Draft
12.2FlexuralStresses 205
M 0 =
(:183)(40)
2
8
= 36: 6 k.ft (12.9-b)
The
exuralstresseswillthus be equalto:
f
w 0
1 ; 2
=
M 0
S 1 ; 2
=
(36:6)(12;000)
1 ; 000
= 439 psi (12.10)
f 1 =
Pi
Ac
1
ec 1
r
2
M 0
S 1
(12.11-a)
= 83 439 = 522 psi (12.11-b)
fti = 3
q
f
0
c= +190
p
(12.11-c)
f 2 =
Pi
Ac
1 +
ec 2
r
2
M 0
S 2
(12.11-d)
= 1 ; 837 + 439= 1 ; 398 psi (12.11-e)
fci = : 6 f
0
c= ^2 ;^400p
(12.11-f)
- PeandM 0. If we have 15%losses,thenthee ective forcePeis equalto (1 0 :15)169=
144 k
f 1 =
Pe
Ac
1
ec 1
r
2
M 0
S 1
(12.12-a)
=
144 ; 000
176
1
(5:19)(12)
68 : 2
439 (12.12-b)
= 71 439 = 510 psi (12.12-c)
f 2 =
Pe
Ac
1 +
ec 2
r
2
M 0
S 2
(12.12-d)
=
144 ; 000
176
1 +
(5:19)(12)
68 : 2
- 439 (12.12-e)
= 1 ; 561 + 439= 1 ; 122 psi (12.12-f)
notethat 71 and 1 ; 561 arerespectivelyequalto (0:85)( 83)and(0:85)( 1 ;837)
respectively.
- PeandM 0 +MDL+MLL
MDL+MLL=
(0:55)(40)
2
8
= 110k.ft (12.13)
andcorrespondingstresses
f 1 ; 2 =
(110)(12;000)
1 ; 000
= 1 ; 320 psi (12.14)
Thus,
f 1 =
Pe
Ac
1
ec 1
r
2
M 0 +MDL+MLL
S 1
(12.15-a)