Stress resolution
An element's stress state is essentially three-dimensional, generally with both normal and shear components in each of the three dimensions. The components are usually deduced by
superposition of load building blocks as in the stress example of the previous section. The two- and one-dimensional cases illustrated here are particularisations of what is essentially three-dimensional.
The element may be rotated about all three axes into a unique
principal orientation in which all shear stresses vanish. The corresponding normal stresses in this principal orientation are termed the
principal stresses. A stress state is characterised most succintly by the principal stresses, say ( σ1, σ2, σ3 ), and failure theories - the next step in failure assessment - are expressed in terms of principal stresses. It is therefore necessary to examine how principal stresses are derived from Cartesian components.
Stress is a
tensor entity, so complex tensor arithmetic must be applied in the three- dimensional case to evaluate stress components as the element rotates. We consider here only the simpler two dimensional case in which one direction is known to be principal, and resolution consists of element rotation only about that principal axis. Thus in the 2-dimensional sketch above, the z-axis is principal and x-y stress components vary as the element rotates about that axis.
Fortunately the great majority of practical cases are two -dimensional since the relatively simple loading in conjunction with the natural choice of axes leads to one of the axes being principal automatically. For example there can be no stress on the free surface of a body, so the surface normal is an obvious choice for one of the three Cartesian axes - as there can be no shear on this surface, the axis is automatically a principal axis.
Although we consider only
"two-dimensional stresses", it is important to remember always that
stress states are essentially three dimensional.In this course the positive convention adopted for the orientation of a plane (characterised by its normal) is shown at (a) below, and for normal and shear stress and strain at (b) - positive shear is counter-clockwise. The two sketches (c) illustrate the consistency of the positive senses for shear stresses and strains; the total shear strain (distortion) is γ = 2( γ/2).
Using these conventions, the plane stress state at 'A' in the example of the previous section is :
σx = -93, σy = 0, τxy = +34, ( τyx = -34) MPa
We now examine the variation of normal and shear stress components as the inclination of the face on which they occur changes.
Consider the elemental unit cube ( size 1x1x1 ) under the known positive 2-dimensional stress components shown in (
i) below - the third dimension (z) is principal.
Rotational equilibrium requires
complementary shear, that is τyx must equal -τxy. This necessity has been incorporated into (
ii), from which it is apparent that the three components ( σx, σy, τxy ) together with the third principal are necessary to define the stress state. The three components are called the
Cartesian triad.We wish to evaluate - in terms of the Cartesian triad - the stress components ( σ, τ) in the general direction θ, so we consider force equilibrium of the wedge element (
iii), one of whose faces is inclined at θ. The height of the wedge remains 1 unit, however the dimension in the x-sense becomes 1.tanθ, and the length of the hypotenuse is 1.secθ.
The force components on each face of the wedge are the stress components multiplied by the face area - these are shown in (
iv). For force equilibrium of the wedge :
in the σ-sense :- σ.secθ - ( σx + τxy.tanθ ) cosθ - ( σy.tanθ + τxy ) sinθ = 0
in the τ-sense :- τ.secθ + ( σx + τxy.tanθ ) sinθ - ( σy.tanθ + τxy ) cosθ = 0
which on simplification give the required resolution equations :-
σ = 1/2 ( σx + σy ) + 1/2 ( σx - σy ) cos 2θ + τxy sin 2θ
τ = - 1/2 ( σx - σy ) sin 2θ + τxy cos 2θ
Similar equations but with different signs are encountered in the literature - sign differences arise from positive conventions other than the above.
The simultaneous occurence of sine and cosine terms in these equations makes it difficult to visualise how the resolved components ( σ, τ) vary as the direction θ changes. More easily interpreted equations result if the stress state is defined by the
basic triad ( σm, σa, θp ) rather than by the Cartesian triad. θp is the inclination of the plane of the
maximum principal stress with respect to the x-reference.
The basic triad is related formally to the Cartesian as follows :-
(
3a) σm = 1/2 ( σx + σy )
σa cos 2θp = 1/2 ( σx - σy ) that is σa = √[ {1/2 ( σx - σy ) }2 + τxy2 ]
σa sin 2θp = τxy θp = 1/2 arctan [ 2 τxy /( σx - σy ) ]
. . . . though these equations seldom need to be implemented.
Making these substitutions leads to resolution equations in the more meaningful form :-
(
4a) σ = σm + σa cos 2 ( θp - θ ) {
nomenclature explanation }
τ = σa sin 2 ( θp - θ )
It is apparent that normal and shear stress components vary sinusoidally with direction θ (not unlike vector components) however the variation is second harmonic - that is stress components are the same along axes which lie at 180o to one another. The two sinusoids are of the same amplitude σa and out of phase by 45o.
