Draft

11.2CrackedSection,UltimateStrengthDesignMethod 191

5. Check equilibriumof forcesin thexdirection(Fx= 0)

a=

Asfs

: 85 f

0

c

b

(11.16)

6. Check assumptionoffsfromthestraindiagram

"s

dc

=

: 003

c

)fs=Es

dc

c

: 003 < fy (11.17)

wherec=

a

(^1)
.

7. Iterateuntil convergenceis reached.

Example11-1: UltimateStrengthCapacity

DeterminetheultimateStrengthof a beamwiththefollowingproperties:b= 10in,d=

23 in,As= 2: 35 in

2

,f

0

c= 4;^000 psiandfy= 60ksi.

Solution:

act =

As

bd

=

2 : 35

(10)(23)

=: 0102

b = : (^851)
fc^0
fy
87
87+fy
= (:85)(:85)
4
60
87
87+60
=: 02885 > act
p
a =
Asfy
: 85 f
0
cb

(2:35)(60)
(:85)(4)(10)
= 4: 147 in
Mn = (2:35)(60)(23
4 : 147
2
) = 2; 950 k.in
MD = Mn= (:9)(2;950)= 2 ; 660 k.in
Notethatfromthestraindiagram
c=
a
0 : 85

4 : 414
0 : 85
= 4: 87 in
Alternative solution
Mn = actfybd
2

1 : 59 act
fy
f
0
c

= Asfyd

1 : 59 act
fy
f
0
c

= (2:35)(60)(23)
h
1 (:59)
60
4
(:01021)
i
= 2; 950 k.in= 245k.ft
MD = Mn= (:9)(2;950)= 2 ; 660 k.in
Example11-2: BeamDesignI