where £ L la the body force vector per unit' mass, p is the volume density, V I is the contravariant velo
city vector and comma denotes the covariant differen
tiation.
Equation of continuity is
V ° (8 .1 .3 )
Xhe boundary conditions are the same as in the ordinary theory of elasticity.
The constitutive equations for incompressible hypoelastic bodies of grade one are
t ‘ + i >
(8 .1 .4 )
where is the extra stress tensor, t^ = ^ + | )
|j being an arbitrary hydrostatic stress and being the ordinary stress tensor when ijx , double the shear modulus, is taken as unit of stress, the dot denote® th«
, material derivative, = 'j{. 0*1, ji + * iCaU r ( f c , k " v k i ) * he10* th« covariant
- 96 -
velocity vector, comma standing for covarinnt differen- tintion, «j|= 1 + ^ t - , , - < > ^ < • the A A ) being dimensionless constants, Equations (8*1*2) and (8.1*3) are the same Tor grsde one, except for the incompressible case when (8.1*3) reduces to
= (8 .1 .5 )
8 .2 . Solid rotating shaft.
We investigate stresses in a right circular cylin
drical solid shaft of an incompressible hypoelastic
material of grade zero, rotating steadily with an angular velocity 65 about its axis. Then remembering that th®
shaft is symmetrically strained, the physical components of velocity field in cylindrical system of coordinates are taken as follows:
W- = j S C M , = a G , oj =.- i j. X , (8 . 2 . 1)
where ji O ) is a function of h, only, k = J xtv ^ ' and °( is a given constant.
■ V The equation ( 8 .1 .3 ) gives
/i = «L fl i f U. = 0 9t U o ,
Thus (3*2*1) can be rewritten in the form
U c / Jt ;
vr
->
00 =. -%<£ %0
(3.2.2)$
tions of h. only, the constitutive equations (3.1*1) and the equations of motion (8.1*2) give
i
f
SlN\ + 4j ; L J + r
( 8 .2 .4 )
•K S * , (8 .2 .S )
A f S6) - - S S = — 7 (8.2.6)
dU ^ 0 ' A- & A.
* j ; ( si ) + s * i - e, ( a - s ' 7 )
(
8.
2.
8)
- 98
f , « ) + K s* ) - ^ ( » » ax <8-a-io)
^ Sz ) +K 5* ) ° ( 8 -2 -u )
where we have taken the definition of rate of stress
given by Truesdell £ S\ J and where 1 is the unit of stress.
The boundary conditions are
S^= o \ at K s. a.j
ft ■
and J ^ 6 ^ ^
~ ° '
(
8.
2.
12)
where a. is the radius of the shaft. Sow integrating ( 8 . 2 . 3 ) , ( 8 . 2 . 6 ) , ( 8 .2 .7 ) and ( 8 . 2 .8 ) , w® have
C4- -r\ **- i
\ -
JJ
A - -L,
(8 .2 .1 3 )Se ~ 3|
ft
- 1.7
(8 .2 .1 4 )( ^ | *■) , (8 .2 .1 5 )
(8 .2 .1 6 )
where D ( , D , M and
C
are constants of integration. Obviously since S® and i 1' cannot be
^ A
Infinite at A = o f we have
^ = ^ = 0 1 (8 .2 .1 7 )
From equations (3 .2 .4 ) and ( 3 .2 . 5 ) , we obtain
^q~ k H 2' 1 ) ’ = (8 .2.18)
where F and are arbitrary functions of their arguments.
Substituting for S and S in (3 .2.10)
& X*
and (8 *2 *1 1 ), we get
F = + (8 .2 .1 9 )
where C and A are constants of integration. Since
h, i.
S. and are not Infinite at A = o , we conclude
ft X
that A stC' = o Hence
S6 =
([^ Iv) -
(f'ty)
* • (8 .2 .2 0 )100 -
How on using equations (8 .2 .1 3 ) and ( 8 .2 .1 4 ), the equa
tion (8 .2 .9 ) gives
(
8.
2.
21)
(3 .2 .2 2 ) Applying the boundary conditions ( 8 .2 .1 2 ) , we find
i> = (-11 ^ ) , (l| ,*<-) 4 . (f (l/1j .
Thus we have the following stress distribution:
S ^ £ ‘(aV j~i) ^ SJ = 1 '^ ^ a 2-—
% \
$2S ( £ a l—
, S =<1
c5 f>^l^f) ■ >
(8 .2 .2 3 )\ ~ z K ° •
We find, on comparison with Seth*s[|]solution for the same
K CK
problem, that S an<3 <-> <*o not vanish in the hypo-
& x
elastic solution.
