# Support Added at the Midspan of Simple Beam to Prevent Excessive Deflection

**Situation**

A simply supported steel beam spans 9 m. It carries a uniformly distributed load of 10 kN/m, beam weight already included.

**Given Beam Properties:**

Area = 8,530 mm

^{2}

Depth = 306 mm

Flange Width = 204 mm

Flange Thickness = 14.6 mm

Moment of Inertia,

*I*= 145 × 10

_{x}^{6}mm

^{4}

Modulus of Elasticity,

*E*= 200 GPa

**1.** What is the maximum flexural stress (MPa) in the beam?

A. 107 | C. 142 |

B. 54 | D. 71 |

**2.** To prevent excessive deflection, the beam is propped at midspan using a pipe column. Find the resulting axial stress (MPa) in the column

**Given Column Properties:**

Outside Diameter = 200 mm

Thickness = 10 mm

Height = 4 m

A. 4.7 | C. 18.8 |

B. 9.4 | D. 2.8 |

**3.** How much is the maximum bending stress (MPa) in the propped beam?

A. 26.7 | C. 15.0 |

B. 17.8 | D. 35.6 |

**Answer Key**

Part (2): [ B ]

Part (3): [ A ]

**Solution**

**Part (1):**

$M = \dfrac{wL^2}{8} = \dfrac{10(9^2)}{8}$

$M = 101.25 ~ \text{kN}\cdot\text{m}$

$f_b = \dfrac{Mc}{I} = \dfrac{101.25(1000^2)(306/2)}{145 \times 10^6}$

$f_b = 106.84 \text{MPa}$ ← *answer*

**Part (2):**

$\dfrac{R_2 L^3}{48EI} = \dfrac{5wL^4}{384EI}$

$R_2 = \frac{5}{8}wL = \frac{5}{8}(1)(9)$

$R_2 = 56.25 ~ \text{kN}$

$f_c = \dfrac{R_2}{A} = \dfrac{56.25(1000)}{\frac{1}{4}\pi (200^2 - 180^2)}$

$f_c = 9.42 ~ \text{MPa}$ ← *answer*

**Part (3):**

$\Sigma F_v = 0$

$2R_1 + R_2 = 10(9)$

$2R_1 + 56.25 = 90$

$R_1 = 16.875 ~ \text{kN}$

$M_2 = 4.5R_1 - 10(4.5)(4.5/2) = 4.5(16.875) - 101.25$

$M_2 = -25.3125 ~ \text{kN}$

$f_b = \dfrac{Mc}{I} = \dfrac{25.3125(1000^2)(306/2}{145 \times 10^6}$

$f_b = 26.71 ~ \text{MPa}$ ← *answer*