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Hello

 

I will try to be as specific as possible.

On my Ti nspire it is possible for me to calculate polar equations like on the buttom picture with the settings on the upper picture. But when I try this in Maple It is not possible. I have worked my way try for two days now and it does not work for me.

Does any body know how to get this solved? 

Regards

Heide

 

hi.please help me for solve algebric equations below?

solve or fsolve dos not any answer?

i think use Newton  Raphson, because solve or fsolve not work.

thankscomparision.mw

restart; mu1 := .1; mu2 := .1; sigma1 := -40; F := 25; upsilon1 := (1/2)*sigma2; upsilon2 := (3/2)*sigma2-sigma1; sigma2 := 100

100

(1)

gamma11 := 2.686901; -1; gamma12 := 7.175339; -1; gamma21 := 2.436735; -1; gamma22 := 12.94855; -1; gamma22 := 12.94855; -1; delta1 := .928207; -1; delta2 := .105073; -1; s11 := .629894; -1; s12 := .217601; -1; s21 := 0.73897e-1; -1; s22 := .805815

.805815

(2)

 

 

 

 
Q1 := -mu1*p1-upsilon1*q1+gamma11*q1*(p1^2+q1^2)+gamma12*q1*(p2^2+q2^2)-delta1*(2*p1*q1*p2-q2*(p1^2-q1^2))-s11*F*q1+s12*F*q2 = 0

-.1*p1-65.747350*q1+2.686901*q1*(p1^2+q1^2)+7.175339*q1*(p2^2+q2^2)-1.856414*p1*q1*p2+.928207*q2*(p1^2-q1^2)+5.440025*q2 = 0

(3)

Q2 := -mu1*q1+upsilon1*p1-gamma11*p1*(p1^2+q1^2)-gamma12*p1*(p2^2+q2^2)-delta1*(2*p1*q1*q2+p2*(p1^2-q1^2))-s11*F*p1-s12*F*p2 = 0

-.1*q1+34.252650*p1-2.686901*p1*(p1^2+q1^2)-7.175339*p1*(p2^2+q2^2)-1.856414*p1*q1*q2-.928207*p2*(p1^2-q1^2)-5.440025*p2 = 0

(4)

Q3 := -mu2*p2-upsilon2*q2+gamma21*q2*(p1^2+q1^2)+gamma22*q2*(p2^2+q2^2)+delta2*(3*p1^2*q1-q1^3)+s21*F*q1 = 0

-.1*p2-190*q2+2.436735*q2*(p1^2+q1^2)+12.94855*q2*(p2^2+q2^2)+.315219*p1^2*q1-.105073*q1^3+1.847425*q1 = 0

(5)

Q4 := -mu2*q2+upsilon2*p2-gamma21*p2*(p1^2+q1^2)-gamma22*p2*(p2^2+q2^2)+delta2*(-p1^3+3*p1*q1^2)-s21*F*p1 = 0

-.1*q2+190*p2-2.436735*p2*(p1^2+q1^2)-12.94855*p2*(p2^2+q2^2)-.105073*p1^3+.315219*q1^2*p1-1.847425*p1 = 0

(6)

 

-.1*q2+((3/2)*sigma2-40)*p2-2.436735*p2*(p1^2+q1^2)-12.94855*p2*(p2^2+q2^2)-.105073*p1^3+.315219*q1^2*p1-1.847425*p1

(7)

NULL

fsolve({Q1, Q2, Q3, Q4}, {p1, p2, q1, q2})

fsolve({-.1*p1-65.747350*q1+2.686901*q1*(p1^2+q1^2)+7.175339*q1*(p2^2+q2^2)-1.856414*p1*q1*p2+.928207*q2*(p1^2-q1^2)+5.440025*q2 = 0, -.1*p2-190*q2+2.436735*q2*(p1^2+q1^2)+12.94855*q2*(p2^2+q2^2)+.315219*p1^2*q1-.105073*q1^3+1.847425*q1 = 0, -.1*q1+34.252650*p1-2.686901*p1*(p1^2+q1^2)-7.175339*p1*(p2^2+q2^2)-1.856414*p1*q1*q2-.928207*p2*(p1^2-q1^2)-5.440025*p2 = 0, -.1*q2+190*p2-2.436735*p2*(p1^2+q1^2)-12.94855*p2*(p2^2+q2^2)-.105073*p1^3+.315219*q1^2*p1-1.847425*p1 = 0}, {p1, p2, q1, q2})

