where [nu] denotes the direction of the outward-drawn normal to the
bounding surface, and X`_[nu] denotes the x-component of the applied
surface traction. The relations between stress-components and
strain-components are expressed by either of the sets of equations (1)
or (3) of S 26. The relations between strain-components and
displacement are the equations (1) of S 11, or the equivalent
conditions of compatibility expressed in equations (1) and (2) of S
16.
39. We may proceed by either of two methods. In one method we
eliminate the stress-components and the strain-components and retain
only the components of displacement. This method leads (with notation
already used) to three partial differential equations of the type
dP /dPu dPv dPw\ /dP^2u dP^2u dP^2u\
([lambda] + [mu]) --- ( --- + --- + --- ) + [mu]( ----- + ----- + ----- ) + [rho]X = 0, (3)
dPx \dPx dPy dPz/ \dPx^2 dPy^2 dPz^2/
and three boundary conditions of the type
_
/dPu dPv dPw\ | dPu
[lambda] cos(x, [nu])( --- + --- + --- ) + [mu] | 2 cos(x, [nu])---
\dPx dPy dPz/ |_ dPx
_
/dPv dPu\ /dPu dPw\ |
+ cos(y, [nu])( -- + -- ) + cos(z, [nu])( -- + -- ) | = X`_[nu], (4)
\dPx dPy/ \dPz dPx/ _|
In the alternative method we eliminate the strain-components and the
displacements. This method leads to a system of partial differential
equations to be satisfied by the stress-components. In this system
there are three equations of the type
dPX_x dPX_y dPX_z
----- + ----- + ----- + [rho]X = 0, (1 _bis_)
dPx dPy dPz
three of the type
dP^2X_x dP^2X_x dP^2X_x 1 dP^2
------- + ------- + ------- + ----------- ----- (X_x + Y_y + Z_z) =
dPx^2 dPy^2 dPz^2 1 + [sigma] dPx^2
[sigma] /dPX dPY dPZ\ dPX
- ---------[rho]( --- + --- + --- ) - 2[rho] ---, (5)
1-[sigma] \dPx dPy dPz/ dPx
and three of the type
dP^2Y_z dP^2Y_z dP^2Y_z 1 dP^2
------- + ------- + ------- + ----------- ------ (X_x + Y_y + Z_z) =
dPx^2 dPy^2 dPz^2 1 + [sigma] dPydPz
/dPZ dPY\
- [rho]( --- + --- ), (6)
\dPy dPz/
the equations of the two latter types being necessitated by the
conditions of compatibility of strain-components. The solutions of
these equations have to be adjusted so that the boundary conditions of
the type (2) may be satisfied.
40. It is evident that whichever method is adopted the mathematical
problem is in general very complicated. It is also evident that, if we
attempt to proceed by help of some intuition as to the nature of the
stress or strain, our intuition ought to satisfy the tests provided by
the above systems of equations. Neglect of this precaution has led to
many errors. Another source of frequent error lies in the neglect of
the conditions in which the above systems of equations are correct.
They are obtained by help of the supposition that the relative
displacements of the parts of the strained body are small. The
solutions of them must therefore satisfy the test of smallness of the
relative displacements.
41. Torsion.--As a first example of the application of the theory we
take the problem of the torsion of prisms. This problem, considered
first by C.A. Coulomb in 1784, was finally solved by B. de Saint-Venant
in 1855. The problem is this:--A cylindrical or prismatic bar is held
twisted by terminal couples; it is required to determine the state of
stress and strain in the interior. When the bar is a circular cylinder
the problem is easy. Any section is displaced by rotation about the
central-line through a small angle, which is proportional to the
distance z of the section from a fixed plane at right angles to this
line. This plane is a terminal section if one of the two terminal
sections is not displaced. The angle through which the section z rotates
is [tau]z, where [tau] is a constant, called the amount of the twist;
and this constant [tau] is equal to G/[mu]I, where G is the twisting
couple, and I is the moment of inertia of the cross-section about the
central-line. This result is often called "Coulomb's law." The stress
within the bar is shearing stress, consisting, as it must, of two sets
of equal tangential tractions on two sets of planes which are at right
angles to each other. These planes are the cross-sections and the axial
planes of the bar. The tangential traction at any point of the
cross-section is directed at right angles to the axial plane through the
point, and the tangential traction on the axial plane is directed
parallel to the length of the bar. The amount of either at a distance r
from the axis is [mu][tau]r or Gr/I. The result that G = [mu][tau]I can
be used to determine [mu] experimentally, for [tau] may be measured and
G and I are known.
