Consider a beam that is in flat direct bending under the action of arbitrary transverse loads in the main plane Ohu(Fig. 7.31, a). We cut the beam at a distance x from its left end and consider the equilibrium of the left side. The influence of the right side in this case must be replaced by the action of the bending moment A / and the transverse force Q y in the section drawn (Fig. 7.31, b). The bending moment L7 in the general case is not constant in magnitude, as was the case with pure bending, but varies along the length of the beam. Since the bending moment M
according to (7.14) is associated with normal stresses o = a x, then the normal stresses in the longitudinal fibers will also change along the length of the beam. Therefore, in the case of transverse bending, the normal stresses are functions of the variables x and y: a x = a x (x, y).
With transverse bending in the beam section, not only normal, but also tangential stresses t act (Fig. 7.31, in), the resultant of which is the transverse force Qy:
Presence of shear stresses x wow accompanied by the appearance of angular deformations y. Shear stresses, like normal stresses, are unevenly distributed over the cross section. Consequently, the angular deformations associated with them by the Hooke law in shear will also be unevenly distributed. This means that in transverse bending, in contrast to pure bending, the beam sections do not remain flat (J. Bernoulli's hypothesis is violated).
The curvature of cross-sections can be clearly demonstrated by the example of the bending of a cantilevered rubber rectangular beam caused by a concentrated force applied at the end (Fig. 7.32). If you first draw straight lines perpendicular to the axis of the beam on the side faces, then after bending these lines do not remain straight. In this case, they are bent so that the greatest shift takes place at the level of the neutral layer.
More accurate studies have established that the effect of cross-sectional distortion on the value of normal stresses is insignificant. It depends on the ratio of the height of the section h to the length of the beam / and at h/ / o x in transverse bending, formula (7.14) is usually used, derived for the case of pure bending.
The second feature of transverse bending is the presence of normal stresses about y, acting in the longitudinal sections of the beam and characterizing the mutual pressure between the longitudinal layers. These stresses occur in areas where there is a distributed load q, and places of application of concentrated forces. Usually these stresses are very small compared to normal stresses. a x. A special case is the action of a concentrated force, in the area of application of which significant local stresses can arise. and u.
Thus, an infinitesimal element in the plane Ohu in the case of transverse bending, it is in a biaxial stress state (Fig. 7.33).
The stresses m and o, as well as the stress o Y , are generally functions of the coordinates* and y. They must satisfy differential equilibrium equations, which for a biaxial stress state ( a z = T yz = = 0) in the absence
volume forces have the following form: 
These equations can be used to determine shear stresses = t and normal stresses OU. The easiest way to do this is for a beam of rectangular cross section. In this case, when determining m, the assumption is made about their uniform distribution over the width of the section (Fig. 7.34). This assumption was made by the famous Russian bridge builder D.I. Zhuravsky. Studies show that this assumption almost exactly corresponds to the actual nature of the distribution of shear stresses in bending for fairly narrow and high beams. (b « AND).
Using the first of the differential equations (7.26) and formula (7.14) for normal stresses a x, we get
Integrating this equation with respect to the variable y, find
where f(x)- an arbitrary function, for the definition of which we use the condition of the absence of shear stresses on the lower face of the beam: 
Taking into account this boundary condition, from (7.28) we find
Finally, the expression for shear stresses acting in the cross sections of the beam takes the following form:
By virtue of the law of pairing of tangential stresses, tangential stresses t, = t also arise in the longitudinal sections
hu uh
beams parallel to the neutral layer.
From formula (7.29) it can be seen that the shear stresses change along the height of the beam cross section according to the law of a square parabola. Shear stresses have the greatest value at points at the level of the neutral axis at y= 0, and in the extreme fibers of the beam at y = ±h/2 they are equal to zero. Using formula (7.23) for the moment of inertia of a rectangular section, we obtain
where F=bh- cross-sectional area of the beam.
Plot t is shown in fig. 7.34.
