Beam reactions, deflection, BMD (30 kN, 20 kN/m)

The beam is solved by the direct stiffness method with three beam elements.

Data and idealisation

$EI=(20\times10^{5}\ \text{N/mm}^2)(5\times10^{6}\ \text{mm}^4)=1\times10^{13}\ \text{N·mm}^2=1\times10^{4}\ \text{kN·m}^2$ (since $1\ \text{kN·m}^2=10^{9}\ \text{N·mm}^2$).

        30 kN     40 kNm        20 kN/m
          |       (clockwise)  ||||||||||||
 ##=======o=========o==============o
  A       2         B              C
  fixed   |<- 5 m ->|    |<- 6 m ->|
 |<- 3 m->|

• Nodes: 1 = A (x = 0, fixed), 2 (x = 3 m, point load), 3 = B (x = 8 m, support, $M=40$ kNm taken clockwise), 4 = C (x = 14 m, roller).
• Elements: 1 (3 m), 2 (5 m), 3 (6 m, carries the UDL from B to C).
• Dofs per node: (deflection $v$ up, rotation $\theta$ anticlockwise). Known: $v_1=\theta_1=0$, $v_3=0$, $v_4=0$. Unknown: $v_2,\theta_2,\theta_3,\theta_4$.

Element stiffness (each element)

$$[k]=\frac{EI}{L^3}\begin{bmatrix}12&6L&-12&6L\\6L&4L^2&-6L&2L^2\\-12&-6L&12&-6L\\6L&2L^2&-6L&4L^2\end{bmatrix}$$

Equivalent nodal loads

Element 3: $wL/2=60$ kN down at each end, $wL^2/12=60$ kNm (clockwise at B, anticlockwise at C). Loads on the structure for $(v_2,\theta_2,\theta_3,\theta_4)$:

$$\{F\}=\begin{bmatrix}-30\\0\\-40-60\\+60\end{bmatrix}=\begin{bmatrix}-30\\0\\-100\\60\end{bmatrix}$$

Reduced equations (after assembly)

$$EI\begin{bmatrix} 0.54044 & -0.42667 & 0.24 & 0 \\ -0.42667 & 2.1333 & 0.4 & 0 \\ 0.24 & 0.4 & 1.4667 & 0.33333 \\ 0 & 0 & 0.33333 & 0.66667 \end{bmatrix}\begin{bmatrix}v_2\\\theta_2\\\theta_3\\\theta_4\end{bmatrix}=\begin{bmatrix}-30\\0\\-100\\60\end{bmatrix}$$

Solution:

$$\begin{aligned}v_2&=\frac{10.657}{EI}=1.066\times10^{-3}\ \text{m}=1.066\ \text{mm},&\theta_2&=\frac{22.551}{EI}=2.255\times10^{-3}\ \text{rad}\\ \theta_3&=-\frac{108.906}{EI}=-10.891\times10^{-3}\ \text{rad},&\theta_4&=\frac{144.453}{EI}=14.445\times10^{-3}\ \text{rad}\end{aligned}$$

(with $EI=10^4$ kN·m$^2$; $v_2$ is upward).

Support reactions ($\{R\}=[K]\{d\}-\{F\}$)

$$R_A=10.298\ \text{kN}\uparrow,\quad M_A=7.930\ \text{kNm (anticlockwise)},\quad R_B=85.627\ \text{kN}\uparrow,\quad R_C=54.076\ \text{kN}\uparrow$$

Check: $\Sigma F_y$: $10.298+85.627+54.076=150.000=30+20\times6$. $\Sigma M_A=0$ also checks.

Deflection at mid-span

Using the element shape functions (and adding the UDL term $wx^2(L-x)^2/24EI$ inside the loaded element):

• Mid-span of A-B ($x=4$ m): $v=4.141\times10^{-3}$ m $=4.141$ mm (upward).
• Mid-span of B-C ($x=11$ m): $v=-25.752$ mm (downward). Without the UDL term the cubic interpolation alone would give $-19.002$ mm.

Bending moment (sagging +), from the reactions

 BMD, sagging up (1 row = 12 kNm, 1 column = 0.35 m)
   84 |                                         
   72 |                              *****      
   60 |                             *     **    
   48 |                            *        *   
   36 |                           *          *  
   24 |        **                *              
   12 |    ****  **             *             * 
    0 |-***--------*---------------------------*
  -12 |*            **         *                
  -24 |               **                        
  -36 |                 *     *                 
  -48 |                  **                     
  -60 |                    **                   
  -72 |                      *                  
  -84 |                                         

The diagram runs from A at the left edge to C at the right edge: hogging at A, sagging under the 30 kN load, deep hogging just left of B, a jump of 40 kNm at B, then sagging in B-C with the maximum at 3.30 m from B, and zero at C.

Answer: $R_A=10.298$, $R_B=85.627$, $R_C=54.076$ kN; $M_A=7.930$ kNm; mid-span deflections $4.141$ mm upward (AB) and $25.752$ mm downward (BC).