(8)

solve(Q1, Q2, Q3, Q4)

Error, invalid input: too many and/or wrong type of arguments passed to solve; first unused argument is -.1*q1+34.252650*p1-2.686901*p1*(p1^2+q1^2)-7.175339*p1*(p2^2+q2^2)-1.856414*p1*q1*q2-.928207*p2*(p1^2-q1^2)-5.440025*p2 = 0

 

Sol := [fsolve(ZZ, omega)]



Download comparision.mw

Here is a code

Maple Worksheet - Error

Failed to load the worksheet /maplenet/convert/EQ.mw .

Download EQ.mw

A fragment of code

for b in extra_bcs do try print(b = 10^(-2)); res[b] := dsolve(dsys4 union {b = 10^(-2)}, numeric, initmesh = 2024, output = listprocedure, approxsoln = [omega2 = 0.1e-2, s(x) = cosh(upsilon*x)-cos(upsilon*x)-(cosh(upsilon)+cos(upsilon))*(sinh(upsilon*x)-sin(upsilon*x))/(sinh(upsilon)+sin(upsilon)), g(x) = sin(((2*n+1)*(1/2))*Pi)], abserr = 0.1e-1) catch: print(lasterror) end try end do; indx := indices(res, nolist); nops([indx]); res[indx[i]]; seq(subs(res[indx[i]](1), omega2(1)), i = 1 .. nops([indx]))

 

 

NULL

SYSTEMATIC APPROACH TO LIFTING EYE DESIGN

Moses

 

restart

with(Optimization)

with(LinearAlgebra)

with(Plots)

Nomenclature

 

Ab  = Required bearing area, sq in. (mm2)

As  = Required shear area at hole, sq in. (mm2)

Aw = Required cheek plate weld area, sq in. (mm2)

b     = Distance from center of eye to the cross section, in. (mm)

C    = Percentage distance of element from neutral axis

D    = Diameter of lifting pin, in. (mm)

e     = Distance between edge of cheek plate and edge of main plate, in. (mm)

Fa   = Allowable normal stress, ksi (kN/mm2)

Fv   = Allowable shear stress, ksi (kN/mm2)

Fw  = Allowable shear stress for weld electrodes, ksi (kN/mm2)

Fy   = Yield stress, ksi (kN/mm2)

fa   =  Computed axial stress, ksi (kN/mm2)

fb   =  Computed bending stress, ksi (kN/mm2)

fmax = Maximum principal stress, ksi(kN/mm2)

fv    = Computed shear stress, ksi (kN/mm2)

g     = Distance between edge of cheek plate and main structure, in. (mm)

h    = Length of lifting eye at any cross section between A-A and C-C, in. (mm)

n   = Total number of lifting eyes used during the lift

P  = Design load per lifting eye, kips (kN)

R  = Radius to edge of lifting eye, in. (mm)

Rh  = Radius of hole, in. (mm)

r    = Radius of cheek plate, in. (mm)

S   = Safety factor with respect to allowable stresses

s  = Cheek plate weld size, in. (mm)

T  = Total plate thickness, in. (mm)

Tp  = Main plate thickness, in (mm)

t    = Thickness of each cheek plate, in (mm)

te  = Cheek plate weld throat, in. (mm)

W  = Total lift weight of structure, kips (kN)

α  =  Angle of taper, deg.

β  =  Angle between vertical and lifting sling, deg.

θ  =  Angle between attaching weldment and lifting sling, deg.

NULL

NULL

The design load for each lifting eye is given by:

restart

P := W*S/(n*cos(beta));

In the above equation, n refers to number of lifting eyes to used for the lift, S is the safety factor with respect to allowable stresses, W is the total weight to be lifted, and β is the angle between the vertical direction and the lifting sling.This analysis applies only to lifting eyes shaped like the one in Fig. 1. For other shapes, the designer should re-evaluate the equations.