42. When the cross-section of the bar is not circular it is clear that
this solution fails; for the existence of tangential traction, near the
prismatic bounding surface, on any plane which does not cut this surface
at right angles, implies the existence of traction applied to this
surface. We may attempt to modify the theory by retaining the
supposition that the stress consists of shearing stress, involving
tangential traction distributed in some way over the cross-sections.
Such traction is obviously a necessary constituent of any stress-system
which could be produced by terminal couples around the axis. We should
then know that there must be equal tangential traction directed along
the length of the bar, and exerted across some planes or other which are
parallel to this direction. We should also know that, at the bounding
surface, these planes must cut this surface at right angles. The
corresponding strain would be shearing strain which could involve (i.) a
sliding of elements of one cross-section relative to another, (ii.) a
relative sliding of elements of the above mentioned planes in the
direction of the length of the bar. We could conclude that there may be
a longitudinal displacement of the elements of the cross-sections. We
should then attempt to satisfy the conditions of the problem by
supposing that this is the character of the strain, and that the
corresponding displacement consists of (i.) a rotation of the
cross-sections in their planes such as we found in the case of the
circle, (ii.) a distortion of the cross-sections into curved surfaces by
a displacement (w) which is directed normally to their planes and varies
in some manner from point to point of these planes. We could show that
all the conditions of the problem are satisfied by this assumption,
provided that the longitudinal displacement (w), considered as a
function of the position of a point (x, y) in the cross-section,
satisfies the equation
dP^2w dP^2w
----- + ----- = 0, (1)
dPx^2 dPy^2
and the boundary condition
/ dPw \ / dPw \
( --- - [tau]y ) cos(x, [nu]) + ( --- + [tau]x ) cos(y, [nu]) = 0, (2)
\ dPx / \ dPy /
where [tau] denotes the amount of the twist, and [nu] the direction of
the normal to the boundary. The solution is known for a great many forms
of section. (In the particular case of a circular section w vanishes.)
The tangential traction at any point of the cross-section is directed
along the tangent to that curve of the family [psi] = const. which
passes through the point, [psi] being the function determined by the
equations
dPw /dP[psi] \ dPw /dP[psi] \
--- = [tau]( ------- + y ), --- = - [tau]( ------- + x ).
dPx \ dPy / dPy \ dPx /
The amount of the twist [tau] produced by terminal couples of magnitude
G is G/C, where C is a constant, called the "torsional rigidity" of the
prism, and expressed by the formula
_ _ _ _
/ / | /dP[psi]\^2 /dP[psi]\^2 |
C = [mu] | | | ( ------- ) + ( ------- ) | dxdy,
_/ _/ |_ \ dPx / \ dPy / _|
the integration being taken over the cross-section. When the coefficient
of [mu] in the expression for C is known for any section, [mu] can be
determined by experiment with a bar of that form of section.
43. The distortion of the cross-sections into curved surfaces is shown
graphically by drawing the contour lines (w = const.). In general the
section is divided into a number of compartments, and the portions that
lie within two adjacent compartments are respectively concave and
convex. This result is illustrated in the accompanying figures (fig. 4
for the ellipse, given by x^2/b^2 + y^2/c^2 = 1; fig. 5 for the
equilateral triangle, given by (x + (1/3)a) [x^2 - 3y^2 - (4/3)ax +
(4/9)a^2] = 0; fig. 6 for the square).
[Illustration: FIG. 4.]
