Engineering Stuffs Archive

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Engineering Professional Practice: Sample Case and a solution



Engineering Professional Practice: Sample Case and a solution

Case: A client came to a designer’s and asked to design a multistoried building. The soil type is found not suitable for that type of structure. The designer hesitated to design a building. The revealed his/her intention as that he/she actually is not going to build that structure but he/she wanted to collect fund from outside sources on behalf of that design and he/she wanted to utilize that money in other business.

Problem: What should a designer do in such situation?


Solution:

Engineering is a profession. A professional is that which is acquired thru a specialized training or education having certain skill those ordinary men does not possess. Because of professional’s knowledge and skill that ordinary people do possess, and because of they use their knowledge and skill for the benefit of men and the society as a whole, professionals do have high recognition in the society. Furthermore, their professional practices are also subjected to public evaluation and hence, a professional’s work cannot be remained as a personal matter. In this background, we have to deal the situation.

A designer is supposed to design a project visiting the site and he/she should look for all the things that are readily apparent on the site that is likely to affect the design. Besides this, a designer is also needed to ascertain the nature and load bearing capacity of the soil underneath ground level. In this case given, it is clearly specified that the type of soil is not suitable for that type of building.

Client’s intention in this regard does not matter for an engineer as a professional designer whether a client is going to build that structure or not. A designer’s main duty is to prepare a design for the project created by a client that is technically feasible to erect on the ground. Non-feasible and non-economical design may lead towards the negligence of a designer or they exhibit the incompetence of the designer. If negligence and incompetence are not the case and if the designer perform such task to fulfill the client’s intention only then the designer is not remained as a professional and his/her behavior becomes unethical and immoral. Professionals committing such unethical/immoral acts are subjected to a disciplinary action if not institutionally (as in the case of Nepal) but by the public and their public image will become very low.

It is also to be noted here that a designer’s duty does not end when he/she has completed his/her design. A designer is under a continuing duty to check that his/her design will work in practice and to correct any errors that emerge during the implementation (construction) of that design. In the case given, as the early stage of starting a design work, it has already been known that the soil type is unsuitable. Thus based on the above discussion as a designer, I will not be involved in designing that structure.


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Case Studies of Engineering Professional Practice for Bachelors of Engineering



Case studies based on Codes of Ethics, Tort, Liability and Negligence in Professional Engineering

  1. Er Satyaman Shrestha was working as a Project Engineer in a road construction project in Chitwan. Er Ramesh Thapa, close friend of Satyaman was also working as a site engineer in the same project for a construction company. Construction work was in full swing. Ramesh and Satyaman were tired every day after the work.Almost every day, Ramesh proposes Satyaman for drinks and good foods in the restaurant after work. After few days of work Satyaman noticed that the ratio of cement mortar used in the stone masonry was 1:6 instead of 1:4 as per design. Similarly, there was 1:3:6 ratio of mixture instead of 1:2:4 as per design in the PCC and RCC works. Satyaman complained of this and asked for the reason. Ramesh told to Satyaman that the strength required for the construction is still safe in the mortar of 1:6 and concrete of 1:3:6 and also this is only the means to cover the overheads of entertainments and other financial benefits, Ramesh did this. Read the rest of this entry »



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Faith of an Engineer – a tale of engineering



The oath of an engineer in his profession, this is a master piece of the work.

I am an ENGINEER. In my profession I take deep pride, but without vainglory; to it owe solemn obligations that I am eager to fulfill.


As an Engineer, I will participate in none but honest enterprise. To him that he has engaged my service, as employer or client, I will give the utmost of performance and fidelity. Read the rest of this entry »


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Graphics Project : Bar Diagram [A Project work in Computer Graphics]



This was a Computer Graphics Project named Bar Diagram required in the coursework. [A Project work in Computer Graphics]

#include
#include
#include
#include
#include
#include

/*FUNCTION FOR READING DATA FROM USER*/

void input(float ,float,float,float,float,float);

/*FUNCTION FOR DRAWING BAR DIAGRAM*/

void bar1 (float);
void bar2 (float);
void bar3 (float);
void bar4 (float);
void bar5 (float);
void bar6 (float);

/* MAIN PROGRAM */

void main()
{
int gd=DETECT,gm;
initgraph(&gd,&gm,”c:\\tc\\bgi”);

float sub1,sub2,sub3,sub4,sub5,sub6;
setbkcolor(9);
setcolor(BLUE);
settextstyle(3,0,2);
outtextxy(getmaxx()/2-130,0,”TRIBHUWAN UNIVERSITY”);
settextstyle(3,0,2);
outtextxy(getmaxx()/2-145,30,”INSTITUTE OF ENGINEERING”);
settextstyle(3,0,3);
outtextxy(getmaxx()/2-155,60,”WESTERN REGION CAMPUS”);
settextstyle(3,0,1);
outtextxy(getmaxx()/2-135,92,”LAMACHAUR-16,POKHARA”);
settextstyle(3,0,2);
outtextxy(getmaxy()/2+50,getmaxy()/2,”A”);
settextstyle(3,0,2);
outtextxy(getmaxy()/2+15,260,”PROJECT”);
settextstyle(3,0,2);
outtextxy(getmaxy()/2+40,getmaxy()/2+40,”ON”);
settextstyle(3,0,1);
outtextxy(getmaxy()/2-50,300,”COMPUTER GRAPHICS”);
outtextxy(getmaxx()-600,getmaxy()-140,”BY”);
outtextxy(getmaxx()-600,getmaxy()-110,”Archana Shrestha”);
outtextxy(getmaxx()-600,getmaxy()-80,”402/BEX/062″);

