This document provides the design calculations for the formwork of a concrete structure. It includes calculations for: - The spacing between sheathing/clamps on the columns based on deflection, bearing and minimum requirements. The design calls for 11 clamps spaced at 30 cm. - The spacing between sheathing/studs on the walls based on similar requirements. The design calls for studs spaced at 35 cm. - The spacing between double wales and ties based on bearing capacity and tie safety requirements. The final design calls for wales spaced at 50 cm and ties at 80 cm.

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