Maple 17 Questions and Posts

These are Posts and Questions associated with the product, Maple 17

Hi!

Everyone,

I want to draw  phase plane of system of three fractional order equations. 

 

Note that 

Also want the  phase portrait when the values of alpha are not same....

Also

Thanks

 

 

 

How do i implement Runge-Kutta of order 6 for a sytem of boundary value problems on maple

,Hello everyone 

...I need code for save maple's project like a video

Thankyou

I  encountered a non-integrable integral in the process of solving the following process, . How to achieve its numerical solution? Such as in a looping   code:

#######
pa[i] := pa[i-1]-(Int(subs(t = tau, Lpa[i-1]+Na1[i-1]-Na2[i-1]), tau = 0 .. t)); 

pw[i] := pw[i-1]-(Int(subs(t = tau, Lpw[i-1]+Nw1[i-1]-Nw2[i-1]), tau = 0 .. t)); u[i] := u[i-1]-(Int(subs(t = tau, Lu[i-1]+Nu1[i-1]+Nu2[i-1]), tau = 0 .. t));

######
Detailed code see annexBC2.mw

The keyboard shortcut sequence for executing an entire worksheet is Alt+e+e+w. Occasionally, I hit Alt+e+w+w by accidence. In that case, Maple sometimes, not always!?, gets completely stuck and can only be shut down through the job list window (Windows Ctrl+Alt+Del). Can anybody else replicate that? And if so, has this erratically odd behaviour been fixed in later versions of Maple?

After running Maple in a shell file, I come up with this error that I do not understand on my Mac,

gap_long := 0.117647058823529 Pi

gap_lat := 0.0588235294117647 Pi

lat_begin := 0.441176470588235 Pi

long_begin := -Pi

lat_begin_0 := 0.441176470588235 Pi

long_begin_0 := -Pi

long_max := 0.882352941176471 Pi

lat_max := -0.441176470588235 Pi

33

Warning, `parameter` is implicitly declared local to procedure `set_par_eff`

distance eff distance_eff
im in has not
im in has not
im in has not
im in has not
im in has not
im in has not
im in has not
im in has not
im in has not
Im in has par
Im in has par
Error, invalid input: eval expects its 2nd argument, eqns, to be of type
{integer, equation, set(equation)}, but received par_eff_post
13

32

31

17

hou := 0

mini := 0

seci := 0

memory used=4.0MB, alloc=32.3MB, time=0.23



If needed, I can attach more files if my question is still a bit too cryptic. Please let me know asap as this is urgent. Thank you so much,
-Z

Dear All,

I am going to solve the following systems of ODEs but get the error: Newton iteration is not converging.
Could you please share your idea with me. In the case of AA=-0.2,0,0.2,0.4,...; I could get the solution.
Thank you in advance.


restart;
with(plots);
Pr := 2; Le := 2; nn := 2; Nb := .1; Nt := .1; QQ := .1; SS := .1; BB := .1; CC := .1; Ec := .1; MM := .2;AA:=-0.4;

Eq1 := diff(f(eta), `$`(eta, 3))+f(eta).(diff(f(eta), `$`(eta, 2)))-2.*nn/(nn+1).((diff(f(eta), eta))^2)-MM.(diff(f(eta), eta)) = 0; Eq2 := 1/Pr.(diff(theta(eta), `$`(eta, 2)))+f(eta).(diff(theta(eta), eta))-4.*nn/(nn+1).(diff(f(eta), eta)).theta(eta)+Nb.(diff(theta(eta), eta)).(diff(h(eta), eta))+Nt.((diff(theta(eta), eta))^2)+Ec.((diff(f(eta), `$`(eta, 2)))^2)-QQ.theta(eta) = 0;
Eq3 := diff(h(eta), `$`(eta, 2))+Le.f(eta).(diff(h(eta), eta))+Nt/Nb.(diff(theta(eta), `$`(eta, 2))) = 0;

bcs := f(0) = SS, (D(f))(0) = 1+AA.((D@@2)(f))(0), theta(0) = 1+BB.(D(theta))(0), phi(0) = 1+CC.(D(phi))(0), (D(f))(etainf) = 0, theta(etainf) = 0, phi(etainf) = 0

Error, (in dsolve/numeric/ComputeSolution) Newton iteration is not converging

This is the coding till i do dhe decryption process. 

