Problem 1009 | Width of aluminum plate reinforcement for the wood section to resist 14 kN-m moment

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 | Finding the width of steel plate reinforcement

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 | Maximum concentrated load at the midspan that the reinforced timber beam can carry

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, fs ≤ 18 ksi and fw ≤ 1200 psi. Show that the neutral axis is 7.1 in. below the top and that INA = 1160 in.4.
 

1005-given-timber-and-steel.gif

 

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.
 

001-equivalent-sections.gif

 

The quantity n is the ratio of the moduli of elasticity of stronger material to the weaker material. In the above case, n = Es / Ew.
 

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