## Analysis and Design of Machine Foundation...

ANALYSIS AND DESIGN OF MACHINE FOUNDATION

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

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 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. In which, G = dynamic shar modulus of he soil and is given  ; 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.

 Sketch

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

 SI No. Direction Spring value Equivalent radius Remarks 1 Vertical This is in vertical Z direction 2 Horizontal This induce sliding in horizontal X 2.1 Horizontal This induce sliding in horizontal Y 3 Rocking This produces roxking about Y axis 3.1 Rocking This produces roxking about X axis 4 Twisting This produces twisting about vertical Z axis

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

SI No.

Direction

Mass ratio (B)

Damping ratio and Damping values

Remarks

1

Vertical

This damping value is in vertical Z direction

2

Horizontal

This damping value is in lateral X direction

2.1

Horizontal

 (5.1)
 (5.2)

This damping value is in lateral Y direction

3

Rocking

 (5.3)
 (5.4)

This damping value is for rocking about Y direction

3.1

Rocking

 (5.5)

This damping value is for rocking about X direction

4

Twisting

 (5.6)

This damping value is valid for twisting about vertical Z axis

 (4)

Vertical Motion Considering damping of the Soil

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

 (6.1)

 (6.2)

 (6.3)

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

 (6.4)

 (6.5)

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

 (6.6)

 (6.7)

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

## Repeat sentence...

H, how i need to write my command to get n times of the sentences i insert.

For example,

>teksbiasa:= `Hello Bob`;

Hello Bob

So, if i want 5 times of `Hello Bob` like

Hello Bob Hello Bob Hello Bob Hello Bob Hello Bob

What should i write?

Thank you ~=]]

## Simple graphical interface in Maple ...

How i need to write

>teksbiasa:=`Hello! Bob`;

in button Action When Clicked at simple graphical interface instead of

Do(teksbiasa=%txtTeksBiasa);

because when i enter Hello! Bob in %txtTeksBiasa, system pop up ERROR

Error

Error in Component button with caption "Botton":

(in unknown) incorrect syntax in parse: missing operator or

`;`(near 7th character of parsed string)

Thank you~=]]

## Calculate message's bits...

>message:=`67A`;

67A

>P:=convert(message, bytes);

[54, 55, 65]

>with(Bits):

>bitP1:=Split(P1);

[0, 1, 1, 0, 1, 1]

>bitP2:=Split(P2);

[1, 1, 1, 0, 1, 1]

>bitP3:=Split(P3);

[1, 0, 0, 0, 0, 0, 1]

>with(Statistics):

>b1:=Count(bitP1);

6

>b2:=Count(bitP2);

6

>b3:=Count(bitP3);

7

>totalBits=b1+b2+b3;

19

Hi, how i need to modify my command so when i write any message with any lenght, i can get the totalBits directly..

Thank you~=]]

## Solve with fsolve...

How I can solve it ? If I want a solution dependent of a. With fsolve? But how?

-x3+ax2-lnx=0

## Karmarkar's algorithm...

I am trying to write a procedure that implements Karmarkar's algorithm for solving linear programming problems.

I am getting a parsing error and I am not sure why. Each of the command in the procedure are working the way I want them to but there is something wrong with the loop controls. The code is attached below:

with(LinearAlgebra);

print(??); # input placeholder
Karmarkar:= proc(A,c)

n:=ColumnDimension(A);

x:=ZeroVector(n);

y:=Vector(n,1/(n));

r:=1/(sqrt(n*(n-1)));

for i from 1 to nops(NullSpace(A))do

x:=x+NullSpace(A)[i];

end do;

C:=1.0;

while C>0.0001 do ;

Diag := DiagonalMatrix(x/Norm(x, 1));

B := `<,>`(A.Diag, Vector[row](n, 1));

p := (IdentityMatrix(n)-Transpose(B).MatrixInverse(B.Transpose(B)).B).Transpose(c.Diag);

p:=evalf(p/(Norm(p,2)));

y:=y-0.9*r*p;

x_new:=evalf((Diag . y)/((Vector[row](n,1) . (Diag . y))));

C:=evalf((c . x_new-c . x)/(c . x));

if C>0.0001 then

x_new:=x;

end if;

end do;

end proc;

Any ideas on how to fix this?

