Upload
byron-phira
View
224
Download
0
Embed Size (px)
Citation preview
7/31/2019 New and Improved Polar Coordinates
http://slidepdf.com/reader/full/new-and-improved-polar-coordinates 1/12
Polar Coordinates
7/31/2019 New and Improved Polar Coordinates
http://slidepdf.com/reader/full/new-and-improved-polar-coordinates 2/12
Strain – Displacement Relations
7/31/2019 New and Improved Polar Coordinates
http://slidepdf.com/reader/full/new-and-improved-polar-coordinates 3/12
7/31/2019 New and Improved Polar Coordinates
http://slidepdf.com/reader/full/new-and-improved-polar-coordinates 4/12
In fig. (a), the line segment ab becomes a’b’ after deformation.
The radial strain εᵣ is
εᵣ=−
=+
−
=u
Tangential strain εθ ,as a result of radial displacement u,
from fig.(a),
εθ
=′′′
=+ −
=
------------------------------- (1)
7/31/2019 New and Improved Polar Coordinates
http://slidepdf.com/reader/full/new-and-improved-polar-coordinates 5/12
Similarly from fig. (b), as a result of tangential displacement v, the
line ac becomes a’c’. The length of ac is rdθ. Thus, the tangential
strain εθ is given as
εθ =−
=+
v
−
=
v
-----------------------------(2)
The resultant tangential strain combining eqtns. (1) & (2),
= εθ + εθ
=
v
7/31/2019 New and Improved Polar Coordinates
http://slidepdf.com/reader/full/new-and-improved-polar-coordinates 6/12
The shear strain ᵣθ can be calculated from the change in the right
angle cab.
Shear strain is given as,
ᵣθ = 1 2
1 and 2 are so small that, tan 1
= 1
tan 2 = 2
From fig. (a),
tan1 =
=
7/31/2019 New and Improved Polar Coordinates
http://slidepdf.com/reader/full/new-and-improved-polar-coordinates 7/12
From fig. (b),
tan = =
tan2 =
= −
=
Total ᵣθ = 1 2
ᵣθ = tan 1 tan 2
ᵣθ =
+
-
7/31/2019 New and Improved Polar Coordinates
http://slidepdf.com/reader/full/new-and-improved-polar-coordinates 8/12
Equilibrium Equations in 2-D
Polar Coordinates
7/31/2019 New and Improved Polar Coordinates
http://slidepdf.com/reader/full/new-and-improved-polar-coordinates 9/12
The equilibrium equation can be derived by a consideration of the
normal and shear stresses acting on an element of a body cut by
two radii and two circular arcs. The position of a point at the
center of the element is given by r and θ.The stresses acting at this point are σᵣ, σθ and τᵣθ.
However the stresses acting on the faces of the element will be
different because of the variation of stress with position. Thus all
normal and shear stresses acting on the element are designatedwith a subscript to associate the stress with the sides of the
element.
The length of line a is radθ, the length of line c is rcdθ and the
length of lines b and d are rc –ra = dr.Also, radial body force is Rr
7/31/2019 New and Improved Polar Coordinates
http://slidepdf.com/reader/full/new-and-improved-polar-coordinates 10/12
Summation of forces in radial direction through the center of the
element gives,
= cos 2
cos 2
sin
2 sin
2 = 0
Now, summation of forces in tangential direction normal to the
radial direction through the center of the element gives,
θ = cos 2
cos 2
sin 2
θ θ sin
2 = 0
7/31/2019 New and Improved Polar Coordinates
http://slidepdf.com/reader/full/new-and-improved-polar-coordinates 11/12
Since dθ/2 is small, sin(dθ/2) = dθ/2 and cos(dθ/2) = 1. Also, dividing
the equations by drdθ gives,
()
2
= 0
And,
2= 0
As the dimensions of the element are made smaller and samller, thefollowing relations will hold within the limit :
()
→
()
=
→
→
→
7/31/2019 New and Improved Polar Coordinates
http://slidepdf.com/reader/full/new-and-improved-polar-coordinates 12/12
→
→ ()
→
→
→
Substitution of these relations in the preceding equations gives the
equilibrium equations in polar coordinates as,
1
= 0
1
2
= 0