98
9 Plane Problem of Elasticity Theory
Even Radial Load The standard [3] for testing of chemically resistant ceramic
materials under the Brazilian method recommends the specimen loading scheme
given in position c in Figs. 9.3 and 9.4. In this case,
N + iT = N − iT =
−q on σ 1 σ 2 and σ 3 σ 4 ,
0 on σ 2 σ 3 and σ 4 σ 1 ,
(9.39)
we assume from the condition of static equivalence of loads at radial distributed
load as follows:
q =
p
2R sin θ
.
(9.40)
By substituting the function (9.39) into formulas (9.30), (9.31), and (9.33) and
calculating the respective integrals, we obtain Muskhelishvili functions for a circle
bearing axisymmetric even load on the arches σ 1 σ 2 ? σ 3 σ 4 :
) = −
q
2πi
ln
σ 2
2 − ζ 2
σ 2
1 − ζ 2
+
qθ
π
,
(ζ ) = −
q
2πi
·
2(σ 2
1 − σ 2
2 )
(σ 2
1 − ζ 2 )(σ 2
2 − ζ 2 )
,
(9.41)
where q is defined by formula (9.40).
Conclusions Figure 9.5 gives curves of normal stresses in diametrical points
located on the axis Ox. The adopted designations are
˜
X x (ζ ) = X x (ζ ) ·
πd
p
; ˜
Y y (ζ ) = Y y (ζ ) ·
πd
p
,
where the stress components X x and Y y are defined by equations (9.15) with known
) and ). In all positions of Fig. 9.5, a continuous straight line depicts the
component ˜
Y y under the action of concentrated forces P , and curves depicted with
dots show the component ˜
X x of the same load. Dashed curves depict the component
˜
Y y when loading the disc by the system of parallel forces (positions a and c) or
radial loads (positions b and d), and dash-and-dot curves show the component ˜
X x
at these loads. For curves in the positions a and b , the arches σ 1 σ 2 and σ 3 σ 4 are
taken equal to 5 0 , and in the positions c and d, they are equal to 10 0 .
The analysis of the curves represented in Fig. 9.5 leads [8] to the following
conclusions:
9 Plane Problem of Elasticity Theory
Even Radial Load The standard [3] for testing of chemically resistant ceramic
materials under the Brazilian method recommends the specimen loading scheme
given in position c in Figs. 9.3 and 9.4. In this case,
N + iT = N − iT =
−q on σ 1 σ 2 and σ 3 σ 4 ,
0 on σ 2 σ 3 and σ 4 σ 1 ,
(9.39)
we assume from the condition of static equivalence of loads at radial distributed
load as follows:
q =
p
2R sin θ
.
(9.40)
By substituting the function (9.39) into formulas (9.30), (9.31), and (9.33) and
calculating the respective integrals, we obtain Muskhelishvili functions for a circle
bearing axisymmetric even load on the arches σ 1 σ 2 ? σ 3 σ 4 :
) = −
q
2πi
ln
σ 2
2 − ζ 2
σ 2
1 − ζ 2
+
qθ
π
,
(ζ ) = −
q
2πi
·
2(σ 2
1 − σ 2
2 )
(σ 2
1 − ζ 2 )(σ 2
2 − ζ 2 )
,
(9.41)
where q is defined by formula (9.40).
Conclusions Figure 9.5 gives curves of normal stresses in diametrical points
located on the axis Ox. The adopted designations are
˜
X x (ζ ) = X x (ζ ) ·
πd
p
; ˜
Y y (ζ ) = Y y (ζ ) ·
πd
p
,
where the stress components X x and Y y are defined by equations (9.15) with known
) and ). In all positions of Fig. 9.5, a continuous straight line depicts the
component ˜
Y y under the action of concentrated forces P , and curves depicted with
dots show the component ˜
X x of the same load. Dashed curves depict the component
˜
Y y when loading the disc by the system of parallel forces (positions a and c) or
radial loads (positions b and d), and dash-and-dot curves show the component ˜
X x
at these loads. For curves in the positions a and b , the arches σ 1 σ 2 and σ 3 σ 4 are
taken equal to 5 0 , and in the positions c and d, they are equal to 10 0 .
The analysis of the curves represented in Fig. 9.5 leads [8] to the following
conclusions: