12Concrete Form Design

12-1 DESIGN PRINCIPLES

The design of concrete formwork that has adequate strength to resist failureand will not deflect excessively when the forms are filled is a problem instructural design. Unless commercial forms are used, this will usually involvethe design of wall, column, or slab forms constructed of wood or plywood. Insuch cases, after the design loads have been established, each of the primaryform components may be analyzed as a beam to determine the maximumbending and shear stresses and the maximum deflection that will occur. Ver-tical supports and lateral bracing are then analyzed for compression and ten-sion loads. The procedures and applicable equations are presented in thischapter.

12-2 DESIGN LOADS

Wall and Column Forms

For vertical forms (wall and column forms), design load consists of the lat-eral pressure of the concrete against the forms. The maximum lateral pres-sure that concrete exerts against a form has been found to be a function ofthe 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) internallyvibrated concrete, the American Concrete Institute (ACI) recommends theuse of the following formulas. to determine the maximum lateral concretepressure.

335

336 Chapter 12

· For all columns and for walls with a vertical rate of placement of 7ft/h (2.1 m/h) or less:

p = 150 + 9000RT

[= 7.2 + 785R

](12-lB)

P T + 18

where p =lateral pressure (lb/sq ft or kPa)R = rate of vertical placement (ft/h or m/h)T = temperature (OF or °C)h = height of form (ft or m)

Maximum pressure =3000 lb/sq ft (143.6 kPa) for columns, 2000 lb/sq ft (95.8kPa) for walls, or 150h (23.6h), whicheveris least. Minimumpressure =600lb/sq ft (28.7 kPa).

(12-lA)

· For walls with a vertical rate of placement of 7 to 10 ft/h (2.1 to 3.0m/h):

_ 150 43,400 2800Rp- + T + T

[

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. Min-imum 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):

(12-2A)

p = 150 h

[p =23.6 h]

Minimum pressure = 600 lb/sq ft (28.7 kPa)

When forms are vibrated externally, it is recommended that a design loadtwice that given by Equations 12-1 and 12-2 be used. When a retarder, poz-zolan, or superplasticizer has been added to the concrete, Equation 12-3should be used. When concrete is pumped into vertical forms from the bot-tom (both column and wall forms), Equation 12-3 should be used and a min-imum additional pressure of 25% should be added to allow for pump surgepressure.

(l2-3A)

(12-3B)

Floor and Roof Slab Forms

The design load to be used for elevated slabs consists of the weight of concreteand reinforcing steel, the weight of the forms themselves, and any live loads(equipment, workers, material, etc.). For normal reinforced concrete, thedesign 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 min-imum 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 liveload should be increased to at least 75 lb/sq ft (3.6 kPa). Any unusual loadswould be in addition to these values. ACI also recommends that a minimumdesign load (dead load plus live load) of 100 lb/sq ft (4.8 kPa) be used. Thisshould 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 bywind, the movement of equipment on the forms, and the placing of concreteinto the forms. Such forces are usually resisted by lateral bracing whosedesign is covered in Section 12-6. The minimum lateral design loads recom-mended for tied wall forms are given in Table 12-1. When form ties are notused, bracing must be designed to resist the internal concrete pressure aswell 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 plusformwork. In determining the value of ws, consider only that part of the slabbeing placed at one time.

TABLE 12-1Recommendedminimum lateral design load forwall forms

Wall Height, h(ft) em]

less than 8 [2.4]

Design Lateral Force Appliedat Top of Form (lb/ft) [kN/m]

hxwf*2

100 [1.46] but

hxwf*at least :-

8 [2.4] or over butless than 22 [6.7]

22 [6.7] or over 7.5 h [0.358 h] but

hxwf*at least :-

*wf =wind force prescribed by local code (lb/sq it) [kPa] butminimum of 15lb/sq it [0.72 kPa]

337

338 Chapter 12

12-3 METHOD OF ANALYSIS

Basis of Analysis

After appropriate design loads have been selected, the sheathing, joists orstuds, and stringers or wales are analyzed in turn, considering each memberto be a uniformly loaded beam supported in one of three conditions (single-span, two-span, or three-span or larger) and analyzed for bending, shear, anddeflection. Vertical supports and lateral bracing must be checked for com-pression and tension stresses. Except for sheathing, bearing stresses must bechecked at supports to ensure against crushing.

Using the methods of engineering mechanics, the maximum valuesexpressed in customary units of bending moment, shear, and deflection devel-oped in a uniformly loaded, simply supported beam of uniform cross sectionare given in Table 12-2.

The maximum fiber stresses (expressed in conventional units) developedin bending, shear, and compression resulting from a specified load may bedetermined from the following equations:

Bending:M

fb=S (12-5)

Shear:

fv = 1.~ V for rectangular wood members (12-6)

fv=~ fi IIb/Q or p ywood

(12-7)

Compression:

pfc or fc.L = A (12-8)

Tension:

Pft=A (12-9)

where fb = actual unit stress for extreme fiber in bending (psi)fc = actual unit stress in compression parallel to grain (psi)

fc.L =actual unit stress in compression perpendicular to grain (psi)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)P =concentrated load (lb)S =section modulus (cu in.)

