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Polar Coordinates

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Strain – Displacement Relations

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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)

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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  

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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 =

=

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From fig. (b),

tan = =

tan2 =

= −

 =

Total ᵣθ = 1 2 

ᵣθ = tan 1 tan 2 

ᵣθ = 

+

-

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Equilibrium Equations in 2-D

Polar Coordinates

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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

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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 

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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 :

()

→

()

=

 →  

→  

→

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→

→ ()

→

 →  

→  

Substitution of these relations in the preceding equations gives the

equilibrium equations in polar coordinates as,

1

= 0 

1

2

= 0