3 . 3 . Propagation or symmetrical disturbance from a transverse cylindrical hole in an infinite piste.
'*fe consider an infinite platef of arbitrary thick
ness ^ t with a transverse cylindrical hole of radius a
the material of the piste being: incompressible hypo- elastic of grade zero* A sudden uniform shearing traction is applied at the hole and then steadily maintained. The problem is to determine the stress
distribution inside the plate. Such a problem is charac*
terized by an axis-symmetric velocity field, appropriate to a purely rotary motion. Thus in cylindrical coordi
nates we write in the deformed state
where \ is a constant. Then the basic equations for grade zero reduce to
(3 .3 .1 )
where denotes time snd h. a , I 'z. 1 ^ ^ 1J- • We take the tentative assumptions as follows:
(8 .3 .3 )
; ( 8 .3 .4 )
- 102 -
‘ t
^ x
) , (8the remaining being identically satisfied. The cross differentiation of (8 .3 .3 ) and (3 .3 .F ) give3
+ .1 __ v = ^ V v
^ K ^ J . ^ £ i . ^
The Initial and boundary conditions are
tfc S. S L- = 0 ,
t
- 0?
k ^a
, alli
tl
- S4 > t > o > a = a , IX 0 ? ■£ > 0 ; h. j
Sj -=> 0 ; t > 0 , 4 -> «>, 8ll I euJLj
Now we introduce dimensionleas variables defined by
1 = k \* ■ , V= t r | t , T = : lb|ftX ' (
(8
and } ,
(8 3.P)
3.6)
.3 .7 )
.3 .8 )
• 3«9)
i - r s *1 ) = i y \
V T ^ 6 ' n U R ( a
v r v n % J '
t ! 4- j_\ v _ v_ a * v
"d'R1- * 'bft i f (a
\) = sj =. o ? T s 0
3 %.
^ ) ? all t and j , V ~ So > T > o , ' R - | ,On defining Laplace transform oO
the equation (8 .3 .1 1 ) reduces to
T V , l >V f f j -nTT
^ r 7 k _ ( -1 + F V = ^ (
.3.11)
1.3.12) .3 .1 0 )
.3 .1 3 )
which is a modified Bessel's equation of order unity,
- 104 -
whose solution is
V = A k, (%Y)>
(8 .3 .1 4 )where K j (.'*) is a modified Bessel'3 function of order one and of second kind, Bessel’ s function of the first kind is excluded to ensure continuity at Infinity and where A is a constant of integration. Laplace trans
form of equations (8*3.10) and (8*3*12) gives
$ 6 =r ! _ ( (8 .3 .1 5 )
and % = ) , (8 .3 .1 6 )
Under conditions ( 8 . 3 . 1«) and ( 8 .3 .1 6 ) , the solution ( 8 .3 .1 4 ) becomes
v = A s . K l c n ) | W R ,
se=
- s . Kt Cty) J j. . (8 .3 .1 7 ), Inverting the Laplace transformations, we get
v - i R . r
A U'l J . — L oo ■ 7. If% k, C'2-)a- ^ '
CH*>
which can also be written as
* T /
U
■e k, CU) 1 1
k,60 ; (8 . 3. 2 0 )V S . * l + S .- * iy € K j C ^ (8 .3 .2 1 )
where ^ ~ — t • ^ ^ +- ‘ ^ 4 i is tha zero of ki ('£/) in the upper half plan® and denotes the real psrt of.
It may be noted that when X = I * i . e . , 5 ^~o , our results coincide with those of Soodier and Jashman £ 5"j J .
8 . 4 . Simple extension of an incompressible right circular solid cylinder under the time dependent axial load.
We consider a body in the form of a right circular solid cylinder, of initial radius f\ , being pulled by
a force L , applied at the plane end, that varies with time and analyse its deformation. I f , O c Z j and (A, &/ z j denote the positions of a typical particle of the cylinder before and after the deforma
tion respectively, we have:
/t*. Kt ( £ ) , 0 = 0o , X a (-At) ■ <8.4.1)
The velocity field i s , then, as follows:
= 0 , U) = —X X 4 ^ (8 .4 .2 )
where £ is a function of tisse
t
to be determined later and £' = d. e [ cU • Using the Lagrangian coordinates \ , 60 , 2.„ x*JL X $ the constitutive equa
tions ( 3 .1 .4 ) of grade one reduce to
(8 .4 .8 )
(8 .4 .6 )
\ ' '*> \
1 t I
'T 7 ‘
at " 7-/ (8 .4 .7 )
A / i
ot
v *8 ’,
I • (8.4*8)
From ( 8 .4 .6 ) f (8 .4 .7 ) rand ( 3 .4 .3 ) , we have:
J, ® i, & / \ 5 j ! ■ "\
V V ^ ' ^ 1 7 ^ 9 ;
where £* (o) , t £ (& ) and (o)
^ ft i
, t ^ and t ^ respectively at r = o . Similarly froa ( 8 .4 .3 ) and ( 8 . 4 .4 ) , we see that
( 8 .4 .9 )
are the values
t*.
