LIST OF SYMBOLS
F, P Hfj, C m
1.5 Principal Stresses as Co-ordinates
STRESS ANALYSIS 21
rx = GX cosflf + t^sin ar, ry = e^ sins + i^cosa'and rt = 0
The shear stress on this plane is found from the substitution of rx, ry and rz into eq(l. 12b) as
^sin2 a+ r^sin 2a)1
0 ( 1 - 2sin2^]
r2 = (azcos &+ c^sin a f + (e^sin a + r^cos df - {e^
= (of - axay + o*) sin2*cos2ar+ rw sin 2«[ax(l -
2 2 2 sin12a)
- rv( o i - 6 0 sin 2arcos
= [^(ax- 00 s i n 2 « - i ^ c o s 2 s ]2
r = % (oi - o p sin 2ar - r^cos 2m
The respective matrix forms for this plane reduction is found from the transformation eq(1.22a): T ' = MTMT in which T and M are now 2 x 2 matrices:
°2l' ff2/
= "^11 ln CTii °i2
hi hz
where, In = cos*, ln = cos(90 - ei) = sin a, 4 i= cos(90 + a) = - sinarand i^ = cosaf. Figure 1.13b shows that the stress state (au', % ' ) applies to the x / - direction and thus the matrix equation reduces to
< =
hi hi
°\\ axi hi hi
where the following associations are made with the engineering notation an = ax, <% = flf, and 0I2 = TV in the co-ordinate directions and <?u' = cand a^ = rfor the oblique plane.
In finding the plane principal stresses (0; and 00, the invariants in eqs(1.25a-e) become:
1 x y, 2 x y t ^ and J3 = 0 and the principal stress cubic eq( 1.24b) reduces to a quadratic
a2 - (ax + a-y)ff+ (trx ay- rv 2) = 0 The solution gives the two principal stresses as roots
a%2 = ¥i(ax+ cry) ± ¥i A (ax - ayf + 4 r ^ ] where the positive discriminant applies to ov
Figure 1.14 Oblique plane set in principal stress axes
Replacing x, y and z in eqsCl.l 1) and (1.12) with 1,2 and 3 respectively and setting TV =
= r^ = 0 gives the following reduced forms;
(1.30a,b,c) (1.31a)
= loj, r2 = mOi, r3 =
The traction across ABC can be written as
from which
r=
The direction cosines for rare, from eqs(1.13),
l
s= (r
1-lff)/T=l(a-
1- a)ft m
s= (r
2- wiff) IT= mlo^ - a) IT
ns = (r3 - «cr) / r = ra (ff3 - <j) IT
(1.31b)
(1.32a) (1.32b) (1.32c) In the mathematical notation, when xt, Xj and *, in Fig. 1.11a are aligned with the principal stress co-ordinates (1,2 and 3), the transformation eq( 1.22b) becomes
"L2
fa
^33
0 0
0
ff22
0 0 0
%
'n hi l
ni
nl
u^
hi hi
hs
This will contain the expressions (1.31 a,b) for the normal and shear stresses on a single oblique plane ABC. Identifying direction cosines /,„ lX2 and ln for the normal to ABC gives its three
stress components as
STRESS ANALYSIS 23
11 7
n
0 0
0 a
n0 0 0
%
in which cr= oj/ and r s V [ (%')
2+ 1.5.1 Maximum Shear Stress
It can be shown from eq(1.31b) that maximum shear stresses act on planes inclined at 45°
to two principal planes and are perpendicular to the remaining plane. For the 1-2 plane in Fig. 1.15, for example, the normal n to the 45° plane shown, has directions I = m = cos 45°
= 1A/2 and n = cos 90° = 0. 3
\
n (I, m, n) Figure 1.15 Maximum 45" shear planeSubstituting /, m and n into eq(l .3 lb), the maximum shear stress for this plane is T
%i = a/fl + a£n - (o|/2 + e^flf = V* (tr? + of - 2<7^) = V4 (^ -
r
M= ± J 4 ( o i - c 5 ) (1.33a) where the subscripts 1, 2 refer to those principal planes to which r i s equally inclined.
Similarly, for the plane inclined at 45° to the 1 and 3 directions {l = n= lMl and m = 0), the maximum shear stress is
i-fli) (1.33b) and, for the plane inclined at 45° to the 2 and 3 directions, where n = m = 1/^2 and I = 0, the maximum shear stress is
ff
s) (1.33c)
The greatest of the three shear stresses, for a system in which a
t> <% > % is *"„„= r
w.
When the 45° shear planes are considered in all four quadrants they form a rhombic
dodecahedron. The normal stress acting upon the planes of maximum shear stress are found from eq(1.31a). For example, with l = m = \H1 and M = 0 for the 45° plane shown in Fig.
