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Bulgara  Ceha slovaca  Croata  Engleza  Estona  Finlandeza  Franceza 
Germana  Italiana  Letona  Lituaniana  Maghiara  Olandeza  Poloneza 
Sarba  Slovena  Spaniola  Suedeza  Turca  Ucraineana 


MECHANICS OF MATERIALS
CYCLE OF LECTURES
COMPOUND STRESSES
CONTENT
2. Position of a neutral axis . . 6
3. Strength requirements for a straight rod . . . . . . . . . . . . . . . . 9
4. Kern of a cross section . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
5. Strength requirements for initially curved beams . . . . . . . . .15
6. Thinwalled pressure vessels . . . . . . . . . . . . . . . . . . . . . . . . 19
1. Stresses at any point
Structural members discussed up to this chapter were loaded very simply. Therefore single internal forces there in cross sections were met: only an axial force (pure tension or compression), only a shearing force (shearing), only a torque (torsion), only a bending moment (pure bending). The only case where we have dealt with two internal forces was bending: a bending moment and shearing force.


If the direction of the force F applied to the cantilever shown in Fig. 1 is not crossing axis z, is perpendicular to this axis and not coinciding with any principal axis (as shown in Fig. 2), axial force N = F_{z} = 0 and torque T = 0, only two bending moments and two shearing forces are acting in all the sections of the cantilever. Such a very often loading is called skew bending or oblique bending (sometimes unsymmetrical bending). The angle under its own weight loading g (Fig. 3) is causing skew bending also because the load direction does coincide with neither principal axis u nor v.
Another simple loading causing composed stresses is eccentric tension or eccentric compression: the only force causes an axial force and bending moments (Fig. 4).
We are able to determine the stress distribution when any internal force component is acting alone. When several components are acting the stress distribution is more complicated. It may be determined by the principle of superposition when this principle is valid (it is applicable when displacements are small and Hooke’s law is valid).
We know that normal stresses in cross sections are related only to the axial force N and the bending moments M_{x}, M_{y}:
_{} (1)
Shearing stresses are related only to three components also:
_{} (2)
Stresses are vectors and should be summed according to the rules of vectorial addition. All the components of the normal stresses are directed parallel to the longitudinal axis z (Fig. 5), therefore the value of the resulting normal stress at any point k is equal to an algebraic sum of all the values obtained by the formulae (1):
_{} (3)
Do not forget that x and y here as well as in the formulae (1) are related to the principal centroidal axes of the cross section.

Shearing stresses related to different internal force components are acting in different directions (Fig. 6). Their values can be calculated by (2) but no algebraic summation is applicable, the resulting stress should be found as a vector sum:
_{} (4)
An algorithm for solving any problem related to compound stresses:
1. Divide all the given loading in two parts – one acting in one principal plane and other acting in another principal plane.
2. Plot the diagrams of shearing force and bending moments for both principal planes. Separately plot the diagrams of torques and axial forces. Determine the dangerous sections by analysing all the diagrams together.
3. Determine the normal and shearing stress distributions in every dangerous section separately from every internal force component. Determine the dangerous points in cross sections by analysing all the distributions together.
4. Determine stress state in every dangerous point and write the strength requirement which is simple for onedimensional state and more complicated (based on hypotheses) for other cases.
5. If stiffness requirements should be satisfied, calculate displacements separately in both principal planes and after that determine a vector sum of all the components of any displacement.
2. Position of a neutral axis

_{}


Neutral axis divides the cross section into two areas: with positive normal stresses (the tensile part) and with negative (the compressive one).
Two cases of the neutral axis position could be analysed: when N = 0 and when N ¹ 0. The first case is termed skew bending or oblique bending. The absolute term in the neutral axis equation is equal zero in this case; it means that the neutral axis is passing the origin of the coordinate system, i. e. the centroid of the cross section. The equation can be written as follows:
_{} (6)

The neutral axis usually is not perpendicular to the load line as it was proved in the case of simple bending: now ½j½ ³ ½a½. The neutral axis is nearer to the coordinate axis of the minimum moment of inertia.

