OneWay Wide Module (Skip) Joist Concrete Floor System Design
A typical floor plan of a 5story office building is shown below. Widemodule joists, or “skip” joists, are similar to standard oneway joists, except the pans are 53 in. or 66 in. wide. For the 53 in. pans, the pan depth varies from 16 in. to 24 in., and for 66 in. pans, the range is 14 in. to 24 in. Widemodule systems are economical for long spans with heavy loads and improved vibration resistance (see references). The gravity loads treatment is shown in this example and the lateral load effects are resisted by reinforced concrete shear walls. The design procedures shown in ACI 31814 are illustrated in detail in this example. The hand solution is also used for a detailed comparison with the analysis and design results of the engineering software program spBeam.
Figure 1 – OneWay Wide Module Joist Concrete Floor Framing System
Contents
1.1. Preliminary slab thickness and joist dimensions
2. Design of Structural Members
2.1.1. Determination of span loads
2.1.2. Determination of design moment and shear
2.1.6. Computer Program Solution
2.1.7. Summary and Comparison of Results
2.1.8. Conclusions and Observations
2.2.1. Determination of span loads
2.2.2. Determination of design moment and shear
2.2.6. Computer Program Solution
2.2.7. Summary and Comparison of Results
2.2.8. Conclusions and Observations
2.3. Design of Beam along Grid B (Interior Frame)
2.3.1. Determination of span loads
2.3.2. Determination of design moment and shear
2.3.6. Computer Program Solution
2.3.7. Summary and Comparison of Results
2.3.8. Conclusions and Observations
2.4. Design of Beam along Grid A (Exterior Frame)
2.4.1. Determination of span loads
2.4.2. Determination of design flexural moment, shear, and torsional moment
2.4.3. Flexural, Shear, and Torsion Design
2.4.5. Computer Program Solution
2.4.6. Summary and Comparison of Results
2.4.7. Conclusions and Observations
2.5. Design of Interior, Edge, and Corner Columns
2.5.1. Determination of factored loads
2.5.2. Column Capacity Diagram (AxialMoment Interaction)
Code
Building Code Requirements for Structural Concrete (ACI 31814) and Commentary (ACI 318R14)
Minimum Design Loads for Buildings and Other Structures (ASCE/SEI 710)
International Code Council, 2012 International Building Code, Washington, D.C., 2012
References
Notes on ACI 31811 Building Code Requirements for Structural Concrete, Twelfth Edition, 2013 Portland Cement Association.
Concrete Floor Systems (Guide to Estimating and Economizing), Second Edition, 2002 David A. Fanella
Simplified Design of Reinforced Concrete Buildings, Fourth Edition, 2011 Mahmoud E. Kamara and Lawrence C. Novak
Design Data
FloortoFloor Height = 12 ft (provided by architectural drawings)
w_{c} = 150 pcf
f_{c}’ = 5,000 psi
f_{y} = 60,000 psi (For flexural reinforcement)
f_{yt} = 60,000 psi (For shear and torsional reinforcement)
Superimposed dead load, SDL = 20 psf framed partitions, wood studs plaster 2 sides
ASCE/SEI 710 (Table C31)
Typical Floor Level, Live load, L_{o} = 80 psf (Office building) ASCE/SEI 710 (Table 41)
Roof Live Load, L_{o} = 20 psf (Ordinary flat roofs) ASCE/SEI 710 (Table 41)
Required fire resistance rating = 2 hours
Solution
In this example deflection will be calculated and checked to satisfy project deflection limits. Minimum member thickness and depths from ACI 31814 will be used for preliminary sizing.
a. Oneway Slab
Using minimum thickness for solid oneway slabs in Table 7.3.1.1 for the solid slab spanning between the ribs.
End Spans: _{}in ACI 31814 (Table 7.3.1.1)
