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SECTION II. DESIGN

DESIGN CONSIDERATIONS

4-22. Forms are the molds that hold the plastic concrete and support it until it hardens. They must keep deflections within acceptable limits, meet dimensional tolerances, prevent paste leakage, and produce a final product that meet appearance needs. Therefore, strength, rigidity, and watertightness are the most important form design considerations. In addition, forms must support all weights and stresses to which they are subject, including the dead load (DL) of the forms; the weight of people, equipment, and materials that transfers to the forms; and any impact due to vibration. Although these factors vary with each project, do not neglect any of them when designing a form. Ease of erection and removal are other important factors in economical form design. Sometimes you may want to consider using platforms and ramp structures that are independent of the formwork because they prevent form displacement due to loading, as well as impact shock from workers and equipment.

BASIS OF FORM DESIGN

4-23. Concrete is in a plastic state when placed in the designed form; therefore, it exerts hydrostatic pressure on the form. The basis of form design is to offset the maximum pressure developed by the concrete during placing. The pressure depends on the rate of placing and the ambient temperature. The rate of placing affects pressure because it determines how much hydrostatic head builds up in the form. The hydrostatic head continues to increase until the concrete takes its initial set, usually in about 90 minutes. However, because the initial set takes more time at low ambient temperatures, consider the ambient temperature at the time of placement. Knowing these two factors (rate of placement and ambient temperature) plus the specified type of form material, calculate a tentative design.

PANEL-WALL FORM DESIGN

4-24. It is best to design forms following a step-by-step procedure. Use the following steps to design a wood form for a concrete wall.

Step 1. Determine the materials needed for sheathing, studs, wales, braces, and tie wires or tie-rods.
Step 2. Determine the mixer output by dividing the mixer yield by the batch time. Batch time includes loading all ingredients, mixing, and unloading. If you will use more than one mixer, multiply the mixer output by the number of mixers.

Step 3. Determine the area enclosed by the form.

Plan area (sq ft) = Length (L) x Width (W)

Step 4. Determine the rate (vertical feet per hour) of placing (R) the concrete in the form by dividing the mixer output by the plan area.

NOTE: If using wood for an economical design, try to keep R < 5 feet per hour.

Step 5. Estimate the placing temperature of the concrete.

Step 6. Determine the maximum concrete pressure (MCP) using the rate of placing (see Figure 4-3 or the formula above the figure). Draw a vertical line from the rate of placing until it intersects the correct concrete-temperature line. Read left horizontally from the point of intersection to the left margin of the graph and determine the MCP in 100 pounds per square foot.

Figure 4-3. MCP graph

Step 7. Find the maximum stud spacing (MSS) in inches. Use Table 4-1 for board sheathing or Table 4-2 for plywood sheathing. Refer to the column headed maximum concrete pressure (MCP) and find the value you have for the MCP. If the value falls between two values in the column, round up to the nearest given value. Move right to the column identified by the sheathing thickness. (Always use the strong way for plywood when possible). This number is the MSS in inches.

Table 4-1. Maximum stud (joist) spacing for board sheathing

Maximum Concrete
Pressure, in
Pounds per Square
Foot
Nominal Thickness of S4S Boards, in Inches
1 1 1/4 1 1/2 2
75 30 37 44 50
100 28 34 41 47
125 26 33 39 44
150 25 31 37 42
175 24 30 35 41
200 23 29 34 39
300 21 26 31 35
400 18 24 29 33
500 16 22 27 31
600 15 20 25 30
700 14 18 23 28
800 13 17 22 26
900 12 16 20 24
1,000 12 15 19 23
1,100 11 15 18 22
1,200 11 14 18 22
1,400 10 13 16 20
1,600 9 12 15 18
1,800 9 12 14 17
2,000 8 11 14 16
2,200 8 10 13 16
2,400 7 10 12 15
2,600 7 10 12 14
2,800 7 9 12 14
3,000 7 9 11 13

 

Table 4-2. Maximum stud (joist) spacing for plywood sheathing, in inches

Maximum
Concrete
Pressure,
in Pounds
per
Square
Foot
Strong Way - 5-Ply Sanded
Face, Grain Perpendicular to
the Stud
Weak Way - 5-Ply Sanded Face,
Grain Parallel to the Stud
1/2 5/8 3/4 1
(7 ply)
1/2 5/8 3/4 1
(7 ply)
75 20 24 26 31 13 18 23 30
100 18 22 24 29 12 17 22 28
125 17 20 23 28 11 15 20 27
150 16 19 22 27 11 15 19 25
175 15 18 21 26 10 14 18 24
200 15 17 20 25 10 13 17 24
300 13 15 17 22 8 12 15 21
400 12 14 16 20 8 11 14 19
500 11 13 15 19 7 10 13 18
600 10 12 14 17 6 9 12 17
700 10 11 13 16 6 9 11 16
800 9 10 12 15 5 8 11 15
900 9 10 11 14 4 8 9 15
1,000 8 9 10 13 4 7 9 14
1,100 7 9 10 12 4 6 8 12
1,200 7 8 10 11 - 6 7 11
1,300 6 8 9 11 - 5 7 11
1,400 6 7 9 10 - 5 6 10
1,500 5 7 9 9 - 5 6 9
1,600 5 6 8 9 - 4 5 9
1,700 5 6 8 8 - 4 5 8
1,800 4 6 8 8 - 4 5 8
1,900 4 5 8 7 - 4 4 7
2,000 4 5 7 7 - - 4 7
2,200 4 5 6 6     4 6
2,400 - 4 5 6     4 6
2,600 - 4 5 5     - 5
2,800 - 4 4 5     - 5
3,000 - - 4 5     - 5
Step 8. Determine the uniform load on a stud (ULS) by multiplying the MCP by the stud spacing divided by 12 inches per foot.