Do(plaintext=GetProperty("message",value));
Do(plaintext=convert(GetProperty("message",value),name));
Do(plaintextInt = convert(plaintext, bytes));
Do(plaintextBin = `~`[convert](plaintextInt, binary));
Do(plaintextBin2 = map2(nprintf, "%07d", plaintextBin));
Do(n0 = plaintextBin2[]);
Do(length1 = length(n0));
Do(plaintextCode = cat("", plaintextBin2[]));
Do(length2 = length(plaintextCode));
Do(z = convert(plaintextCode, decimal, binary));
Do(z1 = z+1);
Do(z2 = z1+%sk1);
Do(z3 = convert(z2, base, 2));
Do(b = cat("", z3[]));
Do(z4 = length(b));
Do(z5 = [Bits:-GetBits(-z2, -1 .. 0, bits = z4)]);
Do(z6 = cat("", z5[]));
Do(z7 = convert(z6, decimal, binary));
Do(%C = `mod`(Power(z7, %txte), %txtN));
Do(%C1 = `mod`(Power(%sk1, %txte), %txtN));

Do(%m = `mod`(Power(%C, %d), %N));

Do(%sk2=`mod`(Power(%C1,%d),%N));

Then nw i need to decrypt back to the original message with the coding:

Do(z8 = [Bits:-GetBits(-%m,-1 .. 0, bits = z4)]);
Do(c = cat("", z8[]));
Do(z9 = convert(c, decimal, binary));
Do(z10 = z9-sk2);
Do(z11 = z10-1);
Do(z12 = [Bits:-GetBits(z11, -1 .. 0, bits = length2)]);
Do(d = cat("", z12[]));
Do(plaintextBin2 = [StringTools:-LengthSplit(d, length1)]);
Do(plaintextInt2 = `~`[convert](plaintextBin2, decimal, binary));
Do(%message1 = convert(plaintextInt, bytes));

when i execute the program it shows the error

so how should I solve this as although i think that it should be problem of parsing the number z4 in the sentence that i highlighed, but whenever i correct it it still can't work.Thus anyone who know please help.Thanks.

 

restart:with(plots):
eq:=(diff(f(eta),eta$2))-a*f(eta)+b*(1+diff(f(eta),eta)^2)^(-1/2)=0;
bc:=f(1)=0,D(f)(0)=0;
ans := dsolve(eq);

 

ANALYSIS AND DESIGN OF MACHINE FOUNDATION

 


restart

Loading Optimization

 

Loading LinearAlgebra  

 

Loading plots  

with(ScientificConstants):

Loading DynamicSystems  

with(Units:-Standard)

with(Units:-Natural)

with(StringTools)

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

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

(1)

NULL

Introduction

 

This document deals with vibration analysis and design of machine foundations subjected to dynamic load.

NULL

NULL

NULL

NULL

GetConstant(g);

standard_acceleration_of_gravity, symbol = g, value = 9.80665, uncertainty = 0, units = m/s^2

(2)

g__SI := evalf(Constant(g, system = SI, units))

9.80665*Units:-Unit(('m')/('s')^2)

(3)

NULL

Richart and Lysmer's Model

 

Richart et al. (1970) idealised the foundation as a lumped mass supported on soil which is idealised as frequency independent springs which he described in term of soil parameter

dynamic shear modulus or shear wave velocity of the soil for circular footing when footings having equivalent circular radius. The Tables below shows the different values of spring and damping vlaues as per Richart and Lysmer.

