Design of bracing

Design of bracing

1. CB523
BRACING DESIGN
ENG. ESRAAHUSSEINMOHAMED

2. BRACING:
•ISA STRUCTURALMEMBERUSEDTOSUPPORTANOTHER.
•ITUSEDTORESISTTHEMOVEMENTOFFORMWORKANDWINDLOADS.
Bracer

3. STEPS OF SOLUTIONS:-
•L=ℎ′!+#′!
L
l’
h’
h
Where:-
h : height of form (ft) [m]
h’: height of top of strut (ft) [m]
L : length of strut (ft) [m]
l’: horizontal distance form the bottom of strut (ft) [m]

4. Strut Load per foot of form
•P’= "∗!∗$
%!∗!&
Where:-
P : strut load per foot of form (lb/ft) [kN/m]
H : lateral load at top of form (lb/ft) [kN/m]
h : height of form (ft) [m]
h’: height of top of strut (ft) [m]
L : length of strut (ft) [m]
l’: horizontal distance from form to bottom of strut (ft) [m]

5. •!
"< 50
•!
"> 50
Safe
Un Safe
StrutLnew= "
#
L/2
L/2
Check of Buckling:-
D
L

6. Allowable Compressive Stress:
•FC’=".$%
(!
")#“PSI” <FC<FTSafe

7. Maximum Allowable Compressive Force:
•P = B*D*FC’
Maximum Strut Spacing:
•S = ⁄(()
“Lb”
“ft”

8. EXAMPLE:-
vDetermine the maximum spacing of 2”x4” lateral braces for the wall form of 8 ft
high. Lateral wood bracing will be attached at a height of 6ft above the form
bottom and anchored 5ft away from the bottom of the form. Design wind load
is 25 psf.
-Wood charactesrstics: Fc = 850 PSI, Ft = 725 PSI,
E =1,400,000 PSI
h= 8ft
h’= 6ft
l’= 5ft
Solution:-
L=ℎ′!+%′!= 6!+5!= 7.81ft
P’= "∗$∗%
&!∗$'= ())∗*∗+.*(
-∗.= 208.27 lb/ft
L
5ft
6ft
8ft

9. h= 8ft àFrom (Table 13-3) H=100lb/ft at least $∗/0
!= *∗!-
!= 100lb/ft
P’= "∗$∗%
&!∗$'= ())∗*∗+.*(
-∗.= 208.27 lb/ft
%
1= +.*(∗(!
(.-= 62.48 > 50
Lnew= +.*(∗(!
!= 46.86 in
%!"#
1= 2..*.
(.-= 31.24 < 50
Un Safe
Safe

10. -Wood charactesrstics: Fc= 850 PSI, Ft = 725 PSI, E =1,400,000 PSI
•Fc’= ).3∗(,2)),)))
("#.%#
&.')(= 430.355 PSI
<Fc= 850PSI
<Ft=725PSI
•P = 3.5 * 1.5 * 430.355=2259.264 Lb
•S = 7
7'= !!-8.!.2
!)*.!+= 10.85 ft
“ Safe ”

11. THANK YOU