Page 98 - 《爆炸与冲击》2026年第10期

第 46 卷 胡 刚，等： 斜交节理岩体中纵波传播特性的理论研究 第 10 期

sinα 1 c P
= 。
式中： α 2 = γ 1 +γ 2 −α 1 β 3 = γ 1 +γ 2 −β 1 ，
，
sinβ 1 c S
( ) ( )
X = 1 0 Y = 0 1 ，分别对式 (6) 进行左乘可以得到：
设矩阵 、矩阵
Å ã
v J1
J1
−1
−1
v TP(i) = XC A 1 v IP(i) + XC B 1 RP(i) (8)
J1
1
1
v
RS(i)
Å J1 ã
v
−1
−1
v J1 = YC A 1 v IP(i) +YC B 1 RP(i) (9)
TS(i)
1
J1
1
v RS(i)
v J2 = v J1 v J2 = v J1 ，所以将式 (8) 代入式 (1)～(2)，式 (9) 代入式 (3)～(4)，同时考虑节理的非
由于 IP(i) TP(i) ， IS(i) TS(i)
[ ( )] 2
k ni + σ (i) /d max
K n = ，可以得到：
线性，设
k ni
Å ã Å Å ãã Å ã
v J2 v J1 v J2
RP(i) −1 −1 −1 RP(i) −1 TP(i) =
1
2
1
2
v J2 =− B A 2 XC A 1 v IP(i) + XC B 1 v J1 + B C 2 v J2
RS(i) RS(i) TS(i)
Å ã Å ã
v J1 v J2
−1 −1 −1 −1 RP(i) −1 TP(i)
− B A 2 XC A 1 v IP(i) − B A 2 XC B 1 + B C 2 (10)
2 1 2 1 v J1 2 v J2
RS(i) TS(i)
Å ã Å Å ãã Å ã Å ã
v J2 v J1 v J2 v J2
TP(i+1) −1 −1 −1 RP(i) −1 RP(i) −1 TP(i) =
1
2
2
v J2 =G D 2 XC A 1 v IP(i) + XC B 1 v J1 +G E 2 v J2 +G F 2 v J2
1
2
TS(i+1) RS(i) RS(i) TS(i)
Å J1 ã Å J2 ã Å J2 ã
v v v
−1 −1 −1 −1 RP(i) −1 RP(i) −1 TP(i)
G D 2 XC A 1 v IP(i) +G D 2 XC B 1 J1 +G E 2 J2 +G F 2 J2 (11)
2 1 2 1 v 2 v 2 v
RS(i) RS(i) TS(i)
Å ã Å ã Å ã
J2 J2 J2
z P cos2β 2 K ∆tcosα 2 −K ∆tcosα 2 K ∆tsinβ 2
n
n
n
式 中 ： A 2 = ， D 2 = J2 ， E 2 = J2 J2 ， F 2 =

z P sin2β 2 tanβ 2 cotα 2 k ∆tsinα 2 k ∆tsinα 2 k ∆tcosβ 2
s
s
s
Å ã Å ã
J2 J2
−K ∆tcosα 2 +z P cos2β 2 −K ∆tsinβ 2 +z S sin2β 2 z P cos2β 2 z S sin2β 2
n n
J2 J2 ， G 2 = 。其
−k ∆tsinα 2 +z P sin2β 2 tanβ 2 cotα 2 k ∆tcosβ 2 −z S cos2β 2 z P sin2β 2 tanβ 2 cotα 2 −z S cos2β 2
s
s
余符号同式 (6)。
Å J2 ã Å Å J1 ãã Å J2 ã
v RP(i) −1 −1 −1 v RP(i) −1 v TP(i) =
3
3
1
1
v J2 =− B A 3 YC A 1 v IP(i) +YC B 1 v J1 + B C 3 v J2
RS(i)
TS(i)
RS(i)
Å J1 ã Å J2 ã
v v
−1 −1 −1 −1 RP(i) −1 TP(i)
− B A 3 YC A 1 v IP(i) − B A 3 YC B 1 J1 + B C 3 J2 (12)
3 1 3 1 v 3 v
RS(i) TS(i)
Å J2 ã Å Å ãã Å J2 ã Å J2 ã
v v J1 v v
TP(i+1) −1 −1 −1 RP(i) −1 RP(i) −1 TP(i) =
1
3
v J2 =G D 3 YC A 1 v IP(i) +YC B 1 v J1 +G E 3 v J2 +G F 3 v J2
3
3
1
TS(i+1) RS(i) RS(i) TS(i)
Å J1 ã Å J2 ã Å J2 ã
v v v
−1 −1 −1 −1 RP(i) −1 RP(i) −1 TP(i)
G D 3 YC A 1 v IP(i) +G D 3 YC B 1 J1 +G E 3 J2 +G F 3 J2 (13)
3 1 3 1 v 3 v 3 v
RS(i) RS(i) TS(i)
Å ã Å ã
J2
z S sin2β 3 K ∆tsinβ 3
n
A 3 = D 3 = ，其余符号同式 (6)、(10)～(11)。
式中： ， J2
−z S cos2β 3 −k ∆tcosβ 3
s
根据叠加原理 [22] ，将式 (10) 和式 (12) 叠加，式 (11) 和式 (13) 叠加，可得纵波入射非线性斜交节理（下
标 o 代表 oblique）的传播方程为：
Å ã Å ã Å ã
v J2 v J1 v J2
RP(i) RP(i) TP(i) (14)
v J2 = −A o v IP(i) − B o v J1 +C o v J2
RS(i) RS(i) TS(i)
Å J2 ã Å J1 ã Å J2 ã Å J2 ã
v TP(i+1) v RP(i) v RP(i) v TP(i) (15)
v J2 = D o v IP(i) + E o v J1 + F o v J2 +G o v J2
TS(i)
TS(i+1)
RS(i)
RS(i)
( ) ( )
−1
−1
−1
−1
−1
−1
−1
−1
−1
A o = B A 2 X + B A 3 Y C A 1 B o = B A 2 X + B A 3 Y C B 1 C o = B C 2 + B C 3 D o = (G D 2 X+
式中： ， ， ，
2 3 1 2 3 1 2 3 2
)
(
G D 3 Y)C A 1 E o = G D 2 X +G D 3 Y C B 1 F o = G E 2 +G E 3 G o = G F 2 +G F 3 。
−1
−1
−1
−1
−1
−1
−1
−1
−1
，
3
3
2
1
，
1
3
3
，
2
2
当入射纵波参数及斜交节理岩体初始条件已知时，反射波和透射波可以通过式 (14)～(15) 计算得
出，由此定义纵波在非线性斜交节理传播时反射系数和透射系数分别为（下标 c 代表 coefficient）：

max v Tm max v Rm
J2
J1
T mc = ，R mc = (16)
max|v IP | max|v IP |
103201-5