Most general rigid 2D displacement in polars
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Short working note by Phil dated 4.14.12, in the Lai Continuum Mechanics support documents. It works through four questions: small rotation in Cartesians, converting it to polar components (ur = 0, uθ = rk with k = sinψ), a translation in polars, and the combination of the two. The result is ur = G cosθ + H sinθ and uθ = -G sinθ + H cosθ + Fr, matching terms in equations 5.29.15 and 5.29.16. Equations are partly lost in the text extraction.
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Most general rigid 2D displacement in polars PhL 4.14.12
Question 1: Consider only rotations for the moment, and do Cartesian only. Consider then x = R(ψ)X:
=
= - = ≈
where we assume ψ is small because all displacements are supposed to be small.
Question 2: How would you express the above result in polar coordinates? I think this is correct:
= = Rz(-θ) since = Rz(+θ)
=
Then we get
= = r sinψ
= r sinψ
This says that ur is unchanged by this rotation but uθ is and we get
ur = 0 uθ = rk k = sinψ
This then seems to account for the "Fr" term in uθ in 5.29.16.
Question 3: What does a translation do in polar coordinates
In Cartesians it does this
= + =
= - =
Then in polars we get
= =
ur = ax cosθ + aysinθ = G cosθ + H sinθ G = ax H = ay
uθ = -ax sinθ + aycosθ = - Gsinθ + Hcosθ
and this accounts for the other terms in 5.29.15,16,
Question 4: How do you handle rotation and translation at the same time?
Let's try first translate the rotate. The translate part results in
' = +
Now rotate to get
= [ + ]
= - = [ + ] -
= +
≈ + = +
where the first term is the same as in Question 1, and in the second we drop order ε2 terms.
Now as before we have
=
and since this is linear, we get the sum of the two results we got in the separate questions:
= +
= r sinψ +
And so
ur = G cosθ + H sinθ
uθ = - Gsinθ + Hcosθ + Fr