We will use forces per unit width (stress resultants) and moments per unit width (stress couples) of the plate edge on which they act instead of forces Px , etc. in the various directions in order to...

Aluminum 6061-T6; 6061-T651
68.9
0.33
276MPa
Titanium Grade 12, Annealed
480 MPa 103 GPa
0.28
http:
www.matweb.com/tools/unitconverter.aspx?fromID=108&fromValue=480
http:
www.matweb.com/tools/unitconverter.aspx?fromID=45&fromValue=103
Materials Used
Materials used for this project are steel, aluminum alloy and titanium alloy.
The modulus of elasticity, yield strength and Poisson’s ratio of steel are 200 GPa, 250 MPa and 0.3 respectively.
Aluminum 1199-H18 is used. The material properties are taken from the following page.
http:
www.matweb.com/search/DataSheet.aspx?bassnum=MA1199H18&ckck=1
The modulus of elasticity, yield strength and Poisson’s ratio of steel are 62 GPa, 110 MPa and 0.33 respectively.
Titanium Grade 12, Annealed is used. The material properties are taken from the following page.
http:
www.matweb.com/search/datasheet.aspx?bassnum=MTU121
The modulus of elasticity, yield strength and Poisson’s ratio of steel are 103 GPa, 480 MPa and 0.28 respectively.
Problem 1
http:
www.matweb.com/search/DataSheet.aspx?bassnum=MA1199H18&ckck=1
http:
www.matweb.com/search/datasheet.aspx?bassnum=MTU121
Analysis for buckling
The differential equation governing buckling is
     4 4 4
, , ,4 2 2 4
2 2 0x xx y yy xy xy
w w w
D N w N w N w
x x y y
  
  
   
      
    
(1)
Here, w is the displacement along the Z direction, Nx, Ny are normal forces per unit length along X and Y direction and Nxy is
the shear force in the XY plane.
Here, we have compressive force along x and tensile force along y. So, we can write
x xN N  , y yN N and Nxy=0
Where, xN and yN are the applied compressive force per unit length and applied tensile force per unit length along the X and Y
directions.
Let us assume the buckling mode shape as
sin sin
m x n y
w w
a
 
  , where m and n are integers.
Hence, Eq. 1 can be written as:
2
2 2 2 2
0x y
m n m n
D N N
a b a
           
                    
(2)
From Eq. 2, it is clear that the variation of xN and yN form a straight line in the XY plane.
We assumes 9 different modes for m=1 to 3 and n=1 to 3 and plotted xN vs. yN .
Von Misses stress analysis
Here, normal stress along Y direction
xx
xx
N
h



Normal stress along Y direction
yy
yy
N
h
 

h is the thickness of the plate.

Tensile stress has been takes as positive and the compressive has been taken as negative.
Using the von Misses criterion 2 2 2
xx xx yy yy Y     
We can write
2 2
2Nx Nx Ny Ny Y
h h h h
   
     
   
(3)
Here Y is the yield strength.
We assumed different values of
yN and solved the quadratic equation (Eq. 3) to find out xN .
After that, we plotted
yN vs. xN .
Below are the plots for different materials and different thickness values.
Each graph has 10 plots – one is from von Misses stress criterion and 9 are from the buckling criterion for 9 modes.
In each plot the areas co
esponding to the allowable combinations of xN and yN are shown by the a
ows.
For each case, there are two figures. The first one is the graph showing the 10 plots mentioned above. Second one is an enlarged
view near the area of allowable xN and yN .
Steel
Thickness = 5mm
0 2 4 6 8 10 12 14 16
0
0.5
1
1.5
2
2.5
Compressive Force Along X (KN/mm)
T
e
n
s
il
e
F
o
c
e
A
lo
n
g
Y
(
K
N
m
m
)
Thickness = 7mm
0 2 4 6 8 10 12 14 16 18 20
0
0.5
1
1.5
2
2.5
Compressive Force Along X (KN/mm)
T
e
n
s
il
e
F
o
c
e
A
lo
n
g
Y
(
K
N
m
m
)

Thickness = 9mm
Aluminum Alloy
0 5 10 15 20 25
0
0.5
1
1.5
2
2.5
Compressive Force Along X (KN/mm)
T
e
n
s
il
e
F
o
c
e
A
lo
n
g
Y
(
K
N
m
m
)
Thickness = 5mm
0 5 10 15
0
0.5
1
1.5
2
2.5
Compressive Force Along X (KN/mm)
T
e
n
s
il
e
F
o
c
e
A
lo
n
g
Y
(
K
N
m
m
)
Thickness = 7mm
Thickness = 9mm
0 2 4 6 8 10 12 14 16
0
0.5
1
1.5
2
2.5
Compressive Force Along X (KN/mm)
T
e
n
s
il
e
F
o
c
e
A
lo
n
g
Y
(
K
N
m
m
)
Titainum Alloy
0 2 4 6 8 10 12 14 16 18
0
0.5
1
1.5
2
2.5
Compressive Force Along X (KN/mm)
T
e
n
s
il
e
F
o
c
e
A
lo
n
g
Y
(
K
N
m
m
)

Thickness=5
0 2 4 6 8 10 12 14 16
0
0.5
1
1.5
2
2.5
Compressive Force Along X (KN/mm)
T
e
n
s
il
e
F
o
c
e
A
lo
n
g
Y
(
K
N
m
m
)
0 5 10 15 20 25 30
0
0.5
1
1.5
2
2.5
3
3.5
4
Compressive Force Along X (KN/mm)
T
e
n
s
il
e
F
o
c
e
A
lo
n
g
Y
(
K
N
m
m
)
0 5 10 15 20 25 30 35
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
Compressive Force Along X (KN/mm)
T
e
n
s
il
e
F
o
c
e
A
lo
n
g
Y
(
K
N
m
m
)

Problem 2
Buckling analysis
The differential equation governing buckling is
     
4 4 4
, 0, , 0, , 0,4 2 2 4
2 2 0x xx xx y yy yy xy xy xy
w w w
D N w w N w w N w w
x x y y
   
         
    
(1)
Here, w is the displacement along the z direction, w0 is the initial imperfection, Nx, Ny are normal forces per unit length along X
and Y direction and Nxy is the shear force in the XY plane.
Here, we have compressive force along x and tensile force along y. So, we can write
x xN N  , y yN N and Nxy=0
Where, xN and yN...