Interior Spans: _{}in ACI 31814 (Table 7.3.1.1)
The minimum slab thickness for widemodule joists for 2hour fire rating is 4.6 in.
IBC 2012 (Table 720.1(3))
Therefore, select a slab thickness of 5 in. for all spans.
b. Oneway Joist
The widemodule joist systems do not meet the limitations of ACI 31814, 9.8.1.1 through 9.8.1.4. Therefore, the structural members of this type of joist construction shall be designed with standard provisions for slabs and beams.
ACI 31814 (9.8.1.8)
Using minimum thickness for nonprestressed beams in Table 9.3.1.1. For the ribs (part of the joists) supporting the solid slab.
End Span: _{}in (governs) ACI 31814 (Table 9.3.1.1)
Interior Span: _{}in ACI 31814 (Table 9.3.1.1)
Therefore, select rib depth of 16 in. for a total joist depth of 21 in.
Figure 2 – Slab and Joist Dimensions
a. Interior Columns
Select a preliminary size based on the axial load demand. Determine interior column loads as follows:
The governing load combination: _{} ACI 31814 (Eq. 5.3.1b)
Where:
D = Dead Load; L= Live Load; L_{r}= Roof Live Load
Typical Floor Level Loads
# of Floors = 4
Dead Loads, D
Selfweight of widemodule joist system (see Figure 2):
Joist average thickness = _{}
Weight of the joist = 0.5524 x 150 pcf = 82.83 psf.
Superimposed dead load = 20 psf
Live Load, L: Calculate the live load reduction per ASCE/SEI 710
_{} ASCE/SEI 710 (Eq. 41)
Where:
L = reduced design live load per ft^{2} of area supported by the member
L_{o} = unreduced design live load per ft^{2} of area supported by the member = 80 psf
K_{LL} = live load element factor ASCE/SEI 710 (Table 42)
A_{T} = tributary area _{}ft^{2}
_{}psf
Which satisfies _{}requirement for members supporting two or more floors.
ASCE/SEI 710 (4.7.2)
Roof Level Loads
Dead Loads, D
Selfweight of widemodule joist system (see Figure 2):
Joist average thickness = _{}
Weight of the joist = 0.5524 x 150 pcf = 82.83 psf.
No superimposed dead load at the roof
Roof Live Load, L_{r}: Calculate the roof live load reduction
_{}; _{} ASCE/SEI 710 (Eq 42)
Where:
_{}psf
_{}since _{} ft^{2} _{} ft^{2}
_{}for flat roof
_{}psf
Total Factored Load on 1^{st} story interior column (@ 1^{st} interior support)
Total Floor Load _{} _{}kips
Total Roof Load_{} lb_{}kips
Assume 24 in square column with 4 – # 11 vertical bars with design axial strength, _{}of
_{} ACI 31814 (22.4.2)
_{} lb
_{}kips
Column Selfweight _{}kips
Total Reaction @ 1^{st} interior support _{} kips _{} kips.
Therefore, the preliminary interior column size of 24 in. x 24 in. is adequate.
b. Edge (Exterior) Columns
Select a preliminary size based on the axial load demand. Therefore, the load takedown for an edge column is done as follows:
The governing load combination: _{} ACI 31814 (Eq. 5.3.1b)
Typical Floor Level Loads
# of Floors = 4
Dead Loads, D
Selfweight of widemodule joist system (see Figure 2):
Weight of the joist = 82.83 psf.
Superimposed dead load = 20 psf
Live Load, L: Calculate the live load reduction per ASCE/SEI 710
_{} ASCE/SEI 710 (Eq. 41)
Where:
L = reduced design live load per ft^{2} of area supported by the member
L_{o} = unreduced design live load per ft^{2} of area supported by the member = 80 psf
K_{LL} = live load element factor = 4 ASCE/SEI 710 (Table 42)
A_{T} = tributary area _{}ft^{2}
_{}psf
Which satisfies _{}requirement for members supporting two or more floors.
ASCE/SEI 710 (4.7.2)
Roof Level Loads
Dead Loads, D
Weight of the joist = 82.83 psf.
No superimposed dead load at the roof
Roof Live Load, L_{r}: Calculate the roof live load reduction
_{}; _{} ASCE/SEI 710 (Eq 42)
Where:
_{}psf
