1. **Problem Statement:** Calculate the reaction forces $R_a$ and $R_b$ at the beam supports, then determine the shear force $V(x)$ and bending moment $M(x)$ diagrams for the beam subjected to: - A downward point load of 200 N at $x=0$ m. - A uniformly distributed load (UDL) of 10 N/m from $x=10$ m to $x=20$ m. - A downward point load of 100 N at $x=20$ m. - A clockwise moment of 3000 N\cdot m at $x=40$ m. The beam length is 50 m (5 segments of 10 m each). 2. **Step 1: Calculate Equivalent Loads and Reactions** - UDL total load: $wL = 10 \times 10 = 100$ N acting at midpoint of UDL segment, $x=15$ m. - Sum of vertical forces for equilibrium: $$ R_a + R_b = 200 + 100 + 100 = 400 \text{ N} $$ - Sum of moments about $A$ (taking clockwise positive): $$ \sum M_A = 0 = -200 \times 0 - 100 \times 15 - 100 \times 20 - 3000 + R_b \times 50 $$ $$ \Rightarrow R_b \times 50 = 200 \times 0 + 100 \times 15 + 100 \times 20 + 3000 $$ $$ R_b \times 50 = 0 + 1500 + 2000 + 3000 = 6500 $$ $$ R_b = \frac{6500}{50} = 130 \text{ N} $$ - Using vertical force equilibrium: $$ R_a = 400 - 130 = 270 \text{ N} $$ 3. **Step 2: Shear Force Diagram $V(x)$** Define $V(x)$ from left to right: - For $0 \leq x < 10$ m: $$ V(x) = R_a - 200 = 270 - 200 = 70 \text{ N} $$ - For $10 \leq x < 20$ m (UDL region): $$ V(x) = 70 - 10(x-10) $$ - At $x=20$ m just before point load: $$ V(20^-) = 70 - 10(10) = 70 - 100 = -30 \text{ N} $$ - Just after 100 N load at $x=20$ m: $$ V(20^+) = -30 - 100 = -130 \text{ N} $$ - For $20 < x < 50$ m: $$ V(x) = -130 \text{ (constant)} $$ - At $x=50$ m (right end): $$ V(50) = -130 + R_b = -130 + 130 = 0 $$ 4. **Step 3: Bending Moment Diagram $M(x)$** Integrate $V(x)$ with respect to $x$: - For $0 \leq x < 10$ m: $$ M(x) = \int V(x) dx = 70x + C_1 $$ At $x=0$, $M(0)=0$, so $C_1=0$. - For $10 \leq x < 20$ m: $$ M(x) = M(10) + \int_{10}^x (70 - 10(t-10)) dt $$ Calculate $M(10)$: $$ M(10) = 70 \times 10 = 700 \text{ Nm} $$ Integrate: $$ \int_{10}^x (70 - 10(t-10)) dt = 70(x-10) - 10 \frac{(x-10)^2}{2} = 70(x-10) - 5(x-10)^2 $$ So: $$ M(x) = 700 + 70(x-10) - 5(x-10)^2 $$ - For $20 \leq x < 40$ m: $$ M(x) = M(20) + \int_{20}^x (-130) dt = M(20) - 130(x-20) $$ Calculate $M(20)$: $$ M(20) = 700 + 70(10) - 5(10)^2 = 700 + 700 - 500 = 900 \text{ Nm} $$ - At $x=40$ m, apply moment $-3000$ Nm (clockwise): $$ M(40^+) = M(40^-) - 3000 $$ Calculate $M(40^-)$: $$ M(40^-) = 900 - 130(20) = 900 - 2600 = -1700 \text{ Nm} $$ So: $$ M(40^+) = -1700 - 3000 = -4700 \text{ Nm} $$ - For $40 \leq x \leq 50$ m: $$ M(x) = M(40^+) - 130(x-40) = -4700 - 130(x-40) $$ - At $x=50$ m: $$ M(50) = -4700 - 130(10) = -4700 - 1300 = -6000 \text{ Nm} $$ 5. **Step 4: Degree of Curve and Curvature** - The curvature $\kappa(x)$ relates to bending moment by: $$ \kappa(x) = \frac{M(x)}{EI} $$ where $E$ is Young's modulus and $I$ is the moment of inertia. - The slope (degree of curve) $\theta(x)$ is: $$ \theta(x) = \frac{d w}{dx} = \int \kappa(x) dx = \frac{1}{EI} \int M(x) dx $$ - Without $E$ and $I$ values, curvature and slope are expressed in terms of $M(x)/EI$. 6. **Summary:** - Reaction forces: $$ R_a = 270 \text{ N}, \quad R_b = 130 \text{ N} $$ - Shear force $V(x)$ piecewise: $$ V(x) = \begin{cases} 70, & 0 \leq x < 10 \\ 70 - 10(x-10), & 10 \leq x < 20 \\ -130, & 20 \leq x \leq 50 \end{cases} $$ - Bending moment $M(x)$ piecewise: $$ M(x) = \begin{cases} 70x, & 0 \leq x < 10 \\ 700 + 70(x-10) - 5(x-10)^2, & 10 \leq x < 20 \\ 900 - 130(x-20), & 20 \leq x < 40 \\ -4700 - 130(x-40), & 40 \leq x \leq 50 \end{cases} $$ - Curvature and slope depend on $E$ and $I$ as $\kappa(x) = \frac{M(x)}{EI}$ and $\theta(x) = \frac{1}{EI} \int M(x) dx$. This completes the solution for reactions, shear and moment diagrams, and curvature relations.