σm is the constant component of the normal stress.
The principal and maximum shear stresses follow immediately as :-
(
5) σmax = σm + σa ( at θp )
σmin = σm - σa ( at θp - π/2 )
τmax = σa ( at θp - π/4 )
These relations are often expressed graphically via Mohr's stress circle, in which σm and σa represent the circle's centre location and radius respectively. The
conventions require that angles on the circle, reckoned from the X-radius, are
double the corresponding angles on the element (which are measured from the x-reference), and in the
opposite sense. The reader should confirm that this construction satisfies equation (
4a).
This example demonstrates typical stress resolution using the simple trigonometry of Mohr's circle rather than the formal resolution equations derived above. The example also shows clearly the variation of stress components with orientation, θ, as predicetd by (
4a).
This script resolves the Cartesian components of a two-dimensional stress state into the principal components.
It is important to remember that
a stress-strain state is always essentially three- dimensional, involving three principals. We have addressed only resolution in two dimensions to obtain a single Mohr's circle involving the principals in these two dimensions - however two other circles must exist relating these two principals with the third.
As noted above the third (eg. automatic) principal is usually deduced from the nature of the problem. Two common situations arise when the state is either one of :
plane stress : If there is no stress orthogonal to the 1-2 resolution plane then σ3 = 0
plane strain : If there is no strain orthogonal to the 1-2 resolution plane then ε3 = 0 and it follows from (
2) that σ3 = ν ( σ1 + σ2 )
Fig F completes the Mohr's circles for the example above (Fig D) assuming that the element is loaded in plane stress, that is the three principals are ( -600, -100, 0 ) MPa.
Fig G illustrates the three principals and three Mohr's circles for a completely unrelated stress state where two-dimensional resolution happens to relate to the largest (3-2) circle and σ1 is the principal stress orthogonal to the resolution plane.
The outcome of stress resolution at an element must be a set of
three principals - all three must be known before the element's safety can finally be assessed by application of an appropriate
failure theory.
Strain resolutionResolution of strain is generally unnecessary when assessing the safety of common engineering components, however the following description is given for completeness.
If a material behaves in a linear elastic manner then the directions of principal strains are identical to the directions of principal stresses, and all the preceding equations, and Mohr's circles, may be expressed in strain terms - provided that everywhere in the stress equations :
normal stress, σ, is replaced by normal strain, ε, and
shear stress, τ, is replaced by half the shear strain, ie. by γ/2.
The strain analogues of the foregoing are therefore :-
(
3b) εm = 1/2 ( εx + εy ) ; εa = 1/2 [ ( εx - εy )2 + γxy2 ]1/2 ; θp = 1/2 arctan [ γxy /( εx - εy ) ]
(
4b) ε = εm + εa cos 2 ( θp - θ ) ; γ/2 = εa sin 2 ( θp - θ )
Experimentally, surface stresses are found from strain gauges attached to accessible surfaces. It is useful therefore to be able to quickly interrelate stresses and strains for a plane stress state, for which, since the principal stress and strain directions coincide, it follows from (
2) and (
3) that :-
(
6) E εm = ( 1 - ν ) σm ; E εa = ( 1 + ν ) σa
which relate the centres and radii of the two circles. If these
circles are drawn to scale, and it is required that their circumferences coincide for ease of drawing, then it may be shown that :
(
7) E $ε = ( 1 + ν ) $σ ; ( 1 + ν ) Cε = ( 1 - ν ) Cσ where
$ε is the scale of the strain circle ( strain units/mm ) $σ is the scale of the stress circle ( MPa/mm ) Cε is the distance (mm) of the strain circle centre from the shear axis Cσ is the distance (mm) of the stress circle centre from the shear axis
Although the use of scaled Mohr's circles is not necessarily advocated, it is strongly recommended that the circles are at least sketched free-hand as an aid to interpretation of the equations. It is important that skill is developed in visualising the interplay between components and principals.
The following example demonstrates application of strain-to-stress transformation.
The 45o lines are each associated with a particular value of stress difference, ie. of equivalent stress from ( 9), and hence of safety factor. No point can appear to the left of the equality line on this graph, as this would imply a reversal of the minimum / maximum roles. The equivalent stress varies from zero on the equality line, through to the strength, S, on the failure locus (n=1).
(read "the change in energy equals work done"). Work done results in a change in energy somewhere in the system and a change in energy requires work to be done by or against some agent..
In the diagram at left no work is done moving an object along a horizontal direction when there is no friction (recall Galileo's principle of inertia. No force is required to keep an object moving. A small amount of work is necessary to start it moving and an equal amount is "given back" when it is stopped.)
10.3. In fact there are no other combinations which conserve energy
