SS beam 6 m: maximum and point deflection

Reading of the question: the text states one 20 kN load at 4 m from the left support, so a single load is used. $EI$ is not given, so the answer is in terms of $EI$ (kNm²). $L = 6$ m, $a = 4$ m, $b = 2$ m.

          20 kN
            v
 A ^---------------^---- B
   |<---4 m--->|<2m>|

Reactions

$R_A = 20\times 2/6 = 6.67$ kN, $R_B = 13.33$ kN.

Position of maximum deflection

The maximum deflection lies in the longer segment (A to load), where slope is zero. Measured from A:

$$x = \sqrt{\frac{L^2 - b^2}{3}} = \sqrt{\frac{36-4}{3}} = 3.266\ \text{m}$$

This is less than 4 m, so it lies in the longer segment.

Maximum deflection

$$y_{max} = \frac{P\,b\,x\,(L^2 - b^2 - x^2)}{6EIL} = \frac{20\times2\times3.266\,(36-4-10.667)}{6\times6\,EI} = \frac{77.42}{EI}\ \text{(kNm}^3\text{)}$$

Deflection under the load

$$y_C = \frac{P\,a^2 b^2}{3EIL} = \frac{20\times16\times4}{3\times6\,EI} = \frac{71.11}{EI}\ \text{(kNm}^3\text{)}$$

Answer: (i) maximum deflection $77.42/EI$ at 3.266 m from the left support; (ii) deflection under the load $71.11/EI$ (with $EI$ in kNm², result in m).

If the figure is read as two 20 kN loads at 2 m and 4 m, the beam is symmetrical: maximum deflection at mid-span $= Pa(3L^2-4a^2)/24EI = 153.33/EI$, and under each load $= 133.33/EI$.