Design of wall and columns formwork ENG. ESRAA HUSSEIN

1. CB523
TERM PROJECT
ENG. ESRAA HUSSEIN MOHAMED

2. Addextra(X*10)cmtoeachdimensionforall
spacingbetweenX-Axis.
Add extra (Y*10) cm to each dimension for all
spacing between Y-Axis
W2
W1
C1
C1
C1
C2
C2C3C2
C2
C3 C3 C3
C3

3. ID: 14104622
X
Y Z
M• Concrete Unit Weight
W = 2300 + (10*M)
• Average Temperature
T = 33 + M ◦C
• Columns and Walls surfaces should have Rough Face
Lumer
• Slab surface should have Fair Face
Plywood

4. • FOR X= 0, 1, 2, 3, 4
Douglas Fir-Larch with lumber sizes equal 1x4”, 1x6”, 2x6”, 2x8”, and
4x4” for formwork members
• FOR X= 5, 6, 7, 8, 9
Southern Pine with lumber sizes equal 1x4”, 1x6”, 2x4”, 2x8”, 4x4”, and
6x6” for formwork members.
• FOR M= 0, 1, 2, 3, 4
Class I plywood sheets with ( 4 + M ) / 8 in thickness and 120 x 240
cm in size
• FOR M= 5, 6, 7, 8, 9
Class I plywood sheets with (M)/8 in thickness and 120 x 240 cm in size

5. • Rate of Placement
R = 3 + ( M * 0.1 ) m/hr
• Wind Force
Wf = ( 16 + M ) lb/ft2
• Concrete will be vibrated internally
Live loads is Manual
• Deflection l/360
• Ties 18KN Capacity with 120mm wide wedges bearing on wales

6. ID: 14104622
X
Y Z
M
• W = 2300 + (10 * M)
= 2300 + (10 * 2) = 2320 Kg/m3 è Cw= 1
• T = 33 + M = 33 + 2 = 35 ◦C
• R = 3 + (M * 0.1) = 3 + (2 * 0.1) = 3.2 m/hr
• P = CwCc (7.2+(
("#$%)
('()#)
))
• = 1*1*(7.2+(
("#$∗+.-)
(+$()#)
)) = 54.6 KPa
• Pmin= 28.7 Cw =28.7 KPa
• Pmax= wh = 2320 * 300 * 10-4 = 69.6 KPa
à Pchosen= 54.6 KPa
• DESIGN OF COLUMN

7. • DESIGN OF SHEATHING ( YOKE OR CLAMP SPACING)
b = 1ft b= 1m = 1000mm d = 1” = 19 mm
A= b* d = 1000*19 = 19000 mm2
S=
.∗/!
0
=
)111∗)2!
0
= 60166.67 mm3
I=
.∗/"
)-
=
)111∗)2"
)-
= 571583.33mm4
Lb=
31."
)111
d(
4!∗.
5
)1/2 =
31."
)111
*19 (
!!!"∗$%%%
54.6
)1/2 = 330.91 mm

8. Lsh=
).))
)111
4"∗6
5
+ 2d =
).))
)111
*
$&'(∗$!%%%
54.6
+ (2 ∗ 19) =530.87 mm
Ldef=
"+.#
)111
(
7∗8
5
)1/3 =
"+.#
)111
(
$$.'∗$%!∗571583.33
*+.(
)1/3 = 366.51 mm
Lclamp= 330.91mm = 33.09 cm = 30 cm
No. of Clamps /side =
!"#.%&'()* +,-./
#!"#$%
+1 =
011+,-./
01
+1 = 10.5 = 11 Clamp
Total No. of Clamps = 11*4 = 44 Clamps

9. DESIGN OF WALLS
• W = 2320 PSF è Cw= 1
• T = 35 ◦C
• R = 3.2 m/hr
• P = CwCc (7.2 + (
))$3
('()#)
) +
(-33%)
('()#)
)
• = 1*1*(7.2+ (
))$3
(+$()#)
) +
(-33∗+.-)
(+$()#)
) = 43.71 KPa
• Pmin= 28.7 Cw =28.7 KPa
• Pmax= wh = 2320 * 300 * 10-4 = 69.6 KPa
à Pchosen= 43.71 KPa

10. • DESIGN OF SHEATHING ( STUD SPACING)
b = 1ft b= 1m = 1000mm d = 1” = 19 mm
A= b* d = 1000*19 = 19000 mm2
S=
.∗/!
0
=
)111∗)2!
0
= 60166.67 mm3
I=
.∗/"
)-
=
)111∗)2"
)-
= 571583.33mm4
Lb=
31."
)111
d(
4!∗.
5
)1/2 =
31."
)111
*19 (
!!!"∗$%%%
43.71
)1/2 = 369.84 mm