44. The distribution of the shearing stress over the cross-section is
determined by the function [psi], already introduced. If we draw the
curves [psi] = const., corresponding to any form of section, for
equidifferent values of the constant, the tangential traction at any
point on the cross-section is directed along the tangent to that curve
of the family which passes through the point, and the magnitude of it is
inversely proportional to the distance between consecutive curves of the
family. Fig. 7 illustrates the result in the case of the _equilateral_
triangle. The boundary is, of course, one of the lines. The "lines of
shearing stress" which can thus be drawn are in every case identical
with the lines of flow of frictionless liquid filling a cylindrical
vessel of the same cross-section as the bar, when the liquid circulates
in the plane of the section with uniform spin. They are also the same as
the contour lines of a flexible and slightly extensible membrane, of
which the edge has the same form as the bounding curve of the
cross-section of the bar, when the membrane is fixed at the edge and
slightly deformed by uniform pressure.
[Illustration: FIG. 5.]
[Illustration: FIG. 6.]
[Illustration: FIG. 7.]
45. Saint-Venant's theory shows that the true torsional rigidity is in
general less than that which would be obtained by extending Coulomb's
law (G = [mu][tau]I) to sections which are not circular. For an elliptic
cylinder of sectional area [omega] and moment of inertia I about its
central-line the torsional rigidity is [mu][omega]^4/4[pi]^2I, and this
formula is not far from being correct for a very large number of
sections. For a bar of square section of side a centimetres, the
torsional rigidity in C.G.S. units is (0.1406)[mu]a^4 approximately,
[mu] being expressed in dynes per square centimetre. How great the
defect of the true value from that given by extending Coulomb's law may
be in the case of sections with projecting corners is shown by the
diagrams (fig. 8 especially no. 4). In these diagrams the upper of the
two numbers under each figure indicates the fraction which the true
torsional rigidity corresponding to the section is of that value which
would be obtained by extending Coulomb's law; and the lower of the two
numbers indicates the ratio which the torsional rigidity for a bar of
the corresponding section bears to that of a bar of circular section of
the same material and of equal sectional area. These results have an
important practical application, inasmuch as they show that
strengthening ribs and projections, such as are introduced in
engineering to give stiffness to beams, have the reverse of a good
effect when torsional stiffness is an object, although they are of great
value in increasing the resistance to bending. The theory shows further
that the resistance to torsion is very seriously diminished when there
is in the surface any dent approaching to a re-entrant angle. At such a
place the shearing strain tends to become infinite, and some permanent
set is produced by torsion. In the case of a section of any form, the
strain and stress are greatest at points on the contour, and these
points are in many cases the points of the contour which are nearest to
the centroid of the section. The theory has also been applied to show
that a longitudinal flaw near the axis of a shaft transmitting a
torsional couple has little influence on the strength of the shaft, but
that in the neighbourhood of a similar flaw which is much nearer to the
surface than to the axis the shearing strain may be nearly doubled, and
thus the possibility of such flaws is a source of weakness against which
special provision ought to be made.
[Illustration: FIG. 8.--Diagrams showing Torsional Rigidities.
(1) Rectilineal square. .84346. .88326.
(2) Square with curved corners and hollow sides. .8186. .8666.
(3) Square with acute angles and hollow sides. .7783. .8276.
(4) Star with four rounded points, being a curve of the eighth degree.
.5374. .6745.
(5) Equilateral triangle. .60000. .72552.]
[Illustration: FIG. 9.]
46. _Bending of Beams._--As a second example of the application of the
general theory we take the problem of the flexure of a beam. In this
case also we begin by forming a simple intuition as to the nature of the
strain and the stress. On the side of the beam towards the centre of
curvature the longitudinal filaments must be contracted, and on the
other side they must be extended. If we assume that the cross-sections
remain plane, and that the central-line is unaltered in length, we see
(at once from fig. 9) that the extensions (or contractions) are given by
the formula y/R, where y denotes the distance of a longitudinal filament
from the plane drawn through the unstrained central-line at right-angles
to the plane of bending, and R is the radius of curvature of the curve
into which this line is bent (shown by the dotted line in the figure).