delay(2500);
cleardevice();
settextstyle(7,0,3);
outtextxy(getmaxx()/2-250,getmaxy()/2,”PROGRAM TO DRAW A BAR DIAGRAM”);
delay(1500);
input(sub1,sub2,sub3,sub4,sub5,sub6);
getch();
}
/*END OF MAIN PROGRAM*/

/*FUNCTION FOR READING DATA FROM USER */

void input(float sub1,float sub2 ,float sub3,float sub4,float sub5,float sub6)
{
cleardevice();
cout<”);

setcolor(12);
line(50,50,50,400);
line(50,400,600,400) ;
settextstyle(11,0,1);
outtextxy(70,430,”SUBJECT——->”);
setcolor(YELLOW);
settextstyle(2,0,2);
outtextxy(40,400,”00″);
outtextxy(40,380,”10″);
outtextxy(40,360,”20″);
outtextxy(40,340,”30″);
outtextxy(40,320,”40″);
outtextxy(40,300,”50″);
outtextxy(40,280,”60″);
outtextxy(40,260,”70″);
outtextxy(40,240,”80″);
outtextxy(40,220,”90″);
outtextxy(40,200,”100″);
setcolor(BLUE);
/* FUNCTION CALL TO DRAW BARDIAGRAM */
bar1(sub1);
bar2(sub2);
bar3(sub3);
bar4(sub4);
bar5(sub5);
bar6(sub6);
delay(3000);

cleardevice();
setbkcolor(1);
setcolor(5);

for (int j=0;j{
settextstyle(4,0,6);
setcolor(j);
delay(200);
outtextxy(getmaxx()/2-200,getmaxy()/2,” THANK-YOU “) ;
cout<

for(float i=0;i{
line(485,400-i,555,400-i);
delay(60);
}
}


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Write an assembly program to READ YOUR NAME AND DISPLAY IT IN NEWLINE



Write an assembly program to READ YOUR NAME AND DISPLAY IT IN NEWLINE

title read and display name

dosseg

.model small

.stack 100H

.code

main proc

mov ax, @data ; initialize ds register

mov ds, ax

mov ah, 09h ; display message1

mov dx, offset msg1

int 21h

mov ah, 0ah ; read string

mov dx, offset string

int 21h

mov alt, 09h ; your name is

mov dx, offset msg2

int 21h

mov ah, 09h ; string output

mov dx, offset string

int 21h

mov ax, 4C00H ; return to DOS

int 21H

main endp

end main

.data

msg1 db “Enter your name $”

msg2 db “OAH, ODH, your name is $”

max db 20

len db ?

string db 20 DUP(‘$’)


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Write an assembly program to REVERSE THE GIVEN DIGITS



Write an assembly program to REVERSE THE GIVEN DIGITS

title to reverse the given digits

dosseg

.model small

.stack 100H

.code

main proc

mov ax, @data ; initialize ds register

mov ds, ax

mov ax, value1 ; move number to ax

mov cx, 0000h ; remainder r = 0

mov bx, 0010h ; for multiplication by 10h

BACK: mov dx, 0000h ; clear upper 16-bits

div bx ; divide the number ie dx= remainder, ax = quotient

push ax ; push remainder and quotient onto the stack

push dx

mov ax, cx ; r = 0

mul bx ; r = r * 10

pop dx

add ax , dx ; r = r * 10 + n mod 10

mov cx, ax ; keep reverse digits in cx register

pop ax ; or ax, ax ie number = 0

jnz BACK

mov ax, 4C00H ; return to DOS

int 21H

main endp

.data

value1 dw 789A h

list dw 4 dup (?)

end main

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Write an assembly program to COPY A BLOCK OF DATA FROM ONE MEMORY TO ANOTHER



Write an assembly program to COPY A BLOCK OF DATA FROM ONE MEMORY TO ANOTHER

title copy a block of data from one memory to another

dosseg

.model small

.stack 100H

.code

main proc

mov ax, @data ; initialize ds register

mov ds, ax

mov SI, offset array1 ; source index

mov di, offset array2 ; destination index

mov cx, 06H ; for 6 data ie cx is implicit counter

BACK: mov al, [SI]

mov [di], al ; move [di], [si]

inc si

inc di

loop BACK ; loop always looks for counter cx

mov ax, 4C00H ; return to DOS

int 21H ; interrupt

main endp

.data

array1 db 01, 02, 03, 04, 05, 06

array2 db 6 dup (?) ; or dup(0) to initialize

result db?

