Timoshenko Beam
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Timoshenko beam.
Displacements
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u 1 = u 0 ( x ) + z φ x u 2 = 0 u 3 = w 0 ( x ) {\displaystyle {\begin{aligned}u_{1}&=u_{0}(x)+z\varphi _{x}\\u_{2}&=0\\u_{3}&=w_{0}(x)\end{aligned}}}
ε x x = ε x x 0 + z ε x x 1 γ x z = γ x z 0 {\displaystyle {\begin{aligned}\varepsilon _{xx}&=\varepsilon _{xx}^{0}+z\varepsilon _{xx}^{1}\\\gamma _{xz}&=\gamma _{xz}^{0}\end{aligned}}} ε x x 0 = d u 0 d x + 1 2 ( d w 0 d x ) 2 ε x x 1 = d φ x d x γ x z 0 = φ x + d w 0 d x {\displaystyle {\begin{aligned}\varepsilon _{xx}^{0}&={\cfrac {du_{0}}{dx}}+{\frac {1}{2}}\left({\cfrac {dw_{0}}{dx}}\right)^{2}\\\varepsilon _{xx}^{1}&={\cfrac {d\varphi _{x}}{dx}}\\\gamma _{xz}^{0}&=\varphi _{x}+{\cfrac {dw_{0}}{dx}}\end{aligned}}} Principle of Virtual Work
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Constitutive Relations
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σ x x = E ε x x ; σ x z = G γ x z {\displaystyle \sigma _{xx}=E\varepsilon _{xx}~;\qquad \sigma _{xz}=G\gamma _{xz}} Then,
N x x = A x x ε x x 0 + B x x ε x x 1 M x x = B x x ε x x 0 + D x x ε x x 1 Q x = S x x γ x z 0 {\displaystyle {\begin{aligned}N_{xx}&=A_{xx}~\varepsilon _{xx}^{0}+B_{xx}~\varepsilon _{xx}^{1}\\\\M_{xx}&=B_{xx}~\varepsilon _{xx}^{0}+D_{xx}~\varepsilon _{xx}^{1}\\\\Q_{x}&=S_{xx}~\gamma _{xz}^{0}\end{aligned}}} where
S x x = K s ∫ A G d A ← shear stiffness {\displaystyle S_{xx}=K_{s}\int _{A}G~dA\qquad \leftarrow \qquad {\text{shear stiffness}}} Equilibrium Equations
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d d x { A x x [ d u 0 d x + 1 2 ( d w 0 d x ) 2 ] } + f = 0 d d x { S x x ( d w 0 d x + φ x ) + A x x d w 0 d x [ d u 0 d x + 1 2 ( d w 0 d x ) 2 ] } + q = 0 d d x ( D x x d φ x d x ) + S x x ( d w 0 d x + φ x ) = 0 {\displaystyle {\begin{aligned}{\cfrac {d}{dx}}\left\{A_{xx}\left[{\cfrac {du_{0}}{dx}}+{\frac {1}{2}}\left({\cfrac {dw_{0}}{dx}}\right)^{2}\right]\right\}+f&=0\\{\cfrac {d}{dx}}\left\{S_{xx}\left({\cfrac {dw_{0}}{dx}}+\varphi _{x}\right)+A_{xx}{\cfrac {dw_{0}}{dx}}\left[{\cfrac {du_{0}}{dx}}+{\frac {1}{2}}\left({\cfrac {dw_{0}}{dx}}\right)^{2}\right]\right\}+q&=0\\{\cfrac {d}{dx}}\left(D_{xx}{\cfrac {d\varphi _{x}}{dx}}\right)+S_{xx}\left({\cfrac {dw_{0}}{dx}}+\varphi _{x}\right)&=0\end{aligned}}} Weak Form