Radius of liftimg eye hole will depend upon the diameter of the pin, D, used in the lifting shackle. It is recommeded that the hole diameter not greater than 1 / 16 in. (2 mm) larger than tha pin diameter. The required bearing area for the pin is

A__b >= P/(.9*F__y);

where Fy is the yield stress. This equation is based on allowable stresses as definde in Ref. 1, which considers stress concentrations in the vicinity of the hole. The designer may choose to use a technique which determines the stresses at the hole and should appropriately adjust the allowable stresses. The total plate thickness is then given by

T >= A__b/D;

At this point, if the thickness, T, is too large to be economically feasible, it may be desirable to use cheek plates (Fig.2) around the hole in order to sustain the bearing stresses. In this case, the above thickness, T, is divided into a main plate of thickness Tp and two cheek plates each of thickness t:

eqn1 := T = T__p+2*t;

It is recommended that t be less than Tp to avoid excessive welding. The radius to the edge of lifting eye plate and the radii of the cheek plates, if they are used, are governed by the condition that the pin cannot shear through these plates. The required area for shear is

A__s >= P/(.4*F__y);

It is possible to compute the required radii by equating the shaering area of the cheek plates plus the shearing area of the main plate to the total shear area. Theis a degree of uncertianty in choosing the appropriate shearing area. Minimum areas are used in the following equation, therefore, leading to conservative values for the radius of the main plate, R, and the radius of the cheek plate, r,

equ2 := (4*(r-R__h))*t+(2*(R-R__h))*T__p = A__s;

equ3 := R = r+e__cheek;

where Rh is the radius of the hole and e is the distance between the edge of the cheek plate and the edge of the main plate (Fig. 2). This difference should be large enough to allow space for welding the cheet plate to the main plate. A reasonable value for e is 1.5*t. It should be noted that the above equations assume there are two cheek plates. If cheek plates are not used, then simply let t equal zero and use Eq. 6 to determine R.

It is not necessary to check tension on this net section, since the allowable stress for shear is 0.4*Fy (Eq. 5); whereas the allowable stress for tension on a net section at a pin hole is given as 0.45*Fy (Ref. 1) which is greater than for shear. Size of weld between the cheek plates and the main plate can be determined as follows. The necessary weld area per cheek plate is

equ4 := A__w = P*t/(F__w*T);

where Fw is the allowable shear stress for the welding electrodes. The weld thickness, te is given by

equ5 := t__e = A__w/(2*Pi*r);

For a manual weld the size, s is given by

s := t__e*sqrt(2);

To assure that this weld size is large enough to insure fusion and minimize distortion, it should be greater than the AISC suggested Minimum Fillet Weld Sizes (Ref. 1).

The axial stress due to uniform tension along a section is

equ6 := f__a = P*sin(theta)/(T__p*h);

where h is the length of the section. The elemental bending stress which is distributed linearly along the section may be expressed as

equ7 := f__b = 12*P*C*(b*cos(theta)-.5*h*sin(theta)+R*sin(theta))/(T__p*h^2);

where C represents the distance of an element from the neutral axis and b is the distance from the center of the eye to the cross section. The shearing stress varies parabolically for section between A-A and B-B and is given as

equ8 := f__v = 1.5*P*cos(theta)*(-4*C^2+1)/(T__p*h);

It is felt that Eq. 13 (i.e., parabolic shear stress distribution) is applicable to the cross sections between A-A and B-B and does not apply to the cross sections between B-B and C-C in the area of the taper. The taper creates discontinuities on the shear plane, which result in significantly large shear stress concentratons along the edge of the taper coincident to point of maximum bending stress. This problem will be addressed ina subsequent section of this article.

