Beam Shear Bending Ef0F99

1. **Problem Statement:** We have a beam with the following loads and supports: - A 25 kN upward point load at the left end (x=0). - A 10 kN downward point load at 2 m from the left end. - A roller support at 5 m from the left end (after 2 m + 3 m spans). - A 25 kN/m downward uniformly distributed load (UDL) over 4 m starting at the roller support (from 5 m to 9 m). - An 11 kN/m downward UDL over the rightmost 5 m span (from 7 m to 12 m), overlapping the previous UDL. - A pin support at the right end (12 m). We need to draw the shear force and bending moment diagrams. 2. **Step 1: Calculate reactions at supports** Let the beam length be $$L = 12\,m$$. Supports: - Roller at $$x=5\,m$$ (vertical reaction $$R_B$$). - Pin at $$x=12\,m$$ (vertical reaction $$R_C$$). Sum of vertical forces = 0: $$R_B + R_C + 25 - 10 - (25 \times 4) - \text{load from 11 kN/m} = 0$$ Calculate the load from 11 kN/m over 5 m: $$11 \times 5 = 55\,kN$$ Sum vertical forces: $$R_B + R_C + 25 - 10 - 100 - 55 = 0$$ Simplify: $$R_B + R_C - 140 = 0 \implies R_B + R_C = 140$$ 3. **Step 2: Take moments about one support to find reactions** Taking moments about the pin support at $$x=12\,m$$ (counterclockwise positive): - Moment from $$R_B$$ at 5 m: $$R_B \times (12 - 5) = R_B \times 7$$ - Moment from 25 kN upward load at 0 m: $$25 \times (12 - 0) = 25 \times 12 = 300$$ (counterclockwise) - Moment from 10 kN downward load at 2 m: $$-10 \times (12 - 2) = -10 \times 10 = -100$$ - Moment from 25 kN/m UDL over 4 m starting at 5 m: Total load = $$25 \times 4 = 100\,kN$$ acting at midpoint of 7 m to 9 m span, i.e., at $$5 + 2 = 7\,m$$ from 5 m, so from left end at $$5 + 2 = 7\,m$$. Distance from pin support: $$12 - 7 = 5\,m$$ Moment: $$-100 \times 5 = -500$$ - Moment from 11 kN/m UDL over 5 m starting at 7 m: Total load = $$11 \times 5 = 55\,kN$$ acting at midpoint of 7 m to 12 m, i.e., at $$7 + 2.5 = 9.5\,m$$. Distance from pin support: $$12 - 9.5 = 2.5\,m$$ Moment: $$-55 \times 2.5 = -137.5$$ Sum moments about pin support: $$R_B \times 7 + 300 - 100 - 500 - 137.5 = 0$$ Simplify: $$7 R_B - 437.5 = 0 \implies 7 R_B = 437.5 \implies R_B = \frac{437.5}{7} = 62.5\,kN$$ 4. **Step 3: Calculate $$R_C$$** From vertical force equilibrium: $$R_C = 140 - R_B = 140 - 62.5 = 77.5\,kN$$ 5. **Step 4: Construct shear force diagram (SFD)** - Start at left end with upward 25 kN reaction: shear = $$+25\,kN$$. - At 2 m, downward 10 kN load: shear = $$25 - 10 = 15\,kN$$. - At 5 m, roller support reaction $$R_B = 62.5\,kN$$ upward: shear = $$15 + 62.5 = 77.5\,kN$$. - From 5 m to 9 m, 25 kN/m downward UDL: Shear decreases linearly by $$25 \times 4 = 100\,kN$$ over 4 m. At 9 m: $$77.5 - 100 = -22.5\,kN$$ - From 7 m to 12 m, 11 kN/m downward UDL overlaps: Between 7 m and 9 m, both UDLs act, total intensity: $$25 + 11 = 36\,kN/m$$ Between 9 m and 12 m, only 11 kN/m acts. Calculate shear at 7 m: From 5 m to 7 m (2 m), shear decreases by $$25 \times 2 = 50\,kN$$: $$77.5 - 50 = 27.5\,kN$$ From 7 m to 9 m (2 m), shear decreases by $$36 \times 2 = 72\,kN$$: $$27.5 - 72 = -44.5\,kN$$ From 9 m to 12 m (3 m), shear decreases by $$11 \times 3 = 33\,kN$$: $$-44.5 - 33 = -77.5\,kN$$ At 12 m, pin support reaction $$R_C = 77.5\,kN$$ upward: Shear jumps from $$-77.5 + 77.5 = 0$$. 6. **Step 5: Construct bending moment diagram (BMD)** - Moment at left end $$x=0$$ is zero. - Calculate moments at key points by integrating shear or summing moments. At 2 m: Moment due to 25 kN upward at 0 m and 10 kN downward at 2 m: $$M(2) = 25 \times 2 - 10 \times 0 = 50\,kN\cdot m$$ At 5 m: Moment from loads and reactions: $$M(5) = 25 \times 5 - 10 \times 3 + R_B \times 0 = 125 - 30 + 0 = 95\,kN\cdot m$$ At 9 m: Calculate moment considering UDLs: Moment decreases due to UDLs from 5 m to 9 m: Moment at 9 m = Moment at 5 m - area under shear curve from 5 to 9 m. Area under shear from 5 to 9 m is trapezoid: $$\frac{(77.5 + (-22.5))}{2} \times 4 = 55 \times 4 = 220\,kN\cdot m$$ So, $$M(9) = 95 - 220 = -125\,kN\cdot m$$ At 12 m: Moment at end is zero due to pin support. 7. **Summary:** - Reactions: $$R_B = 62.5\,kN$$, $$R_C = 77.5\,kN$$. - Shear force diagram starts at +25 kN, jumps and decreases with loads, ends at zero. - Bending moment diagram starts and ends at zero, with maximum and minimum values at calculated points. This completes the shear and bending moment diagrams for the beam.