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∫ x a x b A x x d ( δ u 0 ) d x [ d u 0 d x + 1 2 ( d w 0 d x ) 2 ] d x = ∫ x a x b f δ u 0 d x + [ N x x δ u 0 ] x a x b ∫ x a x b A x x d ( δ w 0 ) d x d w 0 d x [ d u 0 d x + 1 2 ( d w 0 d x ) 2 ] d x + ∫ x a x b S x x d ( δ w 0 ) d x ( d w 0 d x + φ x ) d x = ∫ x a x b q δ u 0 d x + [ ( N x x d w 0 d x + Q x ) δ w 0 ] x a x b − ∫ x a x b S x x δ φ x ( d w 0 d x + φ x ) d x + ∫ x a x b D x x d ( δ φ x ) d x d φ x d x d x = [ M x x δ φ x ] x a x b {\displaystyle {\begin{aligned}\int _{x_{a}}^{x_{b}}A_{xx}{\cfrac {d(\delta u_{0})}{dx}}\left[{\cfrac {du_{0}}{dx}}+{\frac {1}{2}}\left({\cfrac {dw_{0}}{dx}}\right)^{2}\right]~dx&=\int _{x_{a}}^{x_{b}}f\delta u_{0}~dx+\left[N_{xx}\delta u_{0}\right]_{x_{a}}^{x_{b}}\\\\\int _{x_{a}}^{x_{b}}A_{xx}{\cfrac {d(\delta w_{0})}{dx}}{\cfrac {dw_{0}}{dx}}\left[{\cfrac {du_{0}}{dx}}+{\frac {1}{2}}\left({\cfrac {dw_{0}}{dx}}\right)^{2}\right]~dx&+\int _{x_{a}}^{x_{b}}S_{xx}{\cfrac {d(\delta w_{0})}{dx}}\left({\cfrac {dw_{0}}{dx}}+\varphi _{x}\right)~dx=\\&\int _{x_{a}}^{x_{b}}q\delta u_{0}~dx+\left[\left(N_{xx}{\cfrac {dw_{0}}{dx}}+Q_{x}\right)\delta w_{0}\right]_{x_{a}}^{x_{b}}\\\\-\int _{x_{a}}^{x_{b}}S_{xx}\delta \varphi _{x}\left({\cfrac {dw_{0}}{dx}}+\varphi _{x}\right)~dx&+\int _{x_{a}}^{x_{b}}D_{xx}{\cfrac {d(\delta \varphi _{x})}{dx}}{\cfrac {d\varphi _{x}}{dx}}~dx=\left[M_{xx}\delta \varphi _{x}\right]_{x_{a}}^{x_{b}}\end{aligned}}} Finite element model
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Shear Locking
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Example Case
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Linear u 0 {\displaystyle u_{0}} , Linear w 0 {\displaystyle w_{0}} , Linear φ x {\displaystyle \varphi _{x}} .
d w 0 d x = constant . {\displaystyle {\cfrac {dw_{0}}{dx}}=~~{\text{constant}}.} But, for thin beams,
d w 0 d x = slope = − φ x ← ( linear! ) {\displaystyle {\cfrac {dw_{0}}{dx}}=~~{\text{slope}}~~=-\varphi _{x}~~\leftarrow ~~({\text{linear!}})} If constant φ x {\displaystyle \varphi _{x}}
d φ x d x = 0 {\displaystyle {\cfrac {d\varphi _{x}}{dx}}=0} Also
Q x = S x x φ x ≠ 0 ⟹ {\displaystyle Q_{x}=S_{xx}\varphi _{x}\neq 0\implies } Non-zero transverse shear.
M x x = D x x d φ x d x = 0 ⟹ {\displaystyle M_{xx}=D_{xx}{\cfrac {d\varphi _{x}}{dx}}=0\implies } Zero bending energy.Result : Zero displacements and rotations ⟹ {\displaystyle \implies } Shear Locking!
