# Simply Supported Beam with Support Added at Midspan to Prevent Excessive Deflection

**Situation**

A simply supported beam has a span of 12 m. The beam carries a total uniformly distributed load of 21.5 kN/m.**1.** To prevent excessive deflection, a support is added at midspan. Calculate the resulting moment (kN·m) at the added support.

A. 64.5 | C. 258.0 |

B. 96.8 | D. 86.0 |

**2.** Calculate the resulting maximum positive moment (kN·m) when a support is added at midspan.

A. 96.75 | C. 108.84 |

B. 54.42 | D. 77.40 |

**3.** Calculate the reaction (kN) at the added support.

A. 48.38 | C. 161.2 |

B. 96.75 | D. 80.62 |

**Answer Key**

Part (2): [ B ]

Part (3): [ C ]

**Solution**

$R_2 = \frac{5}{8}wL = \frac{5}{8}(21.5)(12)$

$R = 161.25 ~ \text{kN}$ ← *Answer for Part (3)*

$\Sigma F_v = 0$

$2R_1 + R_2 = 21.5(12)$

$2R_1 + 161.25 = 258$

$R_1 = 48.375 ~ \text{kN}$

$M_2 = 48.375(6) - 21.5(6)(3)$

$M_2 = -96.75 ~ \text{kN}\cdot\text{m}$ ← *Answer for Part (1)*

$V_x = 48.375 - 21.5x = 0$

$x = 2.25 ~ \text{m}$

$M_x = 48.375x - 21.5x\left( \dfrac{x}{2} \right)$

$M_x = 48.375(2.25) - 10.75(2.25^2)$

$M_x = 54.42 ~ \text{kN}\cdot\text{m}$ ← *Answer for Part (2)*

- Log in to post comments