NULL

In which, G = dynamic shar modulus of he soil and is given G = `ρ__s`*V__s^2 ; ν = Piosson's ratio of the soil; ρs = mass density of the soil; Vs = shear wave velocity of the soil obtained

from soil testing; g = acceleration due to gravity; m = mass of the machine and foundation; J = mass moment of inertia of the machine and foundation about the appropriate axes; K = equivalent spring stiffness of the soil; C = damping value of the soil; B = interia factor contributing to the damping factor; D = damping ratio of the soil; r = equivalent radius of a circular foundation; L = length of foundation, and B = width of the foundation.

NULL

NULL

NULL

Sketch

 

NULL

NULL

NULL

NULL

nu := .25

Table : Values of soil springs as per Richart and Lysmer (1970) model

 

NULL

NULL

SI No.

Direction

Spring value

Equivalent radius

Remarks

1

Vertical

K__z = 4*G*r__z/(1-nu)"(->)"

r__z = sqrt(L*B/Pi)"(->)"

This is in vertical Z direction

2

Horizontal

K__x = (32*(1-nu))*G*r__x/(7-8*nu)
"(->)"

r__x = sqrt(L*B/Pi)"(->)"

This induce sliding in horizontal X

2.1

Horizontal

K__y = (32*(1-nu))*G*r__y/(7-8*nu)
"(->)"

r__y = sqrt(L*B/Pi)"(->)"

This induce sliding in horizontal Y

3

Rocking

`K__φx` = 8*G*`r__φx`^3/(3*(1-nu))"(->)"

`r__φx` = (L*B^3/(3*Pi))^(1/4)"(->)"

This produces roxking about Y axis

3.1

Rocking

`K__φy` = 8*G*`r__φy`^3/(3*(1-nu))"(->)"

`r__φy` = (L^3*B/(3*Pi))^(1/4)"(->)"

This produces roxking about X axis

4

Twisting

`K__ψ` = 16*G*`r__ψ`^3*(1/3)"(->)"

`r__ψ` = ((B^3*L+B*L^3)/(6*Pi))^(1/4)
"(->)"

This produces twisting about vertical Z axis

 

NULL

NULL

NULL

NULL

NULL

Table : Values of soil damping as per Richart and Lysmer (1970) model

 

NULL

SI No.

Direction

Mass ratio (B)

Damping ratio and Damping values

Remarks

1

Vertical

B__z = .25*m__U*(1-nu)*g__SI/(`ρ__s`*r__z^3)
"(->)"

`ζ__z` = .425/sqrt(B__z)"(->)"C__z = 2*`ζ__z`*sqrt(K__z*m__U)"(->)"

This damping value is in vertical Z direction

2

Horizontal

B__x = (7-8*nu)*m__U*g__SI/((32*(1-nu))*`ρ__s`*r__x^3)
"(->)"

`ζ__x` = .288/sqrt(B__x)"(->)"

C__x = 2*`ζ__x`*sqrt(K__x*m__U)"(->)"

This damping value is in lateral X direction

2.1

Horizontal

B__y = (7-8*nu)*m__U*g__SI/((32*(1-nu))*`ρ__s`*r__y^3)
"(->)"

`ζ__y` = .288/sqrt(B__y)"(->)"

`ζ__y` = .288/((2.145204688-2.451662500*nu)*m__U*Units:-Unit(('m')/('s')^2)/((1-nu)*`ρ__s`*(L*B/Pi)^(3/2)))^(1/2)

(5.1)

NULLError, invalid left hand side in assignmentError, invalid left hand side in assignment

`ζ__ψ` = .5/(1+117.6798000*`J__ψ`*Units:-Unit(('m')/('s')^2)*6^(1/4)/(`ρ__s`*((B^3*L+B*L^3)/Pi)^(5/4)))

(5.2)

C__y = 2*`ζ__y`*sqrt(K__y*m__U)"(->)"

