Subjects structural engineering

Beam Sfd Bmd 3059Ce

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1. **Problem Statement:** Calculate the Shear Force Diagram (SFD) and Bending Moment Diagram (BMD) for a beam with: - Uniformly distributed load (UDL) of 20 kN/m over 4 m - Reaction forces: $Y_A = 30$ kN (left), $Y_B = 70$ kN (right) - Point load of 20 kN downward located 2 m to the right of $Y_B$ 2. **Step 1: Calculate total UDL load and its position** Total UDL load $W = 20 \times 4 = 80$ kN acting at the midpoint of the 4 m span, i.e., 2 m from the left end. 3. **Step 2: Shear Force Calculations** - At left end (just right of $Y_A$): $V = +30$ kN (upward reaction) - Just left of UDL start: $V = 30$ kN - Just right of UDL end (4 m from left): $V = 30 - 80 = -50$ kN - Just left of $Y_B$: $V = -50$ kN - Just right of $Y_B$: $V = -50 + 70 = 20$ kN - Just left of 20 kN point load (6 m from left): $V = 20$ kN - Just right of 20 kN point load: $V = 20 - 20 = 0$ kN 4. **Step 3: Bending Moment Calculations** - At left end ($x=0$): $M=0$ - At $x=2$ m (center of UDL): $$M = Y_A \times 2 - 20 \times 2 \times \frac{2}{2} = 30 \times 2 - 20 \times 2 \times 1 = 60 - 40 = 20 \text{ kNm}$$ - At $x=4$ m (end of UDL): $$M = Y_A \times 4 - 20 \times 4 \times \frac{4}{2} = 30 \times 4 - 20 \times 4 \times 2 = 120 - 160 = -40 \text{ kNm}$$ - At $x=6$ m (right end, just before point load): $$M = Y_A \times 6 - 20 \times 4 \times 3 - 20 \times 2 = 30 \times 6 - 80 \times 3 - 20 \times 2 = 180 - 240 - 40 = -100 \text{ kNm}$$ - At right end ($x=6$ m, after point load): $M=0$ (free end) 5. **Step 4: Summary and Interpretation** - Shear force starts at +30 kN, decreases linearly by 20 kN/m over 4 m due to UDL, jumps by +70 kN at $Y_B$, then drops by 20 kN at the point load, ending at zero. - Bending moment is zero at supports, peaks near the UDL center, and becomes negative near the right end due to loads. Final answer: - Shear Force Diagram values: $V(0^+) = 30$, $V(4^-) = -50$, $V(4^+) = -50$, $V(4^+) + Y_B = 20$, $V(6^-) = 20$, $V(6^+) = 0$ - Bending Moment values: $M(0) = 0$, $M(2) = 20$, $M(4) = -40$, $M(6) = 0$ These values can be used to plot the SFD and BMD.