_{}, since 200 ft^{2} < A_{t} = 480 ft^{2} < 600 ft^{2}
_{}for flat roof
_{}psf
Total Factored Load on 1^{st} story edge column (@ 1^{st} interior support)
Total Floor Load _{} _{}kips
Total Roof Load_{} lb_{}kips
Assume 20 in square column with 4 – # 11 vertical bars with design axial strength, _{}of
_{} ACI 31814 (22.4.2)
_{} lb
_{}kips
Column Selfweight _{}kips
Total Reaction @ 1^{st} interior support _{} kips _{}kips.
Therefore, the preliminary edge column size of 20 in. x 20 in. is adequate.
c. Corner Columns
Select a preliminary size based on the axial load demand. Therefore, the load takedown for a corner column is done as follows:
The governing load combination: _{} ACI 31814 (Eq. 5.3.1b)
Typical Floor Level Loads
# of Floors = 4
Dead Loads, D
Selfweight of widemodule joist system (see Figure 2):
Weight of the joist = 82.83 psf.
Superimposed dead load = 20 psf
Live Load, L: Calculate the live load reduction per ASCE/SEI 710
_{} ASCE/SEI 710 (Eq. 41)
Where:
L = reduced design live load per ft^{2} of area supported by the member
L_{o} = unreduced design live load per ft^{2} of area supported by the member = 80 psf
K_{LL} = live load element factor = 4 ASCE/SEI 710 (Table 42)
A_{T} = tributary area _{}ft^{2}
_{}psf
Which satisfies _{}requirement for members supporting two or more floors.
ASCE/SEI 710 (4.7.2)
Roof Level Loads
Dead Loads, D
Weight of the joist = 82.83 psf.
No superimposed dead load at the roof
Roof Live Load, L_{r}: Calculate the roof live load reduction
_{}; _{} ASCE/SEI 710 (Eq 42)
Where:
_{}psf
_{}, since 200 ft^{2} < A_{t} = 240 ft^{2} < 600 ft^{2}
_{}for flat roof
_{}psf
Total Factored Load on 1^{st} story corner column (@ exterior support)
Total Floor Load _{} _{}kips
Total Roof Load_{} lb_{}kips
Assume 20 in square column with 4 – # 11 vertical bars with design axial strength, _{}of
_{} ACI 31814 (22.4.2)
_{} lb
_{}kips
Column Selfweight _{}kips
Total Reaction @ exterior support _{} kips _{}kips.
Therefore, the preliminary edge column size of 20 in. x 20 in. is adequate.
The design of the following structural members is performed and compared with results of the engineering software program spBeam:
2.1. OneWay Slab
2.2. OneWay Joist
2.3. Interior Beam
2.4. Exterior Beam
2.5. Interior Column
A unit strip of 1 ft is considered for the design of slab spanning between ribs. Note that ACI 31814 does not allow live load reduction for oneway slabs.
Figure 2.1 – Partial plan view illustrating slab design strip
Slab design involves the following steps:
2.1.1. Determination of span loads
2.1.2. Determination of design moments and shears
2.1.3. Flexural Design
2.1.4. Shear Design
2.1.5. Deflections
2.1.6. Computer Program Solution
2.1.7. Summary and comparison of design results
2.1.8. Conclusions and observations
The following gravity load combinations are considered:
_{} ACI 31814 (Eq. 5.3.1a)
_{} kips/ft per ft
_{} ACI 31814 (Eq. 5.3.1b)
_{} kips/ft per ft
Span loads are governed by the second load combination.
The factored moment and shear can be determined using the simplified method if the requirements are satisfied: ACI 31814 (6.5.1)
ü Members are prismatic.
ü Loads are uniformly distributed.
ü L ≤ 3D (0.08 kips/ft per ft ≤ 3 x 0.0825 kips/ft per ft)
ü There are at least two spans.
ü The longer of two adjacent spans does not exceed the shorter by more than 20 percent.
Thus, the approximate coefficients can be used. The factored moments and shears are determined and summarized in the following tables. ACI 31814 (Table 6.5.2 and Table 6.5.3)
Table 2.1.2.1 – OneWay Slab Design Moment Values 

