Question 18.P.8: Obtain the equations of motion for free vibrations of the ha...

Dynamics of Structures

Obtain the equations of motion for free vibrations of the haunched beam shown in Figure P18.8 by dividing its length into three segments and applying the finite element method with a consistent mass formulation. Assume that in the haunched portion, the flexural rigidity varies linearly from $3 E I$ to $E I$ while the mass per unit length varies from $2 m$ to $m$ .

p18.8

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\begin{aligned}& \mathbf{M}=\frac{m}{53,760}\left[\begin{array}{cccccc}13 L^{3} & -9 L^{3} & 152 L^{2} & 0 & 0 & 0 \\-9 L^{3} & 75 L^{3} & 472 L^{2} & -48 L^{3} & 416 L^{2} & 0 \\152 L^{2} & 472 L^{2} & 16,128 L & -416 L^{2} & 3456 L & 0 \\0 & -48 L^{3} & -416 L^{2} & 75 L^{3} & -472 L^{2} & -9 L^{3} \\0 & 416 L^{2} & 3456 L & -472 L^{2} & 16,128 L & -152 L^{2} \\0 & 0 & 0 & -9 L^{3} & -152 L^{2} & 13 L^{3}\end{array}\right] \\& \mathbf{K}=E I\left[\begin{array}{cccccc}\frac{40}{L} & \frac{16}{L} & \frac{-224}{L^{2}} & 0 & 0 & 0 \\\frac{16}{L} & \frac{32}{L} & \frac{-136}{L^{2}} & \frac{4}{L} & \frac{-24}{L^{2}} & 0 \\\frac{-224}{L^{2}} & \frac{-136}{L^{2}} & \frac{1632}{L^{3}} & \frac{24}{L^{2}} & \frac{-96}{L^{3}} & 0 \\0 & \frac{4}{L} & \frac{24}{L^{2}} & \frac{32}{L} & \frac{136}{L^{2}} & \frac{16}{L} \\0 & \frac{-24}{L^{2}} & \frac{-96}{L^{3}} & \frac{136}{L^{2}} & \frac{1632}{L^{2}} & \frac{224}{L^{2}} \\0 & 0 & 0 & \frac{16}{L} & \frac{224}{L^{2}} & \frac{40}{L}\end{array}\right]\end{aligned}

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