level 11
贴吧用户_QU3NtA5
楼主
代码:
Solve[{Integrate[1/
Sqrt[(2 P)/(
E0 Iy) (Cos[\[Theta]A] -
Cos[\[Theta]])], {\[Theta], \[Theta]A, \[Theta]B}] == L/2(*AB*),
Integrate[-(1/
Sqrt[(2 P)/(E0 Iy) (1 - Cos[\[Theta]]) +
d\[Theta]C^2]), {\[Theta], \[Theta]B, 0}] == \[Pi] r1/2(*BC*),
Sqrt[(2 P)/(E0 Iy) (Cos[\[Theta]A] - Cos[\[Theta]B])] +
Sqrt[(2 P)/(E0 Iy) (1 - Cos[\[Theta]B]) + d\[Theta]C^2] ==
1/r1(*B*),
Integrate[Cos[\[Theta]]/
Sqrt[(2 P)/(
E0 Iy) (Cos[\[Theta]A] -
Cos[\[Theta]])], {\[Theta], \[Theta]A, \[Theta]B}] +
Integrate[-(Cos[\[Theta]]/
Sqrt[(2 P)/(E0 Iy) (1 - Cos[\[Theta]]) +
d\[Theta]C^2]), {\[Theta], \[Theta]B, 0}] ==
1/2 \[Epsilon] Lb}, {P, \[Theta]A, \[Theta]B, d\[Theta]C}]
2019年04月21日 06点04分
1
Solve[{Integrate[1/
Sqrt[(2 P)/(
E0 Iy) (Cos[\[Theta]A] -
Cos[\[Theta]])], {\[Theta], \[Theta]A, \[Theta]B}] == L/2(*AB*),
Integrate[-(1/
Sqrt[(2 P)/(E0 Iy) (1 - Cos[\[Theta]]) +
d\[Theta]C^2]), {\[Theta], \[Theta]B, 0}] == \[Pi] r1/2(*BC*),
Sqrt[(2 P)/(E0 Iy) (Cos[\[Theta]A] - Cos[\[Theta]B])] +
Sqrt[(2 P)/(E0 Iy) (1 - Cos[\[Theta]B]) + d\[Theta]C^2] ==
1/r1(*B*),
Integrate[Cos[\[Theta]]/
Sqrt[(2 P)/(
E0 Iy) (Cos[\[Theta]A] -
Cos[\[Theta]])], {\[Theta], \[Theta]A, \[Theta]B}] +
Integrate[-(Cos[\[Theta]]/
Sqrt[(2 P)/(E0 Iy) (1 - Cos[\[Theta]]) +
d\[Theta]C^2]), {\[Theta], \[Theta]B, 0}] ==
1/2 \[Epsilon] Lb}, {P, \[Theta]A, \[Theta]B, d\[Theta]C}]