Items tagged with output

Can you tell me the signification of the number 5 in output below.

Input: maximize(x^2+3,x=-1..2,location)

Output: 7,{[{x=2},5]}




what would i write if i wanted to display the full list of variables that be specified by calling kernelopts? like as output in the interface, i naively attempted ops(kernelopts)  which didnt work of course.

Some years ago a postdoc working with me insisted to use Maple, since he was used to it. He left me some *.mws files with some calculations we had published together. I'd like to re-use the code now, preferably transposed to C or another programming language (there is not much use of Computer algebra in the code). Unfortunately it appears not so easy. I found the free Maple reader, and at least can now see the code. However, I did not find a way to copy-paste code from the Maple reader to a text file, or any other way to store the code in readable format (avoiding the Maple reader). I try to store the code by printing from the Maple reader with a kind of pdf writer, and a process started which kept my Laptop busy for more than an hour now without any output.

At the moment it looks like the two years of Postdoc work is lost, thanks to Maple and its strange restriction policy.

Can somebody let me know how to save the Maple input code of a *.mws file in a text file?



I have Maple output that extends page width. I can of course see the entire output when I scroll to the right. But since I want to make a screenshot of the output, I need to have the output on one page. Is there a possibility to have the output printed on one page, not extending page width?

Any comments would be greatly appreciated!


I apologize, as I'm still very new. I've flipped through a lot of pages, but I'm unsure of how Maple code translates to Maple TA.


I have randomly assigned an integer for the value of a year in Maple TA. I'd like it to appear as "2013" for example, but it will instead appear as "2,013". Is there a way to set the output formatting of a variable individually within a problem in TA?



I have this out put, let me put it simply as a  single variable(call it A) having multiple outputs such that when i print(A) I get         a
How do i put all the values in a single list to get [a,b,c,d]?

This is what I did: aa:=[]: for i in A do aa:=[op(aa),i]:od:
The output is [a]

How do I get [a,b,c,d] without doing a lot of op???


HI all,


I have 

> sol2 := dsolve({odesys, H(0) = 4995, R(0) = 65000, W(0) = 102000, l(0) = 96000}, numeric, method = rosenbrock);
print(`output redirected...`); 
proc(x_rosenbrock) ... end;


I want to have a list of my solutions, t, H(t), R(t), W(t), l(t) that I can put into a spreadsheet (.csv, .txt, etc.), for 600 timesteps. 


There are some answers out there, but I am confused by them, and have not been able to make it work.




Suppose I type certain math expression like as follow:




But when I enter this expression, Maple gives totally different look to this expression, can I force Maple to print similar looking expression as I typed in command line ? I mean without change of  position of intermediates and coefficients.




I would like to know if it is possible to change the color for the result from blue to black or red. Maybe by editing the text that state the colors.

And the same goes for the color palette, can I make my own colors in the color palette text. It should work by default.















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.



The design load for each lifting eye is given by:


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.



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;





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












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.





Here is my Maple 16 code:

 I expected to get outuput

a [a,b,c]

a [a,c,b]

But I get no output.





How do i make output (that blue part i get after pressing Enter) go in one line? Sometimes it's easier to look at one line of answer (even if it is longer than all other parts of the documents) than to look through row after row of equations.

And also - is there a way to hide output? There are many not important equations and i want to see only a few plots and a result. How do i hide outputs - because when i just delete them - maple deletes associated variables.


I wonder if there is an operator to skip inputs and outputs of functions. Consider this example:

The output would be:

However my system is much bigger and I only need the second output of the command and I dont want to waste memory on the first.

In Matlab I would write

~,ans := GenerateMatrix(...)Is there a similar shortcut availible in Maple?


Thanks in advance!


Hi, does anyone know how to choose the variables that populate the DAE Variables box when you use "equation extraction"?

I want the result to be in terms of the voltage source and the voltage drop across the capacitor for a RLC circuit.  I want to be able to choose the input-output variables for the final equation.



Does [Length of output exceeds limit of 1000000] mean that the function couldn't be solved and I have to modify the mathematical model?
If the results are just too long to be displayed, how can I see the results?

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