Draft

11.2CrackedSection,UltimateStrengthDesignMethod 189

11.2.3 Analysis

GivenAs,b,d,f

0

c, andfydeterminethedesignmoment:

1. act=

As

bd

2. 
   

   b= (:85) (^1)
   f
   0
   c
   fy
   87
   87+fy

   
   


3. 
   

   Ifact< b(thatis failureis triggeredby yieldingof thesteel,fs=fy)

   
   

a =

Asfy

: 85 fc^0 b

FromEquilibrium

MD = Asfy

d

a

2

)

MD= Asfy



d 0 : 59

Asfy

f

0

cb



| {z }

Mn

Combiningthislastequationwith=

As

bd

yields

MD= fybd

2



1 : 59 

fy

f

0

c



(11.9)

4. yIf act > bis notallowed by thecode asthiswould be anover-reinforcedsection

which wouldfailwithnopriorwarning.However,if such a sectionexists,andwe needto

determineitsmoment carryingcapacity, thenwe have two unknowns:

(a)Steelstrain"s(which was equalto"yin thepreviouscase)

(b)Locationof theneutralaxisc.

We have two equationsto solve thisproblem

Equilibrium:of forces

c=

Asfs

: 85 f

0

cb^1

(11.10)

Straincompatibility:sincewe know thatat failurethemaximumcompressive strain

"cis equalto 0.003.Thus fromsimilartriangleswe have

c

d

=

: 003

: 003 +"s

(11.11)

Thosetwo equationscanbe solvedby eitheroneof two methods:

(a)Substituteinto onesingleequation

(b)Byiteration

Oncecandfs=E"saredeterminedthen

MD= Asfs



d

1 c

2



(11.12)