle of rotation do not coincide.
Consideration of the differential equation (1.18) and all boundary conditions shows that they are linear with respect to both the deflections and their derivatives included in them, and the loads acting on the beam. Linearity is a consequence of the assumptions about the validity of Hooke's law and the smallness of beam deflections.
Rice. 1.9. A beam, both ends of which are elastically supported and elastically embedded ( a);
forces in elastic supports and elastic seals corresponding to positive
directions of bending moment and shear force ( b)
When several loads act on a beam, each beam bending element (deflection, angle of rotation, moment and shear force) is the sum of the bending elements from the action of each of the loads separately. This very important provision, called the principle of superposition, or the principle of summation of the action of loads, is widely used in practical calculations and, in particular, to reveal the static indeterminacy of beams.
1.3. Initial Parameters Method
The general integral of the beam bending differential equation can be used to determine the elastic line of a single-span beam when the beam load is a continuous function of the coordinate throughout the span. If the load contains concentrated forces, moments or a distributed load acts on parts of the beam length (Fig. 1.10), then the expression (1.24) cannot be used directly. In this case, it would be possible, by denoting the elastic lines in sections 1, 2 and 3 through w 1 , w 2 , w 3 , write out for each of them the integral in the form (1.24) and find all arbitrary constants from the boundary conditions at the ends of the beam and the conjugation conditions at the boundaries of the sections. The conjugation conditions in the case under consideration are expressed as follows:
at x=a 1
at x=a 2
at x=a 3
It is easy to see that such a way of solving the problem leads to a large number of arbitrary constants, equal to 4 n, where n- the number of sections along the length of the beam.
Rice. 1.10. Beam, on some sections of which loads of different types are applied
It is much more convenient to represent the elastic line of the beam in the form
where the terms behind the double line are taken into account when x³ a 1, x³ a 2 etc.
Obviously, δ 1 w(x)=w 2 (x)−w 1 (x); δ2 w(x)=w 3 (x)−w 2 (x); etc.
Differential equations for determining the corrections to the elastic line δ iw (x) based on (1.18) and (1.32) can be written as
General integral for any correction δ iw (x) to the elastic line can be written in the form (1.24) for x a = a i . At the same time, the parameters N a ,M a ,θ a , w a the changes (jump) make sense, respectively: in the shear force, bending moment, angle of rotation and deflection arrow at the transition through the section x=a i . This technique is called the method of initial parameters. It can be shown that for the beam shown in Fig. 1.10, the elastic line equation will be
Thus, the method of initial parameters makes it possible, even in the presence of discontinuity in loads, to write the equation of an elastic line in a form containing only four arbitrary constants N 0 , M 0 , θ 0 , w 0 , which are determined from the boundary conditions at the ends of the beam.
Note that for a large number of variants of single-span beams encountered in practice, detailed bending tables have been compiled that make it easy to find deflections, angles of rotation and other bending elements.
1.4. Determination of shear stresses during beam bending
The hypothesis of flat sections accepted in the theory of beam bending leads to the fact that the shear deformation in the beam section turns out to be equal to zero, and we do not have the opportunity, using Hooke's law, to determine the shear stresses. However, since, in the general case, shear forces act in the beam sections, the shear stresses corresponding to them should arise. This contradiction (which is a consequence of the accepted hypothesis of flat sections) can be avoided by considering the equilibrium conditions. We will assume that when a beam composed of thin strips is bent, the shear stresses in the cross section of each of these strips are uniformly distributed over the thickness and directed parallel to the long sides of its contour. This position is practically confirmed by the exact solutions of the theory of elasticity. Consider a beam of an open thin-walled I-beam. On fig. 1.11 shows the positive direction of shear stresses in the belts and the profile wall during bending in the plane of the beam wall. Select the longitudinal section I-I and two cross sections element length dx (Fig. 1.12).
