Moment area: slopes at A, B, deflection at D

Given data

Simply supported beam AB, $L = 12$ m, 60 kN at C (4 m from A). D is 6 m from B (6 m from A). $EI$ constant.

       60 kN
 A ------v--------D-------- B
 ^  4 m  C   2 m     6 m    ^

Reactions and bending moments

$$R_A = \frac{60\times8}{12} = 40\ \text{kN},\qquad R_B = 20\ \text{kN}$$ $$M_C = 40\times4 = 160\ \text{kNm},\qquad M_D = 20\times6 = 120\ \text{kNm}$$

The $M/EI$ diagram is a triangle A-C-B with peak $160/EI$ at C.

Slope at A

Deviation of B from the tangent at A (Theorem II): area of the triangle $= \tfrac12\times12\times160 = 960$; its centroid is $(0 + 4 + 12)/3 = 5.333$ m from A, i.e. 6.667 m from B.

$$\Delta_{B/A} = \frac{960\times6.667}{EI} = \frac{6400}{EI}$$ $$\theta_A = \frac{\Delta_{B/A}}{L} = \frac{6400}{12\,EI} = \frac{533.33}{EI}\ \text{(clockwise)}$$

Slope at B (Theorem I)

$$\theta_B = \theta_A - \text{Area}_{AB} = \frac{533.33 - 960}{EI} = -\frac{426.67}{EI}$$

so $\theta_B = 426.67/EI$ (anticlockwise).

Deflection at D

Deviation of D from the tangent at A (moment of the $M/EI$ area between A and D about D):

• Triangle A to C: area $= \tfrac12\times4\times160 = 320$, centroid 3.333 m from D: $320\times3.333 = 1066.67$
• Rectangle C to D: $2\times120 = 240$, centroid 1 m from D: $240$
• Triangle (extra height 40 at C, zero at D): $\tfrac12\times2\times40 = 40$, centroid 1.333 m from D: $53.33$

$$\Delta_{D/A} = \frac{1066.67 + 240 + 53.33}{EI} = \frac{1360}{EI}$$

Tangent offset at D: $\theta_A\times6 = \dfrac{533.33\times6}{EI} = \dfrac{3200}{EI}$.

$$y_D = \frac{3200 - 1360}{EI} = \frac{1840}{EI}\ \text{(downward)}$$

Answer: $\theta_A = 533.3/EI$ (clockwise), $\theta_B = 426.7/EI$ (anticlockwise), deflection at D $= 1840/EI$ downward (kNm$^3$/EI, i.e. in metres with $EI$ in kNm$^2$).