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EXAMPLE PROBLEMS 



Example 1 (for Fj^ = 18 inches) 



Given: DL (dead load) = 20 Ibs./ft? 



LL (Uve load) = 25 Ibs./ft? 

 F^ = 18 in. 

 ^(D+L) ~ ^^ ^"' ^^^ seawater) 



Find: 



^(D+L) 



Solution: Enter left side of Figure 78 at ^(})+ii — 12 in. and intersect with curve 



LL = 25 Ibs./ft? 

 Read P on bottom scale (P = 78 percent) 

 From intersect with LL = 25 Ibs./ft? , move vertically to intersect with 



Total Load curves to : 



DL + LL = 45 Ibs./ft? and read ^/d+d = 10.8 in. on left scale. 

 DL = 20 Ibs./ft? and read S^, = 4.8 in. on left scale. 



Answers: P = 78 percent; S/^j^^i = 10.8 in.; S^^ = 4.8 in. 



Example 2 (for Fj^ =^ 18 inches) 



Given: DL = 20 Ibs./ft? 



LL = 251bs./ft? 

 F^ = 20 in. 

 ^(D+L) ~ 12 in. (in seawater) 



Find: S^ 



^(D+L) 

 P 



Solution: ^ (D+L) must first be adjusted because dead load deck level (Fj)) "^ 18 in. 

 (See text) 



^mi) = "^(D+L) - [Fd-18] = 12 in. - [20 in. -2 ft.] 

 ^(D+L) = l^i"- 



Using F/^^^i, find S^,, ?>(j)+i)-, and P in a similar manner to 

 Example 1. 



Answers: P = 58 percent; Sfj)+i) = 14.4 in.; S^^ = 6.4 in. 

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