Let us denote the shear stress in the indicated longitudinal section as τ, and the normal forces in the initial cross section as T. Normal forces in the final section will have increments. Consider only linear increments, then .
Rice. 1.12. Longitudinal forces and shear stresses
in beam girdle element
The condition of static equilibrium of the element selected from the beam (equality to zero of the projections of forces on the axis OX) will
where ; f- the area of the part of the profile cut off by the line I-I; δ is the thickness of the profile at the section site.
From (1.36) it follows:
Since the normal stresses σ x are defined by formula (1.8), then
In this case, we assume that the beam has a section that is constant along the length. Static moment of a part of the profile (cut-off line I-I) relative to the neutral axis of the beam section OY is an integral
Then from (1.37) for the absolute value of stresses we obtain:
Naturally, the resulting formula for determining shear stresses is also valid for any longitudinal section, for example II -II(see Fig. 1.11), and the static moment S ots is calculated for the cut-off part of the beam profile area relative to the neutral axis, without taking into account the sign.
Formula (1.38), according to the meaning of the derivation, determines the shear stresses in the longitudinal sections of the beam. From the theorem on the pairing of shear stresses, known from the course of the strength of materials, it follows that the same shear stresses act at the corresponding points of the beam cross section. Naturally, the projection of the main shear stress vector onto the axis oz must be equal to the shear force N in this section of the beam. Since in the girdle beams of this type, as shown in Fig. 1.11, shear stresses are directed along the axis OY, i.e. normal to the plane of action of the load, and are generally balanced, the shear force must be balanced by shear stresses in the beam web. The distribution of shear stresses along the height of the wall follows the law of change in the static moment S cut off part of the area relative to the neutral axis (with a constant wall thickness δ).
Consider a symmetrical section of an I-beam with a girdle area F 1 and wall area ω = hδ (Fig. 1.13).
Rice. 1.13. Section of an I-beam
The static moment of the cut-off part of the area for a point separated by z from the neutral axis, will
As can be seen from dependence (1.39), the static moment changes from z according to the law of a quadratic parabola. Highest value S ots , and consequently, shear stresses τ , will turn out at the neutral axis, where z= 0:
The greatest shear stress in the beam web at the neutral axis
Since the moment of inertia of the section of the considered beam is equal to
then the greatest shear stress will be
Attitude N/ω is nothing but the average shear stress in the wall, calculated assuming a uniform distribution of stresses. Taking, for example, ω = 2 F 1 , by formula (1.41) we obtain
Thus, for the beam under consideration, the largest shear stress in the wall at the neutral axis is only 12.5% exceeds the average value of these stresses. It should be noted that for the majority of beam profiles used in the ship's hull, the excess of the maximum shear stresses over the average is 10–15%.
If we consider the distribution of shear stresses during bending in the cross section of the beam shown in Fig. 1.14, it can be seen that they form a moment relative to the center of gravity of the section. In the general case, the bending of such a beam in the plane XOZ will be accompanied by twisting.
Beam bending is not accompanied by twisting if the load acts in a plane parallel to XOZ passing through a point called the center of the bend. This point is characterized by the fact that the moment of all tangential forces in the beam section relative to it is equal to zero.
Rice. 1.14. Tangential stresses during channel beam bending (point BUT - bend center)
Denoting the distance of the center of the bend BUT from the axis of the beam web through e, we write down the condition of equality to zero of the moment of tangential forces relative to the point BUT:
where Q 2 - tangential force in the wall, equal to the shear force, i.e. Q 2 =N;
Q 1 =Q 3 - force in the girdle, determined on the basis of (1.38) by the dependence
The shear strain (or shear angle) γ varies along the height of the beam web in the same way as the shear stresses τ , reaching its greatest value at the neutral axis.
As shown, for beams with corbels, the change in shear stresses along the wall height is very insignificant. This allows further consideration of some average shear angle in the beam web
Shear deformation leads to the fact that the right angle between the plane of the cross section of the beam and the tangent to the elastic line changes by the value γ cf. A simplified diagram of the shear deformation of a beam element is shown in fig. 1.15.
