Draft
14.2BuildingsStructures 235
- Thecompressive stressof 740psicaneasilybe sustainedby concrete,as to thetensile
stressof 460psi,it wouldhave to be resistedby somesteelreinforcement.
10.Giventhatthosestressesareservicestressesandnotfactoredones,we adopttheWSD
approach, anduseanallowablestressof 20 ksi,which in turnwillbe increasedby 4=3 for
seismicandwindload,
all=
4
3
(20)= 26: 7 ksi (14.16)
11.Thestressdistributionis linear,compressionat oneend,andtensionat theother. The
lengthof thetensionareais givenby (similartriangles)
x
460
=
20
460 + 740
)x=
460
460 + 740
(20)= 7: 7 ft (14.17)
12.Thetotaltensileforceinsidethistriangularstressblock is
T=
1
2
(460)ksi(7: 7 12)in(12)in
| {z }width
= 250k (14.18)
13.Thetotalamount of steelreinforcement neededis
As=
(250)k
(26:7)ksi
= 9 : 4 in
2
(14.19)Thisamount of reinforcement shouldbe providedat bothendsof thewallsincethewind
or eartquake canactin any direction.In addition,thefoundationsshouldbe designedto
resisttensileupliftforces(possiblyusingpiles).
14.2.1.2 Example: TrussedShearWall
From (Linand Stotesbury 1981)
17 We considerthesameproblempreviouslyanalysed,butusea trussedshearwallinsteadof
a concreteone,Fig.14.6.
- Usingthemaximummoment of 5; 760 kip-ft(Eq.14.8-b), we cancomputethecompression
andtensionin thecolumnsfora lever armof 20 ft.
F=
(5;760)k.ft
(20)ft
= 288 k (14.20)
- If we now addthee ectof the 400 kipverticalload,theforceswouldbe
C =
(400)k
2
288 = 488 k (14.21-a)
T =
(400)k
2
- 288= 88 k (14.21-b)
- Theforcein thediagonalwhich mustresista baseshearof 96 kipis (similartriangles)
F
96
=
p
(20)
2
+ (24)2
20
)F=
p
(20)
2
+ (24)2
20
(96)= 154 k (14.22)
- Thedesigncouldbe modi edto have notensileforcesin thecolumnsby increasingthe
widthof thebase(currentlyat 20 ft).