1. **Problem statement:** Find the rate of change of deflection (the slope) with respect to distance $x$ for the cantilever beam in Fig. 29-8, where deflection is given by $$y = \frac{w x^2}{24 E I} (x^2 + 6L^2 - 4L x)$$ with $E$, $I$, $w$, and $L$ constants. 2. **Formula and rules:** To find the rate of change of deflection, we need the derivative $\frac{dy}{dx}$. Use the product rule: if $y = f(x) g(x)$, then $$\frac{dy}{dx} = f'(x) g(x) + f(x) g'(x)$$ 3. **Identify functions:** Let $$f(x) = \frac{w x^2}{24 E I}, \quad g(x) = x^2 + 6L^2 - 4L x$$ 4. **Compute derivatives:** $$f'(x) = \frac{w}{24 E I} \cdot 2x = \frac{2 w x}{24 E I} = \frac{w x}{12 E I}$$ $$g'(x) = 2x - 4L$$ 5. **Apply product rule:** $$\frac{dy}{dx} = f'(x) g(x) + f(x) g'(x) = \frac{w x}{12 E I} (x^2 + 6L^2 - 4L x) + \frac{w x^2}{24 E I} (2x - 4L)$$ 6. **Simplify terms:** First term: $$\frac{w x}{12 E I} (x^2 + 6L^2 - 4L x) = \frac{w x}{12 E I} x^2 + \frac{w x}{12 E I} 6L^2 - \frac{w x}{12 E I} 4L x = \frac{w x^3}{12 E I} + \frac{6 w L^2 x}{12 E I} - \frac{4 w L x^2}{12 E I}$$ Simplify coefficients: $$= \frac{w x^3}{12 E I} + \frac{w L^2 x}{2 E I} - \frac{w L x^2}{3 E I}$$ Second term: $$\frac{w x^2}{24 E I} (2x - 4L) = \frac{w x^2}{24 E I} 2x - \frac{w x^2}{24 E I} 4L = \frac{2 w x^3}{24 E I} - \frac{4 w L x^2}{24 E I} = \frac{w x^3}{12 E I} - \frac{w L x^2}{6 E I}$$ 7. **Add terms:** $$\frac{dy}{dx} = \left(\frac{w x^3}{12 E I} + \frac{w L^2 x}{2 E I} - \frac{w L x^2}{3 E I}\right) + \left(\frac{w x^3}{12 E I} - \frac{w L x^2}{6 E I}\right) = \frac{w x^3}{6 E I} + \frac{w L^2 x}{2 E I} - \frac{w L x^2}{2 E I}$$ 8. **Final expression for slope:** $$\boxed{\frac{dy}{dx} = \frac{w}{E I} \left( \frac{x^3}{6} + \frac{L^2 x}{2} - \frac{L x^2}{2} \right)}$$ --- 1. **Problem statement:** Find the rate of change of deflection (the slope) of the elastic curve for the simply supported beam in Fig. 29-9 at $x = \frac{L}{4}$, where $$y = \frac{w x}{24 E I} (L^3 - 2 L x^2 + x^3)$$ 2. **Formula and rules:** Again, find $\frac{dy}{dx}$ using the product rule. 3. **Identify functions:** $$f(x) = \frac{w x}{24 E I}, \quad g(x) = L^3 - 2 L x^2 + x^3$$ 4. **Compute derivatives:** $$f'(x) = \frac{w}{24 E I}$$ $$g'(x) = 0 - 4 L x + 3 x^2 = -4 L x + 3 x^2$$ 5. **Apply product rule:** $$\frac{dy}{dx} = f'(x) g(x) + f(x) g'(x) = \frac{w}{24 E I} (L^3 - 2 L x^2 + x^3) + \frac{w x}{24 E I} (-4 L x + 3 x^2)$$ 6. **Simplify terms:** First term: $$\frac{w}{24 E I} (L^3 - 2 L x^2 + x^3)$$ Second term: $$\frac{w x}{24 E I} (-4 L x + 3 x^2) = \frac{w x}{24 E I} (-4 L x) + \frac{w x}{24 E I} (3 x^2) = -\frac{4 w L x^2}{24 E I} + \frac{3 w x^3}{24 E I} = -\frac{w L x^2}{6 E I} + \frac{w x^3}{8 E I}$$ 7. **Combine all:** $$\frac{dy}{dx} = \frac{w}{24 E I} L^3 - \frac{2 w L x^2}{24 E I} + \frac{w x^3}{24 E I} - \frac{w L x^2}{6 E I} + \frac{w x^3}{8 E I}$$ Simplify coefficients: $$- \frac{2 w L x^2}{24 E I} = - \frac{w L x^2}{12 E I}$$ Add with $- \frac{w L x^2}{6 E I}$: $$- \frac{w L x^2}{12 E I} - \frac{w L x^2}{6 E I} = - \frac{w L x^2}{12 E I} - \frac{2 w L x^2}{12 E I} = - \frac{3 w L x^2}{12 E I} = - \frac{w L x^2}{4 E I}$$ Add $\frac{w x^3}{24 E I} + \frac{w x^3}{8 E I} = \frac{w x^3}{24 E I} + \frac{3 w x^3}{24 E I} = \frac{4 w x^3}{24 E I} = \frac{w x^3}{6 E I}$ 8. **Final simplified derivative:** $$\frac{dy}{dx} = \frac{w L^3}{24 E I} - \frac{w L x^2}{4 E I} + \frac{w x^3}{6 E I}$$ 9. **Evaluate at $x = \frac{L}{4}$:** Calculate each term: $$x^2 = \left(\frac{L}{4}\right)^2 = \frac{L^2}{16}$$ $$x^3 = \left(\frac{L}{4}\right)^3 = \frac{L^3}{64}$$ Substitute: $$\frac{dy}{dx} = \frac{w L^3}{24 E I} - \frac{w L \cdot \frac{L^2}{16}}{4 E I} + \frac{w \cdot \frac{L^3}{64}}{6 E I} = \frac{w L^3}{24 E I} - \frac{w L^3}{64 E I} + \frac{w L^3}{384 E I}$$ Find common denominator 384: $$\frac{w L^3}{24 E I} = \frac{16 w L^3}{384 E I}$$ $$\frac{w L^3}{64 E I} = \frac{6 w L^3}{384 E I}$$ $$\frac{w L^3}{384 E I} = \frac{w L^3}{384 E I}$$ Sum: $$\frac{dy}{dx} = \frac{16 w L^3}{384 E I} - \frac{6 w L^3}{384 E I} + \frac{w L^3}{384 E I} = \frac{(16 - 6 + 1) w L^3}{384 E I} = \frac{11 w L^3}{384 E I}$$ 10. **Final answer:** $$\boxed{\frac{dy}{dx} \bigg|_{x=\frac{L}{4}} = \frac{11 w L^3}{384 E I}}$$