Question 6.9: A doubly symmetric hollow box beam (Fig. 6-46) of elastoplas...

The cross section of the beam and the distribution of the normal stresses are shown in Figs. 6-47a and b, respectively. From the figure, we see that the stresses in the webs increase linearly with distance from the neutral axis and the stresses in the flanges equal the yield stress \sigma_{Y}. Therefore, the bending moment M acting on the cross section consists of two parts:
(1) a moment M_{1} corresponding to the elastic core, and
(2) a moment M_{2} produced by the yield stresses \sigma_{Y} in the flanges.
The bending moment supplied by the core is found from the flexure formula (Eq. 6-74) with the section modulus calculated for the webs alone; thus,

M_{Y}=\frac{\sigma_{Y}I}{c}=\sigma_{Y}S            (6-74)

S_{1}=\frac{(b-b_{1})h^{2}_{1}}{6}                (6-91)

and

M_{1}=\sigma_{Y}S_{1}=\frac{\sigma_{Y} (b-b_{1})h^{2}_{1}}{6}                        (6-92)

To find the moment supplied by the flanges, we note that the resultant force F in each flange (Fig. 6-47b) is equal to the yield stress multiplied by the area of the flange:

F=\sigma_{Y}b\left(\frac{h-h_{1}}{2}\right)                (h)

The force in the top flange is compressive and the force in the bottom flange is tensile if the bending moment M is positive. Together, the two forces create the bending moment M_{2}:

M_{2}=F\left(\frac{h+h_{1}}{2}\right)=\frac{\sigma_{Y}b(h^{2}-h^{2}_{1})}{}                (6-93)

Therefore, the total moment acting on the cross section, after some rearranging, is

M=M_{1}+M_{2}=\frac{\sigma_{Y}}{12}\left[3bh^{2}-(b+2b_{1})h^{2}_{1}\right]                      (6-94)

Substituting the given numerical values, we obtain

M = 1330 k-in.

Note: The yield moment M_{Y} and the plastic moment M_{P} for the beam in this example have the following values (determined in Problem 6.10-13):

M_{Y} = 1196 k-in.          M_{P} = 1485 k-in.

The bending moment M is between these values, as expected.