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12
Concrete Form Design
12-1
DESIGN PRINCIPLES
The design of concrete formwork that has adequate strength to resist failure
and will not deflect excessively when the forms are filled is a problem in
structural design. Unless commercial forms are used, this will usually involve
the design of wall, column, or slab forms constructed of wood or plywood. In
such cases, after the design loads have been established, each of the primary
form components may be analyzed as a beam to determine the maximum
bending and shear stresses and the maximum deflection that will occur. Vertical supports and lateral bracing are then analyzed for compression and tension loads. The procedures and applicable equations are presented in this
chapter.
12-2
DESIGN LOADS
Wall and Column Forms
For vertical forms (wall and column forms), design load consists of the lateral pressure of the concrete against the forms. The maximum lateral pressure that concrete exerts against a form has been found to be a function of
the type of concrete, temperature of the concrete, vertical rate of placing,
and height of the form. For ordinary (150 lb/cu ft: 2403 kg/m3) internally
vibrated concrete, the American Concrete Institute (ACI) recommends the
use of the following formulas. to determine the maximum lateral concrete
pressure.
335
336
Chapter
·
12
For all columns and for walls with a vertical rate of placement of 7
ft/h (2.1 m/h) or less:
p
[P
= 150
+ 9000R
T
(12-lA)
= 7.2
+ 785R
(12-lB)
T + 18]
where p = lateral pressure (lb/sq ft or kPa)
R
T
h
= rate of vertical placement
= temperature (OF or °C)
= height of form (ft or m)
Maximum pressure
(ft/h or m/h)
= 3000 lb/sq ft (143.6 kPa)
for columns, 2000 lb/sq ft (95.8
kPa) for walls, or 150 h (23.6h), whicheveris least. Minimum pressure = 600
lb/sq ft (28.7 kPa).
·
For walls with a vertical rate of placement of 7 to 10 ft/h (2.1 to 3.0
m/h):
_ 150 43,400 2800R
p+ T
+ T
(12-2A)
1154
244R
[ P = 7.2 + T + 18 + T + 18 ]
(12-28)
Maximum pressure = 2000 lb/sq ft (95.8 kPa) or 150 h, whichever is less. Minimum pressure = 600 lb/sq ft (28.7 kPa)
·
For walls with a vertical rate of placement greater than 10 ft/h (3.1 m/h):
= 150 h
[p = 23.6 h]
pressure = 600 lb/sq ft
p
Minimum
(l2-3A)
(12-3B)
(28.7 kPa)
When forms are vibrated externally, it is recommended that a design load
twice that given by Equations 12-1 and 12-2 be used. When a retarder, pozzolan, or superplasticizer has been added to the concrete, Equation 12-3
should be used. When concrete is pumped into vertical forms from the bottom (both column and wall forms), Equation 12-3 should be used and a minimum additional pressure of 25% should be added to allow for pump surge
pressure.
Floor and Roof Slab Forms
The design load to be used for elevated slabs consists of the weight of concrete
and reinforcing steel, the weight of the forms themselves, and any live loads
(equipment, workers, material, etc.). For normal reinforced concrete, the
design load for concrete and steel is based on a unit weight of 150 lb/cu ft
12-2
Design Loads
(2403 kg/m3). The American Concrete Institute (AC!) recommends that a minimum live load of 50 lb/sq ft (2.4 kPa) be used for the weight of equipment,
materials, and workers. When motorized concrete buggies are utilized, the live
load should be increased to at least 75 lb/sq ft (3.6 kPa). Any unusual loads
would be in addition to these values. ACI also recommends that a minimum
design load (dead load plus live load) of 100 lb/sq ft (4.8 kPa) be used. This
should be increased to 125lb/sq ft (6.0 kPa) when motorized buggies are used.
Lateral Loads
Formwork must be designed to resist lateral loads such as those imposed by
wind, the movement of equipment on the forms, and the placing of concrete
into the forms. Such forces are usually resisted by lateral bracing whose
design is covered in Section 12-6. The minimum lateral design loads recommended for tied wall forms are given in Table 12-1. When form ties are not
used, bracing must be designed to resist the internal concrete pressure as
well as external loads.
For slab forms, the minimum lateral design load is expressed as follows:
H = 0.02 x dl x ws
(12-4)
where H
= lateral
force applied along the edge of the slab (lb/ft) [kN/m];
minimum value = 100 lb/ft [1.46 kN/m]
dl = design dead load (lb/sqft) [kPa]
ws = width of slab perpendicular to form edge (ft) [m]
In using Equation 12-4, design dead load includes the weight of concrete plus
formwork. In determining the value of ws, consider only that part of the slab
being placed at one time.
TABLE 12-1
Recommendedminimum lateral design load for
wall forms
Wall Height, h
(ft) em]
Design Lateral Force Applied
at Top of Form (lb/ft) [kN/m]
less than 8 [2.4]
hxwf*
2
8 [2.4] or over but
less than 22 [6.7]
100 [1.46] but
hxwf*
at least
:-
22 [6.7] or over
7.5 h [0.358 h] but
at least
hxwf*
:-
*wf = wind force prescribed by local code (lb/sq it) [kPa] but
minimum of 15lb/sq it [0.72 kPa]
337
338
Chapter
12-3
12
METHOD OF ANALYSIS
Basis of Analysis
After appropriate design loads have been selected, the sheathing, joists or
studs, and stringers or wales are analyzed in turn, considering each member
to be a uniformly loaded beam supported in one of three conditions (singlespan, two-span, or three-span or larger) and analyzed for bending, shear, and
deflection. Vertical supports and lateral bracing must be checked for compression and tension stresses. Except for sheathing, bearing stresses must be
checked at supports to ensure against crushing.
Using the methods of engineering mechanics, the maximum values
expressed in customary units of bending moment, shear, and deflection developed in a uniformly loaded, simply supported beam of uniform cross section
are given in Table 12-2.
The maximum fiber stresses (expressed in conventional units) developed
in bending, shear, and compression resulting from a specified load may be
determined from the following equations:
Bending:
M
(12-5)
fb=S
Shear:
fv = 1.~ V for rectangular wood members
(12-6)
fi
I
fv=~
Ib/Q or p ywood
(12-7)
Compression:
p
fc or fc.L
(12-8)
=A
Tension:
P
ft=A
(12-9)
where fb = actual unit stress for extreme fiber in bending (psi)
fc
= actual
unit stress in compression parallel to grain (psi)
= actual unit stress in compression perpendicular
ft = actual unit stress in tension (psi)
fv = actual unit stress in horizontal shear (psi)
A = section area (sq in.)
M = maximum moment (in.-Ib)
fc.L
P = concentrated load (lb)
S = section modulus (cu in.)
to grain (psi)
12-3
TABLE 12-2
Maximum bending, shear, and deflection in a uniformly
Support
Type
Conditions
2 Spans
3 Spans
wZ2
96
M=- wZ2
wZ2
M= 120
V = 5wZ
96
wZ4
Ll= 2220EI
V= wZ
20
wZ4
Ll = 1740EI
M=-
Shear (lb)
V= wZ
24
5wZ4
Ll
loaded beam
1 Span
Bending moment (in.-Ib)
Deflection (in.)
