Three moment theorem: SFD and BMD of beam

Reading of the figure: A fixed; span AB = 8 m with 100 kN at 4 m from A; span BC = 10 m with 60 kN/m; overhang CD = 2 m with 20 kN at D. B and C are rollers and $EI$ is constant.

Moment at C (overhang)

$M_C = -20\times2 = -40$ kNm (hogging).

Load terms

• Span AB: free BM peak $\dfrac{100\times4\times4}{8} = 200$ kNm, $a_1 = \tfrac12(8)(200) = 800$, $\bar x_1 = 4$ m: $\dfrac{6a_1\bar x_1}{L_1} = \dfrac{6(800)(4)}{8} = 2400$
• Span BC: $\dfrac{wL_2^3}{4} = \dfrac{60(10)^3}{4} = 15000$

Three moment equations

At A (fixed end): $2M_A(8) + M_B(8) = -2400$

$$2M_A + M_B = -300$$

At B: $M_A(8) + 2M_B(8+10) + M_C(10) = -2400 - 15000$

$$8M_A + 36M_B + 10(-40) = -17400 \ \Rightarrow\ 8M_A + 36M_B = -17000$$

Substituting $M_A = \dfrac{-300 - M_B}{2}$: $-1200 - 4M_B + 36M_B = -17000$, so $32M_B = -15800$:

$$M_B = -493.750\ \text{kNm},\qquad M_A = 96.875\ \text{kNm}$$

($M_A$ is positive: the fixed end is sagging for this loading.)

Reactions and shear

Span AB ($M_B = M_A + 8R_A - 400$):

$$R_A = \frac{400 + M_B - M_A}{8} = \frac{400 - 493.75 - 96.875}{8} = -23.828\ \text{kN}$$

(negative: A is pulled down). Shear just left of B $= -23.828 - 100 = -123.828$ kN.

Span BC ($M_C = M_B + 10V_{BR} - 60\frac{10^2}{2}$):

$$V_{BR} = \frac{3000 + M_C - M_B}{10} = \frac{3000 - 40 + 493.75}{10} = 345.375\ \text{kN}$$ $$R_B = 123.828 + 345.375 = 469.203\ \text{kN},\qquad R_C = 600 - 345.375 + 20 = 274.625\ \text{kN}$$

Check: $R_A + R_B + R_C = 720.00$ kN $= 100 + 600 + 20 = 720$ kN.

Shear force (kN)

The SF is zero in BC at 5.76 m from B.

Bending moment (kNm)

Answer: $M_A = 96.87$ kNm, $M_B = -493.75$ kNm, $M_C = -40$ kNm; $R_A = -23.828$ kN, $R_B = 469.203$ kN, $R_C = 274.625$ kN; maximum sagging moment in BC $= 500.28$ kNm.