Where:

MCP = maximum concrete pressure, in pounds per square foot
MSS = maximum stud spacing
ULS = uniform load stud, in pounds per linear foot
Step 9. Determine the maximum wale spacing (MWS) using Table 4-3. Refer to the column headed uniform load and find the value for the ULS. If the value falls between two values in the column, round it up to the nearest given value. Move right to the column identified by the size of the stud you are using. This number is the MWS in inches.

Table 4-3. Maximum spacing, in inches, for whales, ties, stringers,
and 4" x 4 or larger shores where member to be supported is a SINGLE member

Uniform Load in
Pounds per
Linear Foot
Supported Members Size (S4S)*
2 x 4 2 x 6 3 x 6 4 x 4 4 x 6
100 60 95 120 92 131
125 54 85 440 80 124
150 49 77 100 75 118
175 45 72 93 70 110
200 42 67 87 65 102
225 40 62 82 61 97
250 38 60 77 58 92
275 36 57 74 55 87
300 35 55 71 53 84
350 32 50 65 49 77
400 30 47 61 46 72
450 28 44 58 43 68
500 27 41 55 41 65
600 24 38 50 37 59
700 22 36 46 35 55
800 21 33 43 32 51
900 20 31 41 30 48
1,000 19 30 38 29 46
1,200 17 27 35 27 42
1,400 16 25 33 25 39
1,600 15 23 31 23 36
1,800 14 22 29 22 34
2,000 13 21 27 21 32
2,200 13 20 26 20 31
2,400 12 19 25 19 30
2,600 12 19 24 18 28
2,800 11 18 23 17 27
3,000 11 17 22 17 26
3,400 10 16 21 16 25
3,800 10 15 20 15 23
4,500 9 14 18 13 21
*S4S indicates surfaced four sides.
Step 10. Determine the uniform load on a wale (ULW) by multiplying the MCP by the MWS and dividing by 12 inches per foot.

Where:

MCP = maximum concrete pressure, in pounds per square foot
MWS = maximum wale spacing, in inches, divided by 12 inches per foot
ULW = uniform load on wale, in pounds per linear foot
Step 11. Determine the tie spacing based on the ULW using Table 4-3 or Table 4-4 (depending on type of wale, either single or double). Refer to the column headed Uniform Load and find the value for the ULW. If the value falls between two values in the column, round it up to the nearest given value. Move right to the column identified by the size of wale you are using. This number is the maximum tie spacing, in inches, based on wale size.

Table 4-4. Maximum spacing, in inches, for ties and 4" x 4" or larger shores where member to be supported is a double member

Uniform
Load, in
Pounds per
Linear Foot
Supporting Member Size (S4S)
2 x 4 2 x 6 3 x 6 4 x 4 4 x 6
100 85 126 143 222 156
125 76 119 135 105 147
150 70 110 129 100 141
175 64 102 124 96 135
200 60 95 120 92 131
225 57 89 116 87 127
250 54 85 109 82 124
275 51 84 104 78 121
300 49 77 100 75 118
350 46 72 93 70 110
400 43 67 87 65 102
450 40 63 82 61 97
500 38 60 77 58 92
600 35 55 71 53 81
700 32 51 65 49 77
800 30 47 61 46 72
900 28 44 58 43 68
1,000 27 43 55 41 65
1,200 25 39 50 38 59
1,400 23 36 46 35 55
1,600 21 34 43 33 51
1,800 20 32 41 31 48
2,000 19 30 39 29 46
2,200 18 29 37 28 44
2,400 17 27 36 27 42
2,600 17 26 34 26 40
2,800 16 25 33 25 39
3,000 15 24 32 24 38
3,400 14 23 30 22 35
3,800 14 21 28 21 33
4,500 12 20 25 19 30
Step 12. Determine the tie spacing based on the tie strength by dividing the tie breaking strength by the ULW. If the breaking strength of the tie is unknown, use Table 4-5 to find the breaking loads for a double-strand wire and tie-rods.

Where:

Tie wire or tie-rod = strength = average breaking load (see Table 4-5)
ULW = uniform load on wales, in pounds per square foot

NOTE: If the result does not equal a whole number, round the value down to the next whole number.