In the case of beams with a non-rectangular cross section (Fig. 7.35), it is difficult to determine the shear stresses m from the equilibrium equation (7.27), since the boundary condition for m is not known at all points of the cross section contour. This is due to the fact that in this case, shear stresses m act in the cross section, which are not parallel to the transverse force Q y . Indeed, it can be shown that at points near the contour of the cross section, the total shear stress m is directed tangentially to the contour. Consider, in the vicinity of an arbitrary point of the contour (see Fig. 7.35), an infinitely small area dF in the plane of the cross section and a platform perpendicular to it dF" on the side of the beam. If the total stress m at the contour point is not directed tangentially, then it can be decomposed into two components: xvx in the direction of the normal v to the contour and X in the direction of the tangent t to the contour. Therefore, according to the law of pairing of shear stresses on the site dF" should-
but act shear stress x equal to x vv . If the side surface is free from tangential loads, then the component x vv = zvx = 0, that is, the total shear stress x must be directed tangentially to the cross-sectional contour, as shown, for example, at points L and AT contour.
Consequently, the shear stress x both at the points of the contour and at any point of the cross section can be decomposed into components x of them.
To determine the components x of the shear stress in beams of non-rectangular cross section (Fig. 7.36, b) suppose that the section has a vertical axis of symmetry and that the component x of the total shear stress x, as in the case of a rectangular cross section, is uniformly distributed over its width.
Using a longitudinal section parallel to the plane Oxz and passing at a distance at from it, and two cross sections xx + dx mentally cut out from the bottom of the beam an infinitesimal element of length dx(Fig. 7.36, in).
We assume that the bending moment M varies within length dx considered element of the beam, and the transverse force Q constant. Then in cross sections x and x + dx the beams will act with the same shear stresses x, and the normal stresses arising from bending moments MzmMz+ dM, will be equal respectively a and a + da. Along the horizontal face of the selected element (in Fig. 7.36, in it is shown in axonometry) according to the law of pairing of shear stresses, stresses x v „ \u003d x will act.
hu uh
resultant R and R+dR normal stresses o and o + d applied to the ends of the element, taking into account the formula (7.14) are equal to
where 
cut-off static moment F(in Fig. 7.36, b shaded) relative to the neutral axis Oz y, - auxiliary variable, changing within at
The resultant shear stress t applied
hu
to the horizontal edge of the element, taking into account the introduced assumption about the uniform distribution of these stresses over the width b(y) can be found by the formula
The equilibrium condition for an element? X=0 gives
Substituting the values of the resultant forces, we obtain 
From here, taking into account (7.6), we obtain a formula for determining shear stresses:
This formula in the domestic literature is called formula D.I. Zhuravsky.
In accordance with formula (7.32), the distribution of shear stresses m along the height of the section depends on the change in the width of the section b(y) and the static moment of the cut-off part of the section S OTC (y).
Using formula (7.32), shear stresses are most simply determined for the rectangular beam considered above (Fig. 7.37).
The static moment of the cut-off cross-sectional area F qtc is equal to
Substituting 5° TC into (7.32), we obtain the previously derived formula (7.29).
Formula (7.32) can be used to determine shear stresses in beams with a stepwise constant section width. Within each section with a constant width, shear stresses change along the height of the section according to the law of a square parabola. In places of abrupt change in the width of the section, shear stresses also have jumps or discontinuities. The nature of the diagram m for such a section is shown in Fig. 7.38.
Rice. 7.37
Rice. 7.38
Consider the distribution of shear stresses in an I-section (Fig. 7.39, a) when bending in a plane Ohu. An I-section can be represented as a conjugation of three narrow rectangles: two horizontal shelves and a vertical wall.
When calculating m in the wall in formula (7.32), one must take b(y) - d. As a result, we get
where S° 1C calculated as the sum of static moments about the axis Oz shelf area F n and parts of the wall F, shaded in Fig. 7.39, a:
The shear stresses t have the greatest value at the level of the neutral axis at y= 0:
where is the static moment of the half-section area relative to the neutral axis:
For rolling I-beams and channels, the value of the static moment of half the section is given in the assortment.
Rice. 7.39
At the level where the wall adjoins the flanges, shear stresses 1 ? equal
where S"- static moment of the sectional area of the flange relative to the neutral axis:
The vertical shear stresses t in the flanges of an I-beam cannot be found by the formula (7.32), as due to the fact that b :» t, the assumption of their uniform distribution over the width of the shelf becomes unacceptable. On the top and bottom faces of the shelf, these stresses must be equal to zero. Therefore, t in
wow
shelves are very small and not of practical interest. Of much greater interest are the horizontal shear stresses in the shelves m, to determine which we consider the equilibrium of an infinitesimal element selected from the bottom shelf (Fig. 7.39 , b).
