problem 8.10 - find and plot the shear (v) and bending moment (m) equations as functions of x.

Explanation:

Step1: Define the distributed load function

The load $w(x)$ varies linearly from $2\text{ kN/m}$ at $[Client Connection Error]x=0$ to $4\text{ kN/m}$ at $x=12$. $$w(x) = 2 + \frac{4-2}{12}x = 2 + \frac{1}{6}x \text{ kN/m}$$

Step2: Calculate support reactions

Sum moments about $A$ to find $R_B$, then sum vertical forces for $R_A$. $$\sum M_A = 0 \Rightarrow R_B(6) - \int_{0}^{12} w(x)(x-6) dx = 0 \Rightarrow R_B = 18\text{ kN}$$ $$\sum F_y = 0 \Rightarrow R_A + R_B - \int_{0}^{12} w(x) dx = 0 \Rightarrow R_A = 18\text{ kN}$$

Step3: Derive shear force equations

Integrate $w(x)$ and account for point reactions at $x=6$. For $0 \le x < 6$: $V(x) = -\int_{0}^{x} (2 + \frac{1}{6}t) dt = -2x - \frac{1}{12}x^2$ For $6 < x \le 12$: $V(x) = -2x - \frac{1}{12}x^2 + R_A = 18 - 2x - \frac{1}{12}x^2$

Step4: Derive bending moment equations

Integrate $V(x)$ with boundary condition $M(0)=0$. For $0 \le x \le 6$: $M(x) = \int V(x) dx = -x^2 - \frac{1}{36}x^3$ For $6 \le x \le 12$: $M(x) = \int (18 - 2x - \frac{1}{12}x^2) dx + C = 18x - x^2 - \frac{1}{36}x^3 - 108$

Answer:

For $0 \le x < 6\text{ m}$: $V(x) = -2x - \frac{1}{12}x^2 \text{ kN}$ $M(x) = -x^2 - \frac{1}{36}x^3 \text{ kN}\cdot\text{m}$

For $6 < x \le 12\text{ m}$: $V(x) = 18 - 2x - \frac{1}{12}x^2 \text{ kN}$ $M(x) = 18x - x^2 - \frac{1}{36}x^3 - 108 \text{ kN}\cdot\text{m}$