Table 4-5. Average breaking load of tie material, in pounds

Steel Wire
Size of Wire Gage
Number
Minimum Breaking Load
Double Strand, in Pounds
8 1,700
9 1,420
10 1,170
11 930
Barbwire
Size of Wire Gage
Number
Minimum Breaking Load
12 1/2 950
13* 660
13 1/2 950
14 650
15 1/2 850
Tie-rod
Description Minimum Breaking Load, in Pounds
Snap ties 3,000
Pencil roads 3,000
*Single-strand barbwire
Step 13. Select the smaller of the tie spacing as determined in steps 11 and 12.
Step 14. Install tie wires at the intersection of studs and wales. Reduce the stud spacing (step 7) or the tie spacing (step 13) to conform with this requirement. tie-rods may be placed along the wales at the spacing (determined in step 13) without adjusting the studs. Place the first tie at one-half the maximum tie spacing from the end of the wale.
Step 15. Determine the number of studs on one side of a form by dividing the form length by the MSS. Add one to this number and round up to the next integer. The first and last studs must be placed at the ends of the form, even though the spacing between the last two studs may be less than the maximum allowable spacing.

NOTE: The length of the form is in feet. The stud spacing is in inches.

Step 16. Determine the number of wales for one side of a form by dividing the form height by the MWS, and round up to the next integer. Place the first wale one-half of the maximum space up from the bottom and place the remaining wales at the MWS.
Step 17. Determine the time required to place the concrete by dividing the height of the form by the rate of placing.

SAMPLE FORM DESIGN PROBLEM NUMBER ONE

4-25. Design a form for a concrete wall 40 feet long, 2 feet thick, and 10 feet high. An M919 concrete mobile mixer is available and the crew can produce and place a cubic yard of concrete every 10 minutes. The concrete placing temperature is estimated at 70 F. The form materials are 2 by 4s, 1-inch board sheathing, and number 9 black annealed wire. The solution steps are as follows:

Step 1. Lay out available materials.

  • Studs: 2 x 4 (single)
  • Wales: 2 x 4 (double)
  • Sheathing: 1-inch board
  • Ties: Number 9 wire

Step 2. Calculate the production rate.

Step 3. Calculate the plan area of the form.

Step 4. Calculate the rate of placing.

Step 5. Determine the concrete-placing temperature.

Temperature = 70 F

Step 6. Determine the MCP (refer to Figure 4-3).

MCP = 400 pounds per square foot

Step 7. Determine the MSS (refer to Table 4-1).

MSS = 18 inches

Step 8. Calculate the ULS.

Step 9. Determine the MWS (refer to Table 4-3).

MWS = 24 inches

Step 10. Calculate the ULW.

Step 11. Determine the tie-wire spacing based on wale size (refer to Table 4-4).
Tie-wire spacing = 30 inches.
Step 12. Determine the tie-wire spacing based on wire strength (referring to Table 4-5, breaking strength of number 9 wire = 1,420 pounds).

Step 13. Determine the maximum tie spacing.

Tie spacing = 21 inches

Step 14. Reduce the tie spacing to 18 inches since the maximum tie spacing is greater than MSS; tie at the intersection of each stud and double wale.
Step 15. Calculate number of studs per side.

Step 16. Calculate the number of double wales per side.

Step 17. Estimate the time required to place concrete.

SAMPLE FORM DESIGN PROBLEM NUMBER TWO

4-26. Design the form for a concrete wall 40 feet long, 2 feet thick, and 10 feet high. An M919 concrete mobile mixer is available and the crew can produce and place a cubic yard of concrete every 7 minutes. The concrete placing temperature is estimated to be 70 F. The materials for constructing the form are 2 by 4s, 3/4-inch thick plywood, and 3,000-pound (breaking strength) snap ties. The following steps will be used:

Step 1. Lay out available needed materials.

  • Studs: 2 x 4 (single)
  • Wales: 2 x 4 (doubled)
  • Sheathing: 3/4-inch plywood (strong)
  • Ties: Snap ties (3,000 pounds)

Step 2. Calculate the production rate.

Step 3. Calculate the plan area of form.

Step 4. Calculate rate the of placing.

Step 5. Determine the concrete placing temperature.

Temperature = 70 F

Step 6. Determine the MCP (refer to Figure 4-3).

MCP = 500 pounds per square foot

Step 7. Determine the MSS (refer to Table 4-2).

MSS = 15 inches

Step 8. Determine the ULS.

Step 9. Determine the MWS (refer to Table 4-3 using 700 pounds per foot load).

MWS = 22 inches

Step 10. Determine the ULW.

Step 11. Determine the tie spacing based on wale size (refer to Table 4-4, using 1,000 pounds per load).

Tie spacing = 27 inches

Step 12. Determine the tie spacing based on tie-rod strength (refer to Table 4-5).

Step 13. Determine the maximum tie spacing.

Tie spacing = 27 inches

Step 14. Insert the first tie-rod one-half the spacing from the end and one full spacing thereafter. Because tie-rods are being used, it is not necessary to adjust the tie/stud spacing.