Location 
Design Moment Value 

End Spans 
Exterior Support Negative 
_{}ftkips/ft 
Midspan 
_{}ftkips/ft 

Interior Support Negative 
_{}ftkips/ft 

Interior Spans 
Midspan Positive 
_{}ftkips/ft 
Support Negative 
_{}ftkips/ft 
Table 2.1.2.2 – OneWay Slab Design Shear Values 

Location 
Design Shear Value 
End Span at Face of First Interior Support 
_{} kips/ft 
At Face of all other Supports 
_{} kips/ft 
For the oneway slab of a widemodule joist system, a single layer of longitudinal reinforcement is provided. The first interior support negative moment governs the design as tabulated in Table 2.1.2.1. Therefore, it is favorable to place the single layer reinforcement closer to the top fiber of the concrete slab. The required reinforcement shall be calculated for the first interior support negative moment first. The required reinforcement for the end span positive moment shall also be calculated as the low effective depth due to the reinforcement location may govern the required reinforcement amount. Finally, the required reinforcement for design shall be checked against the minimum shrinkage and temperature reinforcement requirement per ACI 31814 (24.4.3.2).
Calculate the required reinforcement to resist the first interior support negative moment:
_{} ftkips/ft
Use welded wire fabric reinforcement, 6 x 6W5.5 x W5.5 with 1.5 in. concrete cover. The distance from extreme compression fiber to the centroid of longitudinal tension reinforcement, d, is calculated below:
_{}in.
To determine the area of steel, assumptions have to be made whether the section is tension or compression controlled, and regarding the distance between the resultant compression and tension forces along the slab section (jd). In this example, tensioncontrolled section will be assumed so the reduction factor_{}is equal to 0.9, and jd will be taken equal to 0.95d. The assumptions will be verified once the area of steel is finalized.
Assume_{}in.
Unit strip width, b = 12 in.
_{} in.^{2}/ft
Recalculate ‘a’ for the actual A_{s} = 1.31 in.^{2}: _{} in.
_{}in.
_{}
Therefore, the assumption that section is tensioncontrolled is valid.
_{} in.^{2}/ft
Calculate the required reinforcement to resist the positive moment:
_{} ftkips/ft
The distance from extreme compression fiber to the centroid of longitudinal tension reinforcement:
_{} in.
To determine the area of steel, assumptions have to be made whether the section is tension or compression controlled, and regarding the distance between the resultant compression and tension forces along the slab section (jd). In this example, tensioncontrolled section will be assumed so the reduction factor_{}is equal to 0.9, and jd will be taken equal to 0.95d. The assumptions will be verified once the area of steel is finalized.
Assume_{}in.
Unit strip width, b = 12 in.
_{} in.^{2}/ft
Recalculate ‘a’ for the actual A_{s} = 0.070 in.^{2}: _{} in.
_{}in.
_{}
Therefore, the assumption that section is tensioncontrolled is valid.
_{} in.^{2}/ft
Check the shrinkage and temperature reinforcement requirement:
_{}in.^{2} / ft ACI 31814 (Table 24.4.3.2)
Check reinforcement spacing for crack control:
The maximum spacing of the flexural reinforcement closest to the tension face of the slab shall be:
_{}, but not greater than _{} ACI 31814 (Table 24.3.2)
Where:
_{}Maximum reinforcement spacing for crack control, in
_{}Calculated stress in reinforcement closest to the tension face at service load, ksi
_{}The least distance from surface of reinforcement to the tension face, in
_{}psi ACI 31814 (24.3.2.1)
_{}in. for reinforcement resisting negative moment at supports (i.e. tension at the top)
_{}in. for reinforcement resisting positive moment at midspan (i.e. tension at the bottom)
Thus,
At supports
_{}in (governs @ support)
But not greater than _{}in.
At midspan
_{}in (governs @ midspan)
But not greater than _{}in.
Therefore, for this oneway slab, the shrinkage and temperature reinforcement requirement per ACI 31814 (Table 24.4.3.2) governs the required reinforcement area (_{}in^{2}/ft) and crack control requirement per ACI 31814 (Table 24.3.2) governs the reinforcement spacing (_{}in.).