Even in the case of any complicated loading we can apply the only force loading image for the analysis of normal stresses: having values N, M_{x} and M_{y}, we can determine the value of the resultant load force F and the coordinates of its application point x_{f} = M_{y}/N, y_{f} = M_{x}/N. Applying these expressions the equation of the neutral axis is presented as follows:
_{} (7)
or (applying expression of squared radius of gyration I_{x}/A = i_{})
_{} (8)
The expression between brackets is equal to zero because F/A ¹
_{} (9)
or
_{} (10)
where
_{} _{} (11)
are coordinates of two points at which the neutral axis is crossing coordinate axes (as shown in Fig. 11). The sign of a_{x} is contraverse to the sign of x_{f} because _{} is positive. Therefore
the neutral axis is crossing the quadrant which is crosswise opposite to the quadrant of the resultant load force application. 
10.3. Strength requirements for a straight rod
When no torque and no shearing force are acting in a cross section, no shearing stress there is. The state of stresses is onedimensional. Shearing stresses caused by shearing forces are small and negligible. The only torque causes shearing stresses large significantly, especially at boundary points. Such shearing stresses should be considered, and in this case the state of stresses is multidimensional. So, the strength requirements are different for the next two cases:
onedimensional state of stresses 
multidimensional state of stresses 

t or t (T = 0) 
t ¹ (T ¹ 
Onedimensional state (T = 0)


Strength requirements for the shown section are two restrictions – for s_{max }at the point k and for s_{min} at the point j:
_{} (13)
When you have determined the coordinates of the extreme points k and j, applying strength requirements (12) and (13) is simple.
Simplier strength requirements are for cross sections which shape is rectangle or another figure with four symmetrical corners (Fig. 13). The coordinates of these corners are equal but with different signs: h/2 and b x_{max} and y_{max})._{}



_{} (15)
These expressions of the strength requirements are for the four corners of the rectangular cross section (inspite of the absense of the coordinates of these corners).
Twodimensional state (T ¹
When both normal and shearing stresses are acting at any point of a cross section, state of stresses is twodimensional. Usually shearing stresses caused by shearing forces are negligible, and only torques make to include shearing stresses into strength requirement formulations.
Hypotheses of failure and plasticity are applied when formulating strength requirements for the twodimensional state of stresses. Having normal stress s_{k} and shearing stress t_{k} at some decisive (the most dangerous) point of a cross section, strength requirement can be expressed according to the energy hypothesis as
_{} (16)
or according to the maximum shearing stress hypothesis – as
_{} (17)
Do not forget: both normal and shearing stress values in the expressions (16) and (17) should be determined for the same point k. This the most dangerous point can be found after calculating and comparing normal stresses caused by axial force N and bending moments M_{x}, M_{y}, shearing stresses caused by torque T and shearing forces Q_{x}, Q_{y}.

The shearing stresses caused by the torque T have their extreme values at the boundary points too, therefore the points a and b are the most dangerous. When axial force N = 0, values of the extreme normal stresses at these points are equal: s_{a} s_{b} ; it means that the value of normal stress at any of these points is
_{} (18)
Shearing stress at any of these points (having an expression of polar section modulus W_{p} = W_{x} + W_{y} = 2W):
_{} (19)
Then
_{} (20)
and


(energy hypothesis)
or
t_{max} hypothesis)
There W is axial section modulus (W = pd^{ }
These formulae are very handy for design, therefore they often are applied even in the cases when N, Q_{x} and Q_{y} should be considered: the required section modulus W is determined at first without any consideration of these N and Q, after that stresses s and t are calculated with the consideration and the strength requirement _{}checked.

10.4. Kern of a cross section
Masonry and some other structural materials resist compression very well but do not resist tension. An engineer should exclude any tensile stresses in masonry; all the normal stresses should be negative there. This statement means that the neutral axis should not cross the area of a cross section.
When the neutral axis is at significant distance from the centroid of the crosssectional area, the coordinates of the intersection a_{x}, a_{y} are large, and therefore the coordinates of the loading point are small (_{}). So, the designer should apply the load close to the centroid if he seeks not to meet tension. The load should be applied within the small area surrounding the centroid.
This zone of the crosssectional area applying within which a compressive load does not cause tensile stress anywhere in the section is called the kern (or core) of a section.
Boundary points of the kern can be determined by the following formulae:
_{} _{} (23)

As an example a kern of a rectangular cross section is determined (Fig. 17). There is shown calculation of coordinates of one corner point of the kern boundary; other corners are determined analogously.


As the example has shown, the kern of any rectangular cross section has a shape of diamond. It is easy to prove that the kern of a circular cross section with diameter D is also circular with diameter d = D/4.

Do not forget the main purpose of the kern: it determines the part of the crosssectional area in which you may apply a load without causing different (positive and negative) normal stresses in the whole cross section.
10.5. Strength requirements for initially curved beams
There are many structures consisting of initially curved beams: arches (Fig. 19, a), rims of wheels (b), hooks (c) etc There in this item only plane curved beams will be investigated; the axis of a plane beam and all its loads are lying in a plane of symmetry (the investigation will not cover structures similar to that one shown there in Fig. 19, d). We will discuss the beams which cross sections have dimensions comparable with the radius of curvature (a r, see Fig. 20) – it means that this radius is very small and the curvature is large.


