Beam deflection and stresses at B (FEM)

Assumptions (figure data): A is fixed, C is a roller (simple support), $EI$ is constant. Three beam elements: AB (4 m, 10 kN/m UDL), BD (2 m) and DC (2 m, no load); the 40 kN downward load acts at node D. Positive: $v$ upward, $\theta$ counter-clockwise. Units: kN, m.

Element stiffness

Each element uses

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

(Upward forces and displacements, and counter-clockwise moments and rotations, are positive.) with $L=4$ m (AB), $L=2$ m (BD, DC). Nodes: 1 = A, 2 = B, 3 = D, 4 = C.

Equivalent nodal loads

Element AB: $wL/2=20$ kN, $wL^2/12=13.333$ kN m, so the equivalent loads are $-20$ kN and $-13.333$ kN m at A, and $-20$ kN and $+13.333$ kN m at B. The 40 kN load is a direct nodal load at D.

Assembled equation (free DOFs $v_B,\theta_B,v_D,\theta_D,\theta_C$)

With $v_A=\theta_A=0$ and $v_C=0$:

$$EI\,\begin{bmatrix}1.688 & 1.125 & -1.500 & 1.500 & 0 \\ 1.125 & 3.000 & -1.500 & 1.000 & 0 \\ -1.500 & -1.500 & 3.000 & 0 & 1.500 \\ 1.500 & 1.000 & 0 & 4.000 & 1.000 \\ 0 & 0 & 1.500 & 1.000 & 2.000\end{bmatrix}\begin{Bmatrix}v_B\\\theta_B\\v_D\\\theta_D\\\theta_C\end{Bmatrix}=\begin{Bmatrix}-20.000 \\ 13.333 \\ -40.000 \\ 0 \\ 0\end{Bmatrix}$$

Solving:

$$\begin{aligned} v_B&=-\frac{230.00}{EI}\ \text{m},&\theta_B&=-\frac{34.167}{EI}\ \text{rad}\\ v_D&=-\frac{207.083}{EI}\ \text{m},&\theta_D&=\frac{63.958}{EI}\ \text{rad},\quad \theta_C=\frac{123.333}{EI}\ \text{rad} \end{aligned}$$

Deflection at B = 230.0/EI m downward (with $EI$ in kN m$^2$; e.g. $EI=2\times10^4$ kN m$^2$ gives 11.5 mm).

Element end forces $\{f\}=[k]\{d\}-\{F_{eq}\}$

Reactions: $R_A=50.312$ kN (up), $M_A=82.50$ kN m (counter-clockwise), $R_C=29.687$ kN. Check: $50.3125+29.6875=80$ kN = $10\times4+40$.

Bending moment (sagging +) and shear force

• A: $-82.50$ (hogging); at 2 m from A: $-1.875$; B: $+38.75$; D: $+59.375$; C: $0$.
• Shear: $+50.31$ at A, $+10.31$ at B (just before and after), $-29.69$ right of D.

 A(fixed)      B          D         C(roller)
 |-------------|----------|---------|
 BMD: -82.5   +38.75     +59.375     0   (kN m)
 SFD: +50.31 -> +10.31 (A to D) -> -29.69 (D to C)

Stresses at B

At B, $M_B=38.75$ kN m (sagging) and $V_B=10.31$ kN. The bending stress at a fibre at distance $y$ from the neutral axis is

$$\sigma_B=\frac{M_B\,y}{I}=\frac{38.75\times10^6\ y}{I}\ \text{N/mm}^2$$

tension at the bottom and compression at the top. The section size is not given, so for a rectangular section $b\times h$ use $\sigma_{max}=6M_B/(bh^2)$; the average shear stress is $\tau=V_B/A$ (maximum $1.5V_B/A$ for a rectangle).

Answer: $v_B=-230.00/EI$ m (downward), $\theta_B=-34.167/EI$, $M_B=38.75$ kN m; $\sigma_B=M_By/I$.