12-3 Method of Analysis

TABLE 12-2Maximum bending, shear, and deflection in a uniformly loaded beam

Notation:l =length of span (in.)w =uniform load per foot of span Ob/ft)

E =modulus of elasticity <p,si)I = moment of inertia (in. )

v =maximum shear (lb)Ib / Q =rolling shear constant (sq in./ft)

Since the grain of a piece of timber runs parallel to its length, axial com-pressive 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 com-pressive stresses parallel to the grain (fe)in the member. Loads applied to thetop 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 thebeam. Equating allowable unit stresses in bending and shear to the maxi-mum unit stresses developed in a beam subjected to a uniform load of wpounds per linear foot [kN/m] yields the bending and shear equations ofTables 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 therequired size of the member. Design properties for Plyform (plywood espe-cially engineered for use in concrete formwork) are given in Table 12-4 andsection properties for dimensioned lumber and timber are given in Table12-5. The properties of plywood, lumber, and timber are described in Sec-tion 13-2. However, typical allowable stress values for lumber are given inTable 12-6. The allowable unit stress values in Table 12-6 (but not modulusof elasticity values) may be multiplied by a load duration factor of 1.25 (7-dayload) when designing formwork for light construction and single use or verylimited reuse of forms. However, allowable stresses for lumber sheathing (notPlyform) should be reduced by the factors given in Table 12-6 for wet condi-tions. The values for Plyform properties presented in Table 12-4 are based onwet strength and 7-day load duration, so no further adjustment in these val-ues is required.

339

Support Conditions

Type 1 Span 2 Spans 3 Spans

Bending moment (in.-Ib)wZ2 wZ2 wZ2

M=- M=-M= 12096 96

Shear (lb) V= wZ V = 5wZ V= wZ24 96 20

Deflection (in.)5wZ4 wZ4 wZ4

Ll = 4608EI Ll= 2220EI Ll= 1740EI

340 Chapter 12

TABLE 12-3Concrete form design equations

Design Condition

Bending

Wood

Plywood

Shear

Wood

Plywood

Deflection

If~=h8o

If ~ = Y240

If ~ = Y360

Compression

Tension

Notation:l =length of span, center to center of supports (in.)Fb =allowable unit stress in bending (psi)FbKS =plywood 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)Fv =allowable unit stress in horizontal shear (psi)fc = actual unit stress in compression parallel to grain (psi)feJ. =actual unit stress in compression perpendicular to grain (psi)ft =actual unit stress in tension (psi)A = area of section (in.2)*

E = modulus of elasticity (p,si)I = moment of inertia (in. )*P = applied force (compression or tension) (lb)S =section modulus (in.3)*~ =deflection (in.)b =width of member (in.)d =depth of member (in.)w =uniform load per foot of span (lb/ft)*For a rectangular member: A = bd, S = bd2/6,I = bd3/12.

Support Conditions

1 Span 2 Spans 3 or More Spans

1 = 4.0d (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

1= 16 FvA + 2d 1= 12.8 FvA + 2d 1= 13.3 FvA + 2dw w w

1= 24 F.Ib/Q + 2d 1= 19.2 F.Ib/Q + 2d 1= 20 F.Ib/Q + 2dw w w

1= 5.51 (E t4 1= 6.86 (E t4 1= 6.46 (E 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 1=1.83(t3 1= 1.69(t3P

te or tel.= AP

ft =-A

TABLE 12-3AMetric (SI) concrete form design equations

12-3 Method of Analysis 341

Design Conditions

Bending

1 Span

Wood 1= 36.5 d (Fbb

)112

1000 W

1= 89.9 (FbS

)112

1000 W

1=2.83(F~St2Plywood

Shear

Wood 1= 1.34 FvA 2d1000 w +

1=2.00 Fslb/Q + 2dw

1= 526 (EI!::I.

)1I4

1000 w

1= 75.1 (EI

)1I3

1000 w

= 68.5 (EI

)1I3

1 1000 w

= 59.8 (EI

)1I3

1 1000 wp

tc or tel. = AP

!e=-A

Plywood

Deflection

If ~ =hao

If ~ =Y240

If ~ =Y360

Compression

Tension

Support Conditions

2 Spans

1= 36.5 d (Fbb

)112

1000 w

1= 89.9 (Fbb

)112

1000 w

1=2.83(F~St2

1= 1.07 FvA + 2d1000 w

1= 1.60 Fslb/Q + 2dw

1= 655 (EI~

)1I4

1000 w

= 101 (EI

)1I3

1 1000 w

= 91.7 (EI

)1I3

1 1000 w

= 79.9 (EI

)1I3

1 1000 w

Notation:I =length of span, center to center of supports (mm)Fb =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 (kPa)FJb /Q=plywood section capacity in rolling shear (N/m)tv =allowable unit stress in horizontal shear (kPa)te = actual unit stress in compression parallel to grain (kPa)tel. =actual unit stress in compression perpendicular to grain (kPa)ft =actual unit stress in tension (kPa)A =area of section (mm2)*E =modulus of elasticity (kPa)I =moment of inertia (mm4)*EI =plywood stiffness capacity (kPamm4/m)P =applied force (compression or tension) (N)S =section modulus (mm3)*li =deflection (mm)b =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 = bd3/12

