作业帮用matlab的简单计算
来源：学生作业帮 编辑：拍题作业网作业帮 分类：综合作业 时间：2024/06/04 20:35:00
用matlab的简单计算
syms ua1 sr1 J u1a b z sc1 n a0 r0 a u E1s Es e1 e2 sr2 H s2
q=0
sr1=0.8
J=0.02
ua1=101e3
e1=0.7
H=10
u1a=201e3
b=0.5
u=0.35
Es=8e6
f=sc1*((1-1/(ua1*(1-sr1+sr1*J)/sr1/(u1a*exp(-b*z)+ua1)+1-J))/(1-sr1))^n
d1=diff(f,z)
g=a0*(1-(1-(r0/z)^2)^(-a))-u1a*exp(-b*z)-ua1
d2=diff(g,z)
sr2=1/(ua1*(1-sr1+sr1*J)/sr1/(u1a*exp(-b*z)+ua1)+1-J)
e2=e1*sr1/sr2
E1s=Es*2*(1+e2)^2/(1+e1)/(2+e1+e2)
s2=(1+u)*(1-2*u)/E1s/(1-u)*d1+(1+u)/E1s/(1-u)*d2
for z=1:1:10
q=q+s2*0.0001
end
为什么结果中含有z,如何让结果没有z
syms ua1 sr1 J u1a b z sc1 n a0 r0 a u E1s Es e1 e2 sr2 H s2
q=0
sr1=0.8
J=0.02
ua1=101e3
e1=0.7
H=10
u1a=201e3
b=0.5
u=0.35
Es=8e6
f=sc1*((1-1/(ua1*(1-sr1+sr1*J)/sr1/(u1a*exp(-b*z)+ua1)+1-J))/(1-sr1))^n
d1=diff(f,z)
g=a0*(1-(1-(r0/z)^2)^(-a))-u1a*exp(-b*z)-ua1
d2=diff(g,z)
sr2=1/(ua1*(1-sr1+sr1*J)/sr1/(u1a*exp(-b*z)+ua1)+1-J)
e2=e1*sr1/sr2
E1s=Es*2*(1+e2)^2/(1+e1)/(2+e1+e2)
s2=(1+u)*(1-2*u)/E1s/(1-u)*d1+(1+u)/E1s/(1-u)*d2
for z=1:1:10
q=q+s2*0.0001
end
为什么结果中含有z,如何让结果没有z
用eval( )这个函数,就是把符号型计算转化为数值型计算
%%修改后不含z的%%%%%%%%%%%%%%%%
syms ua1 sr1 J u1a b z sc1 n a0 r0 a u E1s Es e1 e2 sr2 H s2
q=0
sr1=0.8
J=0.02
ua1=101e3
e1=0.7
H=10
u1a=201e3
b=0.5
u=0.35
Es=8e6
f=sc1*((1-1/(ua1*(1-sr1+sr1*J)/sr1/(u1a*exp(-b*z)+ua1)+1-J))/(1-sr1))^n
d1=diff(f,z)
g=a0*(1-(1-(r0/z)^2)^(-a))-u1a*exp(-b*z)-ua1
d2=diff(g,z)
sr2=1/(ua1*(1-sr1+sr1*J)/sr1/(u1a*exp(-b*z)+ua1)+1-J)
e2=e1*sr1/sr2
E1s=Es*2*(1+e2)^2/(1+e1)/(2+e1+e2)
s2=(1+u)*(1-2*u)/E1s/(1-u)*d1+(1+u)/E1s/(1-u)*d2
for z=1:1:10
q=eval(q+s2*0.0001)
end
%%修改后不含z的%%%%%%%%%%%%%%%%
syms ua1 sr1 J u1a b z sc1 n a0 r0 a u E1s Es e1 e2 sr2 H s2
q=0
sr1=0.8
J=0.02
ua1=101e3
e1=0.7
H=10
u1a=201e3
b=0.5
u=0.35
Es=8e6
f=sc1*((1-1/(ua1*(1-sr1+sr1*J)/sr1/(u1a*exp(-b*z)+ua1)+1-J))/(1-sr1))^n
d1=diff(f,z)
g=a0*(1-(1-(r0/z)^2)^(-a))-u1a*exp(-b*z)-ua1
d2=diff(g,z)
sr2=1/(ua1*(1-sr1+sr1*J)/sr1/(u1a*exp(-b*z)+ua1)+1-J)
e2=e1*sr1/sr2
E1s=Es*2*(1+e2)^2/(1+e1)/(2+e1+e2)
s2=(1+u)*(1-2*u)/E1s/(1-u)*d1+(1+u)/E1s/(1-u)*d2
for z=1:1:10
q=eval(q+s2*0.0001)
end