Rice. 1.15. Beam Element Shear Diagram
Denoting the deflection arrow caused by the shear through w sdv , we can write:
Taking into account the sign rule for the shear force N and find the angle of rotation
Insofar as ,
Integrating (1.47), we obtain
Constant a, included in (1.48), determines the displacement of the beam as a rigid body and can be taken equal to any value, since when determining the total deflection arrow from bending w bend and shear w sdv
the sum of the constants of integration will appear w 0 +a determined from the boundary conditions. Here w 0 - deflection from bending at the origin.
We put in the future a=0. Then the final expression for the elastic line caused by the shear will take the form
The bending and shear components of the elastic line are shown in Figs. 1.16.
Rice. 1.16. Flexural ( a) and shear ( b) components of the elastic line of the beam
In the considered case, the angle of rotation of the sections during shear is equal to zero, therefore, taking into account the shear, the angles of rotation of the sections, bending moments and shear forces are associated only with the derivatives of the elastic line from bending:
The situation is somewhat different in the case of the action of concentrated moments on the beam, which, as will be shown below, do not cause shear deflections, but only lead to an additional rotation of the beam sections.
Consider a beam freely supported on rigid supports, in the left section of which acting moment M. The cutting force in this case will be constant and equal
For the right reference section, respectively, we obtain
.(1.52)
Expressions (1.51) and (1.52) can be rewritten as
The expressions in parentheses characterize the relative addition to the angle of rotation of the section caused by the shear.
If we consider, for example, a freely supported beam loaded in the middle of its span by the force R(Fig. 1.18), then the deflection of the beam under the force will be equal to
Bending deflection can be found from beam bending tables. Shear deflection is determined by formula (1.50), taking into account the fact that 
.
Rice. 1.18. Scheme of a freely supported beam loaded with a concentrated force
As can be seen from formula (1.55), the relative addition to the beam deflection due to shear has the same structure as the relative addition to the angle of rotation, but with a different numerical coefficient.
We introduce the notation
where β is a numerical coefficient depending on the specific task under consideration, the arrangement of supports and the load of the beam.
Let us analyze the dependence of the coefficient k from various factors.
If we take into account that , we obtain instead of (1.56)
The moment of inertia of the beam section can always be represented as
,(1.58)
where α is a numerical coefficient depending on the shape and characteristics of the cross section. So, for an I-beam, according to formula (1.40) with ω = 2 F 1 find I= ωh 2 /3, i.e. α=1/3.
Note that with an increase in the dimensions of the beam corbels, the coefficient α will increase.
Taking into account (1.58), instead of (1.57) we can write:
Thus, the value of the coefficient k significantly depends on the ratio of the span length of the beam to its height, on the shape of the section (through the coefficient α), the device of the supports and the load of the beam (through the coefficient β). The relatively longer the beam ( h/L small), the smaller the effect of shear deformation. For rolled profile beams related to h/L less than 1/10÷1/8, the shift correction can practically not be taken into account.
However, for beams with wide girths, such as, for example, keels, stringers and floors as part of bottom slabs, the effect of shear and at the indicated h/L may be significant.
It should be noted that shear deformations affect not only the increase in beam deflections, but in some cases also the results of the disclosure of the static indeterminacy of beams and beam systems.
For a visual representation of the nature of the deformation of the bars (rods) during bending, the following experiment is carried out. A grid of lines parallel and perpendicular to the axis of the beam is applied to the side faces of the rubber bar of rectangular section (Fig. 30.7, a). Then moments are applied to the bar at its ends (Fig. 30.7, b), acting in the plane of symmetry of the bar, crossing each of its cross sections along one of the main central axes of inertia. The plane passing through the beam axis and one of the main central axes of inertia of each of its cross sections will be called the main plane.
