Beam Reactions 00B196

1. **Problem Statement:** Determine the vertical reaction at the roller support, the maximum shear force, and the maximum bending moment for the beam loaded as described. 2. **Given Data:** - A clockwise moment of $$25\ \text{KN-m}$$ applied at $$x=1\ \text{m}$$. - A uniformly distributed load (UDL) of $$10\ \text{kN/m}$$ from $$x=2\ \text{m}$$ to $$x=6\ \text{m}$$. - Beam supports: a pin support at the left end ($$x=0$$) and a roller support at $$x=5\ \text{m}$$. - Beam segments: $$1\ \text{m}$$, $$1\ \text{m}$$, $$3\ \text{m}$$, and $$1\ \text{m}$$ as per the problem description. 3. **Step 1: Calculate reactions at supports** - Let $$R_A$$ be the vertical reaction at the pin support (left end), and $$R_B$$ be the vertical reaction at the roller support ($$x=5\ \text{m}$$). - Total UDL load magnitude: $$ w = 10\ \text{kN/m} \times (6 - 2)\ \text{m} = 40\ \text{kN} $$ - The UDL acts at the midpoint of the loaded length, which is at $$x=4\ \text{m}$$. - Taking moments about point A (left support): $$ \sum M_A = 0 = -25 + R_B \times 5 - 40 \times 4 $$ $$ R_B \times 5 = 25 + 160 = 185 $$ $$ R_B = \frac{185}{5} = 37\ \text{kN} $$ - Using vertical force equilibrium: $$ \sum F_y = 0 = R_A + R_B - 40 $$ $$ R_A = 40 - 37 = 3\ \text{kN} $$ 4. **Step 2: Determine maximum shear force** - Shear just to the right of A: $$ V = R_A = 3\ \text{kN} $$ - Shear just before the UDL starts at $$x=2\ \text{m}$$: $$ V = 3\ \text{kN} $$ - Shear decreases linearly under the UDL: $$ V(x) = 3 - 10(x - 2)\ \text{kN}, \quad 2 \leq x \leq 6 $$ - Shear at $$x=5\ \text{m}$$ (roller support): $$ V(5) = 3 - 10(5 - 2) = 3 - 30 = -27\ \text{kN} $$ - Just to the right of roller support, shear jumps by $$R_B = 37\ \text{kN}$$: $$ V(5^+) = -27 + 37 = 10\ \text{kN} $$ - Shear at the free end $$x=6\ \text{m}$$: $$ V(6) = 10 - 10(6 - 5) = 10 - 10 = 0\ \text{kN} $$ - Maximum shear magnitude is $$\max(|3|, |27|, |10|) = 27\ \text{kN}$$. 5. **Step 3: Determine maximum bending moment** - Moment at $$x=0$$ is zero (pin support). - Moment at $$x=1$$ just left of moment application: $$ M(1^-) = R_A \times 1 = 3 \times 1 = 3\ \text{kN-m} $$ - Moment at $$x=1$$ just right of moment application: $$ M(1^+) = M(1^-) - 25 = 3 - 25 = -22\ \text{kN-m} $$ - Moment at $$x=2$$ (start of UDL): $$ M(2) = M(1^+) + V(1^+) \times (2 - 1) = -22 + 3 \times 1 = -19\ \text{kN-m} $$ - Moment under UDL varies as: $$ M(x) = M(2) + \int_2^x V(t) dt = -19 + \int_2^x [3 - 10(t - 2)] dt $$ $$ = -19 + \left[3t - 5(t - 2)^2\right]_2^x = -19 + 3(x - 2) - 5(x - 2)^2 $$ - To find maximum moment, differentiate and set to zero: $$ \frac{dM}{dx} = 3 - 10(x - 2) = 0 \Rightarrow x = 2.3\ \text{m} $$ - Calculate $$M(2.3)$$: $$ M(2.3) = -19 + 3(0.3) - 5(0.3)^2 = -19 + 0.9 - 0.45 = -18.55\ \text{kN-m} $$ - Moment at roller support $$x=5$$: $$ M(5) = -19 + 3(3) - 5(3)^2 = -19 + 9 - 45 = -55\ \text{kN-m} $$ - Maximum moment magnitude is $$55\ \text{kN-m}$$ (absolute value). **Final answers:** - Vertical reaction at roller support: $$\boxed{37\ \text{kN}}$$ - Maximum shear force: $$\boxed{27\ \text{kN}}$$ - Maximum bending moment: $$\boxed{55\ \text{kN-m}}$$ **Note:** For the shear diagram photo, please visit the create page to generate visual diagrams as per platform policy.