The maximum principal stress that exists on an element is given by

f__max := .5*(f__a+f__b)+(((f__a+f__b)*(1/2))^.5+f__v^2)^.5;

or after dividing by the maximum allowable normal (i.e., tension) stress, Fa, gives a ratio that must be less than unity, where Fa has been taken as 0.6*Fy. A similar analysis for the maximum shear stress on the element yields

f__vmax := ((f__a+f__b)*(1/2))^2+f__v^2;

F__a := .6*F__y;

F__v := .4*F__y;

Ratio__tension := f__max/F__a;

Ratio__shear := f__vmax/F__v;

The designer should now select several critical elements throughout the plate and apply the restrains of Eq. 16 and 17 to obtain a required minimum length for the selected cross section. Eq 18 through 21 apply for an element at the neutral axis of the section. C will be zero and Eq. 11, 13 and 16 reduce to

P*sin(theta)/(.6*F__y*h*T__p)+(1.5*P*cos(theta)/(.6*F__y*h*T__p))^2 <= 1.0;

h >= P*(sin(theta)+(1+8*cos(theta)^2)^.5)/(1.2*F__y*T__p);

An element at the end of the section will be subjected to bending stresses but not shearing stresses.

For this case C = 0.5 and Eq. 16  becomes

P*sin(theta)/(.6*F__y*h*T__p)+6*P*(b*cos(theta)-.5*h*sin(theta)+R*sin(theta)) <= 1.0;

Using the quadratic formular to solve for h yields

h >= .5*(-2*P*sin(theta)/(.6*F__y*T__p)+(2*P*sin(theta)^2/(.6*F__y*T__p)+24*P*(b*cos(theta)+R*sin(theta))/(2*P*sin(theta)/(.6*F__y*T__p)))^.5);

The largest value of h predicted by Eqs. 19, 21 and 23 can be used as a first estimate for the length of the cross section; however, intermediate elements, that is, between the edge and the center of the cross section, should also be checked to determine the appropriate length, h, of the section under consideration.

Cross sections A-A and B-B should be analyzed using the above approach. The designer should use his own discretion to select other cross sections for analysis.At cross section A-A, the lifting eye is assumed to be welded with complete penetration to the support structure. Once length, h, is determined, the angle of taper, α, should be investigated. It can be shown that normal stress and shear stress are related by

equ9 := f__a+f__b = f__v*tan(alpha);

Minimum required length, h, for cross sections between B-B and C-C can be computed by calculating the maximum shear stress for the most critical element of the cross section, which occurs at the tapered surface. It can be shown that the maximum principal stress would not control the required length, h. Using Eqs. 11 and 12 in conjunction with Eq 24, the maximum shear stress yields the following:

(P*sin(theta)/(.4*F__y*h*T__p)+6*P*(b*cos(theta)-.5*h*sin(theta)+R*sin(theta))/(.4*F__y*T__p*h^2))*(.5^2+cot(alpha)^2)^.5 <= 1.0;

 

If the above inequality is not satisfied, the angle of the taper, α, must be adjusted.

The adequacy of the structure to which the lifting eye is to be attached should be checked to verify that it is capable of sustaining the loads from the lifting eye.

In some instances, it may be justifiable to use a more sophisticated technique for analyzing the lifting eye as well as the supporting structure.

 

NULL

Input Variables

 

W := 120;

n := 6;

P := W/n;

S := 3;

F__y := 300;

F__w := 450;

R := 90;

R__h := 89;

alpha := evalf(convert(45*degrees, radians));

beta := evalf(convert(30*degrees, radians));

theta := evalf(convert(20*degrees, radians));

d__pin := 100;

b := 200;

g := 50;

NULL

NULL

Output

 

solve({equ1, equ2, equ3, equ4, equ5, equ7, equ8, equ9}, {A__s, A__w, C, T, T__p, h, r, t, e__cheek});

NULL

``

NULL

NULL

NULL

NULL

NULL

NULL

NULL

 

Download Lifting_Eye_Design.mw

Good Evening Everybody,

Could any one help me with the attached file. I'm trying to solve 9 equations with 9 unknowns with many constraints, I'm getting no output from Maple. Please help.

 

Regards,

 

Moses

Hello,

 

I tried to rearrange below set of equations to have the equations in term of P[0], P[1], P[2], P[3], P[4], P[5] and P[6]. I used the symbol := for function definition for all Ps except one of them. Thus maple will rearrange that excepted one. However, I got error massage stating "Error, (in P[3]) too many levels of recursion" when I tried to rearranged equations for P[0].