3 or More Spans

1= 1.11 FvA 2d1000 w +

1= 1.67 Fslb/Q + 2dw

1= 617 (EI~

)1I4

1000 w

= 93.0 (EI

)1I3

1 1000 w

= 84.7 (EI

)1I3

1 1000 w

= 73.8 (EI

)1I3

1 1000 w

~~~

TABLE 12-4continued

Face Grain Across Supports Face Grain Parallel to Supports

EI F F ,)b/Q EI F F ,Jb/Q

104 Ibin2 103 Ibin 103 Ib 104 Ibin2 103 Ibin 103 Ibft ft ft ft ft ft

Approx.109 kPamm4 103 Nmm 103 N 109 kPamm4 103 Nmm 103 NThickness Weight

in. (mm) psf (kglm2)m m m m m m

Plyform Structural I

(12.5) 1.5 (7.3) 0.117 0.523 0.501 0.043 0.344 0.278(1102) (194) (7.31) (410) (127) (4.06)

% (16) 1.8 (8.8) 0.196 0.697 0.536 0.067 0.459 0.313(1850) (258) (7.83) (636) (170) (4.57)

% (19) 2.2 (10.7) 0.303 0.896 0.631 0.162 0.807 0.413

(2853) (332) (9.21) (1525) (299) (6.02)% (22) 2.6 (12.7) 0.475 1.208 0.769 0.268 1.117 0.611

(4477) (448) (11.22) (2528) (414) (8.92)1 (25.4) 3.0 (14.6) 0.718 1.596 0.814 0.481 1.679 0.712

(6765) (592) (11.88) (4533) (622) (10.39)1-% (28.5) 3.3 (16.1) 0.934 1.843 0.902 0.711 2.119 0.854

(8798) (683) (13.16) (6694) (785) (12.47)

*All properties adjusted to account for reduced effectiveness of plies with grain perpendicular to applied stress. Stresses adjusted for wet conditions, loadduration, and experience factors. Based on APA data.

It should be noted that the deflection of plywood sheathing is preciselycomputed as the sum of bending deflection and shear deflection. While thedeflection equations of Tables 12-3 and 12-3A consider only bending deflec-tion, the modulus of elasticity values of Table 12-4 include an allowancefor shear deflection. Thus, the deflection computed using these tables issufficiently accurate in most cases. However, for very short spans (lid ratioof less than 15) it is recommended that shear deflection be computed sepa-rately and added to bending deflection. See Reference 3 for a recommendedprocedure.

344 Chapter 12

TABLE 12-5

Section properties of U.S. standard lumber and timber (b = width, d = depth)

NominalSize Actual Size Area of Section Section Modulus Moment of Inertia

(b x d) (S4S) A S Iin. in. mm in.2 103 mm2 . 3 105 mm3 in." 106 mm"ID.

1x3 0.75 x 2.5 19 x 64 1.875 1.210 0.7812 0.1280 0.9766 0.40651x4 0.75 x 3.5 19 x 89 2.625 1.694 1.531 0.2509 2.680 1.1151x6 0.75 x 5.5 19 x 140 4.125 2.661 3.781 0.6196 10.40 4.3281x8 0.75 x 7.25 19 x 184 5.438 3.508 6.570 1.077 23.82 9.9131 x 10 0.75 x 9.25 19 x 235 6.938 4.476 10.70 1.753 49.47 20.591 x 12 0.75 x 11.25 19 x 286 8.438 5.444 15.82 2.592 88.99 37.042x3 1.5 x 2.5 38 x 64 3.750 2.419 1.563 0.2561 1.953 0.81292x4 1.5 x 3.5 38 x 89 5.250 3.387 3.063 0.5019 5.359 2.2312x6 1.5 x 5.5 38 x 140 8.250 5.323 7.563 1.239 20.80 8.6562x8 1.5 x 7.25 38 x 184 10.88 7.016 13.14 2.153 47.63 19.832 x 10 1.5 x 9.25 38 x 235 13.88 8.952 21.39 3.505 98.93 41.182 x 12 1.5 x 11.25 38 x 286 16.88 10.89 31.64 5.185 178.0 74.082 x 14 1.5 x 13.25 38 x 337 19.88 12.82 43.89 7.192 290.8 121.03x4 2.5 x 3.5 64 x 89 8.750 5.645 5.104 0.8364 8.932 3.7183x6 2.5 x 5.5 64 x 140 13.75 8.871 12.60 2.065 34.66 14.433x8 2.5 x 7.25 64 x 184 18.12 11.69 21.90 3.589 79.39 33.043 x 10 2.5 x 9.25 64 x 235 23.12 14.91 35.65 5.842 164.9 68.633 x 12 2.5 x 11.25 64 x 286 28.12 18.14 52.73 8.642 296.6 123.53 x 14 2.5 x 13.25 64 x 337 33.12 21.37 73.15 11.99 484.6 201.73 x 16 2.5 x 15.25 64 x 387 38.12 24.60 96.90 15.88 738.9 307.54x4 3.5 x 3.5 89x89 12.25 7.903 7.146 1.171 12.50 5.2054x6 3.5 x 5.5 89 x 140 19.25 12.42 17.65 2.892 48.53 20.204x8 3.5 x 7.25 89 x 184 25.38 16.37 30.66 5.024 111.1 46.264 x 10 3.5 x 9.25 89 x 235 32.38 20.89 49.91 8.179 230.8 96.084 x 12 3.5 x 11.25 89 x 286 39.38 25.40 73.83 12.10 415.3 172.84 x 14 3.5 x 13.25 89 x 337 46.38 29.92 102.4 16.78 678.5 282.44 x 16 3.5 x 15.25 89 x 387 53.38 34.43 135.7 22.23 1034 430.66x6 5.5 x 5.5 140x 140 30.25 19.52 27.73 4.543 76.25 19.526x8 5.5 x 7.5 140 x 191 41.25 26.61 51.56 8.450 193.4 80.486 x 10 5.5 x 9.5 140x 241 52.25 33.71 82.73 13.56 393.0 163.66 x 12 5.5 x 11.5 140x 292 63.25 40.81 121.2 19.87 697.1 290.16 x 14 5.5 x 13.5 140x 343 74.25 47.90 167.1 27.38 1128 469.46 x 16 5.5 x 15.5 140x 394 85.25 55.00 220.2 36.09 1707 710.4

12-4 Slab Form Design 345

TABLE 12-5continued

NominalSize Actual Size Area of Section Section Modulus Moment of Inertia

(b xd) (S4S) A S Iin. in. mm in.2 103 mm2 in.3 105 mm3 in.4 104 mm4

