The value of equation
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syms x u x3
%x = (0:1:4)
L1 = 8;
L = L1
h1 = 0.1*L
W1 = 17;
f(x)=(4*h1*x/L1)-((4*h1*(x^2))/L1^2);
m1 = diff(f(x));
RAh = 3*W1*L1/8;
RBh = W1*L1/8
HAh = (W1*L1^2)/(16*h1)
HBh = HAh;
H_riser = HAh
VAh = RAh
VBh = RBh
%x = (0:1:L/2);
SFnhac = VAh-W1*x
u1 = m1
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Antworten (1)
Star Strider
am 5 Mai 2021
I have absolutely no clue as to what you want.
Try this —
syms x u x3
%x = (0:1:4)
L1 = 8;
L = L1
h1 = 0.1*L
W1 = 17;
f(x)=(4*h1*x/L1)-((4*h1*(x^2))/L1^2);
m1 = diff(f(x));
RAh = 3*W1*L1/8;
RBh = W1*L1/8
HAh = (W1*L1^2)/(16*h1)
HBh = HAh;
H_riser = HAh
VAh = RAh
VBh = RBh
%x = (0:1:L/2);
SFnhac = VAh-W1*x
u1 = m1
solve_u1_for_x = solve(u1)
.
3 Kommentare
Walter Roberson
am 6 Mai 2021
I am not sure why you are not being consistent about what you substitute.
X = (0:1:L/2);
SFnhac = VAh-W1*x
u1 = m1
u11=atan(u1,L/2)
u1ac = vpa(subs(u1,x,X))
u2 = 1/((1+m1^2)^0.5)
u2ac = vpa(subs(u2, x, 0:1:4))
u3=u1*u2
u3ac = vpa(subs(u3, x, 0:1:4))
Qac = SFnhac*u2-H_riser*u3
subs(Qac, x, X)
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