Can I get help to rearranged them. 

P[0](s) = (P[1](s)*mu[1]+P[2](s)*mu[2]+1)/(s+3*lambda+3*sigma)

P[1](s) = (3*P[0](s)*lambda+3*P[3](s)*mu[1]+P[4](s)*mu[2])/(s+mu[1]+2*lambda+2*sigma)

P[2](s) = (3*P[0](s)*sigma+P[4](s)*mu[2]+P[3](s)*mu[1])/(s+mu[2]+2*lambda+2*sigma)

P[3](s) = 2*lambda(P[1](s)+P[2](s))/(s+2*mu[1]+lambda)

P[4](s) = 2*sigma(P[1](s)+P[2](s))/(s+2*mu[2]+sigma)

P[5](s) = lambda(P[3](s)+P[4](s))/s

P[6](s) = sigma(P[3](s)+P[4](s))/s

Thank you for your help

 

restart

with(LinearAlgebra):

with(StringTools):

FormatTime("%m-%d-%Y, %H:%M")

NULL

Pile spacing of each pile from a refrence pile:

S__p := Matrix(9, 9, {(1, 1) = 0., (1, 2) = 3.30, (1, 3) = 6.60, (1, 4) = 3.30, (1, 5) = 4.67, (1, 6) = 7.38, (1, 7) = 6.60, (1, 8) = 7.38, (1, 9) = 9.33, (2, 1) = 3.30, (2, 2) = 0., (2, 3) = 3.30, (2, 4) = 4.67, (2, 5) = 3.30, (2, 6) = 4.67, (2, 7) = 7.38, (2, 8) = 6.60, (2, 9) = 7.38, (3, 1) = 6.60, (3, 2) = 3.30, (3, 3) = 0., (3, 4) = 7.38, (3, 5) = 4.67, (3, 6) = 3.30, (3, 7) = 9.33, (3, 8) = 7.38, (3, 9) = 6.60, (4, 1) = 3.30, (4, 2) = 4.67, (4, 3) = 7.38, (4, 4) = 0., (4, 5) = 3.30, (4, 6) = 6.60, (4, 7) = 3.30, (4, 8) = 4.67, (4, 9) = 7.38, (5, 1) = 4.67, (5, 2) = 3.30, (5, 3) = 4.67, (5, 4) = 3.30, (5, 5) = 0., (5, 6) = 3.30, (5, 7) = 4.67, (5, 8) = 3.30, (5, 9) = 4.67, (6, 1) = 7.38, (6, 2) = 4.67, (6, 3) = 3.30, (6, 4) = 6.60, (6, 5) = 3.30, (6, 6) = 0., (6, 7) = 7.38, (6, 8) = 4.67, (6, 9) = 3.30, (7, 1) = 6.60, (7, 2) = 7.38, (7, 3) = 9.33, (7, 4) = 3.30, (7, 5) = 4.67, (7, 6) = 7.38, (7, 7) = 0., (7, 8) = 3.30, (7, 9) = 6.60, (8, 1) = 7.38, (8, 2) = 6.60, (8, 3) = 7.38, (8, 4) = 4.67, (8, 5) = 3.30, (8, 6) = 4.67, (8, 7) = 3.30, (8, 8) = 0., (8, 9) = 3.30, (9, 1) = 9.33, (9, 2) = 7.38, (9, 3) = 6.60, (9, 4) = 7.38, (9, 5) = 4.67, (9, 6) = 3.30, (9, 7) = 6.60, (9, 8) = 3.30, (9, 9) = 0.})