Method of Analysis
= 4608EI
96
Notation:
l
w
E
I
= length
of span
(in.)
= uniform load per foot of span
=
of inertia
elasticity(in.<p,si)
= modulus
moment of
)
Ob/ft)
v =maximum
Ib / Q
= rolling
shear (lb)
shear constant (sq in./ft)
Since the grain of a piece of timber runs parallel to its length, axial compressive forces result in unit compressive stresses parallel to the grain. Thus,
a compression force in a formwork brace (Figure 12-4) will result in unit compressive stresses parallel to the grain (fe)in the member. Loads applied to the
top or sides of a beam, such as a joist resting on a stringer (Figure 12-1b),
will result in unit compressive stresses perpendicular to the grain (fc.l)in the
beam. Equating allowable unit stresses in bending and shear to the maximum unit stresses developed in a beam subjected to a uniform load of w
pounds per linear foot [kN/m] yields the bending and shear equations of
Tables 12-3 and 12-3A.
When design load and beam section properties have been specified,
these equations may be solved directly for the maximum allowable span.
Given a design load and span length, the equations may be solved for the
required size of the member. Design properties for Plyform (plywood especially engineered for use in concrete formwork) are given in Table 12-4 and
section properties for dimensioned lumber and timber are given in Table
12-5. The properties of plywood, lumber, and timber are described in Section 13-2. However, typical allowable stress values for lumber are given in
Table 12-6. The allowable unit stress values in Table 12-6 (but not modulus
of elasticity values) may be multiplied by a load duration factor of 1.25 (7-day
load) when designing formwork for light construction and single use or very
limited reuse of forms. However, allowable stresses for lumber sheathing (not
Plyform) should be reduced by the factors given in Table 12-6 for wet conditions. The values for Plyform properties presented in Table 12-4 are based on
wet strength and 7-day load duration, so no further adjustment in these values is required.
339
340
Chapter 12
TABLE 12-3
Concrete form design equations
Support Conditions
Design Condition
Bending
Wood
1 Span
2 Spans
(F;b t2
1 = 4.0d(F;bt2
1= 4.46d (F;bt2
1= 9.8 (FSrl2
1= 9.8 (FSt2
1= 10.95(FSt2
1= 9.8 (Fb:;St2
1= 9.8 (Fb:;St2
1= 10.95(Fb:;St2
Wood
1= 16 FvA + 2d
w
1= 12.8 FvA + 2d
w
1= 13.3 FvA + 2d
w
Plywood
1= 24 F.Ib/Q + 2d
w
1= 19.2 F.Ib/Q + 2d
w
1= 20 F.Ib/Q + 2d
w
1= 5.51 (E
1= 6.86 (E
1= 6.46 (E
1 = 4.0d
Plywood
Shear
Deflection
If~=h8o
If ~ = Y240
If ~ = Y360
Compression
Tension
t4
t4
l = 1.72(t3
1= 2.31 (t3
l = 2.13 (t3
1= 1.57(t3
1= 2.10 (t3
1= 1.94(t3
1= 1.37(t3
P
te or tel.= A
P
ft =-A
1=1.83(t3
1= 1.69(t3
Notation:
= length of span, center to center of supports (in.)
l
Fb
FbKS
= allowable
= plywood
unit stress in bending (psi)
section capacity in bending (lb x in./ft)
Fe
= allowable unit stress in compression parallel to grain (psi)
FcJ.
= allowable unit stress in compression perpendicular to grain (psi)
F.Jb / Q = plywood section capacity in rolling shear (lb/ft)
=allowable unit stress in horizontal shear (psi)
Fv
fc
= actual unit stress in compression parallel to grain (psi)
feJ.
ft
A
E
I
P
S
~
b
d
w
3 or More Spans
= actual unit stress in compression perpendicular
= actual unit stress in tension (psi)
= area of section (in.2)*
=
modulus of elasticity
= moment
of inertia (in.(p,si)
)*
= applied force (compression or tension) (lb)
= section modulus (in.3)*
to grain
=deflection (in.)
= width of member
=depth of member
(in.)
(in.)
= uniform load per foot of span (lb/ft)
*For a rectangular member: A = bd, S = bd2/6,I = bd3/12.
(psi)
t4
12-3
341
Method of Analysis
TABLE 12-3A
Metric (SI) concrete form design equations
Support
Design Conditions
1 Span
Conditions
3 or More Spans
2 Spans
Bending
112
1 = 36.5 d Fbb
1000
W
Wood
112
1 = 36.5 d Fbb
1000
w
112
1 = 89.9 Fbb
1000 w
( )
( )
1 = 89.9 FbS
1000 W
Plywood
Shear
( )
( )
112
1 = 2.83(F~St2
1 = 2.83(F~St2
Wood
1 = 1.34 FvA
Plywood
1 = 2.00 Fslb/Q + 2d
1000 w +
2d
w
1 = 526
1000
If ~
= hao
If ~
= Y240
If ~
= Y360
EI!::I.
1I4
(EIw )
1 = 75.1
1000 (w )
EI
=
1 1000 (w )
EI
=
1 1000 (w )
p
Deflection
1I3
1I3
59.8
Compression
tc or tel.
1 = 1.11 FvA 2d
1000 w +
1= 1.60 Fslb/Q + 2d
w
EI~ 1I4
1 = 655
1000 w
= 101 EI 1I3
1 1000 w
= 91.7 EI 1I3
1 1000 w
1 = 1.67 Fslb/Q + 2d
w
EI~ 1I4
1 = 617
( )
( )
( )
= 79.9 EI
1 1000 (w )
1I3
68.5
1= 1.07 FvA + 2d
1000 w
1I3
=A
P
!e=-A
Tension
Notation:
I
Fb
= length
of span, center to center of supports (mm)
= allowable unit stress in bending (kPa)
FtJ{S
= plywood section capacity in bending (Nmmlm)
Fe
= allowable unit stress in compression parallel to grain (kPa)
FcJ.
= allowable unit stress in compression perpendicular to grain
FJb / Q = plywood section capacity in rolling shear (N/m)
= allowable unit stress in horizontal shear (kPa)
tv
te
tel.
ft
A
E
I
EI
P
S
li
b
= actual
unit
stress
in compression
parallel
to grain
= actual unit stress in compression perpendicular
=actual unit stress in tension (kPa)
= area of section (mm2)*
= modulus of elasticity (kPa)
= moment of inertia (mm4)*
= plywood stiffness capacity (kPamm4/m)
= applied force (compression or tension) (N)
= section modulus (mm3)*
= deflection (mm)
= width of member (mm)
d
= depth of member (mm)
w
= uniform load per meter of span (kPalm)
*For a rectangular member: A = bd, S = bd2/6,I
(kPa)
(kPa)
to grain (kPa)
= bd3/12
(EIw )
= 93.0
1 1000 (w )
= 84.7 EI
1 1000 (w )
= 73.8 EI
1 1000 (w )
1000
1I3
1I3
1I3
TABLE 12-4
continued
Face Grain Across
EI
Approx.
