Beam Reactions C13F8A

1. **Problem Statement:** We have a simply supported beam with a pin support at the left end and a roller support near the right end. The beam is loaded with a clockwise moment of $$25\ \text{kN-m}$$ at $$x=1\ \text{m}$$ from the left support and a uniformly distributed load (UDL) of $$10\ \text{kN/m}$$ acting from $$x=2\ \text{m}$$ to the right end over a length of $$4\ \text{m}$$. The beam segments are $$1\ \text{m}$$, $$1\ \text{m}$$, $$3\ \text{m}$$, and $$1\ \text{m}$$. We need to find: - The vertical reaction at the roller support. - The maximum shear force on the beam. - The maximum bending moment on the beam. 2. **Knowns and Setup:** - Total beam length $$L = 1 + 1 + 3 + 1 = 6\ \text{m}$$. - Pin support at $$x=0$$. - Roller support at $$x=6\ \text{m}$$. - Moment applied at $$x=1\ \text{m}$$: $$M_0 = 25\ \text{kN-m}$$ (clockwise). - UDL of $$w = 10\ \text{kN/m}$$ from $$x=2\ \text{m}$$ to $$x=6\ \text{m}$$ (length $$4\ \text{m}$$). 3. **Reactions:** Let $$R_A$$ be the vertical reaction at the pin (left support) and $$R_B$$ be the vertical reaction at the roller (right support). 4. **Equilibrium Equations:** - Sum of vertical forces: $$R_A + R_B - w \times 4 = 0$$ $$R_A + R_B - 10 \times 4 = 0$$ $$R_A + R_B = 40\ \text{kN}$$ - Sum of moments about the pin support (taking counterclockwise as positive): $$\sum M_A = 0 = -M_0 + R_B \times 6 - w \times 4 \times (2 + \frac{4}{2})$$ The UDL resultant force is $$w \times 4 = 40\ \text{kN}$$ acting at the midpoint of the UDL segment, which is at $$x = 2 + 2 = 4\ \text{m}$$ from the pin. So, $$-25 + 6 R_B - 40 \times 4 = 0$$ $$-25 + 6 R_B - 160 = 0$$ $$6 R_B = 185$$ $$R_B = \frac{185}{6} = 30.8333\ \text{kN}$$ 5. **Find $$R_A$$:** $$R_A = 40 - R_B = 40 - 30.8333 = 9.1667\ \text{kN}$$ 6. **Maximum Shear Force:** Shear force changes at points of load application and supports. - Just left of $$x=1\ \text{m}$$ (moment only, no vertical load): $$V = R_A = 9.1667\ \text{kN}$$ - Just right of $$x=1\ \text{m}$$: no vertical load, so shear unchanged $$9.1667\ \text{kN}$$ - At $$x=2\ \text{m}$$ UDL starts, shear begins to decrease by $$10\ \text{kN/m}$$. - At $$x=6\ \text{m}$$ shear is: $$V = 9.1667 - 10 \times (6 - 2) = 9.1667 - 40 = -30.8333\ \text{kN}$$ Maximum shear magnitude is $$30.8333\ \text{kN}$$. 7. **Maximum Moment:** Maximum moment occurs where shear crosses zero. Set shear $$V = 0$$ in the UDL region: $$0 = 9.1667 - 10(x - 2)$$ $$10(x - 2) = 9.1667$$ $$x - 2 = 0.91667$$ $$x = 2.9167\ \text{m}$$ Calculate moment at $$x=2.9167\ \text{m}$$: Moment from left support: $$M = R_A \times x - M_0 \text{(if } x \geq 1) - w \times (x - 2) \times \frac{x - 2}{2}$$ - For $$x=2.9167 > 1$$, subtract moment $$M_0$$ at $$x=1$$. Calculate each term: - $$R_A \times x = 9.1667 \times 2.9167 = 26.75\ \text{kN-m}$$ - Moment couple $$M_0 = 25\ \text{kN-m}$$ (clockwise, subtract) - UDL moment: Length of UDL segment to $$x$$ is $$2.9167 - 2 = 0.9167\ \text{m}$$ Load magnitude: $$w \times 0.9167 = 9.167\ \text{kN}$$ Moment of this load about left support: $$9.167 \times \frac{0.9167}{2} + 2 = 9.167 \times 2.4583 = 22.53\ \text{kN-m}$$ So, $$M = 26.75 - 25 - 22.53 = -20.78\ \text{kN-m}$$ Check moments at supports and load points: - At $$x=1$$: $$M = R_A \times 1 - M_0 = 9.1667 - 25 = -15.8333\ \text{kN-m}$$ - At $$x=2$$: $$M = R_A \times 2 - M_0 = 18.3334 - 25 = -6.6666\ \text{kN-m}$$ - At $$x=6$$: $$M = 0$$ (roller support moment zero) Maximum moment magnitude is $$25\ \text{kN-m}$$ at the applied moment location. **Final answers:** - Vertical reaction at roller support $$R_B = 30.83\ \text{kN}$$ - Maximum shear force $$= 30.83\ \text{kN}$$ - Maximum moment $$= 25\ \text{kN-m}$$ (at $$x=1\ \text{m}$$)