Beam: deflection, slope under load, BMD

Data: $E=210$ GPa $=210\times10^6$ kN/m$^2$, $I=6\times10^6$ mm$^4=6\times10^{-6}$ m$^4$, so $EI=1260$ kN m$^2$. Nodes: 1 = fixed end (x = 0), 2 = point load (x = 3 m), 3 = first roller (x = 5 m), 4 = last roller (x = 10 m). Elements: 1-2 ($L=3$ m, $EI$), 2-3 ($L=2$ m, $EI$), 3-4 ($L=5$ m, $2EI$, UDL 10 kN/m). Units: kN, m.

Element stiffness

$$[k]=\frac{EI}{L^3}\begin{bmatrix}12&6L&-12&6L\\6L&4L^2&-6L&2L^2\\-12&-6L&12&-6L\\6L&2L^2&-6L&4L^2\end{bmatrix},\quad \{d\}=\{v_i,\theta_i,v_j,\theta_j\}^T$$

(Upward forces and displacements, and counter-clockwise moments and rotations, are positive.) Equivalent loads for the UDL element: $-25$ kN and $-20.833$ kN m at node 3; $-25$ kN and $+20.833$ kN m at node 4. The 20 kN load is a nodal load at node 2.

Assembled equation

Boundary conditions: $v_1=\theta_1=0$, $v_3=0$, $v_4=0$. Free DOFs: $v_2,\theta_2,\theta_3,\theta_4$.

$$\begin{bmatrix}2450 & 1050 & 1890 & 0 \\ 1050 & 4200 & 1260 & 0 \\ 1890 & 1260 & 4536 & 1008 \\ 0 & 0 & 1008 & 2016\end{bmatrix}\begin{Bmatrix}v_2\\\theta_2\\\theta_3\\\theta_4\end{Bmatrix}=\begin{Bmatrix}-20.000 \\ 0 \\ -20.833 \\ 20.833\end{Bmatrix}$$

(kN, m units.) Solving:

$$v_2=-4.329\ \text{mm},\quad \theta_2=0.00309\ \text{rad},\quad \theta_3=-0.00669\ \text{rad},\quad \theta_4=0.01368\ \text{rad}$$

Deflection under the 20 kN load = 4.33 mm downward; slope = 0.00309 rad (3.09 x 10$^{-3}$, counter-clockwise).

End forces and bending moments

Reactions: fixed end 5.018 kN and moment 6.230 kN m; first roller 44.210 kN; last roller 20.772 kN. Sum $=70$ kN = $20+10\times5$ (check).

Bending moment diagram (sagging +)

 x (m):   0        3        5                 7.92      10
          |--------|--------|-----------------|---------|
 BMD:    -6.23    +8.82    -21.14            +21.57      0   (kN m)