Question 5.5: A simply supported wood beam having a span length L = 12 ft ...
A simply supported wood beam having a span length L = 12 ft carries a uniform load q = 420 lb/ft (Fig. 5-19). The allowable bending stress is 1800 psi, the wood weighs 35 lb/ft³, and the beam is supported laterally against sideways buckling and tipping.
Select a suitable size for the beam from the table in Appendix F.
Since we do not know in advance how much the beam weighs, we will proceed by trial-and-error as follows: (1) Calculate the required section modulus based upon the given uniform load. (2) Select a trial size for the beam. (3) Add the weight of the beam to the uniform load and calculate a new required section modulus. (4) Check to see that the selected beam is still satisfactory. If it is not, select a larger beam and repeat the process.
(1) The maximum bending moment in the beam occurs at the midpoint (see Eq. 4-15):
M_{\max}=\frac{qL^{2}}{8}=\frac{(420 lb/ft)(12 ft)^{2}(12 in./ft)}{8}=90,720 lb-in.The required section modulus (Eq. 5-24) is
S=\frac{M_{\max}}{\sigma_{allow}}=\frac{90,720 lb-in.}{1800 psi}=50.40 in.^{3}(2) From the table in Appendix F we see that the lightest beam that supplies a section modulus of at least 50.40 in.³ about axis 1-1 is a 3 × 12 in. beam (nominal dimensions). This beam has a section modulus equal to 52.73 in.³ and weighs 6.8 lb/ft. (Note that Appendix F gives weights of beams based upon a density of 35 lb/ft³.)
(3) The uniform load on the beam now becomes 426.8 lb/ft, and the corresponding required section modulus is
S=(50.40 in.^{3})\left(\frac{426.8 lb/ft}{420 lb/ft}\right)=51.22 in.^{3}(4) The previously selected beam has a section modulus of 52.73 in.³, which is larger than the required modulus of 51.22 in.³
Therefore, a 3 × 12 in. beam is satisfactory.
Note: If the weight density of the wood is other than 35 lb/ft³ , we can obtain the weight of the beam per linear foot by multiplying the value in the last column in Appendix F by the ratio of the actual weight density to 35 lb/ft³.