Weight
psf (kglm2)
Thickness
in. (mm)
Plyform
(12.5)
Structural
Face Grain
Supports
F
F ,)b/Q
EI
Parallel
to Supports
F
F ,Jb/Q
104 Ibin2
ft
103 Ibin
ft
103 Ib
ft
104 Ibin2
ft
103 Ibin
ft
103 Ib
ft
109 kPamm4
m
103 Nmm
m
103 N
m
109 kPamm4
m
103 Nmm
m
103 N
m
0.523
(194)
0.697
(258)
0.896
(332)
1.208
(448)
1.596
(592)
1.843
(683)
0.501
(7.31)
0.536
(7.83)
0.631
(9.21)
0.769
(11.22)
0.814
(11.88)
0.902
(13.16)
0.344
(127)
0.459
(170)
0.807
(299)
1.117
(414)
1.679
(622)
2.119
(785)
0.278
(4.06)
0.313
(4.57)
0.413
(6.02)
0.611
(8.92)
0.712
(10.39)
0.854
(12.47)
I
1.5 (7.3)
% (16)
1.8 (8.8)
% (19)
2.2 (10.7)
% (22)
2.6 (12.7)
1 (25.4)
3.0 (14.6)
1-% (28.5)
3.3 (16.1)
0.117
(1102)
0.196
(1850)
0.303
(2853)
0.475
(4477)
0.718
(6765)
0.934
(8798)
0.043
(410)
0.067
(636)
0.162
(1525)
0.268
(2528)
0.481
(4533)
0.711
(6694)
*All properties adjusted to account for reduced effectiveness of plies with grain perpendicular to applied stress. Stresses adjusted for wet conditions, load
duration, and experience factors. Based on APA data.
~
~
~
344
Chapter
TABLE 12-5
Section properties
Nominal
Size
(b x d)
in.
1x3
1x4
1x6
1x8
1 x 10
1 x 12
2x3
2x4
2x6
2x8
2 x 10
2 x 12
2 x 14
3x4
3x6
3x8
3 x 10
3 x 12
3 x 14
3 x 16
4x4
4x6
4x8
4 x 10
4 x 12
4 x 14
4 x 16
6x6
6x8
6 x 10
6 x 12
6 x 14
6 x 16
12
of U.S. standard
Actual Size
(S4S)
in.
0.75 x 2.5
0.75 x 3.5
0.75 x 5.5
0.75 x 7.25
0.75 x 9.25
0.75 x 11.25
1.5 x 2.5
1.5 x 3.5
1.5 x 5.5
1.5 x 7.25
1.5 x 9.25
1.5 x 11.25
1.5 x 13.25
2.5 x 3.5
2.5 x 5.5
2.5 x 7.25
2.5 x 9.25
2.5 x 11.25
2.5 x 13.25
2.5 x 15.25
3.5 x 3.5
3.5 x 5.5
3.5 x 7.25
3.5 x 9.25
3.5 x 11.25
3.5 x 13.25
3.5 x 15.25
5.5 x 5.5
5.5 x 7.5
5.5 x 9.5
5.5 x 11.5
5.5 x 13.5
5.5 x 15.5
mm
19 x 64
19 x 89
19 x 140
19 x 184
19 x 235
19 x 286
38 x 64
38 x 89
38 x 140
38 x 184
38 x 235
38 x 286
38 x 337
64 x 89
64 x 140
64 x 184
64 x 235
64 x 286
64 x 337
64 x 387
89x89
89 x 140
89 x 184
89 x 235
89 x 286
89 x 337
89 x 387
140 x 140
140 x 191
140 x 241
140 x 292
140 x 343
140 x 394
lumber
and timber
Area of Section
A
in.2
103 mm2
1.875
2.625
4.125
5.438
6.938
8.438
3.750
5.250
8.250
10.88
13.88
16.88
19.88
8.750
13.75
18.12
23.12
28.12
33.12
38.12
12.25
19.25
25.38
32.38
39.38
46.38
53.38
30.25
41.25
52.25
63.25
74.25
85.25
1.210
1.694
2.661
3.508
4.476
5.444
2.419
3.387
5.323
7.016
8.952
10.89
12.82
5.645
8.871
11.69
14.91
18.14
21.37
24.60
7.903
12.42
16.37
20.89
25.40
29.92
34.43
19.52
26.61
33.71
40.81
47.90
55.00
(b
= width,
d
= depth)
Section Modulus
S
. 3
ID.
105 mm3
0.7812
1.531
3.781
6.570
10.70
15.82
1.563
3.063
7.563
13.14
21.39
31.64
43.89
5.104
12.60
21.90
35.65
52.73
73.15
96.90
7.146
17.65
30.66
49.91
73.83
102.4
135.7
27.73
51.56
82.73
121.2
167.1
220.2
0.1280
0.2509
0.6196
1.077
1.753
2.592
0.2561
0.5019
1.239
2.153
3.505
5.185
7.192
0.8364
2.065
3.589
5.842
8.642
11.99
15.88
1.171
2.892
5.024
8.179
12.10
16.78
22.23
4.543
8.450
13.56
19.87
27.38
36.09
Moment of Inertia
I
in."
106 mm"
0.9766
2.680
10.40
23.82
49.47
88.99
1.953
5.359
20.80
47.63
98.93
178.0
290.8
8.932
34.66
79.39
164.9
296.6
484.6
738.9
12.50
48.53
111.1
230.8
415.3
678.5
1034
76.25
193.4
393.0
697.1
1128
1707
0.4065
1.115
4.328
9.913
20.59
37.04
0.8129
2.231
8.656
19.83
41.18
74.08
121.0
3.718
14.43
33.04
68.63
123.5
201.7
307.5
5.205
20.20
46.26
96.08
172.8
282.4
430.6
19.52
80.48
163.6
290.1
469.4
710.4
It should be noted that the deflection of plywood sheathing is precisely
computed as the sum of bending deflection and shear deflection. While the
deflection equations of Tables 12-3 and 12-3A consider only bending deflection, the modulus of elasticity values of Table 12-4 include an allowance
for shear deflection. Thus, the deflection computed using these tables is
sufficiently accurate in most cases. However, for very short spans (lid ratio
of less than 15) it is recommended that shear deflection be computed separately and added to bending deflection. See Reference 3 for a recommended
procedure.