Under the action of moments, the beam experiences a straight clean bend. As a result of deformation, as experience shows, the grid lines parallel to the axis of the beam are bent, while maintaining the same distances between them. When indicated in Fig. 30.7, b in the direction of the moments, these lines lengthen in the upper part of the beam, and shorten in the lower part.
Each line of the grid, perpendicular to the axis of the beam, can be considered as a trace of the plane of some cross-section of the beam. Since these lines remain straight, it can be assumed that the cross sections of the beam, which are flat before deformation, remain flat during deformation.
This assumption, based on experience, is known to be called the hypothesis of flat sections, or the Bernoulli hypothesis (see § 6.1).
The hypothesis of flat sections is used not only for pure, but also for transverse bending. For transverse bending, it is approximate, and for pure bending, it is strict, which is confirmed by theoretical studies carried out by methods of the theory of elasticity.
Let us now consider a straight bar with a cross section symmetrical about the vertical axis, embedded with the right end and loaded at the left end with an external moment acting in one of the main planes of the bar (Fig. 31.7). In each cross section of this beam, only bending moments arise acting in the same plane as the moment
Thus, the timber throughout its length is in a state of direct pure bending. In a state of pure bending, individual sections of the beam can also be in the case of transverse loads acting on it; for example, section 11 of the beam shown in fig. 32.7; in the sections of this section, the transverse force
Let us select from the beam under consideration (see Fig. 31.7) with two cross sections an element with a length. As a result of the deformation, as follows from the Bernoulli hypothesis, the sections will remain flat, but will incline relative to each other by a certain angle. Let us conditionally take the left section as fixed. Then, as a result of turning the right section by an angle, it will take a position (Fig. 33.7).
The lines intersect at some point A, which is the center of curvature (or, more precisely, the trace of the axis of curvature) of the longitudinal fibers of the element. 31.7 in the direction of the moment are lengthened, and the lower ones are shortened. The fibers of some intermediate layer perpendicular to the plane of action of the moment retain their length. This layer is called the neutral layer.
Let us denote the radius of curvature of the neutral layer, i.e., the distance from this layer to the center of curvature A (see Fig. 33.7). Consider some layer located at a distance y from the neutral layer. The absolute elongation of the fibers of this layer is equal to and the relative
Considering similar triangles, we find that Therefore,
In the theory of bending, it is assumed that the longitudinal fibers of the beam do not press on each other. Experimental and theoretical studies show that this assumption does not significantly affect the calculation results.
With pure bending, shear stresses do not arise in the cross sections of the beam. Thus, all fibers in pure bending are in uniaxial tension or compression.
According to Hooke's law, for the case of uniaxial tension or compression, the normal stress o and the corresponding relative strain are related by the dependence
or based on formula (11.7)
From formula (12.7) it follows that the normal stresses in the longitudinal fibers of the beam are directly proportional to their distances y from the neutral layer. Consequently, in the cross section of the beam at each point, the normal stresses are proportional to the distance y from this point to the neutral axis, which is the line of intersection of the neutral layer with the cross section (Fig.
34.7, a). It follows from the symmetry of the beam and the load that the neutral axis is horizontal.
At the points of the neutral axis, the normal stresses are equal to zero; on one side of the neutral axis they are tensile, and on the other they are compressive.
The stress diagram o is a graph bounded by a straight line, with the largest absolute value of stresses for points furthest from the neutral axis (Fig. 34.7, b).
Let us now consider the equilibrium conditions for the selected beam element. The action of the left part of the beam on the section of the element (see Fig. 31.7) is represented as a bending moment, the remaining internal forces in this section with pure bending are equal to zero. Let us represent the action of the right side of the beam on the section of the element in the form of elementary forces about the cross section applied to each elementary area (Fig. 35.7) and parallel to the axis of the beam.
