Question 5.5: A simply supported wood beam having a span length L = 3 m ca...

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}                      ( Eq. 4-15)

=\frac{(4  kN/m)(3  m)^2}{8 } =4.5  kN.m

The required section modulus (Eq. 5-24) is

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

=\frac{4.5  kN.m}{12  MPa}= 0.375 \times 10^6  mm^3

(2) From the table in Appendix F we see that the lightest beam that supplies a section modulus of at least 0.375 × 10^6  mm³ about axis 1-1 is a 75 × 200 mm beam (nominal dimensions). This beam has a section modulus equal to 0.456  × 10^6  mm³ and weighs 77.11 N/m (Note that Appendix F (available online) gives weights of beams based upon a density of 5.4 kN/m³.)

(3) The uniform load on the beam now becomes 4.077 kN/m, and the corresponding required section modulus is

S=(0.375 \times 10^6  mm^3)(\frac{4.077}{4.0} )= 0.382 \times 10^6  mm^3

(4) The previously selected beam has a section modulus of 0.456 × 10^6  mm³, which is larger than the required modulus of 0.382 × 10^6  mm³

Therefore, a 75 × 200 mm beam is satisfactory.

Note: If the weight density of the wood is other than 5.4 kN/m³, 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 5.4 kN/m³.