Draft
14.2BuildingsStructures 237
~ 20 ’
~ 20 ’N.A.
w = 0.8 k/ftH = 96 k60 ’120 ’20 ’ 20 ’Figure14.7:DesignExampleof a TubularStructure,(LinandStotesbury1981)
- Themaximum
exuralstresses:
f l=
M C
I
=
(5;760)k.ft(20=2)ft
(4;600)ft
4
= 12 : 5 ksf= 87 psi (14.25)
- Theaverageshearstressis
=
V
A
=
(96)k
2(20)(1)ft
2
= 2: 4 ksf= 17 psi (14.26)
- Theverticalloadof 1,600k producesanaxialstressof
ax=
P
A
=
(1;600)k
(4(20)(1)ft
2
= 20 ksf= 140 psi (14.27)
- Thetotalstressesarethus
= ax+f l (14.28-a)
1 = 140 + 87 = 53 psi (14.28-b)
2 = 140 87 = 227 psi (14.28-c)
thus we donothave any tensilestresses,andthosestressesaremuch betterthanthose
obtainedfroma singleshearwall.
14.2.3 RigidFrames
21 Rigidframescancarrybothverticalandhorizontalloads,however theiranalysisis more
complexthanfortubes.