Step 15. Calculate the number of studs per side.

Step 16. Calculate the number of double wales per side.

Step 17. Estimate the time required to place concrete.

BRACING FOR WALL FORMS

4-27. Braces are used against wall forms to keep them in place and in alignment and to protect them from mishaps due to external forces (wind, personnel, equipment, vibration, accidents). An equivalent force due to all of these forces (the resultant force) is assumed to be acting uniformly along the top edge of the form in a horizontal plane. For most military applications, this force is assumed to be 12.5 times the wall height, in feet. As this force can act in both directions, braces to be used should be equally strong in tension as in compression, or braces should be used on both sides of the wall forms. The design procedure is based on using a single row of braces and assumes that strong, straight, seasoned lumber will be used and that the braces are properly secured against the wall forms and the ground (both ends are secured). Once we know the height of the wall to be built and have selected a material (2 inches or greater) for the braces, we need to determine the maximum safe spacing of these braces (center-to-center) that will keep the formwork aligned.

NOMENCLATURE

4-28. The following terms are used in bracing for a form wall. See Figure 4-4.

LB = total length, in feet, of the brace member from end connection to end connection.
Lmax = the maximum allowable unsupported length of the brace, in feet, due to buckling and bending.
For all 2-inch material, Lmax = 6 1/4 feet; for all 4-inch material, L max = 14 1/2 feet.
L = the actual unsupported length, in feet, of the brace used.
h = the overall height, in feet, of the wall form.
y = the point of application of the brace on the wall form, measured in feet from the base of the form.
0 = the angle, in degrees, that the brace makes with the horizontal shoe brace. For the best effect, the angle should be between 20 and 60 degrees.
J = a factor to be applied which includes all constant values (material properties and assumed wind force). It is measured in feet4, see Table 4-6.
Smax = the maximum safe spacing of braces (feet), center-to-center, to support the walls against external forces.
cos = cosine: the ratio of the distance from the stake to the wall (x) divided by the length of the brace (LB.)
sin = sine: the ratio of the height of the brace (y) divided by the length of the brace (LB.)

 

Figure 4-4. Elements of diagonal bracing

Table 4-6. J factor

Material, in Inches J, In Feet4
2 x 4 2,360
2 x 6 3,710
2 x 8 4,890
2 x 10 6,240
2 x 12 7,590
4 x 4 30,010

PROCEDURE

4-29. The design procedure can best be explained by the example problem that follows:

Step 1. Determine the spacing of braces for a wall 10 feet high. Use 2-by 6- inch by 10-foot material, attached 6 feet from the bottom of the form. The following selected materials are given:

J = 3,710 feet4 (from Table 4-6)

Lmax = 6 1/4 feet (because of the 2-inch material)

LB = 10 feet

h = 10 feet (from example problem)

y = 6 feet (from example problem)

Step 2. Determine the angle of placement, 0.

0 = sin-1 (.600) = 37o

Step 3. Determine the L which is the actual unsupported length of brace. Since the Lmax for all 2-inch material is 6 1/4 feet and the brace in this problem is 10 feet long, we will have to use something to support the braces (usually a 1 by 4 or a 1 by 6). The best position for this support would be in the middle of the brace, thus giving L = 5 feet.

Step 4. Determine the Smax from the formula.

4-30. Using 2- by 6- inches by 10-feet brace applied to the top edge of the wall form at y = 6 feet, place these braces no further apart than 7 feet. After the braces are properly installed, connect all braces to each other at the center so deflection does not occur.

NOTE: This procedure determines the maximum safe spacing of braces. There is no doctrine that states the braces must be placed 7 feet apart--they can be less.

DISCUSSION

4-31. To fully understand the procedure, the following points lend insight to the formula.

Derivation of the formula has a safety factor of 3.

For older or green lumber, reduce Smax according to judgment.

For maximum support, attach braces to the top edge of the forms (or as closely as practicable). Also, better support will be achieved when 0 = 45o

Remember to use intermediate supports whenever the length of the brace (LB) is greater than Lmax.

Whenever there are choices of material, the larger size will always carry greater loads.

To prevent overloading the brace, place supports a minimum of 2 feet apart for all 2-inch material and 5 feet apart for all 4-inch material. This is necessary to prevent crushing of the brace.

OVERHEAD SLAB FORM DESIGN

4-32. There may be instances where a concrete slab will have to be placed above the ground such as bunker and culvert roofs. Carefully consider the design of the formwork because of the danger of failure caused by the weight of plastic concrete and the live load (LL) of equipment and personnel on the forms. The overhead slab form design method employs some of the same figures used in the wall-form design procedure.

NOMENCLATURE

4-33. The following terms are used in the form design (see Figure 4-5).

Sheathing. Shapes and holds the concrete. Plywood or solid sheet metal is best for use.

Joists. Supports the sheathing against deflection. Performs the same function as studs in a wall form. Use 2-, 3-, or 4-inch thick lumber.