1.13, the normal stress is
ff=M(0l + fl&) (1.33d) 1.5.2 Octahedral Plane
The octahedral plane is equally inclined to the principal directions. It follows from eq(1.10) that the direction cosines of its normal are I = m = n = 1A/3 {a= fi= y= 54.8°). Substituting these into eq{L31a), gives the octahedral normal stress
ffo = %(Gi + ^+«73) (1.34a)
Since aa is the average of the principal stresses it is also called the mean or hydrostatic stress, am. The octahedral shear stress is found from substituting I = m = n = 1/V3 in eq(1.31b):
- IT,
(a, -
r0 = % /[(oi - ^ )Z + (<r2 - ffj)2 + (oi - ^ )2] (1.34b) Combining eq(1.34b) with eqs(1.33a,b,c) gives &o in terms of the three maximum shear
ro = (2B) ^(r^ + rM2 + rM 2) (1.34c) The direction of ro is found from eqs(1.32a,b,c) as
(1.35a) (1.35b) (1.35c)
"When the eight octahedral planes in all four quadrants are considered they form the faces of the mgular octahedron, shown in Fig. 1.16.
3
Figure 1.16 Octahedral planes
STRESS ANALYSIS 25
Here, tro and r0 act on the eight planes while T%.%, rM and %3 act along their edges. The deformation that arises from a given stress state may be examined on an octahedral basis.
Since ao acts with equal inclination and intensity it causes a recoverable elastic volume change irrespective of the principal stress magnitudes. Superimposed upon this is the distortion produced by %. Since eq( 1.34b) shows that the magnitude of ro depends upon differences between the principal stresses, a critical value of ra will determine whether the deformation will be elastic or elastic-plastic. In Chapter 3 it is shown that a yield criterion may be formulated on this basis.
Example 1 3 Given the principal steesses at = 7.5, at = 3.1 and c% = 1.4 (MPa) find (i) the maximum shear stresses and their directions and (ii) the magnitude and direction of the normal and shear stresses for the octahedral plane. Confirm the answers using a Mohr's stress circle.
r,MPa
A*
Figure 1.17 Mohr's circle showing max shear and octahedral shear planes
(i) From eqs(1.33a,b,c) the three shear stress maxima are:
rM = ± %(7.5 - 3.1) = ± 2.20 MPa, at 45° to the directions 1 and 2 and perpendicular to 3 r2.3 = ± M(3.1 - 1.4) = ± 0.85 MPa, at 45° to the directions 2 and 3 and perpendicular to 1 r,.3 = ± ^4(7.5 - 1.4) = ± 3.05 MPa, at 45° to the directions 1 and 3 and perpendicular to 2 (ii) From eq(1.34a) the normal stress on the octahedral plane is:
aa = % (7.5 + 3.1 + 1.4) = 4.0 MPa equally inclined to directions 1,2 and 3 The octahedral shear stress is, from eq( 1.34b),
r0 = % • [ (7.5 - 3.1)* + (3.1 - 1.4 )2 + (7.5 - 1.4 )2] = 2.57 MPa with direction cosines, from eq(1.35a,b,c):
lo = (7.5 - 4.0 ) / (73 x 2.57} = 0.786 mo = (3.1 - 4.0) / (73 x 2.57) = - 0.202 n0 = (1.4 - 4.0) / (73 x 2.57) = - 0.584
Mohr's circle (see Fig. 1.17) confirms each answer. The maximum positive shear stresses are the vertical radii of the three circles at A, B and C. Complementary negative shear stresses are associated with the reflective points A', B' and C . To locate the octahedral plane, mark off lines ai&=p= 54.7° (I = m = n = 1/73) from oj, e^ and ff3 as shown. With centres CM and CJ_J draw arcs to intersect at a point P whose co-ordinates are &g and TB. When P
is coincident with A this locates a maximum shear plane where tt= fi= 45° and y~ 90°.
Similarly with P at B and C the orientation of two farther shear planes are confirmed.
1.5.3 Reductions to Plane Principal Stress
The plane stress transformation equations are particular cases of the foregoing 3D equations.
In Fig. 1.18a, the applied principal stresses are er, and a2 and the oblique plane is defined with respective directions: / = cosflf, m = cos(90* - a) = sinaand n = 0 for directions 1,2 and 3.
and 3.