12-4
Slab Form Design
345
TABLE 12-5
continued
Nominal
Size
(b xd)
in.
in.
mm
6 x 18
6 x 20
6 x 22
6 x 24
8x8
8 x 10
8 x 12
8 x 14
8 x 16
8 x 18
8 x 20
8 x 22
8 x 24
lOx 10
10 x 12
10 x 14
10 x 16
10 x 18
10 x 20
10 x 22
lOx24
12 x 12
12 x 14
12 x 16
12 x 18
12 x 20
12 x 22
12 x 24
14 x 14
14 x 16
14 x 18
14 x 20
14 x 22
14 x 24
5.5 x 17.5
5.5 x 19.5
5.5 x 21.5
5.5 x 23.5
7.5 x 7.5
7.5 x 9.5
7.5 x 11.5
7.5 x 13.5
7.5 x 15.5
7.5 x 17.5
7.5 x 19.5
7.5 x 21.5
7.5 x 23.5
9.5 x 9.5
9.5 x 11.5
9.5 x 13.5
9.5 x 15.5
9.5 x 17.5
9.5 x 19.5
9.5 x 21.5
9.5 x 23.5
11.5 x 11.5
11.5x 13.5
11.5x 15.5
11.5 x 17.5
11.5x 19.5
11.5x 21.5
11.5x 23.5
13.5 x 13.5
13.5 x 15.5
13.5 x 17.5
13.5 x 19.5
13.5 x 21.5
13.5 x 23.5
140 x 445
140 x 495
140 x 546
140 x 597
191 x 191
191 x 241
191 x 292
191 x 343
191 x 394
191 x 445
191 x 495
191 x 546
191 x 597
241 x 241
241 x 292
241 x 343
241 x 394
241 x 445
241 x 495
241 x 546
241 x 597
292 x 292
292 x 343
292 x 394
292 x 445
292 x 495
292 x 546
292 x 597
343 x 343
343 x 394
343 x 445
343 x 495
343 x 546
343 x 597
12-4
Method
Actual Size
(S4S)
Area of Section
A
in.2
103 mm2
96.25
107.2
118.2
129.2
56.25
71.25
86.25
101.2
116.2
131.2
146.2
161.2
176.2
90.25
109.2
128.2
147.2
166.2
185.2
204.2
223.2
132.2
155.2
178.2
201.2
224.2
247.2
270.2
182.2
209.2
236.2
263.2
290.2
317.2
62.10
69.19
76.29
83.39
36.29
45.97
55.65
65.32
75.00
84.68
94.36
104.0
113.7
58.23
70.48
82.74
95.00
107.3
119.5
131.8
144.0
85.32
100.2
115.0
129.8
144.7
159.5
174.4
117.5
135.0
152.4
169.8
187.3
204.7
Section Modulus
S
in.3
105 mm3
280.7
348.6
423.7
506.2
70.31
112.8
165.3
227.8
300.3
382.8
475.3
577.8
690.3
142.9
209.4
288.6
380.4
484.9
602.1
731.9
874.4
253.5
349.3
460.5
587.0
728.8
886.0
1058
410.1
540.6
689.1
855.6
1040
1243
46.00
57.12
69.44
82.96
11.52
18.49
27.09
37.33
49.21
62.73
77.89
94.69
113.1
23.42
34.31
47.29
62.34
79.46
98.66
119.9
143.3
41.54
57.24
75.46
96.19
119.4
145.2
173.4
67.20
88.58
112.9
140.2
170.4
203.6
Moment of Inertia
I
in.4
104 mm4
2456
3398
4555
5948
263.7
535.9
950.5
1538
2327
3350
4634
6211
8111
678.8
1204
1948
2948
4243
5870
7868
10274
1458
2358
3569
5136
7106
9524
12437
2768
4189
6029
8342
11181
14600
SLAB FORM DESIGN
of Analysis
The procedure for applying the equations of Table 12-3 and 12-3A to the
design of a deck or slab form is first to consider a strip of sheathing of the
specified thickness and 1 ft (or 1 m) wide (see Figure 12-1a). Determine in
turn the maximum allowable span based on the allowable values of bending
stress, shear stress, and deflection. The lower of these values will, of course,
1022
1415
1896
2476
109.8
223.0
395.7
640.1
968.8
1394
1929
2585
3376
282.5
501.2
810.7
1227
1766
2443
3275
4276
594.2
981.4
1485
2138
2958
3964
4276
1152
1744
2510
3472
4654
6077
346
Chapter
12
TABLE 12-6
Typical values of allowable stress for lumber
Allowable Unit Stress (psi)[kPa]
(Moisture Content = 19%)
Species
(No.2 Grade, 4 x 4
[100 x 100 mm] or smaller)
Douglas fir-larch
Hemlock-fir
Southern
pine
California redwood
Eastern
spruce
Reduction factor for
wet conditions
Load duration factor
(7-day load)
Fb
Fv
FcJ.
Fe
Ft
1450
[9998]
1150
185
[1276]
150
385
[2655]
245
1000
[6895]
800
850
[5861]
675
[7929]
[1034]
[1689]
[5516]
[4654]
1400
180
405
975
825
[9653]
[1241]
[2792]
[6723]
[5688]
1400
160
425
1000
800
[9653]
[1103]
[2930]
[6895]
[5516]
1050
140
255
700
625
[7240]
[965]
[1758]
[4827]
[4309]
E
1.7 x 106
[11.7
x 106]
1.4 x 106
[9.7
x 106]
1.6 x 106
[11.0
x 106]
1.3 X 106
[9.0
x 106]
1.2 x 106
[8.3
x 106]
0.86
0.97
0.67
0.70
0.84
0.97
1.25
1.25
1.25
1.25
1.25
1.00
determine the maximum spacing of the supports (joists). For simplicity and
economy of design, this maximum span vaJue is usuaJly rounded down to the
next lower integer or modular vaJue when selecting joist spacing.
Based on the selected joist spacing, the joist itself is analyzed to determine its maximum allowable span. The load conditions for the joist are illustrated in Figure 12-1b. The joist span selected will be the spacing of the
stringers. Again, an integer or moduJar value is selected for stringer spacing.
Based on the selected stringer spacing, the process is repeated to determine the maximum stringer span (distance between vertical supports or
shores). Notice in the design of stringers that the joist loads are actually
applied to the stringer as a series of concentrated loads at the points where
the joists rest on the stringer. However, it is simpler and sufficiently accurate
to treat the load on the stringer as a uniform load. The width of the uniform
design load applied to the stringer is equal to the stringer spacing as shown
in Figure 12-1c. The calcuJated stringer span must next be checked against
the capacity of the shores used to support the stringers. The load on each
shore is equaJ to the shore spacing muJtiplied by the load per unit length of
stringer. Thus the maximum shore spacing (or stringer span) is limited to the
lower of these two maximum values.
Although the effect of intermediate form members was ignored in determining allowable stringer span, it is necessary to check for crushing at the
point where each joist rests on the stringer. This is done by dividing the load at
12-4
Section
I
I
w
I---
Slab Form Design
Elevation
I
I
1 tt --1
(1m)
.
Joists
£1
w = design load (Ib/sq tt) [kN/m2]
w1 = 1 x w = w (Ib/tt) [kN/m]
a. Sheathing
Joists
Stringers
S2 = spacing of joists (tt) [m]
w2 = w X S2(Ib/tt) [kN/m]
b. Joists
Joist
Stringer
Shores
S3 = spacing of stringers (tt) [m]
w3 = w X S3 (Ib/ft) [kN/m]
c. Stringers
FIGURE 12-1
Design analysis of form members.
this point by the bearing area and comparing the resulting stress to the allowable unit stress in compression perpendicular to the grain. A similar procedure
is applied at the point where each stringer rests on a vertical support.
To preclude buckling, the maximum allowable load on a rectangular
wood column is a function of its unsupported length and least dimension (or
1/d ratio). The 1/d ratio must not exceed 50 for a simple solid wood column.