Stringers. Supports the joists against deflection. Performs the same function as wales in a wall form. Use 2-inch thick or larger lumber. Stringers do not have to be doubled as wales are.

Shores. Supports the stringers against deflection. Performs the same functions as ties in a wall form and also supports the concrete at the desired elevation above the ground. Lumber at least as large as the stringer should be used, but never use lumber smaller than 4 by 4s.

Lateral bracing. Bracing may be required between adjacent shores to keep shores from bending under load. Use 1 by 6s or larger material for bracing material.

Cross bracing. Cross bracing will always be required to support the formwork materials.

Wedges. Wood or metal shims used to adjust shore height

Mudsill. Board which supports shores and distributes the load. Use 2-inch thick or larger lumber.

 

Figure 4-5. Typical overhead slab forms

DESIGN PROCEDURE

4-34. Use the following steps in determining form design:

Step 1. Lay out and specify the materials you will be using for the construction of the overhead roof slab. It is important that anyone using your design will know exactly the materials to use for each of the structural members.

Step 2. Determine the maximum total load (TL) the formwork will have to support.

The LL of materials, personnel, and equipment is estimated to be 50 pounds per square foot unless the formwork will support engine-powered concrete buggies or other power equipment. In this case, a LL of 75 pounds per square foot will be used. The LL is added with the DL of the concrete to obtain the maximum TL. The concrete's DL is obtained using the concrete's unit weight of 150 pounds per cubic foot. The formulas are--

TL = LL + DL

LL = 50 lb/sq ft, or 75 lb/sq ft with power equipment

Step 3. Determine the maximum joist spacing. Use Table 4-1, or Table 4-2 and determine the joist spacing based on the sheathing material used. This is the same procedure used in determining the MSS for wall-form design. Use the maximum TL in place of the MCP.

Step 4. Calculate the uniform load on the joists (ULJ). The same procedure is used as determining uniform loads on structural members in wall-form design.

Step 5. Determine the maximum stringer spacing. Use Table 4-3 and the ULJ calculated in step 4. Round this load up to the next higher load located in the left column of the table. Read right to the column containing the lumber material used as the joist. This is the member to be supported by the stringer. The value at this intersection is the on-center (OC) spacing of the stringer.

Step 6. Calculate the uniform load on the stringer (ULSstr).

Step 7. Determine the maximum shore spacing the following two ways:

Determine the maximum shore spacing based on the stringer strength. Use Table 4-3 orTable 4-4 (depending on the type of stringer) and the ULSstr, rounded to the next higher load shown in the left column of the table. Read right to the stringer material column. This intersection is the maximum OC spacing of the shore required to ensure the stringer is properly supported.

Determine the maximum shore spacing based on the allowable load. This determination is dependent on both the shore strength and the end bearing of the stringer on the shore. Using Tables 4-7 and Table 4-8, select the allowable load on the shore both ways as follows:

(NOTE: The unsupported length is equal to the height above the sill--sheathing thickness, joist thickness, and stringer thickness. This length is then rounded up to the next higher table value.)

Table 4-7. Allowable load (in pounds) on wood shores, based on shore strength

Nominal Lumber
Size, in Inches
4 x 4 4 x 6 6 x 6
Unsupported
Length, in Feet
R S4S R S4S R S4S
4 9,900 9,200 15,300 14,400 23,700 22,700
5 9,900 9,200 15,300 14,400 23,700 22,700
6 9,900 9,200 15,300 14,400 23,700 22,700
7 8,100 7,000 12,500 11,000 23,700 22,700
8 6,200 5,400 9,600 8,400 23,700 22,700
9 4,900 4,200 7,600 6,700 23,700 22,700
10 4,000 3,400 6,100 5,000 23,000 21,000
11 3,300 2,800 5,100 4,500 19,000 17,300
12 2,700 2,400 4,300 3,700 16,000 14,600
13 2,300 2,000 3,600 3,200 13,600 12,400
    l/d = 50   l/d = 50    
14 2,000 1,700 3,100 2,800 11,700 10,700
  l/d = 50   l/d = 50      
15 1,800   2,700   10,200 9,300
          9,000 8,200
          7,900 7,300
          7,100 6,500
          6,400 5,800
          5,700 5,200
NOTES:

1. The above table values are based on wood members with the following characteristics strength:
Compression parallel to gram = 750 psi.
E = 1,100,000 psi.
2. R indicates rough lumber.

Table 4-8. Allowable load on specified shores, in pounds, based on bearing stresses
where the maximum shore area is in contact with the supported member

Nominal Lumber
Size, in Inches
4 x 4 4 x 6 6 x 6
C ^ of
Member
Supported
R S4S R S4S R S4S
250 3,300 3,100 5100 4800 7900 7600
350 4600 4300 7100 6700 11100 10600
385 5100 4700 7800 7400 12200 11600
400 5300 4900 8200 7700 12700 12100
NOTES:
  1. When the compression perpendicular to the grain of the member being supported
    is unknown, assume the most critical C is ^ to the grain.
  2. R indicates rough lumber.
  3. S4S indicates surfaced on four sides.

