Beam Deflection 10Fd85

1. **Problem Statement:** We have a simply supported beam with supports at points A and D. There is a point load of $40$ kN at point B, which is $6$ m from A, and a uniformly distributed load (UDL) of $20$ kN/m from point C to D over $10$ m. We need to find the slope and deflection at point C using the Macaulay method in terms of $EI$. 2. **Given Data:** - Point load $P = 40$ kN at $x = 6$ m - UDL $w = 20$ kN/m from $x = 10$ m to $x = 20$ m (since $C$ is at $10$ m and $D$ at $20$ m) - Beam length $L = 20$ m 3. **Step 1: Calculate reactions at supports A and D** Sum of vertical forces: $$ R_A + R_D = 40 + 20 \times 10 = 40 + 200 = 240 \text{ kN} $$ Taking moments about A: $$ R_D \times 20 = 40 \times 6 + 20 \times 10 \times (10 + \frac{10}{2}) $$ $$ R_D \times 20 = 240 + 20 \times 10 \times 15 = 240 + 3000 = 3240 $$ $$ R_D = \frac{3240}{20} = 162 \text{ kN} $$ Then, $$ R_A = 240 - 162 = 78 \text{ kN} $$ 4. **Step 2: Write the bending moment equation using Macaulay brackets** Define $x$ from A. The bending moment $M(x)$ is: $$ M(x) = R_A x - 40 \langle x - 6 \rangle^0 - 20 \int_0^x \langle x - 10 \rangle^0 dx \times (x - 10) $$ More precisely, $$ M(x) = R_A x - 40 \langle x - 6 \rangle^0 - 20 \langle x - 10 \rangle^1 $$ where Macaulay bracket $\langle x - a \rangle^n = (x - a)^n$ if $x > a$, else 0. 5. **Step 3: Express the moment equation explicitly:** $$ M(x) = 78 x - 40 \langle x - 6 \rangle^0 - 20 \langle x - 10 \rangle^1 $$ 6. **Step 4: Use the differential equation of the elastic curve:** $$ EI \frac{d^2 y}{dx^2} = M(x) $$ Integrate twice to find slope $\theta = \frac{dy}{dx}$ and deflection $y$: $$ EI \frac{d^2 y}{dx^2} = 78 x - 40 \langle x - 6 \rangle^0 - 20 \langle x - 10 \rangle^1 $$ Integrate once: $$ EI \frac{dy}{dx} = \int M(x) dx = 39 x^2 - 40 \langle x - 6 \rangle^1 - 10 \langle x - 10 \rangle^2 + C_1 $$ Integrate again: $$ EI y = 13 x^3 - 40 \frac{\langle x - 6 \rangle^2}{2} - 10 \frac{\langle x - 10 \rangle^3}{3} + C_1 x + C_2 $$ 7. **Step 5: Apply boundary conditions to find constants $C_1$ and $C_2$** At support A ($x=0$), deflection $y=0$: $$ EI y(0) = 0 = 0 - 0 - 0 + 0 + C_2 \Rightarrow C_2 = 0 $$ At support D ($x=20$), deflection $y=0$: Calculate Macaulay terms: $$ \langle 20 - 6 \rangle^2 = 14^2 = 196 $$ $$ \langle 20 - 10 \rangle^3 = 10^3 = 1000 $$ So, $$ 0 = 13 \times 20^3 - 40 \times \frac{196}{2} - 10 \times \frac{1000}{3} + C_1 \times 20 $$ Calculate each term: $$ 13 \times 8000 = 104000 $$ $$ 40 \times 98 = 3920 $$ $$ 10 \times \frac{1000}{3} = \frac{10000}{3} \approx 3333.33 $$ Substitute: $$ 0 = 104000 - 3920 - 3333.33 + 20 C_1 $$ $$ 0 = 104000 - 7253.33 + 20 C_1 $$ $$ 0 = 96846.67 + 20 C_1 $$ Solve for $C_1$: $$ 20 C_1 = -96846.67 $$ $$ C_1 = -4842.33 $$ 8. **Step 6: Write final expressions for slope and deflection at point C ($x=10$ m)** Calculate slope: $$ EI \theta_C = 39 \times 10^2 - 40 \langle 10 - 6 \rangle^1 - 10 \langle 10 - 10 \rangle^2 + C_1 $$ Calculate Macaulay terms: $$ \langle 10 - 6 \rangle^1 = 4 $$ $$ \langle 10 - 10 \rangle^2 = 0 $$ Substitute: $$ EI \theta_C = 39 \times 100 - 40 \times 4 + 0 - 4842.33 $$ $$ EI \theta_C = 3900 - 160 - 4842.33 = -1102.33 $$ Calculate deflection: $$ EI y_C = 13 \times 10^3 - 40 \times \frac{\langle 10 - 6 \rangle^2}{2} - 10 \times \frac{\langle 10 - 10 \rangle^3}{3} + C_1 \times 10 $$ Calculate Macaulay terms: $$ \langle 10 - 6 \rangle^2 = 4^2 = 16 $$ $$ \langle 10 - 10 \rangle^3 = 0 $$ Substitute: $$ EI y_C = 13 \times 1000 - 40 \times \frac{16}{2} - 0 - 4842.33 \times 10 $$ $$ EI y_C = 13000 - 40 \times 8 - 48423.3 $$ $$ EI y_C = 13000 - 320 - 48423.3 = -35743.3 $$ **Final answers:** $$ \boxed{EI \theta_C = -1102.33} $$ $$ \boxed{EI y_C = -35743.3} $$ These are the slope and deflection at point C in terms of $EI$.