For 1/d ratios less than 50, the following equation applies:
F'
C
- 0.3 E
-
(l / d)2
c<: F
""
(12-10)
C
347
348
Chapter
12
where Fc = allowable unit stress in compression parallel to the grain (psi)
[kPa]
F~= allowable unit stress in compression parallel to the grain,
adjusted for 1/d ratio (psi) [kPa]
E = modulus of elasticity (psi) [kPa]
1/d = ratio of member length to least dimension
In using this equation, note that the maximum value used for F~ may not
exceed the value of Fc.
These design procedures are illustrated in the following example.
Sheathing design employing plywood is illustrated in Example 12-2.
EXAMPLE 12-1
Design the formwork (Figure 12-2) for an elevated concrete floor slab 6 in.
(152 mm) thick. Sheathing will be nominal I-in. (25-mm) lumber while
2 x 8 in. (50 x 200 mm) lumber will be used for joists. Stringers will be
4 x 8 in. (100 x 200 mm) lumber. Assume that all members are continuous
over three or more spans. Commercial 4000-lb (17.8-kN) shores will be
used. It is estimated that the weight of the formwork will be 5 lb/sq ft
(0.24 kPa). The adjusted allowable stresses for the lumber being used are
as follows:
Sheathing
psi [kPa]
1075 [7412]
174 [1200]
Fb
Fv
FcJ.
Fe
E
1.36 x 106
[9.4 X 106]
Other Members
psi [kPa]
1250
180
405
850
1.40
[9.7
[8619]
[1241]
[2792]
[5861]
X 106
X 106]
Maximum deflection of form members will be limited to 11360.Use the minimum value of live load permitted by ACI. Determine joist spacing, stringer
spacing, and shore spacing.
Solution
Design Load.
Assume concrete density is 150 lb/cu ft (2403 kg/m3)
Concrete
= 1 sq
ft x 6/12 ft x 150 lb/cu ft
=
75 lb/sq ft
Formwork
=
5 lb/sq ft
Live load = 50 lb/sq ft
Design load = 130 lb/sq ft
12-4
FIGURE 12-2
Example 12-1.
Slab Form Design
Slab form,
Sheathing
Plan
Pressure
per m2:
Concrete
= 1 x 0.152 x 9.8 x 2403/1000 = 3.58 kPa
Formwork
= 0.24
Live load
= 2.40 kPa
= 6.22 kPa
Design load
kPa
Deck Design.
Consider a uniformly loaded strip of decking (sheathing) 12
in. (or 1 m) wide placed perpendicular to the joists (Figure 12-1a) and analyze it as a beam. Assume that the strip is continuous over three or more
spans and use the appropriate equations of Table 12-3 and 12-3A.
349
350
Chapter 12
w
[w
= (1 sq ftJlin ft) x (130 lb/sq ft) = 130 lb/ft
= (1 m2/m) x (6.21 kN/m2) = 6.21 kN/m]
(a) Bending:
1 = 4.46 d (F~b)1I2
= (4.46)
d Fbb
1 =~
1000
(19)
1000
in.
112
(w )
(
= (40.7)
[
= 33.3
(0.75) (1071~0(12)rf2
7412) (1000»
1/2
6.22
)
= 844
mm
]
(b) Shear:
1 = 13.3 FvA + 2d
w
_ (13.3) (174) (12) (0.75) + (2) (0.75)
l
130
=~
= 161.7 in.
FvA + 2d
1000 w
[ -
(1.11) (1200) (1000) (19) + (2) (19)
(1000) (6.22)
(c) Deflection:
= 4107
mm
]
Ebd3
1 = 1.69
(~t3 = 1.69 (W 12 )
1/3
1.36 x 106) (12) (0.75)3
(130) (12)
( EI
_~
l- 1000 ( )
_ 73.8
(
[ = 1.69
1/3
W
=
)
Ebd3
=~
1000 (w12 )
) = 703 mm ]
27.7 in.
1/3
(9.4 x 106) (1000) (19)3
1000
1/3
1/3
(12) (6.22)
Deflection governs in this case and the maximum allowable span is 27.7 in.
(703 mm). We will select a 24-in. (610-mm) joist spacing as a modular value
for the design.
Joist design.
Consider the joist as a uniformly loaded beam supporting a
strip of design load 24 in. (610 mm) wide (same as joist spacing; see Figure
12-1b). Joists are 2 x 8 in. (50 x 200 mm) lumber. Assume that the joists are
continuous over three spans.
w = (2 ft) x (1) x (130 lb/sq ft) = 260 lb/ft
[ w = (0.610 m) x (1) x (6.22 kPa) = 3.79 kN/m]
12-4
Slab Form Design
(a) Bending:
(F:YI2
1 = 10.95
= 10.95 ((1250~i~3.14»)1/2= 87.0
1= ...!QQ...
1000
[
=
F bS
(w )
(
(8619) (2.153 x 105)
3.79
100
1000
in.
1/2
) = 2213 mm ]
1/2
(b) Shear:
1 = 13.3 FvA + 2d
w
= (13.3)
= 114.7 in.
(180) (10.88) + (2) (7.25)
1 =1Jl FvA + 2d
1000 w
[ -_
1.11 (1241) (7016) + (2)(184) = 2918 mm
1000
3.79
]
(c) Deflection:
1 = 1.69
(~t3
(1.4 x 106)(47.63»
(
I: 1~:98EI
1/3
(w )
=
1000 (
260
)
113
= 107.4
in.
1000
73.8
[
9.7 x 106) (19.83 X 106»
3.79
)
1/3
= 2732 mm
]
Thus bending governs and the maximum joist span is 87 in. (2213 mm). We
will select a stringer spacing (joist span) of 84 in. (2134 mm).
Stringer Design.
To analyze stringer design, consider a strip of design
load 7 ft (2.13 m) wide (equal to stringer spacing) as resting directly on the
stringer (Figure 12-1c). Assume the stringer to be continuous over three
spans. Stringers are 4 x 8 (100 x 200 mm) lumber. Now analyze the stringer
as a beam and determine the maximum allowable span.
= (7) (130) = 910 lb/ft
[w = (2.13) (1) (6.22) = 13.25
w
kN/m]
351
352
Chapter
12
(a) Bending:
112
FbS
( )
(
)=
: 1;~:5 FbS
1 1000 ( w )
_ 100
13.25
[ - 1000 (
)
1 = 10.95 --;;-
1250) (30.66)
112
71.1 in.
~210
(8619) (5.024 x 105)
112
= 1808 mm
.
]
(b) Shear:
l= 13.3 FvA + 2d
w
= (13.3)
l
=l::!l
1000
[ --
1.11
1000
(180) (25.38) + (2) (7.25)
910
= 81.3
in.
FvA + 2d
w
(1241) (16.37 x 103) + (2) (184)
13.25
= 2070
mm
J
(c) Deflection:
1 = 1.69
(~t3
= 1.69 (1.4 x l~:b (111.1)t3 = 93.8
EI
l =~
1000 w
[
= 73.8
1000
in.
1I3
()
(
9.7 x 106) (46.26 X 106)
13.25
)
113
= 2388
mmJ
Bending governs and the maximum span is 71.1 in. (1808 mm).
Now we must check shore strength before selecting the stringer span
(shore spacing). The maximum stringer span based on shore strength is equal
to the shore strength divided by the load per unit length of stringer.
