1. **State the problem:** Calculate the reactions at supports A and B, then find shear force and bending moment values along the beam. 2. **Given data:** - Point loads: $P_1=10$ kN at 0 m, $P_2=15$ kN at 7 m, $P_3=12$ kN at 12 m - Uniform distributed loads: $q_1=9$ kN/m over 3 m (0 to 3 m), $q_2=12$ kN/m over 5 m (7 to 12 m) - Beam length: 12 m 3. **Calculate total uniformly distributed loads:** - Load from $q_1$: $Q_1 = q_1 \times 3 = 9 \times 3 = 27$ kN acting at 1.5 m from A - Load from $q_2$: $Q_2 = q_2 \times 5 = 12 \times 5 = 60$ kN acting at 9.5 m from A (7 + 2.5) 4. **Sum of vertical forces for equilibrium:** $$ R_A + R_B = P_1 + Q_1 + P_2 + Q_2 + P_3 = 10 + 27 + 15 + 60 + 12 = 124 \text{ kN} $$ 5. **Sum moments about A (taking counterclockwise positive):** $$ \sum M_A = 0 = -P_1 \times 0 - Q_1 \times 1.5 - P_2 \times 7 - Q_2 \times 9.5 - P_3 \times 12 + R_B \times 12 $$ Calculate each moment: - $P_1 \times 0 = 0$ - $Q_1 \times 1.5 = 27 \times 1.5 = 40.5$ - $P_2 \times 7 = 15 \times 7 = 105$ - $Q_2 \times 9.5 = 60 \times 9.5 = 570$ - $P_3 \times 12 = 12 \times 12 = 144$ So, $$ 0 = -0 - 40.5 - 105 - 570 - 144 + 12 R_B $$ $$ 12 R_B = 40.5 + 105 + 570 + 144 = 859.5 $$ $$ R_B = \frac{859.5}{12} = 71.625 \text{ kN} $$ 6. **Calculate $R_A$ using vertical force equilibrium:** $$ R_A = 124 - R_B = 124 - 71.625 = 52.375 \text{ kN} $$ 7. **Shear force calculation:** - At $x=0^+$: $V = R_A = 52.375$ kN - Just after $P_1$ at 0 m: $V = 52.375 - 10 = 42.375$ kN - At $x=3$ (end of $q_1$): subtract $Q_1=27$ kN uniformly over 3 m, so shear decreases linearly from 42.375 to $42.375 - 27 = 15.375$ kN - Just before $P_2$ at 7 m: shear remains 15.375 kN - Just after $P_2$: $15.375 - 15 = 0.375$ kN - At $x=7$ to $x=12$ (over $q_2$): shear decreases linearly by $Q_2=60$ kN over 5 m, so at 12 m shear is $0.375 - 60 = -59.625$ kN - Just after $P_3$ at 12 m: $-59.625 - 12 = -71.625$ kN - At support B, shear must be zero, and $R_B=71.625$ kN balances this. 8. **Bending moment calculation:** - At $x=0$: $M=0$ - From 0 to 3 m (with $q_1$): $$ M(x) = R_A x - 10 x - \frac{9 x^2}{2} $$ - At $x=3$: $$ M(3) = 52.375 \times 3 - 10 \times 3 - \frac{9 \times 3^2}{2} = 157.125 - 30 - 40.5 = 86.625 \text{ kNm} $$ - From 3 to 7 m (no load except $P_2$ at 7 m): $$ M(x) = M(3) + 15.375 (x - 3) $$ - At $x=7$: $$ M(7) = 86.625 + 15.375 \times 4 = 86.625 + 61.5 = 148.125 \text{ kNm} $$ - From 7 to 12 m (with $q_2$): $$ M(x) = M(7) + 0.375 (x - 7) - \frac{12 (x - 7)^2}{2} $$ - At $x=12$: $$ M(12) = 148.125 + 0.375 \times 5 - 6 \times 25 = 148.125 + 1.875 - 150 = 0 \text{ kNm} $$ **Final answers:** - Reactions: $R_A = 52.375$ kN, $R_B = 71.625$ kN - Shear force varies as calculated above - Bending moment varies as calculated above