1. **Problem Statement:** Calculate the support reactions, draw the shear force diagram, and draw the bending moment diagram for a beam with the following data: - Length AB = $2.T4 = 2 \times 3.4 = 6.8$ m - Length BC = $2.S4 = 2 \times 6.4 = 12.8$ m - Length CD = $0.5Q = 0.5 \times 0 = 0$ m (so point D coincides with C) - Uniformly distributed load (UDL) on BC = 2.63 kN/m downward - Point load at C = 3.25 kN downward - Fixed support at A 2. **Given:** - $T=3$, $S=6$, $Q=0$ - $AB=6.8$ m, $BC=12.8$ m, $CD=0$ m - UDL on BC = 2.63 kN/m - Point load at C = 3.25 kN 3. **Step 1: Calculate total loads and their positions** - Total UDL load on BC: $w = 2.63 \times 12.8 = 33.664$ kN downward - Position of UDL resultant from B: midpoint of BC = $12.8/2 = 6.4$ m - Position of UDL resultant from A: $AB + 6.4 = 6.8 + 6.4 = 13.2$ m - Point load at C is at $AB + BC = 6.8 + 12.8 = 19.6$ m from A 4. **Step 2: Support reactions at fixed support A** - Let vertical reaction at A be $R_A$ and moment reaction at A be $M_A$ - Sum of vertical forces: $R_A - 33.664 - 3.25 = 0 \Rightarrow R_A = 36.914$ kN upward - Sum of moments about A (taking clockwise positive): $$M_A - 33.664 \times 13.2 - 3.25 \times 19.6 = 0$$ $$M_A = 33.664 \times 13.2 + 3.25 \times 19.6$$ $$M_A = 444.0448 + 63.7 = 507.7448 \text{ kN}\cdot\text{m}$$ 5. **Step 3: Shear force diagram** - At A: shear $V = R_A = 36.914$ kN upward - From A to B (0 to 6.8 m): no load, shear constant at 36.914 kN - From B to C (6.8 to 19.6 m): shear decreases linearly due to UDL 2.63 kN/m $$V(x) = 36.914 - 2.63 (x - 6.8)$$ for $6.8 \leq x \leq 19.6$ - At C (x=19.6 m), shear just before point load: $$V(19.6^-) = 36.914 - 2.63 (19.6 - 6.8) = 36.914 - 2.63 \times 12.8 = 36.914 - 33.664 = 3.25$$ kN - Shear just after point load at C: $$V(19.6^+) = 3.25 - 3.25 = 0$$ kN 6. **Step 4: Moment diagram** - Moment at A: $M_A = 507.7448$ kN·m - Moment at any point $x$ between A and B (0 to 6.8 m): $$M(x) = M_A - R_A x = 507.7448 - 36.914 x$$ - Moment between B and C (6.8 to 19.6 m): $$M(x) = M_A - R_A x + \text{moment due to UDL}$$ Moment due to UDL from B to $x$: $$= 2.63 \times (x - 6.8) \times \frac{x - 6.8}{2} = 1.315 (x - 6.8)^2$$ So, $$M(x) = 507.7448 - 36.914 x + 1.315 (x - 6.8)^2$$ - At C (x=19.6 m), moment just before point load: $$M(19.6) = 507.7448 - 36.914 \times 19.6 + 1.315 \times (19.6 - 6.8)^2$$ $$= 507.7448 - 723.5344 + 1.315 \times 163.84 = 507.7448 - 723.5344 + 215.36 = -0.4296 \text{ kN}\cdot\text{m}$$ - Point load at C causes a sudden change in shear but moment is continuous. **Final answers:** - Support reactions: $R_A = 36.914$ kN upward, $M_A = 507.745$ kN·m (rounded) - Shear force varies as described above - Moment varies as described above