6 x 18 5.5 x 17.5 140 x 445 96.25 62.10 280.7 46.00 2456 10226 x 20 5.5 x 19.5 140 x 495 107.2 69.19 348.6 57.12 3398 14156 x 22 5.5 x 21.5 140 x 546 118.2 76.29 423.7 69.44 4555 18966 x 24 5.5 x 23.5 140 x 597 129.2 83.39 506.2 82.96 5948 24768x8 7.5 x 7.5 191 x 191 56.25 36.29 70.31 11.52 263.7 109.88 x 10 7.5 x 9.5 191 x 241 71.25 45.97 112.8 18.49 535.9 223.08 x 12 7.5 x 11.5 191 x 292 86.25 55.65 165.3 27.09 950.5 395.78 x 14 7.5 x 13.5 191 x 343 101.2 65.32 227.8 37.33 1538 640.18 x 16 7.5 x 15.5 191 x 394 116.2 75.00 300.3 49.21 2327 968.88 x 18 7.5 x 17.5 191 x 445 131.2 84.68 382.8 62.73 3350 13948 x 20 7.5 x 19.5 191 x 495 146.2 94.36 475.3 77.89 4634 19298 x 22 7.5 x 21.5 191 x 546 161.2 104.0 577.8 94.69 6211 25858 x 24 7.5 x 23.5 191x 597 176.2 113.7 690.3 113.1 8111 3376

lOx 10 9.5 x 9.5 241 x 241 90.25 58.23 142.9 23.42 678.8 282.510 x 12 9.5 x 11.5 241 x 292 109.2 70.48 209.4 34.31 1204 501.210 x 14 9.5 x 13.5 241 x 343 128.2 82.74 288.6 47.29 1948 810.710 x 16 9.5 x 15.5 241 x 394 147.2 95.00 380.4 62.34 2948 122710 x 18 9.5 x 17.5 241 x 445 166.2 107.3 484.9 79.46 4243 176610 x 20 9.5 x 19.5 241 x 495 185.2 119.5 602.1 98.66 5870 244310 x 22 9.5 x 21.5 241 x 546 204.2 131.8 731.9 119.9 7868 3275lOx24 9.5 x 23.5 241 x 597 223.2 144.0 874.4 143.3 10274 427612 x 12 11.5 x 11.5 292 x 292 132.2 85.32 253.5 41.54 1458 594.212 x 14 11.5x 13.5 292 x 343 155.2 100.2 349.3 57.24 2358 981.412 x 16 11.5x 15.5 292 x 394 178.2 115.0 460.5 75.46 3569 148512 x 18 11.5 x 17.5 292 x 445 201.2 129.8 587.0 96.19 5136 213812 x 20 11.5x 19.5 292 x 495 224.2 144.7 728.8 119.4 7106 295812 x 22 11.5x 21.5 292 x 546 247.2 159.5 886.0 145.2 9524 396412 x 24 11.5x 23.5 292 x 597 270.2 174.4 1058 173.4 12437 427614 x 14 13.5 x 13.5 343 x 343 182.2 117.5 410.1 67.20 2768 115214 x 16 13.5 x 15.5 343 x 394 209.2 135.0 540.6 88.58 4189 174414 x 18 13.5 x 17.5 343 x 445 236.2 152.4 689.1 112.9 6029 251014 x 20 13.5 x 19.5 343 x 495 263.2 169.8 855.6 140.2 8342 347214 x 22 13.5 x 21.5 343 x 546 290.2 187.3 1040 170.4 11181 465414 x 24 13.5 x 23.5 343 x 597 317.2 204.7 1243 203.6 14600 6077

12-4 SLAB FORM DESIGN

Method of Analysis

The procedure for applying the equations of Table 12-3 and 12-3A to thedesign of a deck or slab form is first to consider a strip of sheathing of thespecified thickness and 1 ft (or 1 m) wide (see Figure 12-1a). Determine inturn the maximum allowable span based on the allowable values of bendingstress, shear stress, and deflection. The lower of these values will, of course,

determine the maximum spacing of the supports (joists). For simplicity andeconomy of design, this maximum span vaJue is usuaJly rounded down to thenext lower integer or modular vaJue when selecting joist spacing.

Based on the selected joist spacing, the joist itself is analyzed to deter-mine its maximum allowable span. The load conditions for the joist are illus-trated in Figure 12-1b. The joist span selected will be the spacing of thestringers. Again, an integer or moduJar value is selected for stringer spacing.

Based on the selected stringer spacing, the process is repeated to deter-mine the maximum stringer span (distance between vertical supports orshores). Notice in the design of stringers that the joist loads are actuallyapplied to the stringer as a series of concentrated loads at the points wherethe joists rest on the stringer. However, it is simpler and sufficiently accurateto treat the load on the stringer as a uniform load. The width of the uniformdesign load applied to the stringer is equal to the stringer spacing as shownin Figure 12-1c. The calcuJated stringer span must next be checked againstthe capacity of the shores used to support the stringers. The load on eachshore is equaJ to the shore spacing muJtiplied by the load per unit length ofstringer. Thus the maximum shore spacing (or stringer span) is limited to thelower of these two maximum values.

Although the effect of intermediate form members was ignored in deter-mining allowable stringer span, it is necessary to check for crushing at thepoint where each joist rests on the stringer. This is done by dividing the load at

346 Chapter 12

TABLE 12-6Typical 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) Fb Fv FcJ. Fe Ft E

Douglas fir-larch 1450 185 385 1000 850 1.7 x 106

[9998] [1276] [2655] [6895] [5861] [11.7 x 106]Hemlock-fir 1150 150 245 800 675 1.4 x 106

[7929] [1034] [1689] [5516] [4654] [9.7 x 106]

Southern pine 1400 180 405 975 825 1.6 x 106

[9653] [1241] [2792] [6723] [5688] [11.0 x 106]California redwood 1400 160 425 1000 800 1.3 X 106

[9653] [1103] [2930] [6895] [5516] [9.0 x 106]

Eastern spruce 1050 140 255 700 625 1.2 x 106

[7240] [965] [1758] [4827] [4309] [8.3 x 106]

Reduction factor for 0.86 0.97 0.67 0.70 0.84 0.97wet conditions

