1. **Problem Statement:** Draw the shear and moment diagrams for a cantilever beam fixed at the left end (x=0) and extending to the right for 9 ft. The beam has: - A uniform distributed load (UDL) of 2 kip/ft from x=0 to x=6 ft downward. - A linearly decreasing distributed load from 2 kip/ft at x=6 ft to 0 at x=9 ft downward. - A 4 kip downward point load at the free end (x=9 ft). - A clockwise end moment of 20 kip·ft at the free end (x=9 ft). Find the shear force $V(x)$ and bending moment $M(x)$ functions for $0 < x < 6$ ft and $6 < x < 9$ ft. 2. **Step 1: Define coordinate system and loads** Let $x$ be the distance from the fixed end (left) toward the free end (right). 3. **Step 2: Calculate reactions at the fixed end (x=0)** The fixed end must resist all loads and moments. - Total UDL load from 0 to 6 ft: $w_1 = 2$ kip/ft over 6 ft $$W_1 = 2 \times 6 = 12 \text{ kip}$$ Acts at $3$ ft from fixed end. - Linearly decreasing load from 6 to 9 ft: This is a triangular load with base 3 ft and height 2 kip/ft. $$W_2 = \frac{1}{2} \times 3 \times 2 = 3 \text{ kip}$$ Acts at $6 + \frac{2}{3} = 7$ ft from fixed end (centroid of triangle is 1/3 from larger end). - Point load at 9 ft: 4 kip downward. - End moment at 9 ft: 20 kip·ft clockwise (negative moment). 4. **Step 3: Calculate shear force at $x$ for $0 < x < 6$ ft** Only the uniform load acts here. Shear force at distance $x$ from fixed end: $$V(x) = R_V - w_1 x$$ where $R_V$ is the vertical reaction at fixed end. 5. **Step 4: Calculate bending moment at $x$ for $0 < x < 6$ ft** Moment at $x$ is: $$M(x) = R_M - R_V x + \frac{w_1 x^2}{2}$$ where $R_M$ is the moment reaction at fixed end. 6. **Step 5: Calculate shear force at $x$ for $6 < x < 9$ ft** Shear force includes uniform load over 6 ft plus triangular load from 6 to $x$: Triangular load intensity at $x$: $$w(x) = 2 \times \frac{9 - x}{3} = 2 - \frac{2}{3}(x - 6)$$ Load from 6 to $x$: $$W_{tri}(x) = \frac{1}{2} (x - 6) (w(6) + w(x)) = \frac{1}{2} (x - 6) (2 + w(x))$$ Shear force: $$V(x) = R_V - 12 - W_{tri}(x)$$ 7. **Step 6: Calculate bending moment at $x$ for $6 < x < 9$ ft** Moment is: $$M(x) = R_M - R_V x + 12 (x - 3) + M_{tri}(x)$$ where $M_{tri}(x)$ is moment due to triangular load from 6 to $x$ about fixed end. 8. **Step 7: Calculate reactions $R_V$ and $R_M$** Sum of vertical forces: $$R_V = 12 + 3 + 4 = 19 \text{ kip}$$ Sum of moments about fixed end (taking clockwise as negative): $$-R_M + 12 \times 3 + 3 \times 7 + 4 \times 9 - 20 = 0$$ Calculate: $$-R_M + 36 + 21 + 36 - 20 = 0$$ $$-R_M + 73 = 0 \Rightarrow R_M = 73 \text{ kip·ft}$$ 9. **Step 8: Final expressions** For $0 < x < 6$ ft: $$V(x) = 19 - 2x$$ $$M(x) = 73 - 19x + x^2$$ For $6 < x < 9$ ft: Triangular load from 6 to $x$: $$w(x) = 2 - \frac{2}{3}(x - 6)$$ $$W_{tri}(x) = \frac{1}{2} (x - 6)(2 + w(x)) = \frac{1}{2} (x - 6) \left(2 + 2 - \frac{2}{3}(x - 6)\right) = \frac{1}{2} (x - 6) \left(4 - \frac{2}{3}(x - 6)\right)$$ Simplify: $$W_{tri}(x) = \frac{1}{2} (x - 6) \left(4 - \frac{2}{3}(x - 6)\right) = 2(x - 6) - \frac{1}{3} (x - 6)^2$$ Shear force: $$V(x) = 19 - 12 - W_{tri}(x) - 4 = 3 - W_{tri}(x)$$ Moment due to triangular load about fixed end: Centroid of triangular load from 6 to $x$ is at: $$6 + \frac{2}{3} (x - 6)$$ Moment: $$M_{tri}(x) = W_{tri}(x) \times \left(x - \left(6 + \frac{2}{3} (x - 6)\right)\right) = W_{tri}(x) \times \left(x - 6 - \frac{2}{3} (x - 6)\right) = W_{tri}(x) \times \frac{1}{3} (x - 6)$$ Total moment: $$M(x) = 73 - 19x + 12(x - 3) + M_{tri}(x) = 73 - 19x + 12x - 36 + M_{tri}(x) = 37 - 7x + M_{tri}(x)$$ 10. **Summary:** For $0 < x < 6$ ft: $$V(x) = 19 - 2x$$ $$M(x) = 73 - 19x + x^2$$ For $6 < x < 9$ ft: $$W_{tri}(x) = 2(x - 6) - \frac{1}{3} (x - 6)^2$$ $$V(x) = 3 - W_{tri}(x) = 3 - 2(x - 6) + \frac{1}{3} (x - 6)^2$$ $$M_{tri}(x) = W_{tri}(x) \times \frac{1}{3} (x - 6)$$ $$M(x) = 37 - 7x + M_{tri}(x)$$