Draft

14.2BuildingsStructures 235

9. 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.

1. 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)

2. If we now addtheeectof the 400 kipverticalload,theforceswouldbe

C =

(400)k

2

288 = 488 k (14.21-a)

T =

(400)k

2

• 288= 88 k (14.21-b)

3. 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)

4. Thedesigncouldbe modiedto have notensileforcesin thecolumnsby increasingthe

widthof thebase(currentlyat 20 ft).