1.14 Another Form of Hooke’s Law
15
ε x =
1
E
[σ x − ν(σ y + σ z )],
ε y =
1
E
[σ y − ν(σ z + σ x )],
ε z =
1
E
[σ z − ν(σ x + σ y )];
(1.11)
γ xy =
τ xy
G
, γ yz =
τ yz
G
, γ zx =
τ zx
G
.
(1.12)
Formulas (1.11) and (1.12) are applicable in the case of specific limitations
applied to stress values σ x , . . . , τ zx . In engineering designs, these limitations are
often met by satisfying the requirements dictated by the conditions of strength
and/or rigidity.
1.14 Another Form of Hooke’s Law
Let us consider the change of the body element volume due to deformation. If
rib lengths of an infinitely small parallelepiped before deformation are designated
through dx, dy, dz, its volume will be dV = dxdydz. After deformation, rib
lengths will change and become equal to (1 + ε x )dx, (1 + ε y )dy, (1 + ε z )dz. By
designating the element volume after deformation through dV , we can calculate the
element volume increment due to deformation:
) = (1 + ε x )(1 + ε y )(1 + ε z )dxdydz − dxdydz.
In the case of small deformations, we neglect here the products of linear
deformations as infinitely small and of higher order, and we obtain
) = (ε x + ε y + ε z )dxdydz.
The relation of the volume increment to the initial element volume is referred to
as the volumetric expansion. By designating it through , we can write
=
(dV )
dV
= ε x + ε y + ε z .
(1.13)
Let us call the hydrostatic part of stress (σ 0 ) as the average of normal stresses for
three arbitrary orthogonal areas routed through the body point:
σ 0 =
1
3
(σ x + σ y + σ z ).
(1.14)
15
ε x =
1
E
[σ x − ν(σ y + σ z )],
ε y =
1
E
[σ y − ν(σ z + σ x )],
ε z =
1
E
[σ z − ν(σ x + σ y )];
(1.11)
γ xy =
τ xy
G
, γ yz =
τ yz
G
, γ zx =
τ zx
G
.
(1.12)
Formulas (1.11) and (1.12) are applicable in the case of specific limitations
applied to stress values σ x , . . . , τ zx . In engineering designs, these limitations are
often met by satisfying the requirements dictated by the conditions of strength
and/or rigidity.
1.14 Another Form of Hooke’s Law
Let us consider the change of the body element volume due to deformation. If
rib lengths of an infinitely small parallelepiped before deformation are designated
through dx, dy, dz, its volume will be dV = dxdydz. After deformation, rib
lengths will change and become equal to (1 + ε x )dx, (1 + ε y )dy, (1 + ε z )dz. By
designating the element volume after deformation through dV , we can calculate the
element volume increment due to deformation:
) = (1 + ε x )(1 + ε y )(1 + ε z )dxdydz − dxdydz.
In the case of small deformations, we neglect here the products of linear
deformations as infinitely small and of higher order, and we obtain
) = (ε x + ε y + ε z )dxdydz.
The relation of the volume increment to the initial element volume is referred to
as the volumetric expansion. By designating it through , we can write
=
(dV )
dV
= ε x + ε y + ε z .
(1.13)
Let us call the hydrostatic part of stress (σ 0 ) as the average of normal stresses for
three arbitrary orthogonal areas routed through the body point:
σ 0 =
1
3
(σ x + σ y + σ z ).
(1.14)