Boundary conditions for a cantilevered Timoshenko Beam

I am trying to find out the boundary conditions for above given Timoshenko beam. The boundary conditions that I have derived are

• $$x_1 = 0$$ $$\overline{u}_2^L = 0$$

$$\phi_3^L = 0$$

• $$x_1 = \alpha L$$

$$\overline{u}_2^L = \overline{u}_2^R$$

$$\phi_3^L = \phi_3^L$$

$$-K_{22} \left( \dfrac{d\overline{u}_2^L}{d x_1} - \phi_3^L \right) + P + K_{22} \left( \dfrac{d\overline{u}_2^R}{d x_1} - \phi_3^R \right) = 0$$

$$-H_{33}^c \dfrac{d \phi_3^L}{d x_1} + H_{33}^c \dfrac{d \phi_3^R}{d x_1} = 0$$

• $$x_1 = L$$

$$K_{22} \left( \dfrac{d\overline{u}_2^R}{d x_1} - \phi_3^R \right) = 0$$

$$H_{33}^c \dfrac{d \phi_3^R}{d x_1} = 0$$

where,

$$x_i$$: coordinate along $$i$$-th direction

$$\overline{u}_2^L$$: transverse displacement of left portion of beam

$$\overline{u}_2^R$$: transverse displacement of right portion of beam

$$\phi_3^L$$: sectional rotation of left portion of beam

$$\phi_3^R$$: sectional rotation of right portion of beam

$$K_{22}$$: Sectional shear stiffness, equal to $$\dfrac{5}{6}GA$$ for rectangular cross-sections

However, in the book Structural Analysis with Applications to Aerospace Structures'' by Bauchau and Craig, the boundary conditions at $$x = \alpha L$$ are

$$\overline{u}_2^L = \overline{u}_2^R$$

$$\phi_3^L = \phi_3^L$$

$$-K_{22} \left( \dfrac{d\overline{u}_2^L}{d x_1} - \phi_3^L \right) + P = 0$$

$$H_{33}^c \dfrac{d \phi_3^L}{d x_1} = 0$$

Why are shear force and bending moment equations at $$x = \alpha L$$ are different than mine?

Thank you.

• So, what would Timoshenko say? – Solar Mike Mar 8 at 16:41

• I have used continuity of $\overline{u}_2$ and $\phi_3$, force and moment balance at $x = \alpha L$. At $x = \alpha L$, shear force from both side should be included (?), so there will be three terms in expression, two expressions for shear force and one for P. Kindly correct me if I am wrong. – Ali Baig Mar 9 at 3:29