Design of Slab Formwork

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
ID: 14104622
X
Y Z
M4.3 4.14.24.4
3.89
2.06
2.79
1.72
3.25

3. • W = 2300 + (10 * M)
= 2300 + (10 * 2) = 2320 Kg/m3
• W1= w * ts = 2320*10-2 *20 * 10-2 = 4.64 KPa
• W2= 15 PSF * 0.048 = 0.72 KPa
• W3= 50 PSF *0.048 = 2.4 KPa
• Design load (WT) = W1+ W2+ W3 = 4.64 + 0.72 + 2.4
= 7.76 KPa > 4.8 KPa (100 PSF*0.048)
• Sheathing load = Design load * 1m = 7.76 * 1= 7.76 KN/m
• Plywood thickness if (M< = 4) = ( 4 + M ) / 8
if (M> 4) = ( M ) / 8
M= 2 à Plywood thickness ( 4 + 2 ) / 8
Plywood thickness 6/ 8 = ¾
Plywood thickness 6/ 8 = ¾ *2.54 cm *10 mm = 19.05 mm
ID: 14104622
X
Y Z
M

4. Plywood thickness =19.05 mm , Sheathing load = 7.76 KN/m
• FbKS = 326 * 103
• Fslb/Q = 7.55 * 103
• EI = 2810 * 109
• Lbending = 3.16(
FbKS
!
)1/2 = 3.16(
326 ∗ 103
!.!#
)1/2 = 647.69 mm
• Lshear = 1.67(
Fslb/Q
!
) + 2d = 1.67(
7.55 ∗ 103
!.!#
) + (2*19.05) = 1662.91 mm
• Ldefection =
"#.%
&'''
(
EI
!
)1/3 =
"#.%
&'''
(
2810 ∗ 109
!.!#
)1/3 = 526.02 mm
• Ljoist = 526.02 mm /10 = 52.6 cm = 50 cm
Design of Sheathing

5. • Joist load = 7.76 * 0.5 = 3.88 KN/m
• Lbending =
&''
&'''
(
FbS
!
)1/2 =
&''
&'''
(
9998∗123900
$.%%
)1/2 = 1786.8 mm
• Lshear =
&.&&
&'''
(
Fv A
!
) + 2d =
&.&&
&'''
(
1276 ∗ 5323
$.%%
) + (2*140) = 2223.11 mm
• Ldefection =
"#.%
&'''
(
EI
!
)1/3 =
"#.%
&'''
(
11.7∗106 ∗ 8.656∗106
$.%%
)1/3 = 2189.17 mm
• Lstringer= 1786.17 mm /10 = 178.62 cm = 175 cm
Design of Joist 2”*6” Douglas Fir B= 38mm D=140mm

6. • Stringer load = 7.76 * 1.75 = 13.58 KN/m
• Lbending =
&''
&'''
(
FbS
!
)1/2 =
&''
&'''
(
9998∗215300
&$.'%
)1/2 = 1259.01 mm
• Lshear =
&.&&
&'''
(
Fv A
!
) + 2d =
&.&&
&'''
(
1276 ∗ 7016
&$.'%
) + (2*184) = 1099.75 mm
• Ldefection =
"#.%
&'''
(
EI
!
)1/3 =
"#.%
&'''
(
11.7∗106 ∗ 19.83∗106
&$.'%
)1/3 = 1900.75 mm
• Lshore = 1099.75 mm /10 = 109.975 cm = 105 cm
Design of Stringer 2”*8” Douglas Fir B= 38mm D=184mm

7. Ljoist =50 cm Lstringer= 175 cm Lshore = 105 cm
Check of bearing
1. Between Joist and Stringer :
Load= Design load * LJoist * Lstringer = 7.76 * 0.5 * 1.75= 6.79 KN
Area= BJoist * Bstringer = 38 * 38 = 1444 mm2
Stress = Load / Area = 6.79/ (1444*10-6) = 4702.22 KPa
> Fc⏊= 2655 KN/m2
“ Un safe ”
Lstringer new= 0.95 m

8. • Stringer load = 7.76 * 0.95 = 7.372 KN/m
• Lbending =
&''
&'''
(
FbS
!
)1/2 =
&''
&'''
(
9998∗215300
!.$!(
)1/2 = 1708.78 mm
• Lshear =
&.&&
&'''
(
Fv A
!
) + 2d =
&.&&
&'''
(
1276 ∗ 7016
!.$!(
) + (2*184) = 1715.96 mm
• Ldefection =
"#.%
&'''
(
EI
!
)1/3 =
"#.%
&'''
(
11.7∗106 ∗ 19.83∗106
!.$!(
)1/3 = 2330.04 mm
• Lshore = 1708.78 mm /10 = 170.88 cm = 170 cm
Design of Stringer 2”*8” Douglas Fir B= 38mm D=184mm

9. Check of bearing
2. Between Stringer and Shore :
Load= Design load * Lstringer* Lshore = 7.76 * 0.95 * 1.7 = 12.53 KN
Area= Bstringer * Bshore = 38 *89 = 3382 mm2
Stress = Load / Area = 12.53 / (3382*10-6) = 3704.91 KPa
> Fc⏊= 2655 KN/m2
“ Un safe ”
Lshore new= 1.2 m

10. Ljoist = 50 cm Lstringer= 95 cm Lshore = 120 cm
Check shore safety
Load= Design load * Lstringer * Lshore = 7.76 * 0.95 * 1.2= 8.85 KN
Area= Bshore * Bshore = 89 *89 = 7921 mm2
Stress = Load / Area = 8.85/ (7921*10-6) = 1117.28 KPa
< Fc = 6895 KN/m2
“ safe ”
Final Design:-

11. THANK YOU