1. Chapter 13
Concrete Form Design

2. Frame Construction
Framing member in usually of 2 in (5cm) nominal thickness
Then this frame are covering with siding and roof sheathing of plywood or lumber.
The two principals forms of frames are:
Platform frame construction(in this type of platform the subfloors extends to the outside of the building walls) as shown in figure (11-3)
Balloon frame Construction(in this type of platform the exterior wall studs extends all the way from the sill to the top of the second floor wall) as shown in figure (11-4)

5. Foundation and Roof Construction
Platform frame Construction supported by foundation walls as illustrated in figure (11-5)
In this illustration the floor joints are lapped and rest on top of the girder rather than on the ledger as in figure (11-3).
And as in figures(11-7) for slab in grade and figure (11-8) as combined slab foundation

7. Roof Construction
One method of roof Construction called Joist and Rafter Framing as illustrated in Figure (11-11).
The Rafters are notched where are placed on the wall plates and are held in place by nailing them to the wall plates.
The Collar Beam shown is used to assist in resisting wind loads on the roof.
In figure (11-9) all the anchors and joist hangers that are using in Frame Construction

10. Concrete Form Design SLAB FORM DESIGN Method

11. Concrete Form Design Formwork materials can be classified as:
Lumber and Timber (Wood)
Metals
Plastic

12. 1. Timber Formwork

13. 1. Timber Formwork Lumber:
Lumber is commonly available material and has excellent strength, weight and cost factor
Lumber is classified as:
Boards: 1 to 1.5 inches thick, 2 or more inches width
Dimensions: 2 to 4 inch thick, any width.
Timbers: 5 or more inches thickness, 5 or more inches width

14. 1. Timber Formwork b. Plywood:
The use of plywood in concrete forming for form facing has improved the quality of finished concrete.
The relatively large sheets of plywood have reduced the cost of building and at the same time have provided smooth surfaces that reduces cost of finishing of concrete surfaces

15. 2. Metal Formwork

16. 2. Metal Formwork
The initial cost of metal formwork is more than timber formwork but the lumber of reuses of metal formwork is higher than that of timber.
In long run metal formwork can be economical.
In heavy construction works metal formwork may require a lifting mechanism to handle the formwork panels or props.

17. 2. Metal Formwork
Steel sheet formwork has the problem of rusting also. To avoid rusting, in every use the surfaces should be oiled with an appropriate releasing agent.
in metal formwork usage, the metal sheets are prepared as panels of standard sizes. This brings the difficulties of erecting irregular dimensions of formwork.
Steel or aluminum or magnesium is the most widely used metals.

18. 3. Plastic Formwork

19. 3. Plastic Formwork
They have impervious surfaces that usually create a smooth fınish to the concrete.
Plastic formwork could be reinforced or un-reinforced.
Plastic is reinforced by glass fibers.
Reinforced plastics are specially produced for a specific formwork type.
Un-reinforced plastics are produced in sheet form with smooth or textured surfaces.
Plastic formwork is lighter but less durable than metal formwork.

20. Formwork Design Floor and Roof formwork Design:
The design load that acts on the slab form consist of :
self-weight of the reinforced slab plus
the live load and,
the weight of the formwork themselves.

21. In case of motorized concrete buggies are used 3.6kpa
The American concrete institute (ACI) recommended a minimum live load of:
2.4kpa
In case of motorized concrete buggies are used
3.6kpa
ACI recommended a minimum design load (dead plus live) :
4.8kpa
In case of motorized concrete buggies are used :
6.0 kpa

22. Design steps:
Selecting the design load.
Analyzing the sheeting, joist and stringers as beam under uniformly distributed load supported over one of the three conditions (single span - two spans – three spans or larger).
Determining the allowable span for slab from table (13-5& 13-5A) by considering the smallest span based on the value of bending, shear and deflection.

23. Design the sheathing(decking).
Design the joist.
Design the stringers.
Check the stringer spans and shore capacity.
Check the crushing between joist and stringer.

24. Maximum bending moment, shear force and deflection developed by uniformly distributed load can be obtained from table below:

26. The maximum fiber stress developed in bending, shear and compression resulting from a specified load can be determined from the upper equations.

30. Example (13-1)
Determine joist spacing, stringer spacing, and shore spacing of the following Data:
Design the formwork (Figure 13-2) for an elevated concrete floor slab 6 in. (152 mm) thick.
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).
Stringer
Joist
Sheathing
Types of formwork
4x8 inch
(100x200mm)
2x 8 inch(50x200mm)
1 inch(25mm)
Dimensions

31. Example (13-1)
The adjusted allowable stresses for the lumber being used are as follows:
Maximum deflection of form members will be limited to l/360.
Use the minimum value of live load permitted by ACI.

