Beam deflection at B, rotations (2 elements)

The beam is divided into element 1 (A-B) and element 2 (B-C), each 5 m long, with nodes A, B, C. Dofs per node: (deflection $v$, rotation $\theta$). A is fixed ($v_A=\theta_A=0$) and C is a roller ($v_C=0$), so the unknowns are $v_B,\ \theta_B,\ \theta_C$.

  20 kN/m
 +++++++++++++++++++++++
 |===========o==========o   A fixed, C roller
 A           B          C
 |<--- 5 --->|<--- 5 -->|

Element stiffness ($L=5$ m)

$$[k]=\frac{EI}{125}\begin{bmatrix}12&30&-12&30\\30&100&-30&50\\-12&-30&12&-30\\30&50&-30&100\end{bmatrix}$$

Equivalent nodal loads (per element)

$wL/2=20\times5/2=50$ kN, $wL^2/12=20\times25/12=41.667$ kNm.

• Element 1: $(-50,\ -41.667,\ -50,\ +41.667)$
• Element 2: same.

Loads on the free dofs $(v_B,\theta_B,\theta_C)$: $F_{v_B}=-50-50=-100$ kN, $F_{\theta_B}=+41.667-41.667=0$, $F_{\theta_C}=+41.667$ kNm.

Assembled equations for the free dofs

$$EI\begin{bmatrix}0.192&0&0.24\\0&1.6&0.4\\0.24&0.4&0.8\end{bmatrix}\begin{bmatrix}v_B\\ \theta_B\\ \theta_C\end{bmatrix}=\begin{bmatrix}-100\\0\\41.667\end{bmatrix}$$

Solution

$$v_B=-\frac{1041.67}{EI},\qquad \theta_B=-\frac{104.17}{EI}\ \text{rad (clockwise)},\qquad \theta_C=+\frac{416.67}{EI}\ \text{rad (anticlockwise)}$$

Check with reactions and moments

$\{R\}=[K]\{d\}-\{F\}$ gives $R_A=125$ kN, $M_A=250$ kNm and $R_C=75$ kN. $\Sigma F_y$: $125+75=200=20\times10$ and $\Sigma M_A$: $75\times10-200\times5+250=0$ (with $M_A$ anticlockwise). These agree with the propped-cantilever results $R_C=3wL/8=75$ kN, $R_A=5wL/8=125$ kN, $M_A=wL^2/8=250$ kNm.

Answer: deflection at B $=1041.67/EI$ downward; rotations $\theta_B=104.17/EI$ (clockwise) and $\theta_C=416.67/EI$ (anticlockwise).

Bending moment: $M(x)=-250+125x-10x^2$ (sagging +). $M_A=-250$, $M_B=+125$ kNm, zero at $x=2.5$ m, maximum $+140.6$ kNm at $x=6.25$ m, $M_C=0$.