29.2 General Dependencies in Pure Shear
379
−
ϕ[v 0 , β 0 , ξ] sin ξdξ = 0.
Due to the symmetry in pure shear, the resultant of all slips in the slip plane
coincides in direction with the maximum tangential stress in this plane, i.e.
ctg c = − sin v 0 tg 2β 0 .
Taking into account this dependency, the second of formulas (29.4) looks as
follows:
r x x = tg 2β 0
v 2
v 1
F [v 0 , β 0 ] sin
2 v 0 dv 0 .
(29.5)
By applying the mean value theorem to the integrals (29.3) and (29.4), we can
represent as follows:
r x x = tg 2β 0 F [v, β 0 ](v 2 − v 1 ) sin
2 v, v 1 v v 2 ,
r x y = F [ ˜
v, β 0 ](v 2 − v 1 ), v 1 ˜
v v 2 .
The latter dependencies follow that if the segment [v 1 , v 2 ] is small, the component r x x is small and of the second order as compared to r x y and can be neglected,
which proves the formulated theorem.
29.2 General Dependencies in Pure Shear
Let us assume a = b = 0 in formula (25.3) and represent [3] the operator of shear
resistance S νλ ϕ nl as follows:
S νλ ϕ nl = ψ + [g ((r νλ + ε ˙
r νλ ) + c (γ νλ + ε ˙
γ νλ )] − AA νλ − BB.
(29.6)
In the case of pure shear (τ xy = 0),
τ νλ = τ xy (sin ω cos 2β − cos ω cos αα sin 2β),
γ νλ = γ xy (sin ω cos 2β − cos ω cos αα sin 2β).
(29.7)
Shear resistance in the slip area equals the tangential stress: S νλ ϕ nl = τ νλ . When
substituting the expressions (29.6)–(29.7) into this equation at α = ω = π/2,
379
−
ϕ[v 0 , β 0 , ξ] sin ξdξ = 0.
Due to the symmetry in pure shear, the resultant of all slips in the slip plane
coincides in direction with the maximum tangential stress in this plane, i.e.
ctg c = − sin v 0 tg 2β 0 .
Taking into account this dependency, the second of formulas (29.4) looks as
follows:
r x x = tg 2β 0
v 2
v 1
F [v 0 , β 0 ] sin
2 v 0 dv 0 .
(29.5)
By applying the mean value theorem to the integrals (29.3) and (29.4), we can
represent as follows:
r x x = tg 2β 0 F [v, β 0 ](v 2 − v 1 ) sin
2 v, v 1 v v 2 ,
r x y = F [ ˜
v, β 0 ](v 2 − v 1 ), v 1 ˜
v v 2 .
The latter dependencies follow that if the segment [v 1 , v 2 ] is small, the component r x x is small and of the second order as compared to r x y and can be neglected,
which proves the formulated theorem.
29.2 General Dependencies in Pure Shear
Let us assume a = b = 0 in formula (25.3) and represent [3] the operator of shear
resistance S νλ ϕ nl as follows:
S νλ ϕ nl = ψ + [g ((r νλ + ε ˙
r νλ ) + c (γ νλ + ε ˙
γ νλ )] − AA νλ − BB.
(29.6)
In the case of pure shear (τ xy = 0),
τ νλ = τ xy (sin ω cos 2β − cos ω cos αα sin 2β),
γ νλ = γ xy (sin ω cos 2β − cos ω cos αα sin 2β).
(29.7)
Shear resistance in the slip area equals the tangential stress: S νλ ϕ nl = τ νλ . When
substituting the expressions (29.6)–(29.7) into this equation at α = ω = π/2,