# composite beam

## Example 03: Moment Capacity of a Timber Beam Reinforced with Steel and Aluminum Strips

**Problem**

Steel and aluminum plates are used to reinforced an 80 mm by 150 mm timber beam. The three materials are fastened firmly as shown so that there will be no relative movement between them.

Given the following material properties:

Allowable Bending Stress, F_{b}Steel = 120 MPa Aluminum = 80 MPa Wood = 10 MPa |
Modulus of Elasticity, ESteel = 200 GPa Aluminum = 70 GPa Wood = 10 GPa |

Find the safe resisting moment of the beam in kN·m.

## Problem 1003 | Maximum stresses in wood and steel of composite beam

**Problem 1003**

A simply supported beam 4 m long has the cross section shown in Fig. P-1002. It carries a uniformly distributed load of 20 kN/m over the middle half of the span. If n = 15, compute the maximum stresses in the wood and the steel.

## Problem 1004 | Increase in moment capacity due to aluminum plate reinforcement

**Problem 1004**

Repeat Problem 1002 assuming that the reinforcement consists of aluminum plates for which the allowable stress is 80 MPa. Use n = 5.

## Problem 1002 | Increase in moment capacity due to steel plate reinforcement

**Problem 1002**

A timber beam is reinforced with steel plates rigidly attached at the top and bottom as shown in Fig. P-1002. By what amount is moment increased by the reinforcement if n = 15 and the allowable stresses in the wood and steel are 8 MPa and 120 MPa, respectively?

## Beams with Different Materials

From assumption no. (3) in the previous page: The strains of any two adjacent materials at their junction point are equal.

$\epsilon_s = \epsilon_w$

$\dfrac{f_{bs}}{E_s} = \dfrac{f_{bw}}{E_w}$

$\dfrac{f_{bs}}{f_{bw}} = \dfrac{E_s}{E_w}$

## Chapter 10 - Reinforced Beams

Flexure formula do not apply directly to composite beams because it was based on the assumption that the beam was homogeneous. It is therefore necessary to transform the composite material into equivalent homogeneous section. To do this, consider a steel and wood section to be firmly bolted together so that they can act as one unit. Shown below are the composite wood and steel section and the corresponding equivalent in wood and steel sections.

The quantity n is the ratio of the moduli of elasticity of stronger material to the weaker material. In the above case, n = E_{s} / E_{w}.