(8 .4 .1 0 )Whichever sign is given to ^ , atleast one of the shear stresses is indefinitely amplified i f ever it ha*
- 108 -
any non zero value. There i s , thus, an apparent lnsta- bility in the solution we propose to peek.
s
I f , however, we take a non-zero initial value for any one of the shear stresses, irrespective of , the equations of motion ( 3 .1 .2 ) generally become inconsistent. Accordingly we shall explore the nature of solutions free from shear stresses. We shall also assume that t ^ A. uS for alla - S
t
. The equations(8.4.3)
to (8 .4 .8 ) then further reduce to(8 .4 .1 1 )
= - u ( j « ,
(3 .4 .1 2 )
The equations of motion (8*1*2) becoffi®
N / i ^ I \ f \ 'V\
— \) = ^ fJ t ) ' & + S )> (8.4*13)
bk o /
<5 i
j * . 1
(8 .4 .1 S )
It follows from (8 .4 .1 4 ) and the assumption
A 9
* ' A
that and 5 ^ are independent of 0o • We
—r
further take it for granted that S “ and are also independent of B0 . How from (8 .4 .1 3 ) and (8.4.1*?), we get by cross differentiation
/ T- _ f z \ _
k. >
0 0
which gives
(3 .4 .1 6 )
where p and are arbitrary functions of integra
tion. By subtracting (8 .4 .1 2 ) from (8 .4 .1 1 ) and using ( 3 .4 .1 6 ) , we obtain
t ' - ^1
>1
~ “ 9— \x - i 1 j s>- * c 4f
>
(8 .4 .1 7 )
(8.4.1s)
- 110 -
Substituting the values of and in
(3*4.11 ) o p (3 .4 .1 2 ) w® have
V
f -/ / ' (:/ i- 4-■" i , 5 X. -} + '• ' C'i+ J f ^ o )s ' siC. *
— X ) (3X'|+ ^ e) - 0 .
(8 .4 .1 9 ) Splitting (3 .4 .1 9 ) into two parts, we write
i - "-f
b 7
I
where X ~ ^ and H <^fc) *s an arbitrary
V -J * *
function. Solving (3 .4 .2 0 ) and ( 3 .4 .2 1 ) , we get respec
tively:
* and
Mf
(3 .4 .2 3 )
where
() . ■ ' / 1. „ ■ / . i |4-
~ " " ^ T ' <<u -h 0 U^jy
W 1 U >
'» , .. j.'
and i.1^- T £' U . / ) } ^ (8 .4.24)
* /
and - are arbitrary functions. Integrating u
(3 .4 .1 3 ) w. r.to ,1, | we have
SX s * = £fy*/A ° i ' (8 .4 .2 5 )
where
V
Is also an arbitrary function. Using ( 8 .4 .1 5 ) , ( 3 .4 .1 6 ) and (3 .4 .2 5 ) one can calculatet
-'•> ' ^ ) ~ S t ' ‘^ y ' ^ ~Kf ■ r i ^ - - xe’) Jr U tJj (8 .4 .2 6 )where 'kH :) is again arbitrary. This means that (8 .4 .2 5 ) becomes
C*1 !?> ' .*■ 2-4/ •• ,"t , L —4 6, . ’u . . I
~ Ijt /V ' e v* + [jii -i£ )> f-
13
+ 2 + U t ) , . (8 .4 .2 7 )
112
The boundary condition viz. S*= o at i e e>cf (e).