According to the law of pairing of shear stresses on the longitudinal face of this element, parallel to the plane Ohu voltage is acting xxz , equal in magnitude to the stress t acting in the cross section. Due to the small thickness of the I-beam flange, these stresses can be assumed to be uniformly distributed over the thickness of the flange. With this in mind, from the equilibrium equation for the element 5^=0 we will have
From here we find
Substituting into this formula the expression for a x from (7.14) and taking into account that we get 
Given that
where S° TC - static moment of the cut-off area of the shelf (in Fig. 7. 39, a shaded twice) relative to the axis Oz, we finally get 
In accordance with fig. 7.39 , a 
where z- axis-based variable OU.
With this in mind, formula (7.34) can be represented as
This shows that the horizontal shear stresses change linearly along the axis Oz and take the largest value at z = d/2:
On fig. 7.40 shows diagrams of tangential stresses t and t^, as well as the direction of these stresses in the shelves and the wall of the I-beam under the action of a positive transverse force in the beam section Q. Shear stresses, figuratively speaking, form a continuous flow in the section of the I-beam, directed at each point parallel to the contour of the section.
Let's move on to the definition of normal stresses and at in the longitudinal sections of the beam. Consider a beam section with a uniformly distributed load along the upper face (Fig. 7.41). The cross section of the beam is assumed to be rectangular.
We use to determine the second of the differential equilibrium equations (7.26). Substituting in this equation formula (7.32) for shear stresses t uh, taking into account (7.6), we obtain
By integrating over the variable y, find
Here f(x) - an arbitrary function that is defined using a boundary condition. According to the conditions of the problem, the beam is loaded with a uniformly distributed load q along the top face, and the bottom face is free from loads. Then the corresponding boundary conditions are written as
Using the second of these conditions, we get
With this in mind, the formula for stresses and at will take the following form:
It can be seen from this expression that the stresses o change along the height of the section according to the law of a cubic parabola. In this case, both boundary conditions (7.35) are satisfied. Highest value voltage takes on the upper surface of the beam at y=-h/2:
The nature of the plot and at shown in fig. 7.41.
To estimate the magnitude of the greatest stresses o. a, and m and the relationships between them, consider, for example, the bending of a cantilever beam of rectangular cross section with dimensions bxh, under the action of a uniformly distributed load applied to the upper face of the beam (Fig. 7.42). The largest absolute stresses occur in the termination. In accordance with formulas (7.22), (7.30) and (7.37), these stresses are equal to
As usual for beams l/h» 1, then it follows from the obtained expressions that the stresses with x in absolute value exceed the stresses m and, especially, and u. So, for example, when 1/I == 10 we get a x / m xy \u003d 20 ‘, o x / c y \u003d 300.
Thus, the greatest practical interest in the calculation of beams for bending are the stresses a x, beams acting in cross sections. Voltage with y, characterizing the mutual pressure of the longitudinal layers of the beam, are negligible compared to o v .
The results obtained in this example show that the hypotheses introduced in § 7.5 are well founded.
In the case of transverse bending in the sections of the beam, not only a bending moment occurs, but also a transverse force. Consequently, in this case, not only normal, but also tangential stresses arise in the cross sections of the beam.
Since the tangential stresses are generally unevenly distributed over the cross section, then, strictly speaking, the cross sections of the beam do not remain flat during transverse bending. However, at (where h- cross section height, l- the length of the beam) it turns out that these distortions do not noticeably affect the work of the beam in bending. In this case, the hypothesis of flat sections is also acceptable with sufficient accuracy in the case of pure bending. Therefore, the same formula (5) is used to calculate normal stresses.
Consider the derivation of calculation formulas for shear stresses. Let's single out from a bar experiencing transverse bending an element with a length (Fig. 6.28, a).