We compose six conditions for the equilibrium of an element
Here - the sum of the projections of all forces acting on the element, respectively, on the axis - the sum of the moments of all forces about the axes (Fig. 35.7).
The axis coincides with the neutral axis of the section, and the y-axis is perpendicular to it; both of these axes are located in the plane of the cross section
An elementary force does not give projections on the y-axis and does not cause a moment about the axis. Therefore, the equilibrium equations are satisfied for any values of o.
The equilibrium equation has the form
Substitute in equation (13.7) the value of a according to formula (12.7):
Since (a curved beam element is considered, for which ), then
The integral is the static moment of the cross section of the beam relative to the neutral axis. Its equality to zero means that the neutral axis (i.e. axis) passes through the center of gravity of the cross section. Thus, the center of gravity of all cross sections of the beam, and consequently, the axis of the beam, which is the geometric location of the centers of gravity, are located in the neutral layer. Therefore, the radius of curvature of the neutral layer is the radius of curvature of the curved axis of the bar.
Let us now compose the equilibrium equation in the form of the sum of the moments of all forces applied to the beam element, relative to the neutral axis:
Here represents the moment of the elementary internal force about the axis .
Let us denote the area of the part of the cross section of the beam located above the neutral axis - under the neutral axis.
Then it will represent the resultant of elemental forces applied above the neutral axis, below the neutral axis (Fig. 36.7).
Both of these resultants are equal to each other in absolute value, since their algebraic sum on the basis of condition (13.7) is equal to zero. These resultants form an internal pair of forces acting in the cross section of the beam. The moment of this pair of forces, i.e., the product of the value of one of them and the distance between them (Fig. 36.7), is a bending moment in the cross section of the beam.
Substitute in equation (15.7) the value of a according to formula (12.7):
Here is the axial moment of inertia, i.e., the axis passing through the center of gravity of the section. Hence,
Substitute the value from formula (16.7) into formula (12.7):
When deriving formula (17.7), it was not taken into account that with an external moment directed, as shown in Fig. 31.7, according to the accepted sign rule, the bending moment is negative. If we take this into account, then before the right side of the formula (17.7) it is necessary to put a minus sign. Then, with a positive bending moment in the upper zone of the beam (i.e., at ), the values of a will turn out to be negative, which will indicate the presence of compressive stresses in this zone. However, usually the minus sign is not put on the right side of the formula (17.7), but this formula is used only to determine the absolute values of the stresses a. Therefore, the absolute values of the bending moment and the ordinate y should be substituted into formula (17.7). The sign of the stresses is always easily determined by the sign of the moment or by the nature of the deformation of the beam.
Let us now compose the equilibrium equation in the form of the sum of the moments of all forces applied to the beam element, relative to the y axis:
Here is the moment of the elementary internal force about the y-axis (see Fig. 35.7).
Substitute in the expression (18.7) the value of a according to the formula (12.7):
Here the integral is the centrifugal moment of inertia of the cross section of the beam relative to the axes y and . Hence,
But since
As is known (see § 7.5), the centrifugal moment of inertia of the section is zero relative to the main axes of inertia.
In the case under consideration, the y-axis is the axis of symmetry of the cross-section of the beam and, therefore, the y-axes and are the main central axes of inertia of this section. Therefore, condition (19.7) is satisfied here.
In the case when the cross section of the bent beam does not have any axis of symmetry, condition (19.7) is satisfied if the plane of action of the bending moment passes through one of the main central axes of inertia of the section or is parallel to this axis.
If the plane of action of the bending moment does not pass through any of the main central axes of inertia of the cross section of the beam and is not parallel to it, then condition (19.7) is not satisfied and, therefore, there is no direct bending - the beam experiences oblique bending.
Formula (17.7), which determines the normal stress at an arbitrary point of the considered section of the beam, is applicable provided that the plane of action of the bending moment passes through one of the main axes of inertia of this section or is parallel to it. In this case, the neutral axis of the cross section is its main central axis of inertia, perpendicular to the plane of action of the bending moment.