for all Z„ and for all i requires, unless we are ready to choose special values of )i( and , that
it i ^
£ £ 6 ~ 0 y ~ = 0 J
l = - e ,£ Afi,
(8 .4 .2 3 )
Thus we have
S K - A - _ ( h'L a'-\ ^’ ■*- (8>4>a))
Using ( 8 .4 .1 7 ) , (B .4 .2 2 ), ( 8 .4 .2 3 ) , (8 .4 .2 8 ) and ( 8 .4 .2 9 ), we get
*klO
(3 .4 .3 0 )
Therefore fro® (8*4*18) * (8 *4 *22 ), (8 *4 *2 3 ), (8*4*5$3) and ( 8 .4 * 3 0 ) , we obtain
S2 =
-C 3
+St ^ e ) + I f
e2^ 6>2*, (8 .4 .3 1 )Applying the initial conditions : (i ) S z - o t =. o £ s o for all and (i i ) \» o
4: = o , 6 = o for all
kt
, one obtainsat at
& A o ) ~ ^ 0 {(H ^ /*■ ( ^ - ^ ) (V ^
A 8 .4 . 32)
end
^ ( O
33)
Thus we finally get
s \ i » = i f
't & -~'
£ = i L
- A L) e f ( ( e ) (8.4.3ft)
' i -
* ^ -A,) ^ K A<6j j
+
rl
(8 .4 .3 * )
These agree in nature with those of Green / _?• We can get the load versus strain diagram by using the condition
- 114 .heJf (e)
= % IT
he (q) . d [hoi') ,
(8.4.36)
where L is the load applied at any time. Writing this in detail, we get
ITT) -• i L / ) » * » _ llv.l'
Ujicy
_ _ 3 x l /)2 (7>i_ x * i ) r pa1- /
--— -— 1 1 + 1 * . . < .) l L ^ r -kv)
(8.4.37)
8 . P. Expansion or contraction of a cylindrical tube.
We consider the expansion or contraction of a right circular cylindrical tube of initial radii a, and <X under a radial load which changes with time.
I f the particle which occupied the position ^ ,,0 o /Z,j before deformation, goes to the point >
z)
aft®rthe deformation, we haves
A ? A.0 f A 0) ; b - Q0 , X - Z o , ( 8 .5 .1 )
where A is some function of *t and A 0 is its value st . The velocity field i s , then characterised by
«■ - A'(t) A 1 , o , u j (8 .5 .2 )
where dash denotes differentiation with respect to "t The constitutive equations reduce to
£ f t )
(8 .5 .3 )
\.+ atA-*,)
I [tz)
= -1 h\
Lo + & ( A ~( 8 .5 .4 )
( 8 .5 .5 )
] * •Jt;
( 8 .5 .6 )
( 8 .5 .7 )
' V .
£j
( 8 .5 .8 )
- 116
As before, In the last section, we get from (8 .p. 6 ), ( 8 . * . 7 ) and ( 8 . P .8)
(8 .* .9 )
Now it is clear that unless we take aero initial values of the shear stresses, whatever be the special value of
» the equations of motion may not be satisfied in general. Hence we seek a solution of this problem when a ll the shear stresses vanish. The solution follows on exactly the same lines as the one in Art.8 .4 . Details of the solution are given in the paper /
8 . 6 . Pure shear under loading (varying with time).
Referred to fixed Cartesian system of coordinates, the velocity field is
(
8.
6.
1)
• where € is a given function of "t and CL is a given parameter. Th? rate of deformation tenser is
0 0
o —c i 0
- 6 0 0
(
8.
6.
2)
We take the extra stress in the form
i
T S e>
n
0 - T o
o & n
so that T . =■©
In (8..*• 3 ) , S and are functions of t Here
m<= a t c i t
(3 .6 .3 )
only.
(3 .6 .4 ) Out of ( 7 . 4 .1 ) , (8 .1 .2 ) and ( 8 .1 . 3 ) , the equations that are not identically satisfied are
AT
At
= C e ^ V i / T 1
= - i / t s
(it
J t 6£ - - f C t
( 3 . 6 . c)
( 3 .6 .6 )
( 8 .6 .7 )
- 118
%L
I . cy. ( c 6 - £
Now on integration, the equations and (3*5 give
r = £ u ( ^
eand
5 = c 6c\ -j ^
where £> is a constant.
Pro® (8 .R .7 ) and ( 8 . K. 8 ) , w® have
= i L - V
(£ + Cfe1- ) ^ 6 ^ - 0 <j"
Thus the stresses are
S - T + g [ g V c 0 * + ( « * ■ - «
.3
S) ' ~ r + $ L (f'+ ce'l) * +
(
8.
6.
8)
.6 )
(8 .6 .9 )
(
8.
6.
10)
(
8.
6.
11)
(
8.
6.
12)
(8 .6 .1 3 )
(3 .6 .1 4 )
° - c X ? 'J
X (8.6.15)
where T and «S are given by (8 .5 .9 ) and (8 .5 .1 0 ).
