- Consider that a concordant trajectory has been plotted; express
the eccentricity at section B relative to that at section 2
Thus, CiJe 2 = -0.1250/+0.0700 = - 1.786; therefore, eb = - 1.78fc? 2.
- Determine the allowable range of values of fbp at sections 2
and B
Refer to Fig. 56. At section 2Jbp < + 3621 IMn^2 (+24,966.8 kPa), Eq. a; 0.85/^ => 1221
- 972 - 60; therefore,/^ > + 2509 IMn^2 (+17,299.5 kPa), Eq b. At section BJbp >
-2240 lb/in^2 (-15,444.8 kPa), Eq. c; 0.85/J,, < -(218O + 1280) + 2250; fbp < - 1424 lb/in^2
(-9818.5kPa),Eq.</.
- Substitute numerical values in Eq. 56, expressing eb in
terms of e 2
The values obtained are 1/F 1 < (kt + e£/(362lSb), Eq a'; 1/F 1 < (kt + e£/(25Q9Sb), Eq. b';
HFt > (1.786* 2 - kt)/(224QSb), Eq. c'; ItF 1 < (1.786e 2 - &,)/(! 424S 6 ), Eq. J'.
- Obtain the composite Magnel diagram
Considering the relations in step 7 as equalities, plot the straight lines representing them
to obtain the composite Magnel diagram in Fig. 57. The slopes of the lines have these rel-
ative values: ma = 1/3621; mb = 1/2509; mc = 1.786/2240 = 1/1254; md = 1.786/1424 =
1/797. The shaded area bounded by these lines represents the region of permissible sets of
values of e 2 and 1 IF 1.
- Calculate the minimum allowable value of F 1 and the
corresponding value of e 2
In the composite Magnel diagram, this set of values is represented by point A. Therefore,
consider Eqs. b' and c' as equalities, and solve for the unknowns. Or, (10.32 + e 2 )/2509 =
(1.786^ 2 - 10.32)/2240; solving gives e 2 = 21.87 in (555.5 mm) and F 1 = 1,160,000 Ib
(5,159,680.ON).
Initial state Final state
(b) Limiting values of fbp at section B
FIGURE 56
(a) Limiting values of fbp at section 2
In.tiol state Final state
