Beam Cross-Section

The ceiling beam is composed of three parts, two steel plates and a wood member. The plates are rigidly attached to the top and bottom surface of the wood member as shown in the diagram at the left. The beam must support a continuous load of 2 kN/m over a simple support span of 5 m.

The beam is a composite beam of only two materials even though there are three sections. The wood will be treated as one material and the two plates will be treated as a single material with two areas.

The first issue is to find the actual maximum moment. This is needed for the composite beam bending stress equations. Next, the neutral axis needs to be determined. From there, the actual moment of inertia for each material can be found.

Beam Loading and Moment Diagram

Beam Cut at Center

To determine the maximum bending stress, the maximum bending moment is needed. This moment is not a function of the cross section and can be determined by the moment diagram. This is a common beam loading and the maximum moment is at the center.

Cutting the beam at the center and summing the moments gives

     ΣM = 0

     M - 5 (2.5) + 2 (2.5) (2.5/2) = 0

     M = 6,250 N-m

The location of the neutral axis (NA) is one of the most critical steps when solving for bending stresses in composite beams. Both the moment of inertia and the distance to the outside edge of both materials require knowing where the NA is located.

The NA can be determined from the equation

     0 = E_W Q_W + E_S Q_S

The steel section consists of two parts, 1 and 2, giving,

      0 = E_W (y_W A_W) + E_S [(y_S1 A_S1) + (y_S2 A_S2)]

        = 12 (5 + 1 - h) (5) (10) +
               200[ (10 + 1 + 0.5 - 0.25 - h) (5) (0.5) +
                         (-h + 0.5) (5) (1) ]

        = 3,600 - 600h + 5,625 - 500h + 500 - 1000h

     9,725 = 2,100 h

     h = 4.631 cm

Now that the distance h is known, the location of all material area centroids from the NA can be found.

     y_W = 5 + 1 - 4.631 = 1.369 cm

     y_S1 = 10 + 1 + 0.5 - 0.25 - 4.631 = 6.619 cm

     y_S2 = - 4.631 + 0.5 = -4.131 cm

With the neutral axis (NA) located, the moment of inertia about the NA for each of the two material areas can be calculated. Using the basic formula for the moment of inertia of rectangular areas, and the parallel axis theorem, gives,

     I_W = h³ b / 12 + y_W² A_W
         = 10³ (5) / 12 + 1.369² (10) (5)
         = 510.4 cm⁴

     I_S = I_S1 + I_S2
        = h³ b / 12 + y_S1² A_S1 + h³ b / 12 + y_S2² A_S2
         = 0.5³ (5) / 12 + 6.619² (0.5) (5) +
                1³ (5) / 12 + (-4.131)² (1) (5) +
         = 195.3 cm⁴

The bending stress in each of the material areas can now be determined by using the bending stress equations for a two-material composite beam. There are two equations since each material section has different stress.

The y distance is the maximum distance from the neutral axis for each material section.

     y_W-max = (10 + 1 + 0.5) - (0.5 + 4.631) = 6.369 cm

     y_S-max = (10 + 1 + 0.5) - 4.631 = 6.869 cm

Substituting values into both stress equations give,

The beam will fail, even though the bending stress in the wood is below the wood failure stress of 50 MPa since the maximum bending stress in the steel is over 150 MPa. If the beam was only wood, it would still fail since the maximum bending stress would be 75 MPa (calculations not shown) which is beyond the failure stress of 50 MPa.