27 December 2010

Matlab progress 3 - "Bending Program"

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Matlab progress 3
"Bending Program"



This part of the program deals with a new chapter which is the bending, in this part we can calculate the stresses at any point in the cross section area, we are trying to calculate the stress in many different cross section areas, the program calculates the stress by getting firstly the moment of area of the shape (moment of inertia) around its centroid and then by giving it the moment that act on the shape, the program then get the stress, Not only it can get the stress on the shape but also the moment on it if the stress is given. The program can calculated the centroid and the moment of area of the shape if it is not easy to be obtained by the user. Not only the bending happen by moment but also forces can bend shapes, the program can solve also these problems.

This is the part that doesn’t calculate the centroid of the shape because it is easy to be obtained by user like this problem:


Screenshot of the program :

press on image for full size ..

This is the part that calculates the centroid of the shape because it is difficult to be obtained like this problem :
Screenshot of the program :

for full size press on image ..
 

Notes 11 "Shear Force & Bending Moment Diagrams"

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Shear Force & Bending Moment Diagrams


When designing a beam, there is the need to know how the bending moments vary throughout the length of the beam, particularly the maximum and minimum values of these quantities .

1- Finding Internal Reactions
 
• Pick left side of the cut:
– Find the sum of all the vertical forces to the left of the cut, including V.  Solve for shear, V.
– Find the sum of all the horizontal forces to the left of the cut, including N.  Solve for axial force, N.  It’s usually, but not always, 0.
– Sum the moments of all the forces to the left of the cut about the point of the cut.  Include M.  Solve for bending moment, M
 
• Pick the right side of the cut:
Same as above, except to the right of the cut .
2- Beam Shear

2-Beam Shear
• Why?
• necessary to know the maximum value of the shear .
– necessary to locate where the shear changes from positive to negative .
• where the shear passes through zero.
• Use of shear diagrams give a graphical representation of vertical shear throughout 
the length of a beam .

3-Bending Moment

• Bending moment: tendency of a beam to bend due to forces acting on it .
• Magnitude (M) = sum of moments of forces on either side of the section can be determined at any section along the length of the beam .


• shear is dependent on the loads and reactions
– when a reaction occurs; the shear “jumps” by the amount of the reaction .
– when a concentrated load occurs; the shear “jumps” by the amount of the load Moment is dependent upon the shear diagram .
– the area under the shear diagram = change in the moment (i.e. Ashear diagram = ΔM)
straight lines on shear diagrams create sloping lines on moment diagrams sloping lines on shear diagrams create curves on moment diagrams .
 
- positive shear = increasing slope
negative shear = decreasing slope
 
In beam design, only need to know:
–    reactions
–    max. shear
–    max. bending moment

Now let’s solve some problems

Example 1 :



Example 2 :


Example 3 :


Example 4 :


Now let’s draw the moment Diagrams also ..

press on image for full size ..

Draw the Shear and bending moment diagrams for the beam AB .


Solution :
 
Find RA and RB
∑Fy  =  0  i.e.  RA  +  RB  =  (1.8 x 2.6)  +  4 kN  =  8.68 kN
∑MB =  0  i.e.  - 4 RA  +  2.4  x  4  +  ( 1.8  x  2.6)  x  ( 4 -  1.3 )  =  0
4 RA  =  9.6  +  12.63  =  22.23; 
RA  =  5.56 kN
RB  =  8.68  -  5.56  =  3.12 kN



Common Relationships :



App. - Wind Energy

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Wind Energy


Wind Energy
 
One of the most promising renewable energy resources is the use of wind to produce electricity by driving enormous wind turbines (windmills).


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"External / Bending / MIT"

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"External / Bending / MIT"

Bending and Distributed loads :




Bending :




Source : MIT open courses , www.ocw.mit.edu

26 December 2010

App. - Roller Coasters

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Roller Coasters


1912 : the year John Miller designed the first Under friction roller coaster

these colossal, twisted structures provide an exhilarating and frenetic ride. They wed technology to basic and seemingly incompatible emotions, such as panic, courage, fear, joy, vertigo, and amusement . 

Built as if to exclusively prove Newton's theories, the science of roller coasters abounds with all his terminology: acceleration, mass, gravity, movement, and inertia. But in all this, what is really thrilling is the free fall, the attraction of the abyss.