This damping value is in lateral Y direction

3

Rocking

`B__φx` = (.375*(1-nu))*`J__φx`*g__SI/(`ρ__s`*`r__φx`^5)
"(->)"

`ζ__φx` = .15/((1+`B__φx`)*sqrt(`B__φx`))
"(->)"

`ζ__φx` = .15/((1+11.03248125*(1-nu)*`J__φx`*Units:-Unit(('m')/('s')^2)*3^(1/4)/(`ρ__s`*(L*B^3/Pi)^(5/4)))*((11.03248125-11.03248125*nu)*`J__φx`*Units:-Unit(('m')/('s')^2)*3^(1/4)/(`ρ__s`*(L*B^3/Pi)^(5/4)))^(1/2))

(5.3)

Error, invalid left hand side in assignmentError, invalid left hand side in assignmentNULLError, invalid left hand side in assignment

.15/((1+11.03248125*(1-nu)*`J__φx`*Units:-Unit(('m')/('s')^2)*3^(1/4)/(`ρ__s`*(L*B^3/Pi)^(5/4)))*((11.03248125-11.03248125*nu)*`J__φx`*Units:-Unit(('m')/('s')^2)*3^(1/4)/(`ρ__s`*(L*B^3/Pi)^(5/4)))^(1/2)) = .15/((1+11.03248125*(1-nu)*`J__φx`*Units:-Unit(('m')/('s')^2)*3^(1/4)/(`ρ__s`*(L*B^3/Pi)^(5/4)))*((11.03248125-11.03248125*nu)*`J__φx`*Units:-Unit(('m')/('s')^2)*3^(1/4)/(`ρ__s`*(L*B^3/Pi)^(5/4)))^(1/2))

(5.4)

`C__φx` = 2*`ζ__φx`*sqrt(`K__φx`*`J__φx`)"(->)"

NULL

This damping value is for rocking about Y direction

3.1

Rocking

`B__φy` = (.375*(1-nu))*`J__φy`*g__SI/(`ρ__s`*`r__φy`^5)
"(->)"

NULL

`ζ__φy` = .15/((1+`B__φy`)*sqrt(`B__φy`))
"(->)"

.15/((1+`B__φy`)*`B__φy`^(1/2)) = .15/((1+`B__φy`)*`B__φy`^(1/2))

(5.5)

`C__φy` = 2*`ζ__φy`*sqrt(`K__φy`*`J__φy`)"(->)"

NULL

This damping value is for rocking about X direction

4

Twisting

`B__ψ` = `J__ψ`*g__SI/(`ρ__s`*`r__ψ`^5)"(->)"

`ζ__ψ` = .5/(1+2*`B__ψ`)"(->)"NULLError, invalid left hand side in assignmentError, invalid left hand side in assignmentNULLError, invalid left hand side in assignmentNULLError, invalid left hand side in assignment

.5/(1+117.6798000*`J__ψ`*Units:-Unit(('m')/('s')^2)*6^(1/4)/(`ρ__s`*((B^3*L+B*L^3)/Pi)^(5/4))) = .5/(1+117.6798000*`J__ψ`*Units:-Unit(('m')/('s')^2)*6^(1/4)/(`ρ__s`*((B^3*L+B*L^3)/Pi)^(5/4)))

(5.6)

`C__ψ` = 2*`ζ__ψ`*sqrt(`K__ψ`*`J__ψ`)"(->)"NULL

``

NULL

This damping value is valid for twisting about vertical Z axis

 

NULL

NULL

NULLNULL

(4)

``

NULL

Vertical Motion Considering damping of the Soil

 

For vertical direction the equation becomes that of a lumped mass having single degree of freedom when

deq := m__U*(diff(z(t), t, t))+'C__z'*(diff(z(t), t))+'K__z'*`#mi("z")` = P__0*sin(`ω__m`*t)

m__U*(diff(diff(z(t), t), t))+C__z*(diff(z(t), t))+K__z*`#mi("z")` = P__0*sin(`ω__m`*t)