Angle of each pile from a reference pile in radian:

xi := Matrix(9, 9, {(1, 1) = 0., (1, 2) = 0., (1, 3) = 0., (1, 4) = 1.57, (1, 5) = .785, (1, 6) = .464, (1, 7) = 1.57, (1, 8) = 1.11, (1, 9) = .785, (2, 1) = 3.14, (2, 2) = 0., (2, 3) = 0., (2, 4) = 2.36, (2, 5) = 1.57, (2, 6) = .785, (2, 7) = 2.03, (2, 8) = 1.57, (2, 9) = 1.11, (3, 1) = 3.14, (3, 2) = 3.14, (3, 3) = 0., (3, 4) = 2.68, (3, 5) = 2.36, (3, 6) = 1.57, (3, 7) = 2.36, (3, 8) = 2.03, (3, 9) = 1.57, (4, 1) = 1.57, (4, 2) = .785, (4, 3) = .464, (4, 4) = 0., (4, 5) = 0., (4, 6) = 0., (4, 7) = 1.57, (4, 8) = .785, (4, 9) = .464, (5, 1) = 2.36, (5, 2) = 1.57, (5, 3) = .785, (5, 4) = 3.14, (5, 5) = 0., (5, 6) = 0., (5, 7) = 2.36, (5, 8) = 1.57, (5, 9) = .785, (6, 1) = 2.68, (6, 2) = 2.36, (6, 3) = 1.57, (6, 4) = 3.14, (6, 5) = 3.14, (6, 6) = 0., (6, 7) = 2.68, (6, 8) = 2.36, (6, 9) = 1.57, (7, 1) = 1.57, (7, 2) = 1.11, (7, 3) = .785, (7, 4) = 1.57, (7, 5) = .785, (7, 6) = .464, (7, 7) = 0., (7, 8) = 0., (7, 9) = 0., (8, 1) = 2.03, (8, 2) = 1.57, (8, 3) = 1.11, (8, 4) = 2.36, (8, 5) = 1.57, (8, 6) = .785, (8, 7) = 3.14, (8, 8) = 0., (8, 9) = 0., (9, 1) = 2.36, (9, 2) = 2.03, (9, 3) = 1.57, (9, 4) = 2.68, (9, 5) = 2.36, (9, 6) = 1.57, (9, 7) = 3.14, (9, 8) = 3.14, (9, 9) = 0.})

NULL

Angle of each pile from a reference pile in degrees:

theta := Matrix(9, 9, {(1, 1) = 0., (1, 2) = 0., (1, 3) = 0., (1, 4) = 90.0, (1, 5) = 44.8, (1, 6) = 26.6, (1, 7) = 90.0, (1, 8) = 63.6, (1, 9) = 44.8, (2, 1) = 180., (2, 2) = 0., (2, 3) = 0., (2, 4) = 135., (2, 5) = 90.0, (2, 6) = 44.8, (2, 7) = 116., (2, 8) = 90.0, (2, 9) = 63.6, (3, 1) = 180., (3, 2) = 180., (3, 3) = 0., (3, 4) = 153., (3, 5) = 135., (3, 6) = 90.0, (3, 7) = 135., (3, 8) = 116., (3, 9) = 90.0, (4, 1) = 90.0, (4, 2) = 44.8, (4, 3) = 26.6, (4, 4) = 0., (4, 5) = 0., (4, 6) = 0., (4, 7) = 90.0, (4, 8) = 44.8, (4, 9) = 26.6, (5, 1) = 135., (5, 2) = 90.0, (5, 3) = 44.8, (5, 4) = 180., (5, 5) = 0., (5, 6) = 0., (5, 7) = 135., (5, 8) = 90.0, (5, 9) = 44.8, (6, 1) = 153., (6, 2) = 135., (6, 3) = 90.0, (6, 4) = 180., (6, 5) = 180., (6, 6) = 0., (6, 7) = 153., (6, 8) = 135., (6, 9) = 90.0, (7, 1) = 90.0, (7, 2) = 63.6, (7, 3) = 44.8, (7, 4) = 90.0, (7, 5) = 44.8, (7, 6) = 26.6, (7, 7) = 0., (7, 8) = 0., (7, 9) = 0., (8, 1) = 116., (8, 2) = 90.0, (8, 3) = 63.6, (8, 4) = 135., (8, 5) = 90.0, (8, 6) = 44.8, (8, 7) = 180., (8, 8) = 0., (8, 9) = 0., (9, 1) = 135., (9, 2) = 116., (9, 3) = 90.0, (9, 4) = 153., (9, 5) = 135., (9, 6) = 90.0, (9, 7) = 180., (9, 8) = 180., (9, 9) = 0.})

The total number of piles in a pile group:

n__pile := 9

The length of each pile in the pile group in (meters):