(a) (b) Figure 1.18 Reduction to plane stress
Substituting these into eq(1.31a,b) gives the normal and shear stresses for the oblique plane:
a= oi cos2 ar+ ^sin2 a= oi (1 + cos 2a)/2 + aj(l - cos 2s>)/2 er= %(ai + 00 + %(oi - fit;) cos 2a
t2= <T,2COSZ«+ o^sin2flr- (ojeos:!ir+
STEESS ANALYSE
Alternatively, these can be found from the transformation matrix T ' = MTM T. In MI, this takes the plane form
° 2 l ' ff22'.
=
hi hi
'21 *22
0 «,
hi hi In h2.
Referring to Fig. 1.18b, the direction cosines for the xt' axis become: ln = cos a and ln = cos(90 - flf) = s a w and for the %' axis: 4i - cos (90 + as) = - sinar and 122= cosaf. The required stress components «ru' and a%{ for the plane normal to x,' are found from the reduced matrix multiplication:
hi hi hi 1 hi
0
h
hi
where the correspondence is «ru' = trand CTB' = r. Full matrix multiplication will also supply those stress components <%', <%' on the plane normal to %', as shown in Fig. 1.18b.
1.6 Alternative Stress Definitions
The components ffy, of the Cauchy stress tensor T, appearing in eq(l .6), refer to the elemental deformed area da of the surface on which the stress traction vector rw acts. Thus, as the foregoing stress transformations apply to the geometry of the deformed material, the stress components are said to be true stresses. Provided deformations are less than 1%, as with the elasticity of metallic materials, it is unnecessary to distinguish between the initial and current areas. A nominal stress, calculated from the original area, will give the Cauchy stress with acceptable accuracy. For example, in a tensile test on a metal, it will only become necessary to convert nominal stresses to true stresses within the reduced cross-sectional area when deformation advances well into the plastic range. However, it is unacceptable to calculate Cauchy elastic stress components from the original area where large elastic deformations arise in certain non-metals. When the elastic extension in an incompressible rubber reaches 300% the cross-sectional area will have diminished by 67%. This means that the nominal stress is only 1/3 of the true stress! Clearly, the choice between initial and current areas can become critical to defining stress properly. The Cauchy definition may be regarded as a true stress. Alternative definitions of nominal stress, given by Piola and Kirchhoff, employ the original area for convenience.
In Fig. 1.19 the resultant force <SF is shown for both the reference and current configurations, Xt and xt, respectively. The traction vectors rTO and r°* are associated with the vectors N and n, lying normal to the reference and current areas *M and fia. The corresponding stress tensors are T (with Cauchy components <% in*,) and S (with nominal components Sg in X-). Since the force $? is common between the two configurations, it follows from eqs(l.S) and (1.6) that i*0 = i*0, when:
or
(1.36)Sa
xl,X1
Figure 1.19 Tractions in reference and current configurations
To express the relationship between n, N, Sa and SA it is convenient to take, without loss of generality, infinitesimal triangular areas. Let vectors dk0' and Srf® lie along the sides of
Sa and vectors SKf1'* and <SX® lie along the sides of SA as shown. The areas appear within their cross products
or or
(1.37a) (1.37b) where e^ is the alternating unit tensor. The latter gives: (i) unity for clockwise permutations of differently valued indices, i.e. em = e231 = %2 = 1, (ii) minus unity for anticlockwise differently valued indices, i.e. em = em = em = - 1 and (Hi) zero when any two or more indices are the same, e.g. em = %j = 0 etc. Now, since x = x (X) or x, = x, (Xj), it follows:
^ J or <5z = (3xf dX)SK (1.38)
Substituting eq(1.38) into eq( 1.37a) and making appropriate changes to the indices,
n, Sa = %epOV BX^SX™ (dx
kf ax
r)ffl
r»
Multiplying both sides by dx,/ dX^ leads to
nddxJdX^da = Vieljk(dxl/dXp)(dxildXq)(dxkf3Xr)MM « , » (1.39) Use is now made of the identity
% Fj pFi, F4 r (1,40a)
where Ftp = 3JC, tdXp etc, are the deformation gradients and J = det F = |ck, fdXj | is the Jacobian determinant:
STRESS ANALYSIS
dx
lfdX
1dx
3/dX
1Sx
$/dX
2A simple physical interpretation of / is as a ratio between material densities (or volume elements) in the undeformed and deformed configurations. That is, J = p
Blp= d WdF
0. Combining eqs{1.39) and (1.40);
and substituting from eq( 1.37b)
(1.41) or
or The matrix form may further be expressed as
Substituting eq(L41) into eq(1.36) gives
= S
l,
ior
This transpose of the nominal stress tensor (S^ = S) defines the first Piola-Kirchoff stress tensor. It is seen that the components Spi are related to the Cauchy sixess components a^ and the deformation gradients as follows;
p p