Rice. 6.28
With a longitudinal horizontal section drawn at a distance y from the neutral axis, we divide the element into two parts (Fig. 6.28, in) and consider the equilibrium of the upper part, which has a base of width b. At the same time, taking into account the law of pairing of tangential stresses, we obtain that the tangential stresses in the cross section are equal to the tangential stresses arising in the longitudinal sections (Fig. 6.28, b). Taking into account this circumstance and from the assumption that shear stresses are uniformly distributed over the area, using the condition , we obtain:
where is the resultant of normal forces in the left cross section of the element within the shaded area:
Taking into account (5), the last expression can be represented as
where is the static moment of the part of the cross section located above the y coordinate (in Fig. 6.28, b this area is shaded). Therefore, (15) can be rewritten as
As a result of joint consideration of (13) and (16), we obtain
or finally
The resulting formula (17) is named after the Russian scientist DI. Zhuravsky.
Strength condition for shear stresses:
where is the maximum value of the transverse force in the section; - allowable shear stress, it is usually equal to half.
To study the stress state at an arbitrary point of a beam experiencing transverse bending, we select an elementary prism from the composition of the beam around the point under study (Fig. 6.28, G), so that the vertical platform is part of the cross section of the beam, and the inclined platform is an arbitrary angle relative to the horizon. We accept that the selected element has the following dimensions along the coordinate axes: along the longitudinal axis - dz, i.e. along the axis z; along the vertical axis - dy, i.e. along the axis at; along the axis X- equal to the width of the beam.
Since the vertical area of the selected element belongs to the cross section of the beam experiencing transverse bending, the normal stresses on this area are determined by formula (5), and the shear stresses are determined by the D.I. Zhuravsky (17). Taking into account the law of pairing of shear stresses, it is easy to establish that shear stresses on a horizontal platform are also equal. The normal stresses on this site are equal to zero, according to the hypothesis of the bending theory already known to us that the longitudinal layers do not exert pressure on each other.
Let us denote the values of normal and tangential stresses on the inclined area through and , respectively. Taking the area of the inclined platform , for the vertical and horizontal platforms we will have and , respectively.
Composing the equilibrium equations for an elementary cut prism (Fig. 6.28, G), we get:
from where we will have:
Consequently, the final expressions for stresses on an inclined platform take the form:
Let us determine the orientation of the site, i.e. the value at which the voltage reaches its extreme value. According to the rule for determining the extrema of functions from mathematical analysis, we take the derivative of the function from and equate it to zero:
Assuming we get:
From where we will finally have:
According to the last expression, extreme stresses arise on two mutually perpendicular areas, called main , and the stresses themselves - main stresses.
Comparing the expressions and , we have:
whence it follows that the tangential stresses on the main areas are always equal to zero.
In conclusion, taking into account the well-known trigonometric identities:
and formulas,
we determine the principal stresses, expressing from through and:
In the previous section, we saw that only normal stresses arise in pure bending. Accordingly, the internal forces are reduced to a bending moment in the section.
With transverse bending in the cross section of the beam, not only a bending moment arises, but also a shearing force. This force is the resultant of elementary forces lying in the section plane (Fig. 5.8).
Thus, during transverse bending, not only normal, but also tangential stresses arise. The occurrence of tangential stresses is accompanied by the appearance of angular deformations. Therefore, the hypothesis of flat sections is violated. Figure 5.9 shows a typical pattern of cross-sectional curvature.
It has been theoretically and experimentally proven that the distortion of the plane of cross sections does not noticeably affect the magnitude of normal stresses. Thus, normal stresses in transverse bending are calculated using the same formulas as in pure bending
Thus, the hypothesis of flat sections is extended to transverse bending.
Now let us determine approximately the magnitude of shear stresses during transverse bending. Let's select an element of length from the beam (Fig. 5.10).
In case of transverse bending, the moments arising in the left and right sections of the element are not the same and differ by .
With a longitudinal horizontal section drawn at a distance from the neutral layer (Fig. 5.10, b), we divide this element into two parts and consider the equilibrium condition of the upper part. On the right side, the voltage at each point is greater than on the left, because. the bending moment on the right is greater than on the left (Fig. 5.10, b).
The resultant of normal forces in the left section within the shaded area is equal to
or according to formula (5.8)
,
where is the current ordinate of the site (Fig. 5.10, b),
Static moment about the axis of the part of the area located above the longitudinal section.
In the right section, the normal force will be different
.
The difference between these forces in the right and left sections is equal to
.
This difference must be balanced by the tangential forces arising in the longitudinal section of the element (Fig. 5.10, b and c).
As an approximation, we assume that shear stresses are uniformly distributed over the width of the section.
Then 
.
