## Info

294,200 (N-m)

With a bending moment arm of 5 m (the center of mass location is at the cylinder mid-length), we can find the equivalent axial load using Eq. (11-51):

Limit load x Ultimate Factor of Safety = Ultimate Load (11 -62a) or 715,900 X 1.25 = 894,900 N. (ll-62b)

### Sizing for Tensile Strength

The equation for axial stress, o-, is c= P/A. To size the cylinder for tensile strength, we use the ultimate Peq load = 894,900 N, and the material's allowable stress, Ftu = 524 x 106 N/m2, and use A = 2nRt to solve for the required thickness.

Although we won't show you here, we must check for yield conditions in the same way, using a factor of safety of 1.10 with limit load and F^ = 448 x 106 N/m2.

Sizing for Stability (Compressive Strength)

We must now size the cylinder for stability [Ref. Eqs. (11-52) and (11-53)], using the cylinder thickness required for bending stability. The cylinder must withstand an ultimate Peq = 894,900 N.

r 16\ r 16)10.00286 y= 1.0 - 0.901 (1.0 - e~v ) = 0.379

The equation for cylinder buckling stress is o„ =0.6yf = (0.6)(0.379)(71Xl°9i><000286> = 46.16 xl06N/m2

Note that if o^ were greater than the material's proportional limit, we would use additional methods for inelastic buckling. With the cylinder's cross-sectional area, A = 180 cm2, the critical buckling load is

P„=A oa — (0.0180)(46.16 x 106) = 830,9«) N (ultimate)

Thus, the cylinder is not adequate because the applied ultimate load is greater than the critical buckling load. Structural integrity is often shown in terms of the margin of safety (MS), defined as w„ Allowable Load or Stress , . MS = ———:-;-:-7--1.0

Design Load or Stress and must be greater than or equal to zero. For the stability conditions (ultimate), 830,900

—1.0 = -0.07 (7% negative margin of safety) (11-70)

894,900

Results for a small increase in thickness are shown in Table 11-59.

TABLE 11-59. Summary of Sizing the Monocoque Cylinder for Stability. This table summarizes our Initial sizing attempt and the first (and final) iteration for an equivalent axial load of 894,900 N.

iteration

Thickness (cm)

r

Oc

Area (cm2)

Per

0 0