Question

求大神用matlab求解三角函数 非常急！！！！！！！！！万分感谢。

L1=4 L2 =4 求解未知数入

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Answer 1

在matlab中输入
>> L1=4;
>> L2=4;
>> EI1=0.53*10^12;
>> EI2=0.073*10^12;
>> syms lamda;
>> A=-(sinh(L1*lamda)-sin(L1*lamda)-L2*lamda*(cosh(L1*lamda)-cos(L1*lamda)))/(cosh(L1*lamda)+cos(L1*lamda)+L2*lamda*(sin(L1*lamda)-sin(L1*lamda)))
A =

(-sinh(4*lamda)+sin(4*lamda)+4*lamda*(cosh(4*lamda)-cos(4*lamda)))/(cosh(4*lamda)+cos(4*lamda))

y=EI1*(cosh(L1*lamda)+cos(L1*lamda)+A*(sinh(L1*lamda)+sin(L1*lamda)))*(-1.64*(cos(1.64*L2*lamda)/sin(1.64*L2*lamda)+cosh(1.64*L2*lamda)/sinh(1.64*L2*lamda)))+5.3792*EI2*(A*(cosh(L1*lamda)-cos(L1*lamda))+sinh(L1*lamda)+sin(L1*lamda))

y =

(530000000000*cosh(4*lamda)+530000000000*cos(4*lamda)+530000000000*(-sinh(4*lamda)+sin(4*lamda)+4*lamda*(cosh(4*lamda)-cos(4*lamda)))/(cosh(4*lamda)+cos(4*lamda))*(sinh(4*lamda)+sin(4*lamda)))*(-41/25*cos(164/25*lamda)/sin(164/25*lamda)-41/25*cosh(164/25*lamda)/sinh(164/25*lamda))+392681600000*(-sinh(4*lamda)+sin(4*lamda)+4*lamda*(cosh(4*lamda)-cos(4*lamda)))/(cosh(4*lamda)+cos(4*lamda))*(cosh(4*lamda)-cos(4*lamda))+392681600000*sinh(4*lamda)+392681600000*sin(4*lamda)
solve(y)
ans =

-226.83621981532200097501890241426
你看这个和你会不会有出入

追问

应该是有很多解的 需要比零大的最小3个解 麻烦大神再帮帮忙啊 万分感谢

追答

E1I1(column)=0.53*10^12 E2I2(beam)=0.073*10^12里面column 和beam是怎么回事