11. Lsh=
).))
)111
4"∗6
5
+ 2d =
).))
)111
*
$&'(∗$!%%%
43.71
+ (2 ∗ 19) =653.67 mm
Ldef=
"+.#
)111
(
7∗8
5
)1/3 =
"+.#
)111
( $$.'∗$%!∗571583.33
43.71 )1/3 = 394.72 mm
Lstud= 369.84 mm = 36.98 cm = 35 cm
• DESIGN OF STUD ( WALES SPACING)
X = 4 2’’ * 6”
Stud Load = Pchosen* 𝐿 𝑠𝑡𝑢𝑑 = 43.71 * 0.35 = 15.3 KN/m
A= 5323 mm2 S=1.239*105 mm3 I=8.656*106 mm4
B = 38mm D = 140mm

12. Lb=
)11
)111
(
4!∗9
5
)1/2 =
)11
)111
* (
!!!"∗$.&,!∗$%!
15.3
)1/2 = 899.8 mm
Lsh=
).))
)111
4"∗6
5
+ 2d = (
).))
)111
*
$&'( ∗ *,&,
15.3
) + (2 ∗ 140) = 772.76 mm
Ldef=
"+.#
)111
(
7∗8
5
)1/3 =
"+.#
)111
( $$.'∗$%!∗".(*(∗$%"
15.3
)1/3 = 1385.67mm
Lwale= 772.76 mm = 77.28 cm = 75 cm
• DESIGN OF DOUBLE WALES ( TIES SPACING)
Wales Load = Pchosen* 𝐿 𝑤𝑎𝑙𝑒𝑠 = 43.71 * 0.75 = 32.78 KN/m
A= 7016 *2 mm2 S=2.153*105 *2 mm3 I=19.83*106 *2 mm4
2’’ * 8” B = 38mm D = 184mm

13. Lb=
)11
)111
(
4!∗9
5
)1/2 =
)11
)111
* (
!!!"∗&.$*,∗$%!
∗&
32.78
)1/2 = 1146.01 mm
Lsh=
).))
)111
4"∗6
5
+ 2d = (
).))
)111
*
$&'( ∗'%$(∗&
32.78
) + (2 ∗ 184) = 974.3 mm
Ldef=
"+.#
)111
(
7∗8
5
)1/3 =
"+.#
)111
( $$.'∗$%!∗$!.",∗$%"
∗&
32.78
)1/3 = 1785.25 mm
Ltie= 974.3 mm = 97.43 cm = 95 cm
Lstud= 35 cm Lwale= 75 cm Ltie= 95 cm

14. Lstud= 35 cm Lwale= 75 cm Ltie= 95 cm
• CHECK OF BEARING:-
1. BETWEEN STUD & WALES
Load = Pchosen * Lstud * Lwale = 43.71 * 0.35 * 0.75 = 11.47 KN
Area = bstud * bwale * 2 = 38 * 38 * 2 = 2888 mm2
Stress = Load/Area =11.47/(
-###
)111#)= 3971.61 KN/mm2 > Fc⏊= 2655 KN/mm2
è UN SAFE
Lwale New= (2655*(
-###
)111#))/(43.71 * 0.35) = 0.5 m

15. Lb=
)11
)111
(
4!∗9
5
)1/2 =
)11
)111
* (
!!!"∗&.$*,∗$%!
∗&
21.855
)1/2 = 1403.52 mm
Lsh=
).))
)111
4"∗6
5
+ 2d = (
).))
)111
*
$&'( ∗'%$(∗&
&$."**
) + (2 ∗ 184) = 1277.37 mm
Ldef=
"+.#
)111
(
7∗8
5
)1/3 =
"+.#
)111
( $$.'∗$%!∗$!.",∗$%"
∗&
21.855
)1/3 = 2043.55 mm
Ltie= 1277.37 mm = 127.74 cm = 125 cm
Wales Load = Pchosen* 𝐿 𝑤𝑎𝑙𝑒𝑠 = 43.71 * 0.5 = 21.855 KN/m

16. • CHECK OF BEARING:-
2. BETWEEN WALES & TIE WEDGE
Load = Pchosen * Lwale * Ltie = 43.71 * 0.5 * 1.25 = 27.32 KN
Area = bwale * 2 * btie wedge = 38 * 2 * 120 = 9120 mm2
Stress = Load/Area =27.32/(
2)-1
)111#)= 2995.61 KN/mm2 > Fc⏊= 2655 KN/mm2
è UN SAFE
Ltie New= (2655*(
2)-1
)111#))/(43.71 * 0.5) = 1.1 m

17. • CHECK TIE SAFETY:-
Load = Pchosen * Lwale * Ltie = 43.71 * 0.5 * 1.1 = 24.04 KN > 18 KN
è UN SAFE
Ltie New= 0.8 m
Lstud= 35 cm Lwale= 50 cm Ltie= 80 cm
• FINAL DESIGN:-

18. THANK YOU