Load duration factor 1.25 1.25 1.25 1.25 1.25 1.00(7-day load)

Section

I w II I

I--- 1 tt --1(1m) .

w = design load (Ib/sq tt) [kN/m2]w1 = 1 x w = w (Ib/tt) [kN/m]

a. Sheathing

S2 = spacing of joists (tt) [m]w2 = w X S2(Ib/tt) [kN/m]

S3 = spacing of stringers (tt) [m]

w3 = w X S3 (Ib/ft) [kN/m]

c. Stringers

12-4

Elevation

Joists

£1

Joists

Stringers

b. Joists

Slab Form Design

Stringer

Shores

FIGURE 12-1 Design analysis of form members.

Joist

this point by the bearing area and comparing the resulting stress to the allow-able unit stress in compression perpendicular to the grain. A similar procedureis applied at the point where each stringer rests on a vertical support.

To preclude buckling, the maximum allowable load on a rectangularwood column is a function of its unsupported length and least dimension (or1/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' - 0.3E c<:FC - (l / d)2 "" C

(12-10)

347

348 Chapter 12

where Fc =allowable unit stress in compressionparallel 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 notexceed 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 while2 x 8 in. (50 x 200 mm) lumber will be used for joists. Stringers will be4 x 8 in. (100 x 200 mm) lumber. Assume that all members are continuousover three or more spans. Commercial 4000-lb (17.8-kN) shores will beused. 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 areas follows:

Maximum deflection of form members will be limited to 11360.Use the mini-mum value of live load permitted by ACI. Determine joist spacing, stringerspacing, 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 ftLive load = 50 lb/sq ft

Design load =130lb/sq ft

Sheathing Other Memberspsi [kPa] psi [kPa]

Fb 1075 [7412] 1250 [8619]Fv 174 [1200] 180 [1241]FcJ. 405 [2792]Fe 850 [5861]E 1.36 x 106 1.40 X 106

[9.4 X 106] [9.7 X 106]

12-4 Slab Form Design 349

FIGURE 12-2 Slab form,Example 12-1.

Sheathing

Plan

Pressure per m2:

Concrete = 1 x 0.152 x 9.8 x 2403/1000 =3.58 kPa

Formwork = 0.24 kPa

Live load =2.40 kPa

Design load = 6.22 kPa

Deck Design. Consider a uniformly loaded strip of decking (sheathing) 12in. (or 1 m) wide placed perpendicular to the joists (Figure 12-1a) and ana-lyze it as a beam. Assume that the strip is continuous over three or morespans and use the appropriate equations of Table 12-3 and 12-3A.

350 Chapter 12

w =(1 sq ftJlin ft) x (130 lb/sq ft) = 130 lb/ft

[w = (1 m2/m) x (6.21 kN/m2) =6.21 kN/m]

(a) Bending:

1=4.46 d (F~b)1I2

= (4.46) (0.75) (1071~0(12)rf2 = 33.3 in.

[

1=~ d (Fbb

)112

]

1000 w

= (40.7) (19) (7412) (1000»

)1/2 =844

1000 6.22 mm

(b) Shear:

1= 13.3 FvA + 2dw

_ (13.3) (174) (12) (0.75) + (2) (0.75) = 161.7 in.- 130

[

l =~ FvA + 2d

]

1000 w

- (1.11) (1200) (1000) (19) + (2) (19) = 4107 mm- (1000) (6.22)

(c) Deflection: Ebd3

)1= 1.69 (~t3 = 1.69 (W 12 1/3

(1.36 x 106) (12) (0.75)3

)1/3 =27.7 in.= 1.69 (130) (12)

[

_ ~ (EI

)1/3 =~ (Ebd3

)1/3

]

l- 1000 W 1000 w12

_ 73.8 ((9.4 x 106) (1000) (19)3

)1/3 = 703 mm- 1000 (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 valuefor the design.

Joist design. Consider the joist as a uniformly loaded beam supporting astrip of design load 24 in. (610 mm) wide (same as joist spacing; see Figure12-1b). Joists are 2 x 8 in. (50 x 200 mm) lumber. Assume that the joists arecontinuous 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:

1= 10.95 (F:YI2

= 10.95 ((1250~i~3.14»)1/2= 87.0 in.

[

1= ...!QQ... (F bS

)1/2

]

1000 w

= 100 ((8619) (2.153 x 105)

)1/2= 2213 mm1000 3.79

(b) Shear:

1 = 13.3 FvA + 2dw

= (13.3) (180) (10.88) + (2) (7.25) = 114.7 in.

[

1=1Jl FvA + 2d

]

1000 w

_ 1.11 (1241) (7016) + (2)(184) =2918mm- 1000 3.79

(c) Deflection:

1= 1.69 (~t3

((1.4 x 106)(47.63»

)113= 107.4 in.

[

I: 1~:98(EI

)1/3 260

]

1000 w

= 73.8 (9.7 x 106) (19.83 X 106»

)1/3 = 2732 mm1000 3.79

Thus bending governs and the maximum joist span is 87 in. (2213 mm). Wewill select a stringer spacing (joist span) of 84 in. (2134 mm).

Stringer Design. To analyze stringer design, consider a strip of designload 7 ft (2.13 m) wide (equal to stringer spacing) as resting directly on thestringer (Figure 12-1c). Assume the stringer to be continuous over threespans. Stringers are 4 x 8 (100 x 200 mm) lumber. Now analyze the stringeras a beam and determine the maximum allowable span.

w =(7) (130) = 910 lb/ft

[w =(2.13) (1) (6.22) = 13.25 kN/m]

351

352 Chapter 12

(a) Bending:

(FbS

)112

1= 10.95 --;;-

(1250) (30.66)

)112 = 71.1 in.

[

: 1;~:5 (FbS)

~210

]

1 1000 w

_ 100. (