From (5.11)
This formula allows you to calculate the stresses in the longitudinal sections of the beam. The stresses in the cross sections are equal to them according to the law of pairing.
Thus, the formula makes it possible to calculate shear stresses at any points along the height of the cross section.
Consider the distribution of shear stresses for some types of cross sections.
Rectangular section (Fig. 5.11).
Let's take an arbitrary point , spaced from the neutral axis at a distance . Draw a section through this point parallel to the axis; the width of this section is .
The static moment of the cut off (shaded) part is equal to
; 
,
Consequently,
.
As is known,
Substituting the obtained values into formula (5.11), we have
(5.12)
Formula (5.12) shows that shear stresses along the height of the section change according to the law of a square parabola. For we get , and for we have 
.
I-section (Fig. 5.12). A characteristic feature of this section is a sharp change in the width of the section at the transition from the I-beam wall to its flange. Basically, the transverse force is perceived by the wall, and a small amount falls on the share of the shelves.
Consider an arbitrary point (Figure 5.12). Draw a line parallel to the axis through this point. The static moment of the area of the upper cut-off part (shaded in Fig. 5.12) can be found as the sum of the static moments of the areas and:
.
This formula is valid when the point is within the vertical wall, i.e. as long as the value is within . The diagram of shear stresses for a vertical wall has the form shown in fig. 5.12.
.
.
Pure oblique bend
A bend is called oblique if the plane of acting forces passes through the axis of the beam, but does not coincide with any of the main axes of the section.
It is most convenient to consider it as a simultaneous bending of the beam in two main planes and (Fig. 5.13).
For this, the bending moment is decomposed into components relative to the axes and:
, 
.
Thus, an oblique bend is reduced to two flat bends about the axes, and . Bending moments are considered positive if they cause tension in the first quadrant.
Normal stresses at a point having coordinates and will be equal to the sum of stresses from , i.e. ,
The slope of the neutral line is
.
Because in the general case, then the condition of perpendicularity of lines, known from analytic geometry, is not observed, since
Therefore, the neutral line is not perpendicular to the plane of the moment, but is somewhat turned towards the minimum moment of inertia. The beam "prefers" bending not in the plane of the bending moment, but in some other plane, where the bending plane will be smaller.
Because diagram of normal stresses in the cross section of the ruler, then the maximum stresses occur at the point furthest from the neutral line. Let the coordinates of this point be then:
. (5.15)
The strength condition can be written as:
. (5.16)
If the section has a simple shape, then the most distant points are found immediately, if it is complex, then by drawing the section on a scale (Fig. 5.14), the position of the neutral line is plotted, and the most distant point is graphically located (Fig. 5.14).
In case of flat transverse bending, when the bending moment also acts in the sections of the beam M and shear force Q, not only normal 
, but also shear stresses 
.
Normal stresses in transverse bending are calculated using the same formulas as in pure bending:
; 
.(6.24)
P
Fig.6.11. flat bend
Shear stresses acting at the same distance at from the neutral axis, constant along the width of the beam;
Tangential stresses are everywhere parallel to the force Q.
Let us consider a cantilever beam under conditions of transverse bending under the action of a force R. Let's build diagrams of internal forces O y, and M z .
On distance x from the free end of the beam, we select an elementary section of the beam with a length dx and a width equal to the width of the beam b. Let us show the internal forces acting on the faces of the element: on the faces cd there is a transverse force Q y and bending moment M z, but on the verge ab- also transverse force Q y and bending moment M z +dM z(because Q y remains constant along the length of the beam, and the moment M z changes, Fig. 6.12). On distance at cut off part of the element from the neutral axis abcd, we will show the stresses acting on the faces of the obtained element mbcn, and consider its equilibrium. There are no stresses on the faces that are part of the outer surface of the beam. On the side faces of the element from the action of the bending moment M z, normal stresses arise:
;
(6.25)
.
(6.26)
In addition, on these faces, from the action of a transverse force Q y, shear stresses arise 
, the same stresses arise according to the law of pairing of tangential stresses on the upper face of the element.
Let's compose the balance equation of the element mbcn, projecting the resultant stresses under consideration onto the axis x:
. (6.29)
The expression under the integral sign is the static moment of the side face of the element mbcn about the axis x, so we can write
.