Recall
d d x { A x x [ d u 0 d x + 1 2 ( d w 0 d x ) 2 ] } + f = 0 {\displaystyle {\cfrac {d}{dx}}\left\{A_{xx}\left[{\cfrac {du_{0}}{dx}}+{\frac {1}{2}}\left({\cfrac {dw_{0}}{dx}}\right)^{2}\right]\right\}+f=0} or,
d d x { A x x ε x x 0 } + f = 0 {\displaystyle {\cfrac {d}{dx}}\left\{A_{xx}\varepsilon _{xx}^{0}\right\}+f=0} If f = 0 {\displaystyle f=0} and A x x = {\displaystyle A_{xx}=} constant
A x x d d x ( ε x x 0 ) = 0 ⟹ ε x x 0 = constant . {\displaystyle A_{xx}{\cfrac {d}{dx}}(\varepsilon _{xx}^{0})=0\qquad \implies \qquad \varepsilon _{xx}^{0}=~{\text{constant}}.} If there is only bending but no stretching,
ε x x 0 = 0 = d u 0 d x + 1 2 ( d w 0 d x ) 2 {\displaystyle \varepsilon _{xx}^{0}=0={\cfrac {du_{0}}{dx}}+{\frac {1}{2}}\left({\cfrac {dw_{0}}{dx}}\right)^{2}} Hence,
d u 0 d x ≈ ( d w 0 d x ) 2 {\displaystyle {\cfrac {du_{0}}{dx}}\approx \left({\cfrac {dw_{0}}{dx}}\right)^{2}} Also recall:
d d x { S x x ( d w 0 d x + φ x ) + A x x d w 0 d x [ d u 0 d x + 1 2 ( d w 0 d x ) 2 ] } + q = 0 {\displaystyle {\cfrac {d}{dx}}\left\{S_{xx}\left({\cfrac {dw_{0}}{dx}}+\varphi _{x}\right)+A_{xx}{\cfrac {dw_{0}}{dx}}\left[{\cfrac {du_{0}}{dx}}+{\frac {1}{2}}\left({\cfrac {dw_{0}}{dx}}\right)^{2}\right]\right\}+q=0} or,
d d x { S x x γ x z 0 + A x x d w 0 d x ε x x 0 } + q = 0 {\displaystyle {\cfrac {d}{dx}}\left\{S_{xx}\gamma _{xz}^{0}+A_{xx}{\cfrac {dw_{0}}{dx}}\varepsilon _{xx}^{0}\right\}+q=0} If q = 0 {\displaystyle q=0} and S x x = {\displaystyle S_{xx}=} constant, and no membrane strains
S x x d d x ( γ x z 0 ) = 0 ⟹ γ x z = constant = d w 0 d x + φ x {\displaystyle S_{xx}{\cfrac {d}{dx}}(\gamma _{xz}^{0})=0\qquad \implies \qquad {\gamma _{xz}=~{\text{constant}}}~={\cfrac {dw_{0}}{dx}}+\varphi _{x}} Hence,
φ x ≈ d w 0 d x {\displaystyle \varphi _{x}\approx {\cfrac {dw_{0}}{dx}}} Shape functions need to satisfy:
d u 0 d x ≈ ( d w 0 d x ) 2 ; and φ x ≈ d w 0 d x {\displaystyle {\cfrac {du_{0}}{dx}}\approx \left({\cfrac {dw_{0}}{dx}}\right)^{2}~;\qquad {\text{and}}\qquad \varphi _{x}\approx {\cfrac {dw_{0}}{dx}}}
Example Case 1
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Linear u 0 {\displaystyle u_{0}} , Linear w 0 {\displaystyle w_{0}} , Linear φ x {\displaystyle \varphi _{x}} .
First condition ⟹ {\displaystyle \implies } constant = {\displaystyle =} constant. Passes! No Membrane Locking.
Second condition ⟹ {\displaystyle \implies } linear = {\displaystyle =} constant. Fails! Shear Locking. Example Case 2
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Linear u 0 {\displaystyle u_{0}} , Quadratic w 0 {\displaystyle w_{0}} , Linear φ x {\displaystyle \varphi _{x}} .
First condition ⟹ {\displaystyle \implies } constant = {\displaystyle =} quadratic. Fails! Membrane Locking.
Second condition ⟹ {\displaystyle \implies } linear = {\displaystyle =} linear. Passes! No Shear Locking. Example Case 3
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Quadratic u 0 {\displaystyle u_{0}} , Quadratic w 0 {\displaystyle w_{0}} , Linear φ x {\displaystyle \varphi _{x}} .
First condition ⟹ {\displaystyle \implies } linear = {\displaystyle =} quadratic. Fails! Membrane Locking.
Second condition ⟹ {\displaystyle \implies } linear = {\displaystyle =} linear. Passes! No Shear Locking. Example Case 4
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Cubic u 0 {\displaystyle u_{0}} , Quadratic w 0 {\displaystyle w_{0}} , Linear φ x {\displaystyle \varphi _{x}} .
First condition ⟹ {\displaystyle \implies } quadratic = {\displaystyle =} quadratic. Passes! No Membrane Locking.
Second condition ⟹ {\displaystyle \implies } linear = {\displaystyle =} linear. Passes! No Shear Locking. Overcoming Shear Locking
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