(8619) (5.024 x 105)

)112 = 1808 mm- 1000 13.25

(b) Shear:

13.3 FvA + 2dl=w

(13.3) (180) (25.38) + (2) (7.25) =81.3 in.= 910

[

l =l::!l FvA + 2d

J

1000 w

- 1.11 (1241) (16.37 x 103) + (2) (184) = 2070 mm- 1000 13.25

(c) Deflection:

1= 1.69 (~t3

= 1.69 (1.4 x l~:b (111.1)t3 = 93.8 in.

[

l =~ (EI

)1I3

J

1000 w

= 73.8 (9.7 x 106) (46.26 X 106)

)113 =23881000 13.25 mm

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 equalto 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 353

Before completing our design, we should check for crushing at the pointwhere each joist rests on a stringer. The load at this point is the load per unitlength 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) = 5.25 sq in.[A=(38) (89) =3382 mm2 ]

fc.L =~ = ;~:~ = 347 psi < 405 psi (Fc.L)

[

8.09 x 106

]fc.L= 3382 2392 kPa < 2792 kPa (fc.L)

OK

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) spacingShore: 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 forslab forms substituting studs for joists, wales for stringers, and ties forshores. First, the maximum lateral pressure against the sheathing is deter-mined from the appropriate equation (Equation 12-1, 12-2, or 12-3). If thesheathing thickness has been specified, the maximum allowable span for thesheathing based on bending, shear, and deflection is-the maximum stud spac-ing. If the stud spacing is fixed, calculate the required thickness of sheathing.

Next, calculate the maximum allowable stud span (wale spacing) basedon stud size and design load, again considering bending, shear, and deflection.If the stud span has already been determined, calculate the required size ofthe stud. After stud size and wale spacing have been determined, determinethe maximum allowable spacing of wale supports (tie spacing) based on walesize and load. If tie spacing has been preselected, determine the minimumwale size. Double wales are commonly used (see Figure 12-3) to avoid thenecessity of drilling wales for tie insertion.

Next, check the tie's ability to carry the load imposed by wale and tiespacing. The load (lb) [kN] on each tie is calculated as the design load (lb/sqft) [kPa] multiplied by the product of tie spacing (ft) [m] and wale spacing

354 Chapter 12

Wale

~ q...:.I, ')~. . ~

"'O~''':'4,:" I~:

': ',...o.~.::

:~.:: <\'-:':

:~~<.o~ :>.

[itI

:'.':~ ',".:':'~.'.".,'

T

Stud

,>:~'.f,,.»"

.".> c'.. . ~.." .'

. -: :.,:>.. r'

. .Po: ~. : . . 0

.::<.~.:;.

.;:/:f'~":

J..:~ .. 0 .',.'

Tie/

T

T

Studspacing

Tie

IWalespacing

1 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 thetie spacing must be reduced.

The next step is to check bearing stresses (or compression perpendicularto the grain) where the studs rest on wales and where tie ends bear on wales.Maximum bearing stress must not exceed the allowable compression stressperpendicular to the grain or crushing will result. Finally, design lateral brac-ing 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 pouredat a rate of 4 ftJh (1.219 mIh), internally vibrated, at 900 F (320 C) tempera-ture (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 are3000-lb (13.34-kN) capacity with 1-l/2-in. (38-mm) wide wedges bearing onwales. Deflection must not exceed 1/360. Determine stud, wale, and tie spac-ing. Use Plyform section properties and allowable stresses from Table 12-4and lumber section properties from Table 12-5. Allowable stresses for thelumber 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 = 150 + (9000) (4) =550 lb/sq ftT 90

[ =7 2 785R =7 2 (785) (1.219) = 26 3 kNl 2]P . + T + 18 . + 32 + 18 . m

Check: 550 < 150 h (= 1200 lb/sq ft) < 2000 lb/sq ft OK[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 = 1 x 1 x 550 =550 lb/ft

[w = 1 x 1 x 26.3 = 26.3 kN/m ]

(a) Bending:

355

356 Chapter 12

(b) Shear:

Z=20 Fslb/Q + 2dw

_ (20) (0.412 x 103) + (2) (5/8)- 550

= 14.98 + 1.25 = 16.23 in.

Z= 1.67 FJb/Q + 2dw

_ (1.67) (6.01 x 103) + (2) (16)- 26.3= 382 + 32 =414 rom

(c) Deflection:

Z= 1.69 (~t3

=(1.69) (0.19:5~ 106t3 = 12..0 in.

[

z=~ (EI

)1I3

]

1000 w

= 73.8 (1836 x 109

)113 =3041000 26.3 rom

Deflection governs. Maximum span = 12.0 in. (304 mm). Use a 12-in. (304-mm) stud spacing.

Select Wale Spacing (Three or More Spans). Since the stud spacing is12 in. (304 mm), consider a uniform design load 1 ft (304 rom) wide resting oneach stud.

w = 1 x 1 x 550 =550 lb/ft304

[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 in.

[

z=...!QQ..(FbS )

112

]

1000 w

Z= 100 ((12480) (0.5019 x 105)

)1/2 = 885 mm1000 8.0