(6.1)

NULL

t1 := subs(P__0*sin(`ω__m`*t)/m__U = F, expand(deq/m__U));

diff(diff(z(t), t), t)+1.085721853*(G*(L*B/Pi)^(1/2)*m__U/(1-nu))^(1/2)*(diff(z(t), t))/(m__U*(m__U*Units:-Unit(('m')/('s')^2)/(`ρ__s`*(L*B/Pi)^(3/2))-m__U*Units:-Unit(('m')/('s')^2)*nu/(`ρ__s`*(L*B/Pi)^(3/2)))^(1/2))+4*G*(L*B/Pi)^(1/2)*`#mi("z")`/(m__U*(1-nu)) = F

(6.2)

NULL

(6.3)

By algebraically manipulating the expression, the form traditionally used by engineers is derived:

t2 := algsubs('C__z'/m__U = 2*zeta*omega, t1)

diff(diff(z(t), t), t)-(-1.085721853*(-G*(L*B/Pi)^(1/2)*m__U/(nu-1))^(1/2)*(diff(z(t), t))*nu+1.085721853*(-G*(L*B/Pi)^(1/2)*m__U/(nu-1))^(1/2)*(diff(z(t), t))+4.*G*(L*B/Pi)^(1/2)*`#mi("z")`*(-Units:-Unit(('m')/('s')^2)*(nu-1)*m__U/(`ρ__s`*(L*B/Pi)^(3/2)))^(1/2))/((nu-1.)*(-Units:-Unit(('m')/('s')^2)*(nu-1)*m__U/(`ρ__s`*(L*B/Pi)^(3/2)))^(1/2)*m__U) = F

(6.4)

NULL

t3 := algsubs('K__z'/m__U = omega^2, t2)

diff(diff(z(t), t), t)-(-1.085721853*(-G*(L*B/Pi)^(1/2)*m__U/(nu-1))^(1/2)*(diff(z(t), t))*nu+1.085721853*(-G*(L*B/Pi)^(1/2)*m__U/(nu-1))^(1/2)*(diff(z(t), t))+4.*G*(L*B/Pi)^(1/2)*`#mi("z")`*(-Units:-Unit(('m')/('s')^2)*(nu-1)*m__U/(`ρ__s`*(L*B/Pi)^(3/2)))^(1/2))/((nu-1.)*(-Units:-Unit(('m')/('s')^2)*(nu-1)*m__U/(`ρ__s`*(L*B/Pi)^(3/2)))^(1/2)*m__U) = F

(6.5)

This form includes the damping ratio , the natural frequency , and the external forcing term .  Consider only free vibration by setting

gen3 := subs(F = 0, t3)

diff(diff(z(t), t), t)-(-1.085721853*(-G*(L*B/Pi)^(1/2)*m__U/(nu-1))^(1/2)*(diff(z(t), t))*nu+1.085721853*(-G*(L*B/Pi)^(1/2)*m__U/(nu-1))^(1/2)*(diff(z(t), t))+4.*G*(L*B/Pi)^(1/2)*`#mi("z")`*(-Units:-Unit(('m')/('s')^2)*(nu-1)*m__U/(`ρ__s`*(L*B/Pi)^(3/2)))^(1/2))/((nu-1.)*(-Units:-Unit(('m')/('s')^2)*(nu-1)*m__U/(`ρ__s`*(L*B/Pi)^(3/2)))^(1/2)*m__U) = 0

(6.6)