Corresponding to this strain there must be traction acting across the
cross-sections. If we assume that there is no other stress, then the
magnitude of the traction in question is Ey/R, where E is Young's
modulus, and it is tension on the side where the filaments are extended
and pressure on the side where they are contracted. If the plane of
bending contains a set of principal axes of the cross-sections at their
centroids, these tractions for the whole cross-section are equivalent to
a couple of moment EI/R, where I now denotes the moment of inertia of
the cross-section about an axis through its centroid at right angles to
the plane of bending, and the plane of the couple is the plane of
bending. Thus a beam of any form of section can be held bent in a
"principal plane" by terminal couples of moment M, that is to say by a
"bending moment" M; the central-line will take a curvature M/EI, so that
it becomes an arc of a circle of radius EI/M; and the stress at any
point will be tension of amount My/I, where y denotes distance (reckoned
positive towards the side remote from the centre of curvature) from that
plane which initially contains the central-line and is at right angles
to the plane of the couple. This plane is called the "neutral plane."
The restriction that the beam is bent in a principal plane means that
the plane of bending contains one set of principal axes of the
cross-sections at their centroids; in the case of a beam of rectangular
section the plane would bisect two opposite edges at right angles. In
order that the theory may hold good the radius of curvature must be very
large.
47. In this problem of the bending of a beam by terminal couples the
stress is tension, determined as above, and the corresponding strain
consists therefore of longitudinal extension of amount My/EI or y/R
(contraction if y is negative), accompanied by lateral contraction of
amount [sigma]My/EI or [sigma]y/R (extension if y is negative), [sigma]
being Poisson's ratio for the material. Our intuition of the nature of
the strain was imperfect, inasmuch as it took no account of these
lateral strains. The necessity for introducing them was pointed out by
Saint-Venant. The effect of them is a change of shape of the
cross-sections in their own planes. This is shown in an exaggerated way
in fig. 10, where the rectangle ABCD represents the cross-section of the
unstrained beam, or a rectangular portion of this cross-section, and the
curvilinear figure A'B'C'D' represents in an exaggerated fashion the
cross-section (or the corresponding portion of the cross-section) of the
same beam, when bent so that the centre of curvature of the central-line
(which is at right angles to the plane of the figure) is on the line EF
produced beyond F. The lines A'B' and C'D' are approximately circles of
radii R/[sigma], when the central-line is a circle of radius R, and
their centres are on the line FE produced beyond E. Thus the neutral
plane, and each of the faces that is parallel to it, becomes strained
into an _anticlastic surface_, whose principal curvatures are in the
ratio [sigma] : 1. The general appearance of the bent beam is shown in
an exaggerated fashion in fig. 11, where the traces of the surface into
which the neutral plane is bent are dotted. The result that the ratio of
the principal curvatures of the anticlastic surfaces, into which the top
and bottom planes of the beam (of rectangular section) are bent, is
Poisson's ratio [sigma], has been used for the experimental
determination of [sigma]. The result that the radius of curvature of the
bent central-line is EI/M is used in the experimental determination of
E. The quantity EI is often called the "flexural rigidity" of the beam.
There are two principal flexural rigidities corresponding to bending in
the two principal planes (cf. S 62 below).
[Illustration: FIG. 10.]
[Illustration: FIG. 11.]
[Illustration: FIG. 12.]
48. That this theory requires modification, when the load does not
consist simply of terminal couples, can be seen most easily by
considering the problem of a beam loaded at one end with a weight W, and
supported in a horizontal position at its other end. The forces that are
exerted at any section p, to balance the weight W, must reduce
statically to a vertical force W and a couple, and these forces arise
from the action of the part Ap on the part Bp (see fig. 12), i.e. from
the stresses across the section at p. The couple is equal to the moment
of the applied load W about an axis drawn through the centroid of the
section p at right angles to the plane of bending. This moment is called
the "bending moment" at the section, it is the product of the load W and
the distance of the section from the loaded end, so that it varies
uniformly along the length of the beam. The stress that suffices in the
simpler problem gives rise to no vertical force, and it is clear that in
addition to longitudinal tensions and pressures there must be tangential
tractions on the cross-sections. The resultant of these tangential
tractions must be a force equal to W, and directed vertically; but the
direction of the traction at a point of the cross-section need not in
general be vertical. The existence of tangential traction on the
cross-sections implies the existence of equal tangential traction,
directed parallel to the central-line, on some planes or other which are
parallel to this line, the two sets of tractions forming a shearing
stress. We conclude that such shearing stress is a necessary constituent
of the stress-system in the beam bent by terminal transverse load. We
can develop a theory of this stress-system from the assumptions (i.)
