"Notes" Index ..

"Notes" Index

For Easy access to the lectures published in Aero103 blog ,

here is the lectures index :

• Notes 1 : Equilibruim
• Notes 2 : Trusses
• Notes 3 : Hook's law & poisson ratio
• Notes 4 : Stress and strain - Axial loading
• Notes 5 : deformation & thermal strain
• Notes 6 : Torsion (Pt.1)
• Notes 7 : Torsion (Pt.2)
• Notes 8 : Bending (Pt.1)
• Notes 9 : Bending (Pt.2)
• Notes 10 : Bending (Pt.3)
• Notes 11 : Shear Force & Bending Moment Diagrams

Matlab progress 3 - "Bending Program"

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"

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 :