NULL

sol1 := dsolve({gen3, z(0) = P, (D(z))(0) = V}, z(t))

z(t) = -(1000000000/1178791942081753609)*Pi*m__U*exp(-(1085721853/1000000000)*Units:-Unit(('s')/('m')^(1/2))*t/(Pi*((nu-1)^2/(`ρ__s`*L^2*B^2*G))^(1/2)*m__U))*(1085721853*V*Units:-Unit(('s')/('m')^(1/2))*((nu-1)^2/(`ρ__s`*L^2*B^2*G))^(1/2)*`ρ__s`*L^2*B^2-4000000000*(L*B/Pi)^(1/2)*Pi*nu*`#mi("z")`+4000000000*`#mi("z")`*Pi*(L*B/Pi)^(1/2))/(Units:-Unit(('s')/('m')^(1/2))^2*B^2*L^2*`ρ__s`)+(4000000000/1085721853)*`#mi("z")`*G*Pi*(L*B/Pi)^(1/2)*((nu-1)^2/(`ρ__s`*L^2*B^2*G))^(1/2)*t/((nu-1)*Units:-Unit(('s')/('m')^(1/2)))+(1/1178791942081753609)*(1085721853000000000*V*Units:-Unit(('s')/('m')^(1/2))*Pi*((nu-1)^2/(`ρ__s`*L^2*B^2*G))^(1/2)*m__U*`ρ__s`*L^2*B^2+1178791942081753609*Units:-Unit(('s')/('m')^(1/2))^2*P*`ρ__s`*L^2*B^2-4000000000000000000*(L*B/Pi)^(1/2)*Pi^2*m__U*nu*`#mi("z")`+4000000000000000000*`#mi("z")`*Pi^2*(L*B/Pi)^(1/2)*m__U)/(B^2*L^2*`ρ__s`*Units:-Unit(('s')/('m')^(1/2))^2)

(6.7)

NULL

NULL

 

Download Analysis_and_Design_of_Machine_Foundations_1.mw

Good Morning Mapleprime Community,

Would anybody please help in the attached worksheet. I'm trying to use the new function in Maple that is the clicable method, but I was having problem in some of my output such as zeta_y and zeta_phi as this two equations are generating an error message.

 

Regards,

Moses

 

 

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

I'm trying to solve a system of 4 ODE's.

 

 

however I have 4 equations and six unknowns. I dont know how else to describe the functions a,b,c,d

 

cause these just represent vector valued functions at points (x1,y1) and (x2,y2) where i have chosing (x1,y1)=(-1,0) and (x2,y2) = (1,0)

 

I have that

 

dx1/dt = (u,v)

dx2/dt=(f,g)

I know that if i graph these functions I should get vertical lines, but I keep getting circles if I instead consider a(t) to be x(t) and b(t) to be y(t)...

 

I need to solve this system and plot it but i am misinterpreting something somewhere..

I have the following question:

Illustrate how the sequence N->R de fined by n ->n^2/n^2 + 31n + 228 can be shown to be
within a given epsilon > 0 of its limiting value x0.
(a) use an appropriate conditional statement to find N such that abs (xn -􀀀 x0) < epsilon for every n>=N
and produce an appropriate list of the data points (n,xn) to illustrate
this

I found N but without using any CONDITIONAL STATEMENT.Can you help me find N using IF FOR WHILE?

Hi. I'm hacing trouble writing a maple procedure for the question below, can anyone help?

 

Write a maple procedure which takes as its input the vectoeat u1 and u2 and the eigenvectors lambda1 and lambda2 where u1,u2 are element of R^2 and the lambdas are real numbers.

If u1,U2 is linearly independent then the output is the matrix A an element of R^2x2 with the property that Au1= lambda1u1 and AU2=lambda2u2;

if u1,u2 is linearly dependent then the output is the statement "not an eigenbasis".

 

I I then have two inputs which I have to do but I'm not sure on how to write the procedure. Any help will be much appreciated.  

 

Thanks :)

 

 

Dear all,

I developed a program to solve f(x, y) = 0 and g(x, y) = 0, I obtained as results (x=2.726, y=2.126) . running the same program another time it gives (x=2.762, y=1.992). how to explain this?

> fsolve({f(x, y) = 0, g(x, y) = 0}, {x = 0 .. infinity, y= 0 .. infinity});

Thanks in advance.

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