Formula (16.7) shows that with direct pure bending, the curvature of the curved axis of the beam is directly proportional to the product of the modulus of elasticity E and the moment of inertia. The product will be called the bending stiffness of the section; it is expressed in etc.
With pure bending of a beam of constant section, the bending moments and section stiffnesses are constant along its length. In this case, the radius of curvature of the bent axis of the beam has a constant value [see. expression (16.7)], i.e., the beam is bent along a circular arc.
From formula (17.7) it follows that the largest (positive - tensile) and smallest (negative - compressive) normal stresses in the cross section of the beam occur at points farthest from the neutral axis, located on both sides of it. With a cross section symmetrical about the neutral axis, the absolute values of the largest tensile and compressive stresses are the same and can be determined by the formula
where is the distance from the neutral axis to the most distant point of the section.
The value that depends only on the size and shape of the cross section is called the axial section modulus and is denoted
(20.7)
Hence,
Let us determine the axial moments of resistance for rectangular and round sections.
For a rectangular section with width b and height
For a circular section with a diameter d
The moment of resistance is expressed in .
For sections that are not symmetrical about the neutral axis, for example, for a triangle, a brand, etc., the distances from the neutral axis to the outermost stretched and compressed fibers are different; therefore, for such sections there are two moments of resistance:
where are the distances from the neutral axis to the outermost stretched and compressed fibers.
The calculation of a beam for bending "manually", in an old-fashioned way, allows you to learn one of the most important, beautiful, clearly mathematically verified algorithms of the science of the strength of materials. The use of numerous programs such as "entered the initial data ...
...– get the answer” allows the modern engineer today to work much faster than his predecessors a hundred, fifty and even twenty years ago. However, with such a modern approach, the engineer is forced to fully trust the authors of the program and eventually ceases to "feel the physical meaning" of the calculations. But the authors of the program are people, and people make mistakes. If this were not so, then there would not be numerous patches, releases, "patches" for almost any software. Therefore, it seems to me that any engineer should sometimes be able to "manually" check the results of calculations.
Help (cheat sheet, memo) for calculating beams for bending is shown below in the figure.
Let's use a simple everyday example to try to use it. Let's say I decided to make a horizontal bar in the apartment. A place has been determined - a corridor one meter twenty centimeters wide. On opposite walls at the required height opposite each other, I securely fasten the brackets to which the beam-beam will be attached - a bar of St3 steel with an outer diameter of thirty-two millimeters. Will this beam support my weight plus additional dynamic loads that will arise during exercise?
We draw a diagram for calculating the beam for bending. Obviously, the most dangerous external load application scheme will be when I start to pull myself up, clinging to the middle of the crossbar with one hand.
Initial data:
F1 \u003d 900 n - the force acting on the beam (my weight) without taking into account the dynamics
d \u003d 32 mm - the outer diameter of the bar from which the beam is made
E = 206000 n/mm^2 is the modulus of elasticity of the St3 steel beam material
[σi] = 250 n/mm^2 - allowable bending stresses (yield strength) for the material of the St3 steel beam
Border conditions:
Мx (0) = 0 n*m – moment at point z = 0 m (first support)
Мx (1.2) = 0 n*m – moment at point z = 1.2 m (second support)
V (0) = 0 mm - deflection at point z = 0 m (first support)
V (1.2) = 0 mm - deflection at point z = 1.2 m (second support)
Calculation:
1. First, we calculate the moment of inertia Ix and the moment of resistance Wx of the beam section. They will be useful to us in further calculations. For a circular section (which is the section of the bar):
Ix = (π*d^4)/64 = (3.14*(32/10)^4)/64 = 5.147 cm^4
Wx = (π*d^3)/32 = ((3.14*(32/10)^3)/32) = 3.217 cm^3
2. We compose equilibrium equations for calculating the reactions of the supports R1 and R2:
Qy = -R1+F1-R2 = 0
Mx (0) = F1*(0-b2) -R2*(0-b3) = 0
From the second equation: R2 = F1*b2/b3 = 900*0.6/1.2 = 450 n
From the first equation: R1 = F1-R2 = 900-450 = 450 n
3. Let's find the angle of rotation of the beam in the first support at z = 0 from the deflection equation for the second section:
V (1.2) = V (0)+U (0)*1.2+(-R1*((1.2-b1)^3)/6+F1*((1.2-b2)^3)/6)/
U (0) = (R1*((1.2-b1)^3)/6 -F1*((1.2-b2)^3)/6)/(E*Ix)/1,2 =
= (450*((1.2-0)^3)/6 -900*((1.2-0.6)^3)/6)/
/(206000*5.147/100)/1.2 = 0.00764 rad = 0.44˚
4.
