Three moment equation: SFD and BMD of beam

Reading of the figure: A fixed, span AB = 8 m with a 300 kN load at 4 m from A, B an interior support, span BC = 10 m with 50 kN/m over the whole span, C a roller. EI is constant.

The three moment equation is $M_A L_1 + 2M_B(L_1+L_2) + M_C L_2 = -\dfrac{6a_1\bar x_1}{L_1} - \dfrac{6a_2\bar x_2}{L_2}$ with sagging moments positive. The fixed end A is treated by adding an imaginary span of zero length to its left, which makes the slope at A zero. At the roller C the moment is $M_C = 0$.

Load terms

• Span AB: free BM is a triangle with peak $\dfrac{300\times4\times4}{8} = 600$ kNm, area $a_1 = \tfrac12(8)(600) = 2400$ kNm$^2$, $\bar x_1 = 4$ m. $\dfrac{6a_1\bar x_1}{L_1} = \dfrac{6(2400)(4)}{8} = 7200$.
• Span BC: UDL, $\dfrac{wL_2^3}{4} = \dfrac{50(10)^3}{4} = 12500$.

Equations

At A (fixed end):

$$2M_A(8) + M_B(8) = -7200 \quad\Rightarrow\quad 16M_A + 8M_B = -7200$$

At B:

$$M_A(8) + 2M_B(8+10) + 0 = -7200 - 12500 \quad\Rightarrow\quad 8M_A + 36M_B = -19700$$

Solving:

$$M_B = -503.125\ \text{kNm},\qquad M_A = -198.4375\ \text{kNm}$$

Both are hogging.

Reactions

With sagging-positive moments, $M(x) = M_A + R_A x - 300\langle x-4\rangle$ in AB, so $M_B = M_A + 8R_A - 1200$:

$$R_A = \frac{1200 + M_B - M_A}{8} = \frac{1200 + (-503.125) - (-198.4375)}{8} = 111.914\ \text{kN}$$

Shear just left of B $= 111.914 - 300 = -188.086$ kN. In BC, $M(x) = M_B + V_B' x - 25x^2$ with $M_C = 0$ at $x = 10$:

$$V_B' = \frac{2500 - M_B}{10} = 300.3125\ \text{kN},\qquad R_B = 188.086 + 300.312 = 488.398\ \text{kN}$$ $$R_C = 500 - 300.3125 = 199.6875\ \text{kN}$$

Check: $R_A + R_B + R_C = 800.00$ kN $= 300 + 50(10) = 800$ kN.

Shear force (kN)

SF is zero in BC at $x = 3.994$ m from C.

Bending moment (kNm)

Points of contraflexure: 1.77 m from A; 1.33 m beyond E (towards B); and 2.01 m beyond B in BC.

Answer: $M_A = -198.44$ kNm and $M_B = -503.12$ kNm (hogging); $R_A = 111.91$ kN, $R_B = 488.40$ kN, $R_C = 199.69$ kN; peak sagging BM $= 398.75$ kNm in BC.