4000
.
1 = 910 x 12 = 52.7 m.
[l = 1~~285 = 1.343m]
Thus the maximum stringer span is limited by shore strength to 52.7 in.
(1.343 m). We select a shore spacing of 4 ft (1.22 m) as a modular value.
12-5
Wall and Column Form Design
Before completing our design, we should check for crushing at the point
where each joist rests on a stringer. The load at this point is the load per unit
length of joist multiplied by the joist span.
P = (260) (84/12) = 1820 lb
[P = (3.79) (2.134) = 8.09 kN]
Bearing area (A) = (1.5)(3.5)
[A = (38) (89)
fc.L =~
[fc.L=
= ;~:~ = 347
8.09 x 106
3382
= 5.25 sq in.
= 3382 mm2 ]
OK
psi < 405 psi (Fc.L)
2392 kPa < 2792 kPa
(fc.L)
]
Final Design.
Decking: nominal1-in. (25-mm) lumber
Joists: 2 x 8's (50 x 200-mm) at 24-in. (610-mm) spacing
Stringers: 4 x 8's (100 x 200-mm) at 84-in. (2.13-m) spacing
Shore: 4000-lb (17.8-kN) commercial shores at 48-in. (1.22-m) intervals
12-5
WALLAND COLUMN FORM DESIGN
Design Procedures
The design procedure for wall and column forms is similar to that used for
slab forms substituting studs for joists, wales for stringers, and ties for
shores. First, the maximum lateral pressure against the sheathing is determined from the appropriate equation (Equation 12-1, 12-2, or 12-3). If the
sheathing thickness has been specified, the maximum allowable span for the
sheathing based on bending, shear, and deflection is-the maximum stud spacing. If the stud spacing is fixed, calculate the required thickness of sheathing.
Next, calculate the maximum allowable stud span (wale spacing) based
on stud size and design load, again considering bending, shear, and deflection.
If the stud span has already been determined, calculate the required size of
the stud. After stud size and wale spacing have been determined, determine
the maximum allowable spacing of wale supports (tie spacing) based on wale
size and load. If tie spacing has been preselected, determine the minimum
wale size. Double wales are commonly used (see Figure 12-3) to avoid the
necessity of drilling wales for tie insertion.
Next, check the tie's ability to carry the load imposed by wale and tie
spacing. The load (lb) [kN] on each tie is calculated as the design load (lb/sq
ft) [kPa] multiplied by the product of tie spacing (ft) [m] and wale spacing
353
354
Chapter 12
Wale
Stud
Tie/
T
Stud
spacing
,>:~'.f,
,.»"
.".> c'.
. . ~.." .'
. -: :.,:>
..
r'
. .Po:
.: ..0
~
T
Wale
spacing
.;:/:f'~":
J..:~ .
.~
': ',...o.~.::
1
I
:~.:: <\'-:':
:
~~<.o~
:>.
:'.':~
',".:
':'~.'.".,'
T
Tie
"'O~''':
'4,:" I~:
[it
.::<.~.:;.
. 0 .',.'
~ q...:.
I, ')~.
I
Plan
Section
FIGURE 12-3
Wall form, Example 12-2.
(£1)[m] . If the load exceeds tie strength, a stronger tie must be used or the
tie spacing must be reduced.
The next step is to check bearing stresses (or compression perpendicular
to the grain) where the studs rest on wales and where tie ends bear on wales.
Maximum bearing stress must not exceed the allowable compression stress
perpendicular to the grain or crushing will result. Finally, design lateral bracing to resist any expected lateral loads, such as wind loads.
EXAMPLE 12-2
Forms are being designed for an 8-ft (2.44-m) high concrete wall to be poured
at a rate of 4 ftJh (1.219 mIh), internally vibrated, at 900 F (320 C) temperature (see Figure 12-3). Sheathing will be 4 x 8 ft (1.2 m x 2.4 m) sheets of
12-5
Wall and Column Form Design
5/8-in. (16-mm) thick Class I Plyform with face grain perpendicular to studs.
Studs and double wales will be 2 x 4 in. (50 x 100 mm) lumber. Snap ties are
3000-lb (13.34-kN) capacity with 1-l/2-in. (38-mm) wide wedges bearing on
wales. Deflection must not exceed 1/360. Determine stud, wale, and tie spacing. Use Plyform section properties and allowable stresses from Table 12-4
and lumber section properties from Table 12-5. Allowable stresses for the
lumber being used for studs and wales are:
Fb = 1810 psi (12480 kPa)
Ff} = 120 psi (827 kPa)
Fc.L= 485 psi (3340 kPa)
E = 1.7 X 106 psi (11.7 x 106 kPa)
Solution
Design Load.
p = 150 + 9000R
T
[
P
=7 2
785R
. + T + 18
= 150
=7 2
.
+ (9000)
90 (4)
+
= 550 lb/sq ft
(785) (1.219)
32 + 18
= 26 3 kNl
.
2
m
Check: 550 < 150 h (= 1200 lb/sq ft) < 2000 lb/sq ft
[26.3 < 23.6 h (= 57.5 kN/m2) < 95.8 kN/m2 ]
Select Stud Spacing
(Three or More Spans).
Material: 5/8-in. (16-mm) Class I Plyform (Table 12-4)
Consider a strip 12 in. wide (or 1 m wide):
w
[w
(a) Bending:
= 1 x 1 x 550 = 550
=1 x
1 x 26.3
= 26.3
lb/ft
kN/m ]
]
OK
355
356
Chapter 12
(b) Shear:
Z = 20 Fslb/Q
w
+ 2d
x 103) + (2) (5/8)
-_ (20) (0.412
550
= 14.98 + 1.25 = 16.23
Z = 1.67 FJb/Q
in.
+ 2d
w
x 103) + (2) (16)
-_ (1.67) (6.01
26.3
= 382
+ 32
= 414 rom
(c) Deflection:
Z = 1.69
(~t3
= (1.69)
z=~
[
(0.19:5~ 106t3
EI 1I3
1000 w
= 73.8 1836 x 109
1000
26.3
( )
(
Deflection governs. Maximum span
mm) stud spacing.
)
113
= 12..0 in.
= 304
= 12.0 in.
rom
]
(304 mm). Use a 12-in. (304-
Select Wale Spacing (Three or More Spans).
Since the stud spacing is
12 in. (304 mm), consider a uniform design load 1 ft (304 rom) wide resting on
each stud.
w
= 1 x 1 x 550 = 550 lb/ft
304
[w = 1000 x 1 x 26.3 = 8.0 kN/m]
(a) Bending:
Z = 10.95 (F~Srl2
= (10.95) (1810~~~.063»)1/2 = 34.77
z= 1000
...!QQ.. FbS
in.
112
(w )
Z= 100
[ 1000 (
(12480) (0.5019 x 105)
8.0
)
1/2
= 885 mm
]
12-5
Wall and Column Form Design
(b) Shear:
Z = 13.3 FuA + 2d
w
- (13.3) (120) (5.25) + (2) (3.5) = 22.23 in.
