equivalent section

Problem 1009
A timber beam 150 mm wide by 200 mm deep is to be reinforced at the top and bottom by aluminum plates 6 mm thick. Determine the width of the aluminum plates if the beam is to resist a moment of 14 kN·m. Assume n = 5 and take the allowable stresses as 10 MPa and 80 MPa in the wood and aluminum, respectively.
 

Problem 1008
A timber beam 150 mm wide by 250 mm deep is to be reinforced at the top and bottom by steel plates 10 mm thick. How wide should the steel plates be if the beam is to resist a moment of 40 kN·m? Assume that n = 15 and the allowable stresses in the wood and steel are 10 MPa and 120 MPa, respectively.
 

Problem 1005
A timber beam 6 in. by 10 in. is reinforced only at the bottom by a steel plate as shown in Fig. P-1005. Determine the concentrated load that can be applied at the center of a simply supported span 18 ft long if n = 20, f_s ≤ 18 ksi and f_w ≤ 1200 psi. Show that the neutral axis is 7.1 in. below the top and that I_NA = 1160 in.⁴.
 

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

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?
 

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

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

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

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 usually taken as the ratio of the moduli of elasticity of stronger material to the weaker material. In the above case, n = E_s / E_w.