that the tension at any point of the cross-section is related to the
bending moment at the section by the same law as in the case of uniform
bending by terminal couples; (ii.) that, in addition to this tension,
there is at any point shearing stress, involving tangential tractions
acting in appropriate directions upon the elements of the
cross-sections. When these assumptions are made it appears that there is
one and only one distribution of shearing stress by which the conditions
of the problem can be satisfied. The determination of the amount and
direction of this shearing stress, and of the corresponding strains and
displacements, was effected by Saint-Venant and R.F.A. Clebsch for a
number of forms of section by means of an analysis of the same kind as
that employed in the solution of the torsion problem.
[Illustration: FIG. 13.]
49. Let l be the length of the beam, x the distance of the section p
from the fixed end A, y the distance of any point below the horizontal
plane through the centroid of the section at A, then the bending
moment at p is W(l - x), and the longitudinal tension P or X_x at any
point on the cross-section is - W(l - x)y/I, and this is related to
the bending moment exactly as in the simpler problem.
50. The expressions for the shearing stresses depend on the shape of
the cross-section. Taking the beam to be of isotropic material and the
cross-section to be an ellipse of semiaxes a and b (fig. 13), the a
axis being vertical in the unstrained state, and drawing the axis z at
right angles to the plane of flexure, we find that the vertical
shearing stress U or X_y at any point (y, z) on any cross-section is
2W[(a^2 - y^2){2a^2(1 + [sigma]) + b^2} - z^2a^2(1 - 2[sigma])]
---------------------------------------------------------------.
[pi]a^3b(1 + [sigma])(3a^2 + b^2)
The resultant of these stresses is W, but the amount at the centroid,
which is the maximum amount, exceeds the average amount, W/[pi]ab, in
the ratio
{4a^2(1 + [sigma]) + 2b^2}/(3a^2 + b^2)(1 + [sigma]).
If [sigma] = 1/4, this ratio is 7/5 for a circle, nearly 4/3 for a flat
elliptic bar with the longest diameter vertical, nearly 8/5 for a flat
elliptic bar with the longest diameter horizontal.
In the same problem the horizontal shearing stress T or Z_x at any
point on any cross-section is of amount
4Wyz{a^2(1 + [sigma]) + b^2[sigma]}
- -----------------------------------.
[pi]a^3b(1 + [sigma])(3a^2 + b^2)
The resultant of these stresses vanishes; but, taking as before
[sigma] = 1/4, and putting for the three cases above a = b, a = 10b,
b = 10a, we find that the ratio of the maximum of this stress to the
average vertical shearing stress has the values 3/5, nearly 1/15, and
nearly 4. Thus the stress T is of considerable importance when the
beam is a plank.
As another example we may consider a circular tube of external radius
r0 and internal radius r1. Writing P, U, T for X_x, X_y, Z_x, we find
4W
P = - -----------------(l - x)y,
[pi](r0^4 - r1^4)
_
W | /
U = ------------------------------- |(3 + 2[sigma]) (r0^2 + r1^2 - y^2
2(1 + [sigma])[pi](r0^4 - r1^4) |_ \
_
r0^2r1^2 \ |
- ------------- (y^2 - z^2) ) - (1 - 2[sigma])z^2|
(y^2 + z^2)^2 / _|
W
T = - ------------------------------
(1 + [sigma])[pi](r0^4 - r1^4)
_ _
| r0^2r1^2 |
| 1 + 2[sigma] + (3 + 2[sigma]) ------------- | yz;
|_ (y^2 + z^2)^2 _|
and for a tube of radius r and small thickness t the value of P and
the maximum values of U and T reduce approximately to