550
z= 1:.!l
FuA + 2d
1000 w
[ =
(1.11) (827) (3.387 x 103) + (2) (88.9)
(1000) (8.0)
= 566
mm
J
(c) Deflection:
Shear governs, so maximum span (wale spacing) is 22.23 in. (566 mm). Use a
16-in. (406-mm) wale spacing for modular units.
Select Tie Spacing (Three or More Spans, Double Wales).
wale spacing of 16 in. (406 mm):
16
w =12 x 550 = 733.3 lb/ft
406
[w = 304 x 8.0 = 10.68 kN/m ]
(a) Bending:
Z = 10.95 (F~St2
= 10.95
z= .lQQ...
1000
((1810) 7C:3~33.063) t2
FbS
= 42.58
(w )
= 100
1000 (
(12480) (2 x 0.5019 x 105)
[
in.
1/2
10.68
)
112
= 1083
mmJ
Based on a
357
358
Chapter
12
(b) Shear:
I = 13.3 FvA + 2d
w
= (13.3)(120)(2
733.3
x 5.25) + (2)(3.5)
= 29.85
in.
I =lJl FvA + 2d
1000 w
x 3.387
[ = (1.11)(827)(2
(1000)(10.68)
= 760 mm
x 103) + (2)(88.9)
]
(c) Deflection:
1=1.69(~r3
= 1 69 (1.7 x 106)(2 x 5.359) 1/3= 49 32 .
(
.
=~
EI
w
()
= 73.8
[ 1000 (
I
1000
)
733.3
1/3
.
(11. 7 x 106)(2 x 2.231 x 106)
10.68
m.
)
1/3
= 1252
mm]
Shear governs. Maximum span is 29.85 in. (760 mm). Select a 24-in. (610mm) tie spacing for a modular value.
(d) Check tie load:
P
= wale
spacing x tie spacing x p
16 24
= -12 -12
(550)
= 14671b/tIe. < 3000 lb
OK
(406)(610)
[P
= (1000)(1000)
x 26.3
= 6.51
kN < 13.34 kN ]
Check Bearing.
(a) Stud on wales:
Bearing area (A) (double wales)
[A
Load at each panel
point (P)
= loadlft(m)
= (2)(1.5)(1.5) = 4.5 sq in.
= (2)(38)(38) = 2888
of stud x wale spacing (ft/m)
16
= 733 lb
12
406
P = (550) [P
= (8.0)
mm2)
1000 = 3.25 kN ]
12-6
j:
= AP = 733
= 163
4.5
j:
= 3.252888
X 106 = 1125 kPa
Icl.
[I cl.
Design of Lateral Bracing
psi < 485 psi (FcJ)
OK
< 3340 kPa (FCJ)
]
(b) Tie wedges on wales:
Tie load (P)
= 1467 lb [6.51 kN]
Bearing area (A) = (1.5)(1.5)(2) = 4.5 sq in.
[A = (38)(38)(2) = 2888 mm2]
= ~ = 1:.~7 = 326 psi < 485 psi (Fcl.)
6.51 x 106
fc.l=
2888
= 2254 kPa < 3340 kPa (Fcl.)]
[
OK
fc.l-
Final Design.
Sheathing: 4 x 8 ft (1.2 x 2.4 m) sheets of 5/8-in. (16-mm) Class I Plyform
placed with the long axis horizontal.
Studs: 2 x 4's (50 x 100 mm) at 12 in. (304 mm) on center.
Wales: Double 2 x 4's (50 x 100 mm) at 16 in. (406 mm) on center.
Ties: 3000-lb (13.34-kN) snap ties at 24 in. (610 mm) on center.
12-6
DESIGN OF LATERAL BRACING
Many failures of formwork have been traced to omitted or inadequately
designed lateral bracing. Minimum lateral design load values were given in
Section 12-2. Design procedures for lateral bracing are described and illustrated in the following paragraphs.
Lateral
Braces
for Wall and Column Forms
For wall and column forms, lateral bracing is usually provided by inclined
rigid braces or guy-wire bracing. Since wind loads, and lateral loads in general, may be applied in either direction perpendicular to the face of the form,
guy-wire bracing must be placed on both sides of the forms. When rigid braces
are used they may be placed on only one side of the form if designed to resist
both tension and compression forces. When forms are placed on only one side
of a wall with the excavation serving as the second form, lateral bracing must
be designed to resist the lateral pressure of the concrete as well as other lateral forces.
359
360
Chapter
12
Inclined bracing will usually resist any wind uplift forces on vertical
forms. However, uplift forces on inclined forms may require additional consideration and the use of special anchors or tiedowns. The strut load per foot of
form developed by the design lateral load can be calculated by the use of
Equation 12-11. The total load per strut is then P' multiplied by strut spacing.
P' = H x h x 1
h' x l'
1 = (h,2 + l'2)1/2
( 12-11 )
(12-12)
where P' = strut load per foot of form (lb/ft) [kN/m]
H
= lateral
load at top of form (lb/ft) [kN/m]
of form (ft) [m]
h' = height of top of strut (ft) [m]
1 = length of strut (ft) [m]
h
= height
l' = horizontal distance from form to bottom of strut (ft) [m]
If struts are used on only one side of the form, the allowable unit stress for
strut design will be the lowest of the three possible allowable stress values
(Fe, Fe, or Ft).
EXAMPLE 12-3
Determine the maximum spacing of nominal 2 x 4 in. (50 x 100 mm) lateral
braces for the wall form of Example 12-2 placed as shown in Figure 12-4.
Assume that local code wind requirements are less stringent than Table 12-1.
Allowable stress values for the braces are as follows.
Allowable Stress
psi
Fe
Ft
E
850
725
1.4 x 106
kPa
5861
4999
9.7 X 106
Solution
Determine the design lateral force per unit length of form.
H = 100 lb/ft (Table 12-1)
[H = 1.46 kN/m (Table 12-1)]
Determine the length of the strut using Equation 12-12.
1 = (h,2 + l'2)1/2
= (62 + 52)112
= 7.81 it
[l = (1.832 + 1.532)1/2= 2.38 m]
The axial concentrated load on the strut produced by a unit length of form
may now be determined from Equation 12-1.
12-6
Design of Lateral Bracing
H
-------
8ft
(2.44 m)
r
.
6 ft _
(1.83
m)
~5ft(1.53m)-1
FIGURE 12-4
Wall form bracing, Example 12-3.
P' = H x h x I =
= 208 3 lb/ft f fi
.
0 orm
(1008)(8)(7.81)
h' x l'
(6)(5)
p' = (1.46)(2.44)(2.38) = 3 03 kN/
(1.83)(1.53)
.
m]
[
Next, we determine the allowable compressive stress for each strut using
Equation 12-10. To do this, we must determine the lid ratio of the strut.
lid
= (7.8:.~12) = 62.5>
50
[lid
= (2.3~~.~000) = 62.5
> 50]
Since the lid ratio exceeds 50, each strut must be provided lateral bracing to
reduce its unsupported length. Try a single lateral support located at the mid-
point of each strut, reducing I to 46.9 in. (1.19m).
