# Uniform Load

**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,

*H*= 4 m

Modulus of Elasticity,

*E*= 200 GPa

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 |

## Reactions of Tripod Made from Wood Planks

**Situation**

In the figure shown, each plank carries a uniform load of 100 N/m throughout its length. The supports are on the same plane.

- Find the reaction at A.

A. 900 N

B. 800 N

C. 1400 N

D. 2400 N

- Find the reaction at B.

A. 900 N

B. 800 N

C. 1400 N

D. 2400 N

- Find the reaction at C.

A. 900 N

B. 800 N

C. 1400 N

D. 2400 N

## Problem 833 | Reactions of Continuous Beams

**Problem 833**

Refer to Problem 825 for which M_{2} = -980 lb·ft and M_{3} = -1082 lb·ft.

## Problem 823 | Continuous Beam by Three-Moment Equation

**Problem 823**

A continuous beam simply supported over three 10-ft spans carries a concentrated load of 400 lb at the center of the first span, a concentrated load of 640 lb at the center of the third span and a uniformly distributed load of 80 lb/ft over the middle span. Solve for the moment over the supports and check your answers using the results obtained for Problems 819 and 822.

## Problem 822 | Continuous Beam by Three-Moment Equation

**Problem 822**

Solve Prob. 821 if the concentrated load is replaced by a uniformly distributed load of intensity w_{o} over the middle span.

Answers:

$M_2 = -\dfrac{w_o L^2}{4} \cdot \dfrac{1 + 2\beta}{4(\alpha + 1)(1 + \beta) - 1}$

$M_3 = -\dfrac{w_o L^2}{4} \cdot \dfrac{1 + 2\alpha}{4(1 + \alpha)(1 + \beta) - 1}$

## Problem 820 | Continuous Beam by Three-Moment Equation

**Problem 820**

Solve Prob. 819 if the concentrated load is replaced by a uniformly distributed load of intensity w_{o} over the first span.

## 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.

## Problem 734 | Restrained beam with uniform load over half the span

**Problem 734**

Determine the end moments for the restrained beams shown in Fig. P-734.