Question 5.7: A simple beam AB of span length 21 ft must support a uniform...

In this example, we will proceed as follows: (1) Find the maximum bending moment in the beam due to the uniform load. (2) Knowing the maximum moment, find the required section modulus. (3) Select a trial wide-flange beam from Table E-1 in Appendix E (available online) and obtain the weight of the beam. (4) With the weight known, calculate a new value of the bending moment and a new value of the section modulus. (5) Determine whether the selected beam is still satisfactory. If it is not, select a new beam size and repeat the process until a satisfactory size of beam has been found.

Maximum bending moment. To assist in locating the cross section of maximum bending moment, we construct the shear-force diagram (Fig. 5-21b) using the methods described in Chapter 4. As part of that process, we determine the reactions at the supports:

R_A = 188.6 kN                    R_B = 171.4 kN

The distance x_1 from the left-hand support to the cross section of zero shear force is obtained from the equation

V = R_A – qx_1 = 0

which is valid in the range 0 ≤ x ≤ 4 m. Solving for x_1, we get

x_1=\frac{R_A}{q} =\frac{188.6  kN}{60  kN/m} =3.14  m

which is less than 12 ft, and therefore the calculation is valid.

The maximum bending moment occurs at the cross section where the shear force is zero; therefore,

M_{max}=R_Ax_1-\frac{qx_1^2}{2} =296.3  kN.m

Required section modulus. The required section modulus (based only upon the load q) is obtained from Eq. (5-24):

S=\frac{M_{max}}{\sigma_{allow}}        (Eq. 5-24)

=\frac{296.3 \times 10^6  N.mm}{110  MPa} =2.694 \times 10^6  mm^3

Trial beam. We now turn to Table E-1 and select the lightest wide-flange beam having a section modulus greater than 2694 cm³ The lightest beam that provides this section modulus is HE 450A with S = 2896 cm³. This beam weighs 140 kg/m (Recall that the tables in Appendix E are abridged, and therefore a lighter beam may actually be available.)

We now recalculate the reactions, maximum bending moment, and required section modulus with the beam loaded by both the uniform load q and its own weight. Under these combined loads the reactions are

R_A = 193.4 kN                  R_B = 176.2 kN

and the distance to the cross section of zero shear becomes

x_1 = 3.151 m

The maximum bending moment increases to 304.7 kNm, and the new required section modulus is

S=\frac{M_{max}}{\sigma_{allow}}=\frac{304.7 \times 10^6  N.mm}{110  MPa} =2770  cm^3

Thus, we see that the HE 450A beam with section modulus S = 2896 cm³ is still satisfactory.

Note: If the new required section modulus exceeded that of the HE 450A beam, a new beam with a larger section modulus would be selected and the process repeated.