F, , _ 0.3 E _ (0.3)(1.4 X 106)
-
c
D
(lId)2
-
, = (0.3)(9.7
[ L'C
430
(46.9/1.5)2
X 106)
(1190/38)2
2 97 kP
.
a]
As Fc' < Ft < Fc, the value of Fc' governs.
The maximum allowable compressive force per strut is:
P
[p
= (1.5)(3.5)(430) = 2257 lb
=
(38)(8~6~2967)
= 10.03
.
pSI
kN]
361
362
Chapter
12
Thus maximum strut spacing is:
p
2257
s = p' = 208.3 = 10.8ft
10.03
s
=
3.03 = 3.31 m]
[
Keep in mind that this design is based on providing lateral support to each
strut at the midpoint of its length.
Lateral
Braces for Slab Forms
For elevated floor or roof slab forms, lateral bracing may consist of cross
braces between shores or inclined bracing along the outside edge of the form
similar to that used for wall forms. The following example illustrates the
method of determining the design lateral load for slab forms.
EXAMPLE 12-4
Determine the design lateral force for the slab form 6 in. (152 mm) thick, 20
ft (6.1 m) wide, and 100 ft (30.5 m) long shown in Figure 12-5. The slab is to
be poured in one pour. Assume concrete density is 150 lb/cu ft (2403 kg/m3)
and that the formwork weighs 15 lb/sq ft (0.72 kPa).
Solution
Dead load = (1/2)(1)(150) + 15 = 90 lb/sq ft
[dl
H
= (0.152)(~~~~03)(9.8)
= 0.02 x dl x ws
+ 0.72
= 4.30 kpa]
For the 20-ft (6.1-m) face, the width of the slab is 100 ft (30.5 m).
= (0.02)(90)(100) = 180 lb/lin ft
[H = (0.02)(4.30)(30.5) = 2.62 kN/m]
H20
For the 100-ft (30.5-m) face, the width of the slab is 20 ft (6.1 m).
HIOO= (0.02)(90)(20)
= 36 lb/ft < 100 lb/ft (minimum value)
[H = (0.02)(4.29)(6.1) = 0.52 kN/m < 1.46 kN/ml
Therefore, use HIOO= 100 lb/ft (1.46 kN/m) = minimum load.
FIGURE 12-5 Slab form
bracing design, Example 12-4.
-r
r
100 tt (30.5 m)
1 .......
20 tt
(6.1 m)
...L
J
~
~
~
~
~
~
~
H100 Ib/tt (kN/m)
~
~
--
~.......
H20 Ib/tt (kN/m)
Problems
PROBLEMS
1. Calculate the appropriate design load due to concrete pressure for a wall
form 9 ft (2.75 m) high to be poured at a rate of 6.5 ft (2.0 m) per hour at
a temperature of 68°F (20°C).
2. What should be the design load due to concrete pressure for a column
form 15 ft (4.9 m) high that is to be filled by pumping concrete into the
form from the bottom?
3. Determine the proper design load for a slab form 9 in. (229 mm) thick
using motorized buggies for placing concrete. The formwork is estimated
to weigh 10 lb/sq ft (48.8 kg/m2).
/4.
Calculate the maximum allowable span for 3/4-in. (19-mm) Class I Plyform decking with face grain across supports carrying a design load of
150 lb/sq ft (7.2 kPa). Assume that the decking is continuous over three
or more spans and limit deflection to 1/240 of span length.
S. Determine the maximum allowable spacing of nominal 2 x 4 in. (50 x 100
mm) studs for a wall form sheathed with nominal I-in. (25-mm) Hem-fir
lumber. The design load is 500 lb/sq ft (23.9 kPa). Assume that the
sheathing is continuous over three or more spans. Use the allowable
stresses of Table 12-6 with 7-day load and wet conditions. Limit deflection to 1/360 of span length.
6. Determine the maximum allowable span of nominal 2 x 4 in. (50 x 100
mm) wall form studs carrying a design load of 1000 lb/sq ft (47.9 kPa).
The stud spacing is 16 in. (406 mm) on center. Use the allowable stresses
for Douglas Fir from Table 12-6 and 7-day load duration. Assume that
studs are continuous over three or more spans. Limit deflection to 1/360
of span length. Based on the maximum allowable span, check for crushing of studs on double 2 x 4 in. (50 x 100 mm) wales.
7. Design the formwork for a wall 8 ft (2.44 m) high to be poured at the
rate of 5 ft/h (1.53 m/h) at a temperature of 77°F (25°C). Sheathing will
be 3/4-in. (19-mm) Class I Plyform with face grain across supports. All
lumber is Southern Pine. Use nominal 2 x 4 in. (50 x 100 mm) studs,
double 2 x 4 in. (50 x 100 mm) wales, and 3000-lb (13.3-kN) snap ties.
Lateral wood bracing will be attached at a height of 5 ft (1.53 m) above
the form bottom and anchored 4 ft (1.22 m) away from the bottom of the
form. Use the allowable stresses of Table 12-6 adjusted for 7-day load.
Limit deflection to 1/360 of the span length. Design wind load is 25 lb/sq
ft (1.2 kPa).
~8.
Design the formwork for a concrete slab 8 in. (203 mm) thick whose net
width between beam faces is 15 ft (4.58 m). Concrete will be placed using
a crane and bucket. The formwork is estimated to weigh 5 lb/sq ft (24.1
kg/m2). Decking will be 3/4-in (19-mm) Class I Plyform with face grain
across supports. All lumber will be Southern Pine. Joists will be nominal
2-in. (50-mm) wide lumber. One 4-in. (lOO-mm) wide stringer will be
placed between the beam faces. Limit deflection to 1/360 of span length.
Commercial shores of 4500 lb (20 kN) capacity will be used. Lateral support will be provided by the beam forms.
363
364
Chapter 12
9. Derive the equations for bending given in Table 12-3 using the relationships of Table 12-2 and Equation 12-5.
10. Develop a computer program to calculate the maximum safe span of a
wood formwork member as determined by bending, shear, and deflection
based on the equations of Table 12-3 or 12-3A. Input should include support conditions, allowable deflection, allowable stresses, load per unit
length of span, and dimensions of the member.
REFERENCES
1. American Institute Of Timber Construction. Timber Construction Manual,
3rd ed. New York: Wiley, 1985.
2. American Plywood for Concrete Forms. APA-The Engineered WoodAssociation, Tacoma, WA, 1992.
3. Concrete Forming. APA-The Engineered Wood Association, Tacoma, WA,
1994.
4. Design of Wood Formwork for Concrete Structures (Wood Construction
Data No.3). American Forest and Paper Products Association, Washington, D.C., 196!.
5. Guide to Formwork for Concrete (ACI 347R-94). American Concrete Institute, Detroit, MI, 1994.
6. Hurd, M. K. Formwork for Concrete, 6th ed. American Concrete Institute,
Detroit, MI, 1995.
7. National Design Specification for Wood Construction. American Forest and
Paper Products Association, Washington, D.C., 1982.
8. Peurifoy, Robert L., and Garold D. Oberlender. Formwork for Concrete
Structures, 3rd ed. New York: McGraw-Hill, 1996.
9. Plywood Design Specifications. APA-The Engineered Wood Association,
Tacoma, WA, 1995.