Beam reactions and mid-span deflection (2EI)

Data: $EI=2\times10^5\ \text{MPa}\times5\times10^6\ \text{mm}^4=10^{12}$ N mm$^2=1000$ kN m$^2$ (same for both spans). Nodes: 1 = fixed end (x = 0); 2 = intermediate support (x = 7 m) with an applied moment of 25 kN m (taken clockwise); 3 = right roller (x = 12 m). Span 1: $L=7$ m with 25 kN at 3 m from the fixed end; span 2: $L=5$ m with 15 kN/m UDL. 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 nodal loads

Span 1 ($a=3$, $b=4$, $L=7$): $\dfrac{Pb^2(3a+b)}{L^3}=\dfrac{25\times16\times13}{343}=15.16$ kN, $\dfrac{Pab^2}{L^2}=24.49$ kN m, $\dfrac{Pa^2(a+3b)}{L^3}=9.84$ kN, $\dfrac{Pa^2b}{L^2}=18.37$ kN m. Span 2: $wL/2=37.5$ kN, $wL^2/12=31.25$ kN m. Equivalent loads are the negatives of the fixed-end forces. The applied clockwise moment gives $-25$ kN m at $\theta_2$ (counter-clockwise positive).

Assembled equation

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

$$\begin{bmatrix}1371.429 & 400.000 \\ 400.000 & 800.000\end{bmatrix}\begin{Bmatrix}\theta_2\\\theta_3\end{Bmatrix}=\begin{Bmatrix}-37.883 \\ 31.250\end{Bmatrix}$$ $$\theta_2=-0.04568\ \text{rad},\qquad \theta_3=0.06190\ \text{rad}$$

Support reactions $\{f\}=[k]\{d\}-\{F_{eq}\}$

• Fixed end: $R_1=9.567$ kN (up), $M_1=11.439$ kN m (counter-clockwise)
• Intermediate support: $R_2=56.827$ kN (up)
• Right roller: $R_3=33.606$ kN (up)
• Check: $\Sigma R=100.000$ kN $=25+15\times5=100$ kN.

Deflection at mid span

The displacement within a span is the Hermite interpolation of the nodal values plus the fixed-end beam deflection due to the load. Evaluating (checked by refining the mesh with nodes at the mid-points):

• Mid of span 1 (x = 3.5 m): $v=-2.22$ mm (downward)
• Mid of span 2 (x = 9.5 m): $v=-91.65$ mm (downward)

The value is large because $EI=1000$ kN m$^2$ is small for these spans.

Answer: $R_1=9.57$ kN, $M_1=11.44$ kN m, $R_2=56.83$ kN, $R_3=33.61$ kN; mid-span deflections $2.22$ mm (span 1) and $91.65$ mm (span 2), both downward.