Safety Details :

The designers of these extreme machines take into account all possible safety factors to provide as safe an experience as possible.

Riders are made to wear safety belts, and machine parts are inspected on a regular basis to prevent accidents.

Joints and beams are X-rayed for flaws. Safety devices applied to the drive chain before cars reach the top prevent the train of cars from moving backward.

These devices are also installed on some of the hills, where the train slows down in its climb. In the event of wind gusts and sudden decelerations, these preventive measures keep the train in place and stop it from backtracking .


Wheels to keep the trolley on the track :

Three types of wheels are needed:

  • Upper wheels to control the train for most of the route .
  • Lower ones for use on the hills-G forces are sometimes greater than the weight of the train .
  • Lateral wheels to prevent the train from derailing on curves.

Force of Gravity in Action :

Most of the motion in a roller-coaster ride is a response to the Earth's gravitational pull. No engines are mounted on the cars. After the train reaches the top of the first slope—the highest point on the ride— the train rolls downhill and gains speed under the Earth's gravitational pull. 

The speed is sufficient for it to climb over the next hill. This process occurs over and over again until all the train's energy has been lost to friction and the train of cars slows to a stop. If no energy were lost to friction, the train would be able to keep running as long as no point on the track was higher than the first peak.
 
1-POTENTIAL ENERGY

When the wagon reaches the highest point of the roller coaster, it has a great deal of potential energy.
 
2-MECHANICAL ENERGY

At a certain point in the trajectory, both energies (potential and kinetic) cancel each other out .
 
3-KINETIC ENERGY

is energy of motion that is, the energy released by the train every time it descends.

press on image for full size ..

Acceleration :


Powerpoint Presentation:




25 December 2010

Matlab progress 2 "

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Matlab progress 2 
"Stress & Strain Program"


This post is about chapter 2 which deals with (Stress & Strain) , this part from our program is used to solve the problems of stress & strain by getting the deflection in the rod and also calculate both stress and strain in the rod ,For example  in the following picture the rod is divided into 3 sections according to the forces and the lengths and the areas on each section , For example :


Screenshot of the program :

Press on image for full size


The program can get the deflection &(or) stress , strain  not only in the whole rod but also in each section in the rod.
 
It is clear that the program also solve the problems which are affected by thermal effect which have a great effect on the deflection and also the stain, The program is capable of solving a problem of 5 sections not only three.
 
Finally we can say that it is easy to deal with this program, just to enter the input in the problem and it calculated what do you want, But the main use of this program is to be like a calculator the user must think about the problem and divided it in to sections and know the force and length and area of each section only the program calculate the inputs and get the final answer.

21 December 2010

Notes 10 " Bending (Pt.3) "

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Bending "Pt.3"




This beam was forced  so it feels with a tension and a compression  and to know the parts  which feels the tension or compression and also , Draw the bending moment diagram that applies on this system.



Note:  I=0.25πR4

Example(1) :






Bending Moment Diagram
M=PL-PX

Example(2) :

Note : Blue is the reaction forces,moments












Powerpoint Presentation
" Solved Problems Included "

18 December 2010

Matlab progress 1

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Matlab progress 1



Introduction :

  • Through this page, we will show how is the progress of our project (MATLAB SOLVER), its mission is solving our problems in structure analysis whatever its type (reaction forces, truss, bars, shafts ……etc ;).
  • We need to your help by your attempt in using our solver so as to know the faults within our SOLVER. 
  • We can answer your questions about our solver or about MATLAB (the language technical computing) by notes, references and courses from a vary universities.

Member of our group:

1-Ashraf nabil
2-Mostafa abd elhamed
3-Hossam ahmed
4-Hozaifa elsaid

5-Ahmed abd ellatif
6-Ahmed yasser
7-Ahmed yossry

17 December 2010

Notes 9 " Bending (Pt.2) "

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Bending "Pt.2"

"Bending Machine"

Introduction :

in the previous lecture we saw that there is a axial strain and stress due to the bending .

I : is the second moment of area

Example(1) :




Example(2) :



NOTE: If the height of the beam is small compare with the width of it we neglect the I of this beam.
a
Example(3) :


Example(4) :



Example(5) :



Example(6) : "VIP"




PDF  Document :