# Simply Supported Beam

## 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 |

## 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 |

## Problem 1007 | Flexural stresses developed in the wood and steel fibers

**Problem 1007**

A uniformly distributed load of 300 lb/ft (including the weight of the beam) is simply supported on a 20-ft span. The cross section of the beam is described in Problem 1005. If n = 20, determine the maximum stresses produced in the wood and the steel.

## Solution to Problem 691 | Beam Deflection by Method of Superposition

**Problem 691**

Determine the midspan deflection for the beam shown in Fig. P-691. (Hint: Apply Case No. 7 and integrate.)

## Solution to Problem 690 | Beam Deflection by Method of Superposition

## Solution to Problem 673 | Midspan Deflection

**Problem 673**

For the beam shown in Fig. P-673, show that the midspan deflection is δ = (Pb/48EI) (3L^{2} - 4b^{2}).

## Deflections in Simply Supported Beams | Area-Moment Method

The deflection δ at some point B of a simply supported beam can be obtained by the following steps:

## Solution to Problem 585 | Design for Flexure and Shear

**Problem 585**

A simply supported beam of length L carries a uniformly distributed load of 6000 N/m and has the cross section shown in Fig. P-585. Find L to cause a maximum flexural stress of 16 MPa. What maximum shearing stress is then developed?

## Solution to Problem 581 | Design for Flexure and Shear

**Problem 581**

A laminated beam is composed of five planks, each 6 in. by 2 in., glued together to form a section 6 in. wide by 10 in. high. The allowable shear stress in the glue is 90 psi, the allowable shear stress in the wood is 120 psi, and the allowable flexural stress in the wood is 1200 psi. Determine the maximum uniformly distributed load that can be carried by the beam on a 6-ft simple span.