1. When the compression perpendicular to the grain of the member being supported is unknown, assume the most critical C is ^ to the grain.

2. R indicates rough lumber.

3. S4S indicates surfaced on four sides.

- First, determine the allowable load based on the shore strength. Select the shore material dimensions and determine the unsupported length in feet of the shore. See Table 4-7. The allowable load for shore is given in pounds at the intersection. Read down the left column to the unsupported length of the shore, then read right to the column of the size material used as the shore.

- Second, determine the allowable load based on end bearing area. Select the size of the shore material and the compression C perpendicular (^) to the grain of the stringer. If the C ^ to the grain is unknown, use the lowest value provided in Table 4-8. Read down the left column to the C ^ to the grain of the stringer material and then right of the column of the shore material. The allowable load between the stringer and the shore will be in pounds.

- Compare the two loads just determined and select the lower as the maximum allowable load on the shore to be used in the formula below.

Calculate shore spacing by the following formula:

Step 8. Determine the most critical shore spacing. Compare the shore spacing based on the stringer strength in step 7, with the shore spacing based on the allowable load in step 7. Select the smaller of the two spacings.

Step 9. Check the shore bracing.

Verify that the unbraced length (l) of the shore (in inches) divided by the least dimension (d) of the shore does not exceed 50. If l/d exceeds 50, the lateral and cross bracing must be provided. Table 4-7 indicates the l/d > 50 shore lengths and can be used if the shore material is sound and unspliced. It is good engineering practice to provide both lateral and diagonal bracing to all shore members if material is available.

SAMPLE FORM DESIGN PROBLEM NUMBER THREE

4-35. Design the form for the roof of a concrete water tank to be 6 inches thick, 20 feet wide, and 30 feet long. The slab will be constructed 8 feet above the floor (to the bottom of the slab). Available materials are 3/4-inch plywood and 4 by 4 (S4S) (surfaced on four sides) lumber. Mechanical buggies will be used to place concrete. Use the following steps in design:

Step 1. Lay out available materials for construction.

Sheathing:

3/4-inch plywood (strong way)

Joists:

4 x 4 (S4S)

Shores:

4 x 4 (S4S)

Stringers:

4 x 4 (S4S)

Step 2. Determine the maximum TL.

LL = personnel and equipment = 75 lb/sq ft

TL = DL + LL

TL = 75 lb/sq ft + 75 lb/sq ft

TL = 150 lb/sq ft

Step 3. Determine the maximum joist spacing using Table 4-2.

3/4-inch plywood (strong way) and TL = 150 pounds per square foot

Joist spacing = 22 inches

Step 4. Calculate the ULJ.

Step 5. Determine the maximum stringer spacing using Table 4-3.

Load = 275 pounds per foot

Joist material = 4 x 4

Maximum stringer spacing = 55 inches

Step 6. Calculate the ULSstr.

Step 7. Determine the maximum shore spacing based on stringer strength. Use the maximum shore spacing shown in Table 4-3 single member stringers.

Load = 687.5 pounds per foot (round up to 700 pounds per foot)

Stringer material = 4 x 4 (S4S)

Maximum shore spacing = 35 inches, based on stringer strength

Step 8. Determine shore spacing based on allowable load. This determination is based on both the shore strength and end bearing stresses of the stringer to the shore. Determine both ways as follows:

Allowable loads based on shore strength are shown in Table 4-7.

Unsupported length = 8 feet - 3/4 inch - 3 1/2 inches - 3 1/2 inches = 7 feet 4 1/4 inches (round up to 8 feet).

For an 8-foot 4 by 4 (S4S), the allowable load = 5,400 pounds, based on shore strength.

For an allowable load based on end bearing stresses see Table 4-8. Since we do not know what species of wood we are using, we must assume the worst case. Therefore, the (C ^) to the grain = 250, and the allowable load for a 4 by 4 (S4S) = 3,100 pounds based on end bearing stresses.

Select the most critical of the two loads determined.

Since the (C ^) is less than the allowable load on the shore perpendicular to the grain (II), 3,100 pounds is the critical load.

Calculate the shore spacing as follows:

Step 9. Select the most critical shore spacing. The spacing determined in step 7 is less than the spacing determined in step 8; therefore, the shore spacing to be used is 35 inches.

Step 10. Check shore deflection.

Unsupported length = 8 feet - 3/4 inch - 3 1/2 inch - 3 1/2 inches = 7 feet 4 1/4 inches = 88.25 inches

d = least dimension of 4 x 4 (S4S) lumber = 3.5 inches

NOTE: Lateral bracing is not required. Cross bracing is always required.

4-36. Summary of materials needed for construction.

