Question 5.5: A simply supported wood beam having a span length L = 12 ft ...

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}}{σ_{allow}}=\frac{90,720  lb-in.}{1800  psi}=50.40  in.^3

(2) From the table in Appendix G 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\times 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 G available online gives weights of beams based upon a density of  35  lb/ft^3.)

(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)(\frac{426.8  lb/ft}{420  lb/ft})=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 \times 12  in.  beam is satisfactory.

Note: If the weight density of the wood is other than  35  lb/ft^3,  we can obtain the weight of the beam per linear foot by multiplying the value in the last column in Appendix G by the ratio of the actual weight density to  35  lb/ft^3.