Calculate for me shear force calculations and bending moment calculations, then draw for me shear force diagram and bending moment diagram

Step 1: Calculate Reactions at the Fixed Support Let the fixed end be point A (left side) and the free end be point B (right side). The total length of the beam is $L = 4 \, \text{m}$. The uniformly distributed load (UDL) is $w = 3 \, \text{kN/m}$. The point load at the free end is $P = 2 \, \text{kN}$.

The vertical reaction force $R_A$ at the fixed end is the sum of all downward loads: $R_A = wL + P$ $R_A = (3 \, kN/m)(4 \, m) + 2 \, \text{kN}$ $R_A = 12 \, kN + 2 \, \text{kN}$ $R_A = 14 \, \text{kN}$ The reaction moment $M_A$ at the fixed end is the sum of moments due to all loads about point A: $M_A = (wL)\frac{L}{2} + PL$ $M_A = (3 \, kN/m \cdot 4 \, m) \cdot \frac{4 \, m}{2} + 2 \, kN \cdot 4 \, \text{m}$ $M_A = (12 \, kN) \cdot (2 \, m) + 8 \, kN \cdot \text{m}$ $M_A = 24 \, kN \cdot m + 8 \, kN \cdot \text{m}$ $M_A = 32 \, kN \cdot \text{m}$

Step 2: Shear Force Calculations Consider a section at a distance $x$ from the fixed end (left side). Using the standard sign convention (upward forces positive for shear force when looking at the left part of the section): $V(x) = R_A - wx$ $V(x) = 14 \, kN - (3 \, kN/m)x$ At the fixed end ($x=0$): $V(0) = 14 - 3(0) = 14 \, \text{kN}$ At the free end ($x=4 \, \text{m}$): $V(4) = 14 - 3(4) = 14 - 12 = 2 \, \text{kN}$

Step 3: Bending Moment Calculations Consider a section at a distance $x$ from the fixed end (left side). Using the standard sign convention (hogging moments are negative): $M(x) = -M_A + R_A x - (wx)\frac{x}{2}$ $M(x) = -32 \, kN \cdot m + (14 \, kN)x - \frac{(3 \, kN/m)x^2}{2}$ $M(x) = -32 + 14x - 1.5x^2$ At the fixed end ($x=0$): $M(0) = -32 + 14(0) - 1.5(0)^2 = -32 \, kN \cdot \text{m}$ At the free end ($x=4 \, \text{m}$): $M(4) = -32 + 14(4) - 1.5(4)^2$ $M(4) = -32 + 56 - 1.5(16)$ $M(4) = -32 + 56 - 24$ $M(4) = 0 \, kN \cdot \text{m}$

Step 4: Shear Force Diagram (SFD) The shear force diagram starts at a positive value of $14 \, \text{kN}$ at the fixed end ($x=0$). It decreases linearly along the beam due to the uniformly distributed load. At the free end ($x=4 \, \text{m}$), the shear force is $2 \, \text{kN}$. The entire diagram is above the zero line, indicating positive shear force throughout the beam. The equation for the SFD is $V(x) = 14 - 3x \, \text{kN}$.

Step 5: Bending Moment Diagram (BMD) The bending moment diagram starts at a negative value of $-32 \, kN \cdot \text{m}$ at the fixed end ($x=0$), which is the maximum (in magnitude) bending moment. It then increases parabolically (becomes less negative) along the beam. At the free end ($x=4 \, \text{m}$), the bending moment is $0 \, kN \cdot \text{m}$. The entire diagram is below the zero line, indicating negative bending moment (hogging) throughout the beam. The equation for the BMD is $M(x) = -32 + 14x - 1.5x^2 \, \text{kN \cdot m}$.

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