Sheathing:

3/4-inch plywood (strong way)

Joists:

4 x 4 (S4S) lumber spaced 22 inches OC

Stringer:

4 x 4 (S4S) lumber spaced 55 inches OC

Shores:

4 x 4 (S4S) lumber spaced 35 inches OC

Lateral braces:

not required

CONCRETE SLAB ON GRADE THICKNESS DESIGN

4-37. Concrete slabs on grade are the most often constructed concrete projects by engineer units. Many slabs that are constructed fail because the thickness of the slab is not adequate. In other cases, the slab thickness is so excessive that materials and personnel are wasted in the construction. Use the following procedure, tables, and figures for the design of concrete slab thickness in the field to eliminate possible failure or wasted materials.

BASIC ASSUMPTIONS

4-38. The following three points apply to the design thickness:

The minimum slab thickness will be 4 inches for class 1, 2, and 3 floors, as listed in Table 4-9.

Only the load area and the flexural strength required will be considered using this method. Whenever the loaded area exceeds 80 square inches, the soil-bearing capacity must be considered. When the load will be applied at the edges of the concrete slab, either thicken the edge by 50 percent and taper back to normal slab thickness at a slope of not more than 1 in 10 or use a grade beam to support the edge.

The controlling factor in determining the thickness of a floor on ground is the heaviest concentrated load the slab will carry. The load is usually the wheel load plus impact loading caused by the vehicle.

Table 4-9. Concrete floor classifications

Class Usual Traffic Use Special Consideration Concrete Finishing
Technique
1 Light foot Residential or tile-covered Grade for drainage; make plane for tile Medium steel trowel
2 Foot Offices, churches, schools, hospitals

Ornamental residential

Nonslip aggregate; mix in surface

Color shake, special

Steel trowel; special finish for nonslip

Steel trowel, color, exposed aggregate; wash if aggregate is to be exposed

3 Light foot and pneumatic wheels Drives, garage floors, sidewalks for residences Crown; pitch joints
Air entrainment
Float, trowel, and broom
4 Foot and pneumatic wheels* Light industrial commercial Careful curing Hard steel trowel and brush for nonslip
5 Foot and wheel abrasive wear* Single-course industrial, integral topping Careful curing Special hard aggregate float and trowel
6 Foot and steel-tire vehicles - severe abrasion Bonded two-course, heavy industrial Base Textured surface and bond Surface leveled by screeding
Top Special aggregate and/or mineral or metallic surface treatment Special disc-type power floats with repeated steel troweling
7 Same as classes 3, 4, 5, 6 Unbonded topping Mesh reinforcing; bond breaker on old concrete surface; minimum thickness 2 1/2 inches  
*Under abrasive conditions on floor surface, the exposure will be much more severe and a higher quality surface will be required for class 4 and 5; for a class 6, two-course floor, a mineral or metallic aggregate monolithic surface treatment is recommended.

DESIGN PROCEDURE

4-39. Use the following steps in design procedures for concrete slabs.

Step 1. Determine the floor classification. Use Table 4-9 with the usual traffic and use of the slab.

Step 2. Determine the minimum compressive strength. Using Table 4-10, read down the class of floors column until you find the floor classification selected. Read to the right and select the f 'c in psi. Note that this table gives the recommended slump range for the plastic concrete.

Table 4-10. Recommended slumps and compressive strengths

Class of Floors Slump Range, in inches Minimum Compressive
Strength, f 'c in Pounds
per Square Inch*Use
1 2-4 3,000
2 2-4 3,500
3 2-4 3,500
4 1-3 4,000
5 1-3 4,500
6 base 2-4 3,500
6 topping** 0-1 5,000 - 8,000
7 1-3 4,000
NOTE: These recommendations are specially for concrete made with normal-weight aggregate. For structural slabs, the requirements of ACI 318 and the contract documents should be met.

*Refer to the minimum compressive strength of cylinders made and tested according to applicable ASTM standards for 28 days. The average of any five consecutive strength tests of continuously moist-cured specimens representing each class of concrete should be equal to or greater than the specified strength.

**The cement content of the heavy-duty floor topping depends upon the severity of the abrasion. The minimum is 846 pounds per cubic yard.

Step 3. Determine the allowable flexural tensile stress f 't after the concrete f 'c is determined. Use the formula:

Step 4. Determine the equivalent static load (ESL). The expected impact loading is needed for this step. The impact loading is 25 percent more than the static load (SL) for the vehicles.

ESL = (1 + 25%) x SL

Step 5. Evaluate and correct the ESL if f 't is not equal to 300 psi. When calculating the allowable flexural tensile f 'c in step 4 above, an f 't was determined. If this f 't is not equal to 300 psi, then the ESL must be corrected based on a ratio between the standard (300 psi) and the actual f 't. This correction is necessary so that the standard thickness (see Figure 4-6) may be used to determine the required slab thickness. The procedure for correction is as follows:

Figure 4-6. Maximum wheel loads for industrial floors

Step 6. Determine the slab thickness. Using Figure 4-6, read up the left side until the TL is the same as the design CESL from 5 above. Read to the right until the loaded area (in square inches) for the slab design is intersected. The slab thickness can then be determined by interpolating between the slanted slab thickness lines on the figure. Round up to the next 1/4 inch.