I f we neglect Inertia terms, our solution becomes
provided we choose G> such that <S ® o when £ =^° . The relations (8 .5 .1 2 ) to (8 .5 .1 6 ) together with (8 .5 .9 ) suggest that for large values of o(. , as is usually the case, the transition from elastic to plastic region is gradual and that the^ yield stress is reached for small values of the strain in the case of Pure Shear depending on time through £ •
8 . 7 . Solid rotating shaft under simple extension.
Here we consider the problem with the help of
• equations ( 7 .4 .1 ) that are valid for incompressible hypo
elastic work hardening materials. Referred to fixed
cylindrical system of coordinates, we take the physical components of the velocity field as follows:
K C O £ = -a.A z e ,
(8 .7 .1 )
where A is a given constant, « la the constant angular velocity of the shaft and £ is a function of time to be determined. The rate of deformation tensor becomes
i s. r a ^ o o
h i o (8 .7 .2 )
» -khi
We assume the extra stress tensor In the form
ft]
ft]
so that t * +
t V
(a .7. 3)
, t t i 0 and where f J,
ar® functions of A , k and Now out of equa
tions ( 7 . 4 . 1 ) , those which are not identically satisfied
A i di + ( * ! ) - i + u V t ?
r m
/ 0 2.
Solution of (8 .7 .6 ) is
tl- f U L
—1 J
where j> is an arbitrary function. From (8 .7 .4 ) ( 8 .7 * 5 ) , we can write in a particular cas«
Therefore from the condition that t ' t-t -f =-o and from ( 8 . 7 .9 ) , we can say that
Lko( ( k e - f )
>(8.7.S)
(8 .7 .6 )
*(8.7.7)
(8 .7 .8 )
(8.7*9)
and
(8 .7 .1 0 )
122
which does satisfy (8 .7 .4 ) gnd (8.7..*) leaving <j> still undetermined. Integrating (8 .7 .7 ) partially w.r.to h and observing that i ' ( «r t , we get
V - 1 r V = £ L*‘ + (8.7.11)
H u J
c 1
Applying the boundary condition that o = o at
— x * we find the value of P and subs
titute it back in ( 8 .7 .1 1 ), we thus obtain
H a
(3 .7.12)
Thus
I
h i ’ + t L(8 .7 .1 3 ) Substituting this value of j> in ( 8 .7 .8 ) and using
( 8 . 7 . 9 ) , we get
f
r
(8 .7 .1 4 )
This can be satisfied i f
^ f ( -
0
and £ A fc - £ = 0 / (8 . 7. 15)unless ^ is small which is not physically the case.
There fore
i ‘ i
J and £ = ^ V u ) U ^ C C K + V » t >(8 .7 .1 6 ) where (2 8nd jji ar® arbitrary constants that can be determined by initial conditions snd j> ia determined from the load distribution over the plane ends. Thus we arrive at
s ' , s t =
II c ^ ) ( i * t + ty\
Si = ~ ( A e ^ j - i U / U
+ (i ft'-<£) (i-H 4 tj)'Z
(8 .7 .1 7 )
(8 .7 .1 8 )
I f o( is large, it is obvious that the transition from elastic to plastic region is gradual and that the approach
124
to the yield stress is quite rapid in the case of a solid cylindrical shaft rotating with constant angular velocity & under a given load distribution at the ends.
8 . 8 . Remarks.
The observation that we made regarding the shear stresses in Art.8 .4 and Art.8 .5 is worth noting. I f a solution of the basic equations of hypoelasticity Is possible to find in consistence with the equationsof
motion and with non-zero shear stresses, the whole theory goes down since physically in simple extension, shear stresses must die out. But it is difficult to find such a solution-wlth non-zero shear stresses in the case of simple extension and at the same time consistent with the equations of motion. The question is still open.
In other hypoelastic solutions given here, we find additional stresses that are absent in the ordinary elastic solutions. A number of solutions for hypoelastic problems using grade zero have already been given by
Truesdell
L
Al _7» Green I J \ Verma Sinha / 5 T __/ and Paria / S~6 _ /. Grade one solutions are• however few. Infact the one of simple extension, given in this thesis, is the first solution / *3
J
with inertia terms also taken into account. On plasticity side thesehypoelastic equations do help us in exploring the
transition fro® elastic to plastic stage without using any yield criterion. This of course happens in parti
cular cases only. General case is yet to be established.
The theory is in the formative stage *nd is capable of predicting many new effects. The calculation of solu
tion of the non-linear differential equations makes it a little difficult to study. In next chapter, we pro
pose to give the deformation energy of hypoelastic materials and also suggest the lines on which tempera
ture can be introduced in hypoelastic equations.