12-5 Wall and Column Form Design 357

(b) Shear:

Z= 13.3 FuA + 2dw

- (13.3) (120) (5.25) + (2) (3.5) = 22.23 in.- 550

[

z=1:.!l FuA + 2d

J

1000 w

(1.11) (827) (3.387 x 103) + (2) (88.9) = 566 mm= (1000) (8.0)

(c) Deflection:

Shear governs, so maximum span (wale spacing) is 22.23 in. (566 mm). Use a16-in. (406-mm) wale spacing for modular units.

Select Tie Spacing (Three or More Spans, Double Wales). Based on awale spacing of 16 in. (406 mm):

16w =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 ((1810) 7C:3~33.063) t2 = 42.58 in.

[

z=.lQQ... (FbS

)1/2

J

1000 w

= 100 ((12480) (2 x 0.5019 x 105)

)112 = 10831000 10.68 mm

358 Chapter 12

(b) Shear:

I =13.3 FvA + 2dw

(13.3)(120)(2 x 5.25) + (2)(3.5) = 29.85 in.= 733.3

[

I =lJl FvA + 2d

]

1000 w

(1.11)(827)(2 x 3.387 x 103) + (2)(88.9) = 760 mm= (1000)(10.68)

(c) Deflection:

1=1.69(~r3= 169

((1.7 x 106)(2 x 5.359)

)1/3= 49 32

.. 733.3 . m.

[

=~ (EI

)1/3

]

I 1000 w

= 73.8 ((11. 7 x 106)(2 x 2.231 x 106)

)1/3 = 12521000 10.68 mm

Shear governs. Maximum span is 29.85 in. (760 mm). Select a 24-in. (610-mm) tie spacing for a modular value.

(d) Check tie load:

P =wale spacing x tie spacing x p16 24 .= - - (550) =14671b/tIe < 3000 lb OK12 12

[(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) =(2)(1.5)(1.5) =4.5 sq in.[A = (2)(38)(38) = 2888 mm2)

Load at each panel point (P) = loadlft(m) of stud x wale spacing (ft/m)

16P =(550) - =733 lb

12

[

406

]P =(8.0) 1000 = 3.25 kN

12-6 Design of Lateral Bracing

j: = P = 733 = 163 psi < 485 psi (FcJ)Icl. A 4.5

[j: = 3.25 X 106 = 1125 kPa < 3340 kPa (FCJ)

]I cl. 2888

OK

(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]

fc.l- = ~ = 1:.~7 = 326 psi < 485 psi (Fcl.)

[

6.51 x 106

]fc.l-= 2888 = 2254 kPa < 3340 kPa (Fcl.)

OK

Final Design.

Sheathing: 4 x 8 ft (1.2 x 2.4 m) sheets of 5/8-in. (16-mm) Class I Plyformplaced 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 inadequatelydesigned lateral bracing. Minimum lateral design load values were given inSection 12-2. Design procedures for lateral bracing are described and illus-trated in the following paragraphs.

Lateral Braces for Wall and Column Forms

For wall and column forms, lateral bracing is usually provided by inclinedrigid braces or guy-wire bracing. Since wind loads, and lateral loads in gen-eral, 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 bracesare used they may be placed on only one side of the form if designed to resistboth tension and compression forces. When forms are placed on only one sideof a wall with the excavation serving as the second form, lateral bracing mustbe designed to resist the lateral pressure of the concrete as well as other lat-eral forces.

359

360 Chapter 12

Inclined bracing will usually resist any wind uplift forces on verticalforms. However, uplift forces on inclined forms may require additional consid-eration and the use of special anchors or tiedowns. The strut load per foot ofform developed by the design lateral load can be calculated by the use ofEquation 12-11. The total load per strut is then P' multiplied by strut spacing.

P' = H x h x 1 (12-11 )h' x l'

1= (h,2+ l'2)1/2 (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]h =height of form (ft) [m]h' = height of top of strut (ft) [m]

1 = length of strut (ft) [m]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 forstrut 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) lateralbraces 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 kPa

FeFtE

850725

1.4 x 106

58614999

9.7 X 106

SolutionDetermine 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 formmay now be determined from Equation 12-1.

r8ft

(2.44 m) .

6 ft _

(1.83 m)

12-6 Design of Lateral Bracing 361

H--------

~5ft(1.53m)-1

FIGURE 12-4 Wall form bracing, Example 12-3.

P ' = H x h x I = (1008)(8)(7.81) = 208 3 lb/ft f fih' x l' (6)(5) . 0 orm

[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 usingEquation 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 toreduce 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) 430 .c - (lId)2 - (46.9/1.5)2 pSI

[D , = (0.3)(9.7 X 106) 2 97 kP

]L'C (1190/38)2 . a

As Fc' < Ft < Fc, the value of Fc' governs.

The maximum allowable compressive force per strut is:

P =(1.5)(3.5)(430) = 2257 lb

[p = (38)(8~6~2967) = 10.03 kN]

362 Chapter 12

Thus maximum strut spacing is:

p 2257s = 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 eachstrut at the midpoint of its length.

Lateral Braces for Slab Forms