L__p := 12.182   

 

``

Diameter of pile (meter):

`&phi;__pile` := .457

The radius of each pile (meter):

r__0 := evalf((1/2)*`&phi;__pile`, 2)

NULL

`&rho;__A` := 1.0

NULL

D__s := 0.5e-1

NULL

Omega := 46``

``

`&nu;__soil` := .5

NULL

E__p := 3.2*10^10``

NULL

G__s := 5.7*10^7``

NULL

 

V__s := 182.88

NULL

``

NULL

NULL

V__La := 395.845``

NULL

`&alpha;__&nu;` := proc (S__p, xi, n__pile) local `&alpha;__tmp`, flist, i, j, `&alpha;__1`, `&alpha;__2`, `&alpha;__3`; global D__s, V__La, V__s, Omega; description "Calculate the horizontal interaction factor from a refernce pile in a pile group"; flist := []; for i to `n__pile ` do for j to n__pile do `&alpha;__1`[i, j] := evalf[4](.5*(S__p[i, j]/`&phi;__pile`)^.5*exp(-D__s*Omega*S__p[i, j]/V__La)*exp(-I*Omega*S__p[i, j]/V__La)); `&alpha;__2`[i, j] := evalf[4]((3/4)*(S__p[i, j]/`&phi;__pile`)^.5*exp(-D__s*Omega*S__p[i, j]/V__s)*exp(-I*Omega*S__p[i, j]/V__s)/sqrt(2)); `&alpha;__3`[i, j] := evalf[4](`&alpha;__1`[i, j]*cos(xi[i, j])^2+`&alpha;__2`[i, j]*sin(xi[i, j])^2); if i = j then `&alpha;__tmp`[i, i] := 1.0 elif xi[i, j] = 0. then `&alpha;__tmp`[i, j] := `&alpha;__1`[i, j] elif xi[i, j] = 1.57 then `&alpha;__tmp`[i, j] := `&alpha;__2`[i, j] elif xi[i, j] <> 1.57 and xi[i, j] <> 0. then `&alpha;__tmp`[i, j] := `&alpha;__3`[i, j] end if; flist := [op(flist), `&alpha;__tmp`[i, j]] end do end do end proc:

NULL

NULL

NULL

NULLNULL

NULLNULLNULL

`&alpha;__v` := eval(Matrix(n__pile, n__pile, `&alpha;__&nu;`(S__p, xi, n__pile)))``

NULL

``

``

``

NULL

NULL

``

NULL

 

Download Test_pile.mw

Good Evening Maple Community,

Can anybody please help in the attached file. Highlighted in yellow are the three equations I'm trying to program in Maple procedure, but could not get any output due to an error message. Please Help.

 

Boyer 

Hi 

I have 3 equations

eq1 := vs = (Rs+Z)*i1+Z*i3

eq2 := A*vi = Z*i1+(Z+Rf+ro)*i3

eq3:= vo = (Rf+Z)*i3+Z*i1

 

and I want to solve for vo/vs . How to do that ?

the expected solution is 

  Continuation.
  One way to get rolling without slipping animation in 3d. The trajectory and circle are divided into segments of equal length. In the next segment of the trajectory we construct circle, taking into account the fact that it turned on one segment. Rolling sphere or cylinder can be simulated, if we take plottools templates of the same radius, and replace them on the site of our circle.

ROLLING_WITHOUT_3d.mw













The method of solving underdetermined systems of equations, and universal method for calculating link mechanisms. It is based on the Draghilev’s method for solving systems of nonlinear equations. 
When calculating link mechanisms we can use geometrical relationships to produce their mathematical models without specifying the “input link”. The new method allows us to specify the “input link”, any link of mechanism.

