Question 5.7: A simple beam AB of span length 21 ft must support a uniform...
A simple beam AB of span length 21 ft must support a uniform load q = 2000 lb/ft distributed along the beam in the manner shown in Fig. 5-21a.
Considering both the uniform load and the weight of the beam, and also using an allowable bending stress of 18,000 psi, select a structural steel beam of wide-flange shape to support the loads.
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 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:
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}=0which is valid in the range 0 ≤ x ≤ 12 ft. Solving for x_{1}, we get
x_{1}=\frac{R_{A}}{q}=\frac{18,860 lb}{2,000 lb/ft}=9.430 ftwhich 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,
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}}=\frac{(88,920 lb-ft)(12 in./ft)}{18,000 psi}=59.3 in.^{3}Trial beam. We now turn to Table E-1 and select the lightest wide-flange beam having a section modulus greater than 59.3 in.³ The lightest beam that provides this section modulus is W 12 × 50 with S = 64.7 in.³ This beam weighs 50 lb/ft. (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
and the distance to the cross section of zero shear becomes
x_{1}=\frac{19,380 lb}{2,050 lb/ft}= 9.454 ftThe maximum bending moment increases to 91,610 lb-ft, and the new required section modulus is
S=\frac{M_{\max}}{\sigma_{allow}}=\frac{(91,610 lb-ft)(12 in./ft)}{18,000 psi}=61.1 in.^{3}Thus, we see that the W 12 × 50 beam with section modulus S = 64.7 in.³ is still satisfactory.
Note: If the new required section modulus exceeded that of the W 12 × 50 beam, a new beam with a larger section modulus would be selected and the process repeated.