For elevated floor or roof slab forms, lateral bracing may consist of crossbraces between shores or inclined bracing along the outside edge of the formsimilar to that used for wall forms. The following example illustrates themethod 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, 20ft (6.1 m) wide, and 100 ft (30.5 m) long shown in Figure 12-5. The slab is tobe 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).

SolutionDead load =(1/2)(1)(150) + 15 =90 lb/sq ft

[dl = (0.152)(~~~~03)(9.8) + 0.72 =4.30 kpa]H = 0.02 x dl x ws

For the 20-ft (6.1-m) face, the width of the slab is 100 ft (30.5 m).

H20 =(0.02)(90)(100) = 180 lb/lin ft[H = (0.02)(4.30)(30.5) = 2.62 kN/m]

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 < 100lb/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 formbracing design, Example 12-4. r

-r20 tt

(6.1 m)...L

~ ~ ~ ~ ~ ~ ~ ~

100 tt (30.5 m) 1 .......

J -- H20 Ib/tt (kN/m)

~ ~.......

H100 Ib/tt (kN/m)

Problems 363

PROBLEMS

1. Calculate the appropriate design load due to concrete pressure for a wallform 9 ft (2.75 m) high to be poured at a rate of 6.5 ft (2.0 m) per hour ata temperature of 68°F (20°C).

2. What should be the design load due to concrete pressure for a columnform 15 ft (4.9 m) high that is to be filled by pumping concrete into theform from the bottom?

3. Determine the proper design load for a slab form 9 in. (229 mm) thickusing motorized buggies for placing concrete. The formwork is estimatedto weigh 10 lb/sq ft (48.8 kg/m2).

/4. Calculate the maximum allowable span for 3/4-in. (19-mm) Class I Ply-form decking with face grain across supports carrying a design load of150 lb/sq ft (7.2 kPa). Assume that the decking is continuous over threeor 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 100mm) studs for a wall form sheathed with nominal I-in. (25-mm) Hem-firlumber. The design load is 500 lb/sq ft (23.9 kPa). Assume that thesheathing is continuous over three or more spans. Use the allowablestresses of Table 12-6 with 7-day load and wet conditions. Limit deflec-tion to 1/360 of span length.

6. Determine the maximum allowable span of nominal 2 x 4 in. (50 x 100mm) 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 stressesfor Douglas Fir from Table 12-6 and 7-day load duration. Assume thatstuds are continuous over three or more spans. Limit deflection to 1/360of span length. Based on the maximum allowable span, check for crush-ing 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 therate of 5 ft/h (1.53 m/h) at a temperature of 77°F (25°C). Sheathing willbe 3/4-in. (19-mm) Class I Plyform with face grain across supports. Alllumber 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) abovethe form bottom and anchored 4 ft (1.22 m) away from the bottom of theform. 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/sqft (1.2 kPa).

~8. Design the formwork for a concrete slab 8 in. (203 mm) thick whose netwidth between beam faces is 15 ft (4.58 m). Concrete will be placed usinga crane and bucket. The formwork is estimated to weigh 5 lb/sq ft (24.1kg/m2). Decking will be 3/4-in (19-mm) Class I Plyform with face grainacross supports. All lumber will be Southern Pine. Joists will be nominal2-in. (50-mm) wide lumber. One 4-in. (lOO-mm) wide stringer will beplaced between the beam faces. Limit deflection to 1/360 of span length.Commercial shores of 4500 lb (20 kN) capacity will be used. Lateral sup-port will be provided by the beam forms.

364 Chapter 12

9. Derive the equations for bending given in Table 12-3 using the relation-ships of Table 12-2 and Equation 12-5.

10. Develop a computer program to calculate the maximum safe span of awood formwork member as determined by bending, shear, and deflectionbased on the equations of Table 12-3 or 12-3A. Input should include sup-port conditions, allowable deflection, allowable stresses, load per unitlength 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 WoodAsso-ciation, Tacoma, WA, 1992.

3. Concrete Forming. APA-The Engineered Wood Association, Tacoma, WA,1994.

4. Design of Wood Formwork for Concrete Structures (Wood ConstructionData No.3). American Forest and Paper Products Association, Washing-ton, D.C., 196!.

5. Guide to Formwork for Concrete (ACI 347R-94). American Concrete Insti-tute, 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 andPaper Products Association, Washington, D.C., 1982.

8. Peurifoy, Robert L., and Garold D. Oberlender. Formwork for ConcreteStructures, 3rd ed. New York: McGraw-Hill, 1996.

9. Plywood Design Specifications. APA-The Engineered Wood Association,Tacoma, WA, 1995.