We compose equations for constructing diagrams for the first section (0 Shear force: Qy (z) = -R1 Bending moment: Mx (z) = -R1*(z-b1) Rotation angle: Ux (z) = U (0)+(-R1*((z-b1)^2)/2)/(E*Ix) Deflection: Vy (z) = V (0)+U (0)*z+(-R1*((z-b1)^3)/6)/(E*Ix) z = 0 m: Qy (0) = -R1 = -450 n Ux(0) = U(0) = 0.00764 rad Vy(0)=V(0)=0mm z = 0.6 m: Qy (0.6) = -R1 = -450 n Mx (0.6) \u003d -R1 * (0.6-b1) \u003d -450 * (0.6-0) \u003d -270 n * m Ux (0.6) = U (0)+(-R1*((0.6-b1)^2)/2)/(E*Ix) = 0.00764+(-450*((0.6-0)^2)/2)/(206000*5.147/100) = 0 rad Vy (0.6) = V (0)+U (0)*0.6+(-R1*((0.6-b1)^3)/6)/(E*Ix) = 0+0.00764*0.6+(-450*((0.6-0)^3)/6)/ (206000*5.147/100) = 0.003 m The beam will sag in the center by 3 mm under the weight of my body. I think this is an acceptable deflection. 5.
We write the diagram equations for the second section (b2 Shear force: Qy (z) = -R1+F1 Bending moment: Mx (z) = -R1*(z-b1)+F1*(z-b2) Rotation angle: Ux (z) = U (0)+(-R1*((z-b1)^2)/2+F1*((z-b2)^2)/2)/(E*Ix) Deflection: Vy (z) = V (0)+U (0)*z+(-R1*((z-b1)^3)/6+F1*((z-b2)^3)/6)/( E*Ix) z = 1.2 m: Qy (1,2) = -R1+F1 = -450+900 = 450 n Мx (1,2) = 0 n*m Ux (1,2) = U (0)+(-R1*((1,2-b1)^2)/2+F1*((1,2-b2)^2)/2)/(E* ix) = 0,00764+(-450*((1,2-0)^2)/2+900*((1,2-0,6)^2)/2)/ /(206000*5.147/100) = -0.00764 rad Vy (1.2) = V (1.2) = 0 m 6.
We build diagrams using the data obtained above. 7.
We calculate the bending stresses in the most loaded section - in the middle of the beam and compare with the allowable stresses: σi \u003d Mx max / Wx \u003d (270 * 1000) / (3.217 * 1000) \u003d 84 n / mm ^ 2 σi = 84 n/mm^2< [σи] = 250 н/мм^2 In terms of bending strength, the calculation showed a threefold margin of safety - the horizontal bar can be safely made from an existing bar with a diameter of thirty-two millimeters and a length of one thousand two hundred millimeters. Thus, you can now easily calculate the beam for bending "manually" and compare with the results obtained in the calculation using any of the numerous programs presented on the Web. I ask those who RESPECT the work of the author to SUBSCRIBE to the announcements of articles.
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