Method of sections is applied for estimating internal forces. It is illustrated by an example (Fig. 21).
The internal forces are calculated applying equilibrium equations (SF_{z}= S F_{y}= S M_{c}=
_{}
_{}
_{}
Let be F_{1} = 5F and F_{2} = 4F, then internal forces _{} _{} _{} and their diagrams are presented in Fig. 22.
Calculation of the coordinate j_{j} for the position of the M_{extreme}:
_{} _{} _{}
M_{extreme} = M_{j} = Fr(5 – 5cos38.7 – 4 sin38.7 1.40Fr
So we have a simple curved structure and very complicated, nonlinear distributions of internal forces.
Now let us consider normal stresses in a cross section. They may be caused by an axial force and by a bending moment. Normal stresses caused by N are constant in all the area of the cross section (the distribution is the same as in a cross section of a straight rod). The distribution of normal stresses caused by the bending moment is not linear but hyperbolic (Fig. 23).
An expression for the normal stress at any point k:
_{} (24)
There in the formula (24) A is an area of the cross section, r_{k} – a distance between the point k and the centre of curvature, r_{0} – a distance between the neutral axis and the centre of curvature. The latter distance is expressed as follows:
_{} (25)
There in literature expressions of r_{0} for various shapes of cross section (rectangular, triangular, circular a. o.) are presented.
If you want to sattisfy strength in an initially curved beam, you should:
to find the extremum bending moment M_{j};
to find the value of the axial force N_{j} at the section with extremum bending moment;
to estimate the position of the neutral axis in this section (when only bending moment is acting);
to calculate normal stresses caused by N_{j} and M_{j}, to sum them at the most dangerous point (Fig. 24), to check if the strength requirements are sattisfied:
_{} _{}
10.6. Thinwalled pressure vessels

consider only the vessels with symmetrical loading (Fig. 26, a) because in this case a vessel acts as a membrane, i. e. bending of the walls does not take place. When the
A symmetrical thinwalled vessel acted by an internal symmetrical pressure p is under consideration.A segment of the vessel wall (Fig. 27) is isolated from this vessel by passing two planes trough the axis of symmetry z and two circular sections which are perpendicular to the exernal surface. This segment is presented in Fig. 28. There in two meridional cuts (ad and bc) circumferial (or hoop) stresses s_{t} are acting; meridional (or longitudinal) stresses s_{m} are acting in two hoop cuts (ab and dc). No shearing stress is in these cuts, therefore the stresses s_{t} and s_{m} are principle stresses. Stresses in the third direction, in the radial one, are very small, negligible. It means that the stress state is twodimensional.


2) four resultants of normal stresses – two s_{m}rds_{t} and two s_{t}rds_{m}; the lines of these resultants does not coincide, between directions of both equal forces there are small angles (dj and dq
The equilibrium equation (the sum of projections of all the five forces to the axis perpendicular to the surface of the shell is equal zero):
_{}
Angles are very small, therefore
_{} _{}
After substituting these expressions into equation we have:
_{} (26)
There in the equation two unknown quantities are included: s_{t} and s_{m}. Therefore we need one more equation.

_{} (27)
When the pressure p is constant (for example, pressure of gas),
P = pS,


there G is the weight of the liquid contained in the separated part of vessel.
We see that _{} in a cylindrical vessel. Hoop stresses there are twice more dangerous than meridional ones (you can prove it when inflating a cylindrical rubber balloon up to its burst: crack will be meridional – caused by hoop stresses).
In a spherical vessel p = const, _{} and the equation (26) can be solved without supplementary equations; there _{} and
_{}, _{} (32)
The strength requirements for thinwalled pressure vessels are composed by applying any strength hypothesis because the stress state is twodimensional. For example, when the energy hypothesis is applied (_{}, _{})
_{} (33)
Any designer knows that both s_{t} and s_{m} are positive and _{} Therefore a stronger hypothesis often is applied – the hypothesis of maximum normal stress:
_{} (34)
All these formulae are valid for the estimation of stress and strength at points remote from any break in a surface of the vessel (for example, from head of a boiler). At any breaking the strain state is more complicated, the head of the boiler is preventing a uniform deformation, the shell is bended, there can be noticed a socalled fringe effect (or edge effect). Consideration of any fringe effect is complicated but very important. This effect was one of the reasons of the failure of a vessel in Jonava plant “Azotas”.
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