1. **Problem Statement:** Calculate the slope and deflection at point A of a cantilever beam with a moment load of 75 klb·ft at A and a uniformly distributed load of 3.5 klb/ft over the segment BC (10 ft). The beam has length AB = 5 ft and BC = 10 ft, modulus of elasticity $E = 29000$ ksi, and moment of inertia $I = 110$ in$^4$. 2. **Formula and Method:** We use Macaulay’s Step Function method to find slope $\theta(x)$ and deflection $y(x)$. The bending moment $M(x)$ along the beam is expressed using step functions $\langle x - a \rangle^n = (x - a)^n$ if $x > a$, else 0. The beam equation is: $$EI \frac{d^2 y}{dx^2} = M(x)$$ Integrate twice to find slope and deflection: $$EI \frac{dy}{dx} = \int M(x) dx + C_1$$ $$EI y = \int \left( \int M(x) dx + C_1 \right) dx + C_2$$ Boundary conditions at fixed end C ($x=15$ ft): $$y(15) = 0, \quad \theta(15) = 0$$ 3. **Express $M(x)$:** - Moment at A (x=0): $-75$ klb·ft (negative for clockwise moment) - Uniform load $w=3.5$ klb/ft from $x=5$ to $x=15$ Moment at section $x$ (from left end A): $$M(x) = -75 + 3.5 \langle x - 5 \rangle^2 / 2$$ 4. **Integrate $M(x)$:** First integration: $$EI \theta(x) = \int M(x) dx = -75x + \frac{3.5}{2} \int \langle x - 5 \rangle^2 dx + C_1$$ $$= -75x + \frac{3.5}{2} \cdot \frac{\langle x - 5 \rangle^3}{3} + C_1 = -75x + \frac{3.5}{6} \langle x - 5 \rangle^3 + C_1$$ Second integration: $$EI y(x) = \int EI \theta(x) dx = \int \left(-75x + \frac{3.5}{6} \langle x - 5 \rangle^3 + C_1 \right) dx + C_2$$ $$= -\frac{75 x^2}{2} + \frac{3.5}{6} \cdot \frac{\langle x - 5 \rangle^4}{4} + C_1 x + C_2$$ $$= -37.5 x^2 + \frac{3.5}{24} \langle x - 5 \rangle^4 + C_1 x + C_2$$ 5. **Apply boundary conditions at $x=15$ ft:** - Slope zero: $$EI \theta(15) = -75(15) + \frac{3.5}{6} (15-5)^3 + C_1 = 0$$ Calculate: $$(15-5)^3 = 10^3 = 1000$$ $$-75 \times 15 = -1125$$ $$\frac{3.5}{6} \times 1000 = \frac{3.5 \times 1000}{6} = 583.3333$$ So: $$-1125 + 583.3333 + C_1 = 0 \Rightarrow C_1 = 541.6667$$ - Deflection zero: $$EI y(15) = -37.5 (15)^2 + \frac{3.5}{24} (10)^4 + 541.6667 \times 15 + C_2 = 0$$ Calculate: $$-37.5 \times 225 = -8437.5$$ $$(10)^4 = 10000$$ $$\frac{3.5}{24} \times 10000 = 1458.3333$$ $$541.6667 \times 15 = 8125$$ Sum: $$-8437.5 + 1458.3333 + 8125 + C_2 = 0$$ $$1145.8333 + C_2 = 0 \Rightarrow C_2 = -1145.8333$$ 6. **Calculate slope and deflection at A ($x=0$):** - Slope: $$EI \theta(0) = -75 \times 0 + \frac{3.5}{6} \langle 0 - 5 \rangle^3 + 541.6667 = 541.6667$$ Since $\langle 0 - 5 \rangle^3 = 0$ (because $0 < 5$), $$\theta(0) = \frac{541.6667}{EI}$$ Convert $EI$ to consistent units: $$E = 29000 \text{ ksi} = 29000 \times 1000 = 29,000,000 \text{ psi}$$ $$I = 110 \text{ in}^4$$ $$EI = 29,000,000 \times 110 = 3.19 \times 10^9 \text{ lb-in}^2$$ Convert length units from ft to in for slope (since $x$ in ft, multiply by 12 for in): Slope is dimensionless (radians), so: $$\theta(0) = \frac{541.6667}{3.19 \times 10^9} = 1.697 \times 10^{-7} \text{ radians}$$ - Deflection: $$EI y(0) = -37.5 \times 0 + \frac{3.5}{24} \times 0 + 541.6667 \times 0 - 1145.8333 = -1145.8333$$ $$y(0) = \frac{-1145.8333}{3.19 \times 10^9} = -3.59 \times 10^{-7} \text{ ft}$$ Convert to inches: $$-3.59 \times 10^{-7} \times 12 = -4.31 \times 10^{-6} \text{ in}$$ **Final answers:** - Slope at A: $\theta_A = 1.7 \times 10^{-7}$ radians (approx.) - Deflection at A: $y_A = -4.3 \times 10^{-6}$ inches (approx.) These small values are expected for a stiff steel beam under the given loads.