Step 7. Determine the minimum cement content and recommended air content for structures subjected to freeze-thaw cycles, depending upon the maximum size of CA to be used.

4-42. The slab thickness design procedure is now complete, and with the accumulated information you should proceed to the mix design procedure as detailed in Chapter 3 of this manual.

SLAB CONSTRUCTION

4-40. Slab construction is often spoiled by improper construction practices that can be easily prevented. Forms for slabs on grades are relatively simple to construct. See Section III of this chapter. The following practices should always be employed for ensuring that the slab will hold up under service load.

Always use non-frost susceptible (free-draining) material under the slab. Material such as silts and very fine sand expands under frost action if groundwater is present. Removing these types of soil and replacing them with a free-draining material (gravel or coarse sand) to one-half of the expected frost penetration, (in addition to providing adequate drainage at the edges of the structure), will prevent frost heave.

Always use some form of steel reinforcement. Welded-wire mesh is very desirable for preventing excessive cracking and preventing widening of cracks that do develop.

Always keep the slab moist and protect it from excessive heat and freezing for the required curing time. This will ensure the concrete is cured properly.

SAMPLE SLAB CONSTRUCTION PROBLEM

4-41. Determine the slab thickness for the floor of a wheeled-vehicle shop. The total weight on one wheel is 7,500 pounds and the loaded area equals 40 square inches. The surface will be exposed to foot and wheel traffic. Abrasive wear is expected. The maximum-size aggregate will be 1 inch.

Step 1. Determine the floor classification, see Class 5, Table 4-9.

Step 2. Determine the minimum compressive strength from Table 4-10.

Step 3. Determine the allowable flexural tensile stress.

Step 4. Determine the ESL.

ESL = (1 +.25) x SL = 1.25 x 7,500 = 9375 lbs

Step 5. Evaluate and correct the ESL if f 't 300 psi

Step 6. Determine the slab thickness from Figure 4-6.

Loaded area = 40 sq in

Slab thickness = 6.6 in

NOTE: Always round up to the next higher 1/4-inch thickness for convenience. Therefore, the thickness is rounded up to 6 3/4-inches.

4-42.The minimum cement content per cubic yard from, Table 4-11, with 1-inch maximum-size aggregate (MSA) is 520 pounds. Because the structure is a vehicle-shop floor, the air content may be less than the table allowance. The slab design information is now:

f 'c = 4,500 psi

Slab thickness = 6 3/4 inches

Minimum cement content = 520 lb/cu yd

Air content = 6%

Table 4-11. Minimum cement contents and percentages of entrained air

Maximum Size of
Coarse Aggregate
Minimum Cement
Content, in Pounds
per Cubic Yard
Air Content for
Freeze-Thaw
Resistance,
Shown in
Percentage*
1 1/2 470 5 + 1
1 520 6 + 1
3/4 540 6 + 1
1/2 590 7 1/2 + 1
3/8 610 7 1/2 + 1
NOTES: These mixtures are specially for concrete made with normal-weight aggregate; different mixtures may be needed for lightweight aggregate concrete. For structural slabs, the requirements of ACI 318 and the contract document must be met.

*Smaller percentages of entrained air can be used for concrete floors that will not be exposed to freezing and thawing or deicing. This may improve the workability of the concrete and the finish product.

COLUMN FORM DESIGN

4-43. Use the following steps in determining design procedures of a wood form for a concrete column.

Step 1. Determine the materials you will use for sheathing, yokes, and battens. The standard materials for column forms are 2 by 4s and 1-inch sheathing.

Step 2. Determine the column height.

Step 3. Determine the largest cross-sectional column dimension.

Step 4. Determine the maximum yoke spacing by referring to Table 4-12. First, find the column height in feet in the first column. Then move right horizontally to the column heading of the largest cross-sectional dimension in inches of the column you are constructing. The center-to-center spacing between the second yoke and the base yoke is the lowest value in the interval that falls partly in the correct column height line. You can obtain all subsequent yoke spacing by reading up this column to the top. These are maximum yoke spacing; you can place yokes closer together. Adjust final spacing to be at the top of the column.

Table 4-12. Column yoke spacing using 2 by 4s and 1-inch cheathing

SAMPLE COLUMN FORM DESIGN PROBLEM

4-44. Determine the yoke spacing for a 9-foot column whose largest cross-sectional dimension is 36 inches. Construction materials are 2 by 4s and 1-inch sheathing.

Solution Steps:

Step 1. Lay out materials available.

2 x 4s and 1-inch sheathing

Step 2. Determine column height.

9 feet

Step 3. Determine the largest cross-sectional dimension.

36 inches

Step 4 and Step 5. Determine the maximum yoke spacing, refer to Table 4-12. Starting from the base, the yokes are: 8, 8, 10, 11, 12, 15, 17, 17, and 10-inches. The spacing between the top two yokes are reduced due to the limits of the column height. Adjust the final spacing to be at the top of the column.

David L. Heiserman, Editor

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All Rights Reserved

Revised: June 06, 2015