end main

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Hardware Concept in Computer Graphics- Mouse and Pointing Devices



Mouse play a very important role in development of Computer Graphics and to implement graphics in the computer system. Mouse is a pointing input device that points out item(s) by cursor. The cursor can be move from one position to another, this movement of cursor is due to the movement of mouse in the plane surface. The movement of mouse occurs only in two directions- vertical and horizontal. Mouse always measures its relative movement from its last position. Read the rest of this entry »



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Write an assembly language program (Intel 8086) to READ A STRING, CONVERT IT INTO UPPER CASE AND FINALLY DISPLAY THE CONVERTED STRING



Write an assembly language program (Intel 8086) to READ A STRING, CONVERT IT INTO UPPER CASE AND FINALLY DISPLAY THE CONVERTED STRING

Read More

 

title ALP to read a string, convert it into upper case and display the converted string

dosseg

.model small

.stack 100H

.code

main proc far

mov ax, @data ; initialize ds register

mov ds, ax

 

mov ah, 0ah ; read string

lea ax, param

int 21h

 

mov bx, 00 ; convert to upper string

mov cx, act_len

L2:          mov ah, str [bx]

cmp ah, 41h

jb L1

cmp ah, 51h

ja L1

XOR ah, 00100001B

mov str [bx], ah

L1:          inc bx

loop L2

 

mov ah, 02h ; display newline

mov dl, 0a

int 21h

 

mov ah, 09h ; display string

lead dx, str

int 21h

 

mov ax, 4C00H ; return to DOS

int 21H

main endp

.data

param label Byte
max_men db 60
act_len db ?
str db 60  dup(?)

 

end main


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To evaluate ordinary differential equations by RK-4 (Classical) Method



Evaluating ordinary differential equations by RK-4 or classical method both order first and second

//To compute ordinary differential equations by RK-4 or classical method for order first

#include

#include

#include

void main()

{

cout<<”\t first order R-K 4th Classical Method\t”<

double h,n,m1,i,m2,m3,m4,x[100],y[100];

cout<<”Number of intervals “;

cin>>n;

cout<

cout<<”Enter the value of h”<<<”h = “;

cin>>h;

cout<<”Enter the value of x0″<<<”x0 = “;

cin>>x[0];

cout<<”Enter the value of y0″<<<”y0 = “;

cin>>y[0];

cout<

for(i=0;i<=n;i++)

{

x[i+1]=x[i]+h;

m1=h*.5*(1+x[i])*(y[i]*y[i]);

m2=h*.5*(1+(x[i]+.5*h))*(y[i]+m1*.5*h)*(y[i]+m1*.5*h);

m3=h*.5*(1+(x[i]+.5*h))*(y[i]+m2*.5*h)*(y[i]+m2*.5*h);

m4=h*.5*(1+(x[i]+h))*(y[i]+m3*h)*(y[i]+m3*h);

y[i+1]=y[i]+(m1+2.0*m2+2.0*m3+m4)/6;

cout<<”x”<<<”= “<<

cout<<”y”<<<”= “<<

cout<

}

getch();

}

// To compute ordinary differential equations by RK-4 or classical method for order second

#include

#include

#include

void main()

{

cout<<”\t 2nd order order R-K 4th Order Method\t”<

double h,n,m1,i,m2,m3,m4,p1,p2,p3,p4,x[100],y[100],z[100];

cout<<”Number of intervals “;

cin>>n;

cout<

cout<<”Enter the value of h”<<<”h = “;

cin>>h;

cout<<”Enter the value of x0″<<<”x0 = “;

cin>>x[0];

cout<<”Enter the value of y0″<<<”y0 = “;

cin>>y[0];

cout<<”Enter the value of z0″<<<”z0 = “;

cin>>z[0];

cout<

for(i=0;i<=n;i++)

{

x[i+1]=x[i]+h;

m1=h*z[i];

p1=h*(6*x[i]+3*y[i]-2*z[i]);

m2=h*(z[i]+.5*p1);

p2=h*(6*(x[i]+.5*h)+3*(y[i]+m1*.5)-2*(z[i]+.5*p1)) ;

m3=h*(z[i]+.5*p2);

p3=h*(6*(x[i]+.5*h)+3*(y[i]+m2*.5)-2*(z[i]+.5*p2)) ;

m4=h*(z[i]+.5*p3);

p4=h*(6*(x[i]+.5*h)+3*(y[i]+m3*.5)-2*(z[i]+.5*p3)) ;

y[i+1]=y[i]+(m1+2.0*m2+2.0*m3+m4)/6;

z[i+1]=z[i]+(p1+2.0*p2+2.0*p3+p4)/6;

cout<<”x”<<<”= “<<

cout<<”y”<<<”= “<<

cout<<”z”<<<”= “<<

cout<

}

getch();

}

To evaluate ordinary differential equations by RK-4 (Classical) Method in C++ Programming Language for Numerical Methods for Engineering Students


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