Draft

4.4InternalForces 83

INCLINED

INTERNAL

FORCE: N

KNOWN VERTICAL

COMPONENT: H COMPONENT: V

HORIZONTAL

CONSEQUENT

b=18.4^0

V

H

N

b=18.4

FORCE POLYGON

Figure4.7:EielTower,InternalGravity Forces;(BillingtonandMark1983)

12 Gravity loadarerstconsidered,remember thosearecausedby thedeadloadandthelive

load,Fig.4.7:

cos =

V

N

)N=

V

cos

(4.12-a)

N =

11 ; 140 k

cos 18 : 4

o

= 11,730kip (4.12-b)

tan =

H

V

)H=Vtan (4.12-c)

H = 11 ; 140 k(tan 18 : 4

o

) = 3,700kip (4.12-d)

Thehorizontalforceswhich mustbe resistedby thefoundations,Fig.4.8.

H H

3700 k

3700 k

Figure4.8:EielTower,HorizontalReactions;(BillingtonandMark1983)

13 Becausetheverticalloaddecreaseswithheight, theaxialforcewillalsodecreasewithheight.

14 At thesecondplatform,thetotalverticalloadisQ= 1; 100 + 2; 200 = 3; 300 kandat that

height theangleis 11: 6

o

thus theaxialforce(per pairof columns)willbe

Nvert =

3 ; 300 k

2

cos 11 : 6

o

= 1; 685 k (4.13-a)

Hvert =

3 ; 300 k

2

(tan 11 : 6

o

) = 339k (4.13-b)

Notethatthisis aboutseven timessmallerthantheaxialforceat thebase,which fora given

axialstrength,wouldleadthedesignerto reduce(ortaper)thecross-section.