Example.
Three-bar mechanism.  The system of equations linkages in this mechanism is as follows:

f1 := x1^2+(x2+1)^2+(x3-.5)^2-R^2;
f2 := x1-.5*x2+.5*x3;
f3 := (x1-x4)^2+(x2-x5)^2+(x3-x6)^2-19;
f4 := sin(x4)-x5;
f5 := sin(2*x4)-x6;

Coordinates green point x'i', i = 1..3, the coordinates of red point x'i', i = 4..6.
Set of x0'i', i = 1..6 searched arbitrarily, is the solution of the system of equations and is the initial point for the solution of the ODE system. The solution of ODE system is the solution of system of equations linkages for concrete assembly linkage.
Two texts of the program for one mechanism. In one case, the “input link” is the red-green, other case the “input link” is the green-blue.
After the calculation trajectories of points, we can always find the values of other variables for example the angles.
Animation displays the kinematics of the mechanism.
MECAN_3_GR_P_bar.mw 
MECAN_3_Red_P_bar.mw

(if to use another color instead of color = "Niagara Dark Orchid", the version of Maple <17)

Method_Mechan_PDF.pdf






Hello,

I have two equations (1) and (2) and i want to divide (2) with (3). A good point is that Maple understand division with equations. Nevertheless, I didn't obtain a simplified solution.

Here my code :

restart;
eq1:=-sin(alpha0(t))*cos(beta0(t)) = -sin(alpha[1](t))*cos(beta[1](t));
-sin(alpha0(t)) cos(beta0(t)) = -sin(alpha[1](t)) cos(beta[1](t))
eq2:=cos(alpha0(t))*cos(beta0(t)) = cos(alpha[1](t))*cos(beta[1](t));
cos(alpha0(t)) cos(beta0(t)) = cos(alpha[1](t)) cos(beta[1](t))
simplify(eq1/eq2,trig);

Here the result obtained :-sin(alpha0(t))/cos(alpha0(t)) = -sin(alpha[1](t))/cos(alpha[1](t))

Consequently, I would like to obtain tan(alpha0(t))=tan(alpha1(t))

Do you have ideas why I didn't obtain a simplified result ? And How can I obtain the solution with tangents ?

Thank you for your help

Hello dears! Hope everything going fine with you. I have faced problem while solving the system of equations using fsolve command please find the attacment and fixed my problem.

I am very thankful to you for this favour. 

VPM_Help.mw

Mob #: 0086-13001903838

Dear Maple users

In order to solve a problem in Biology, I have come across four equations with three unknowns, making it overdetermined. The equation system does not have any specific solutions, but since the equations can be considered containing errors, I am looking after the best possible solution - which will minimize the coefficients in the original equation system. My question is twofold:

1. Does there exist a specific Mathematical Theory to handle non-linear equation systems of this type?

2. Which command in Maple can solve it?

 

 

Regards,

Erik

 

hi.how i can dsolve couple linear equations with power series solutions or taylor series expantion?

file attached below.

thanks

 

TAYLOR.mw

Hello,

I would like to create a function from an equation.
I have define 4 functions u[1](t), ..., u[4](t).
From these 4 functions, I would like to define 4 equations. I would like to obtain this result:
Equ[1]:=u[1](t)=L/2*cos(w*t+phi[1])
Equ[2]:=u[2](t)=L/2*cos(w*t+phi[2])
Equ[3]:=u[3](t)=L/2*cos(w*t+phi[3])
Equ[4]:=u[4](t)=L/2*cos(w*t+phi[4])

My code is the following:

for i to 4
do
u[i]:=unapply(L/2*cos(w*t+phi[i]),t);
Equ[i]:='u[i](t)'=u[i](t);
end do;

These lines are not working because the left member is not incremented (u[i](t) stays at each iteration u[i](t))

Generally, how can I transform a function to an equation ?
And, in this specific case, how can I obtain the four equations mentioned above ?

Thanks a lot for your help.

 

Hello,

I try his code:

 

for i to 6 do u[i, j] := (u[i-1, j]+u[i+1, j]+lambda*(u[i, j-1]+u[i, j+1])+sigma(h, r[i+1])*(u[i+1, j]-u[i-1, j]))/mu end do;


 
Error, too many levels of recursion.

I want to the  results of the analytic equations, for example:

                                                          
 u[1,j]= (u[0, j] + u[2, j] + lambda (u[1, j - 1] + u[1, j + 1])                                                       

      + sigma(h, r[2]) (u[2, j] - u[0, j]))/mu

.

.

.

.
Regards.

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