## Error, Dsolve error for my first equation...

I have an equation

eq := diff(y(x), x) = ((76.559-179.54*((1/3600)*x+1)+113.15*((1/3600)*x+1)^2-13.661*((1/3600)*x+1)^3+.4621*((1/3600)*x+1)^4)*.88-.15*((76.559-179.54*((1/3600)*x+1)+113.15*((1/3600)*x+1)^2-13.661*((1/3600)*x+1)^3+.4621*((1/3600)*x+1)^4)*.88)*(1-0.45e-2*(y(x)-9.8-273))-(3.8+2.8*3.7+5.76*10^(-8)*(y(x)^2+(273+9.8+(1/46800)*(18.3-9.8)*x)^2)*(y(x)+273+9.8+(1/46800)*(18.3-9.8)*x))*(273+9.8+(1/46800)*(18.3-9.8)*x-y(x)))/(903*(2702*0.2e-2));

Maple gives me a general solution.

then I  tried to get a particular solution using dsolve({eq, y(0) = 283.8});

But Maple dose not return anything.

Thank you for any help.

## 2D Matrix interpolation...

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 (5)

 (6)

 (7)

 (8)

Hello Mapleprime Community,

Would you be kind to help me with the attached Maple worksheet. My goal is to do a 2D interpolation given the value of SR (slenderness ratio) and E_G (Young's modulus / Shear modulus ratio) in the Matrix table Vert_Coeff with header as vector E_G and the other as vector SR. I’m confused as to how to do the interpolation. Please help.

Regards,

Moses

## Solving system with constraints ...

SYSTEMATIC APPROACH TO LIFTING EYE DESIGN

Moses

 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.

The design load for each lifting eye is given by:

 >
 >

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

 >

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

 >

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:

 >

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

 >

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,

 >
 >

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

 >

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

 >

For a manual weld the size, s is given by

 >

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

 >

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

 >

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

 >

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

 >

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

 >
 >
 >
 >
 >

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

 >
 >

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

 >

Using the quadratic formular to solve for h yields

 >

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

 >

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:

 >

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

 >
 >
 >
 >
 >
 >
 >
 >
 >
 >
 >
 >
 >
 >

 Output

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

## How do I solve this system of ODEs in Maple?...

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..

## How to calculate size of data in bytes?...

>tekssifer:="Qnuux*#Huh";

tekssifer:=Qnuux*#Huh

Hi, how i need to continue my command to get the size of tekssifer in bytes?

Thank you~=]]

## Random binary generator...

Wanna ask that how to make maple program to select a random binary number in base 2.

For example, suppose that we have a value 11101100(8 bit which in base 2),then how to let Maple to select a binary number randomly and the binary bit of the random number always shorter than the previous then sum it together?

For example the program will select 1101101(7-bit) and sum it together(11101100+1101101=101011001,in base 2)

Can somebody help me thanks.

## Perform 2's complement in binary form...

Hai everyone, wanna know that how to perform 2's complement method in maple 17?

For instance, 2's complement of 19 is 11101100(in base 2) in 8-bit binary representation.

## Perform binary arithmetic in Maple...

Hello everyone,wanna ask that how to perform binary arithmetic(addition,subtraction,multiplication and division) in Maple.

For instances, how i need to perform:

1101+111=10100(all in base 2)

11000-1011=1101(all in base 2)

1011*1101=10001111(all in base 2)

10010011/1011=1101(all in base 2)

Kindly help,thanks.

## Splitting strings into words...

>teksbiasa:=`Hello!`;

teksbiasa:=Hello!

>nilaiASCII:=convert(teksbiasa,bytes);

nilaiASCII:=[72,101,108,108,111,33]

>E:=nilaiASCII+~nops(nilaiASCII);

E:=[78,107,114,114,117,39]

Hi, how i need to modify my command so the length of each word can be detected if enter a sentence instead of 1 word ?

For example, if i entered >> `Hello! Bob`, so the length of each word is [6, 3], so what i need is

>teksbiasa:=`Hello! Bob`;

teksbiasa:=Hello! Bob

>nilaiASCII:=convert(teksbiasa,bytes);

nilaiASCII:=[72, 101, 108, 108, 111, 33, 32, 66, 111, 98]

and my E will be >> E:=[72+6, 101+6, 108+6, 108+6, 111+6, 33+6, 32, 66+3, 111+3, 98+3] which is >>

E:= [78, 107, 114, 114, 117, 39, 32, 69, 114, 101]

Thanks for help~=]]
Have a nice day~=]]

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