(6.30)
Given that, according to the differential dependencies of D. I. Zhuravsky, when bending,
,
(6.31)
expression for tangents stresses during transverse bending can be rewritten as follows ( Zhuravsky's formula)
.
(6.32)
Let's analyze Zhuravsky's formula.
Q y is the transverse force in the considered section;
J z - axial moment of inertia of the section about the axis z;
b- section width in the place where shear stresses are determined;
is the static moment about the z-axis of the part of the section located above (or below) the fiber where the shear stress is determined:
, (6.33)
where 
and F" - the coordinate of the center of gravity and the area of the considered part of the section, respectively.
6.6 Complete strength test. Dangerous sections and dangerous points
To check the bending strength, according to the external loads acting on the beam, plots of changes in internal forces along its length are built and the dangerous sections of the beam are determined, for each of which it is necessary to carry out a strength test.
With a full strength test, there will be at least three such sections (sometimes they coincide):
The section in which the bending moment M z reaches its maximum modulo value;
The section in which the transverse force Q y, reaches its maximum modulo value;
The section in which and the bending moment M z and shear force Q y reach sufficiently large values in modulus.
In each of the dangerous sections, it is necessary, having built diagrams of normal and shear stresses, to find the dangerous points of the section (strength check is carried out for each of them), which will also be at least three:
The point at which the normal stresses 
, reach their maximum value, - that is, the point on the outer surface of the beam is the most distant from the neutral axis of the section;
The point at which shear stresses 
reach their maximum value, - a point lying on the neutral axis of the section;
The point at which both normal stresses and shear stresses reach sufficiently large values (this check makes sense for sections such as a tee or I-beam, where the width of the section is not constant in height).
As it was established earlier, in the cross sections of the beam during transverse bending, not only normal, but also tangential stresses arise, causing shear deformations. By virtue of the pairing law, the same tangential stresses will also occur in longitudinal sections parallel to the neutral layer. The presence of shear stresses in longitudinal sections is confirmed by the appearance of longitudinal cracks in wooden beams during transverse bending.
Let's move on to the derivation of a formula for calculating shear stresses during transverse bending of rectangular beams. This formula was derived in 1855 by D.I. Zhuravsky. The need for such a formula was caused by the fact that in the last century, wooden structures were widely used in the construction of bridges, and timber beams usually have a rectangular cross section and do not work well for chipping along the fibers.
Consider a rectangular beam bxh (Fig. 6.19). Let in cross section 1 there is a bending moment M k, and in section 2, spaced from the first one by an infinitely close distance d z - bending moment M and + dM". On distance at from the neutral axis we draw a longitudinal section ace and consider the equilibrium of the elementary parallelepiped atps , which has measurements
The resultant of the normal internal forces acting on the face am , denote N u but acting on the edge cn - N 2; we denote the variable normal stresses in these faces by cTi and 02, respectively. In the cross section of the beam, we select an infinitely narrow strip cL4 located at a variable distance at from the neutral axis. Then
Let us assume that shear stresses in the cross section of a rectangular beam are parallel to the transverse force Q and are evenly distributed over the width of the section. Assuming that the tangential stresses m are also uniformly distributed in the longitudinal section, we determine the tangential force d F, operating on the edge ace: d F-xbdz.
Let us compose the equilibrium equation of the parallelepiped atps
:IZ = 0; N x
+ dF-N 2
= 0, whence dF \u003d N 2 - N x,
or 
Rice. 6.19
Expression J ydA there is a static moment shaded
hovannaya square And at sections relative to the neutral axis; let's denote it by S. Then
where 
Since, according to Zhuravsky's theorem,
This equality is called Zhuravsky's formula. 
Zhuravsky's formula reads like this: shear stresses in the cross section of the beam are equal to the product of the transverse force Q and the static moment S with respect to the neutral axis of the part of the section, lying above the layer of fibers under consideration, divided by the moment of inertia I of the entire section about the neutral axis and by the width b of the fiber layer under consideration.
The derived formula gives the value of shear stresses in longitudinal sections, but according to the law of pairing, at the points of the cross section lying on the line of intersection of the longitudinal and transverse planes, shear stresses of the same modulus will act.
Let's define the law of distribution of tangential stresses for a rectangular beam (Fig. 6.20, a). For fiber layer ad:
at at= ±I/ 2 t = 0;
at at= 0 t = t max = 2Q/(2bh)= 3Q/2A= Zx media /2.