32. Figure (13-2) Slab

33. Figure (13-2) Slab

34. Solution

35. 1. Deck Design (Joist Spacing)
Consider a uniformly loaded strip of decking (sheathing) 12 in. (or 1 m) wide placed perpendicular to the joists (Figure 13-1a) and analyze it as a beam.
Assume that the strip is continuous over three or more spans and use the appropriate equations of Table (13-5) and (13-5A) in the following Page.
w =(1 sq. ft/in ft) x (130 lb/sq ft) = 130 lb/ft
[w =(1 m2/in m) x (6.21 kN/m2) =6.21 kN/m]

37. Figure 13-1 Design Analysis for form member

38. 1. Deck Design (Joist Spacing)

39. 1. Deck Design (Joist Spacing)

40. 1. Deck Design (Joist Spacing)

41. 1. Deck Design (Joist Spacing)
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.

42. 2. 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 13-1b):
Joists are 2 x 8 in. (50 x 200 mm) lumber.
Assume that the joists are continuous over three spans.
wdesignload= (wconcrete+wformwork+wLiveload)x Area x Depth from sheathing (the least)
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]

43. Figure 13-1 Design Analysis for form member

46. 2. Joist design

47. 2. Joist design

48. 2. Joist design

49. 2. Joist design
Thus bending governs and the maximum joist span is 87 in. (2213 mm).
We will select a stringer spacing (joist span) of 84 in (7 ft). (2134 mm).

50. 3. Stringer Design (shore spacing)
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 13-1c).
Assume the stringer to be continuous over three spans.
Stringers are 4 x 8 (100 x 200 mm) lumber (given)
Now analyze the stringer as a beam and determine the maximum allowable span.
w =(7) (130) =910 lb/ft
[w =(2.13) (1) (6.22) = kN/m]

51. Figure 13-1c Design Analysis for form member

52. 3. Stringer Design

53. 3. Stringer Design

54. 3. Stringer Design
Bending governs and the maximum span is 71.1 in. (1808 mm).

55. 4. Check Shore Strength
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.
As 4000 is given before and 910 is w=7x130
Thus the maximum stringer span is limited by shore strength to 52.7 in. (1.343 m)< 71.1 in (1808mm) choosing the least span.
We select a shore spacing of 4 ft (1.22 m) as a modular value.

56. 5. Check Shore Crushing
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.
As 2.134m is the span of joist and 3.79 is W of joist given the Force of shore point where joists rest on Stringer.
The Bearing area is the actual area where 2 becomes 1.5 for Joists and 4 becomes 3.5 for stringer.

57. 6. 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

58. Example 2
Design the formwork for the slab. Concrete will be placed, Unit weight of concrete are 2403 kg/m3. Sheathing will be plywood structure I 1in (25.4 mm) with face grain parallel supports. All lumber will be Eastern Spruce.
Joist will be nominal (50*100 mm). Stringer will be nominal (37*100 mm). Formwork weighs 0.26 KN/m2. Commercial 14.5 KN shore capacity will be used
Live Load = 2.40 KN/m2 Maximum allowable deflection is limited to L/240. Assume all members are continuous over three or more spans.

59. Step 1: Load per meter square
Concrete = ( 1 x 0.15 x 9.8 x 2403 ) / 1000 = 3.53KN/m2
Formwork = 0.26 L.L Total+3.53Concrete = 6.19 KN/m2
Design Load = 6.19 KN/m2

60. Step 2: Sheathing Design
Consider a uniformly loaded strip of sheathing 1 m wide placed parallel to the joists;
W = (1m2/m) x (6.19 KN/m2) = 6.19 KN/m2
L Bending = mm
L Shearing= mm
L Deflection= mm
Deflection governs the design. Maximum allowable span of sheathing mm
Number of Spans = 5000𝑚𝑚 𝑚𝑚 =7.54=8𝑠𝑝𝑎𝑛𝑠
Span Length = =712.5𝑚𝑚

61. Step 3: Joist Design W= 712.5 1000 𝑥1𝑥6.19=4.41 𝑘𝑁/𝑚2
L Bending=907.44mm
L Shearing= mm
L Deflection= mm
Bending governs the design. Maximum allowable span of joist is mm
No of Spans= 𝟑𝟕𝟎𝟎 𝟗𝟎𝟕.𝟕𝟒 =𝟒.𝟏 𝒔𝒑𝒂𝒏𝒔=𝟓𝒔𝒑𝒂𝒏𝒔
Span Length = 𝟑𝟕𝟎𝟎 𝟓 =𝟕𝟒𝟎𝒎𝒎

62. Step 4: Stringer Design W= 740 1000 𝑥1𝑥6.19=4.58𝑘𝑛/𝑚2
L Bending= mm
L Shearing= mm
L Deflection= mm
Bending governs the design, Maximum allowable stringer span is mm
No of spans= =4.957=5 𝑆𝑝𝑎𝑛𝑠
Span Length= 5700/5=1140mm

63. Step5: Checking Bearing
Bearing Area =38*64=2432mm2
P=6.19x 𝑥 =3.263 𝐾𝑁
Bearing Stress = 𝑥10−6 =
KN/m2<=1758 KN/m2