Thus, in the upper and lower layers of fibers, shear stresses are equal to zero, and in the fibers of the neutral layer, they reach a maximum value. The laws of distribution of shear stresses over the width and height of a rectangular section are shown in fig. 6.20 a.
With some approximation, the Zhuravsky formula can be used to calculate shear stresses in beams with cross sections of a different shape. Let us consider a cantilever beam of a trough profile, the section of which is shown in fig. 6.20 b, bent by force Y 7 at the end.
plane 1-1 cut off a part of the shelf with an area BUT. Since the bending of the beam is transverse, then in the plane 1-1 longitudinal tangential forces and stresses will act xz(by analogy, see Figure 6.19). According to the law of pairing, tangential stresses will arise in the cross section of the shelf x x of the same value and can be calculated using the Zhuravsky formula
where Q- transverse force in the beam section; Sx- cut-off static moment BUT about the x-axis (neutral axis), S x = AhJ2 ; / - the moment of inertia of the entire section relative to the neutral axis; t- shelf thickness.
Rice. 6.20
If the flange thickness is constant, then shear stresses x x change according to a linear law; then
Resultant Rx tangential stresses in the upper shelf is equal to
The same force acts on the lower shelf R, but directed in the opposite direction. Two forces Ri form a pair with the moment M to = Rhx. Therefore, in the section, along with the vertical transverse force Q = Ri there is also a torque M k, which twists the beam. R2- resultant of shear stresses in the beam web.
To prevent torsion deformation, an external force F should be applied at some point AT on distance a from the middle of the wall and observe the condition Fa \u003d M k. From here a = M K / F. Such a point AT called bend center. If the beam section has two axes of symmetry, then the center of bending coincides with the center of gravity of the section.
Without derivation, we give a formula for determining the maximum shear stresses at round beams:
The shear stresses in the beams correspond to the shear deformation, as a result of which the flat cross-sections during transverse bending do not remain flat, as in pure bending, but are bent (Fig. 6.21).
Rice. 6.21
Most beams are designed only for normal stresses; but three types of beams should also be checked for shear stresses, namely:
- 1) wooden beams, since wood does not work well for chipping;
- 2) narrow beams (for example, I-beams), since the maximum shear stresses are inversely proportional to the width of the neutral layer;
- 3) short beams, since with a relatively small bending moment and normal stresses, such beams can experience significant transverse forces and shear stresses.
The maximum shear stress in an I-section is determined by the Zhuravsky formula. The product range tables show the values of the static moment of the half-section area for I-beams and channels.
Example 6.7
The cantilever beam, rigidly clamped at one end in the embedment, consists of two square-section wooden beams connected at the other end with a bolt (Fig. 6.22). A force is applied to the free end of the beam R= 15 kN. Beam length / = 4 m. Determine the diameter of the bolt shaft if the allowable shear stress [t cf ] = 120 MPa. The size of the cross section of the bars a = 20 cm
Rice. 6.22
Solution. In all cross sections of the beam, in addition to the bending moment, a transverse force arises Q=R= 15 kN and the corresponding tangential shear stresses calculated according to the Zhuravsky formula, and the maximum stresses m max occur on the neutral axis, that is, at the point of contact of the bars. According to the pairing law, the same shear stresses also occur in the longitudinal sections of the beam. Then
where Q - transverse force: Q = 15-10 3 N; S - static moment of the half-section area of the beam relative to the neutral axis: S= a 2 -a / 2= a r /2 ; I- moment of inertia of the entire section about the neutral axis: I - a(2a) 3 /2-2a 4 /3 ; b - section width: b= a.
Substituting these expressions into the Zhuravsky formula, we have m max \u003d 3 () / (4n 2), and substituting numerical values and taking into account the dimensions, we obtain
Shear force F= x max And sd, where is the shear area A sd = al. Consequently F== Htah aI= 0.282 10 6 0.2 4 = 226 10 3 N. Force F, acting at the junction of the beams, tends to cut off the bolt. Find the required diameter d bolt shaft based on its shear: F/A Cf) A cf - cut area equal to the cross-sectional area of \u200b\u200bthe bolt rod: D. p \u003d lx / 2/4
Substituting this expression into the calculation formula, we have,