The most feasible reinforcement solution that meets both requirements mentioned above is to provide welded wire fabric reinforcement, 6 x 6W5.5 x W5.5. Note that the welded wire reinforcement selected provides minimum shrinkage and temperature reinforcement in the slab direction parallel to the joists as well. Alternately, deformed bars can be utilized in lieu of welded wire fabric. It should be noted that two conditions specific to this design contribute to having such a stringent spacing requirement.
These are listed below:
 The 5 in. slab has a single layer reinforcement that is placed near the top surface (i.e. clear cover from the top surface to the reinforcement is 1.5 in. This result in a high c_{c} value for the calculation of reinforcement spacing for crack control due to positive moment.
 The stress in reinforcement closest to the tension face at service load, f_{s}, is taken as 2/3 f_{y} as permitted by ACI 31814 without calculation. It is very likely that under the loading considered, the stress in the steel will be lower than 2/3 f_{y}. The f_{s} value is expected to be in the range of 1/3 f_{y} to 1/2 f_{y}. Even if it is assumed to be 1/2 f_{y} , s value will be 12 in.
From Table 2.1.2.2 above, the shear value in end span at face of first interior support governs.
_{}kips/ft
The design shear at a distance, d, away from the face of support,
_{}kips/ft
Shear strength provided by concrete
_{} ACI 31814 (Eq. 22.5.5.1)
_{} lb/ft _{}kips/ft
_{}kips/ft _{}kips/ft .
Therefore, the slab shear capacity is adequate.
Since the preliminary slab thickness met minimum thickness requirement, the deflection calculations are not required. Unless governed by fire rating requirements; as in this example; lesser thicknesses and consequently cost savings can be achieved through deflection computations. Deflection values are calculated and provided for every model created by spBeam Program and can be used by the engineer to make additional optimization decisions.
spBeam Program can be utilized to analysis and design beams and oneway slab systems. The oneway slab is modeled as 1ft unit strip supported on ribs. The ribs provide some rotational stiffness at the supports. In spBeam solution, the rotational stiffness is assumed as 32,000^{*} kipin/rad for modeling the joist supports. Also, for oneway slab run, the rib widths assumed as 6 in. and modeled through dummy columns of 6 in. x 12 in. with zero height (i.e. column stiffness is zero, but the 6 in. dimension of the column is utilized to push the design moments 3 in. from the support centerline). In this example, userdefined bar size #2 is defined in spBeam to represent welded wire fabric, W5.5, with the crosssectional area of 0.055 in^{2} (see Fig. 2.1.6.1).
Figure 2.1.6.1 – spBeam Reinforcement Database – Userdefined Bar Set
The program calculates the internal forces (shear force and bending moment), moment and shear capacities, immediate and longterm deflections, and required reinforcements. The graphical and text results are provided below for input and output of the spBeam program. The graphical and text results are provided here for both input and output of the spBeam model.
^{*} Refer to spBeam manual (Chapter 2 – Special Considerations for One and TwoWay Joist Systems)
^{*} Refer to spBeam manual (Chapter 4 – Defining Boundary Conditions, Rotational Stiffness)
Figure 2.1.6.2 – spBeam Model – Isometric View of 15 Span – 1ft Wide Unit Strip of OneWay Slab
Figure 2.1.6.3 – spBeam Model – Loads (Including Live Load Patterning) units in lb/ft^{2}
Figure 2.1.6.4 – spBeam Model – Internal Forces (Shear Force Diagram and Bending Moment Diagram)
Figure 2.1.6.5 – spBeam Model – Moment Capacity Diagram
Figure 2.1.6.6 – spBeam Model – Shear Capacity Diagram
Figure 2.1.6.7 – spBeam Model – Immediate Deflection Diagram
Figure 2.1.6.8 – spBeam Model – Reinforcement Diagram
Table 2.1.7.1 – Comparison of Hand Solution with spBeam Solution 

Flexural Design 

Span Location 
Design Moment (ftkips/ft) 
Reinforcement Required for Flexure (in^{2}/ft) 
Minimum Reinforcement (in^{2}/ft) (Shrinkage & Temperature Reinforcement) 

End Span 
Hand Solution 
spBeam Solution 
Hand Solution 
spBeam Solution 
Hand Solution 
spBeam Solution 
Interior Negative 
0.69 
0.63 
0.046 
0.042 
0.108 (2#2) 
0.108 (2#2) 
Positive 
0.49 
0.49 
0.069 
0.069 
0.108 (2#2) 
0.108 (2#2) 
Shear Design 

Span Location 
V_{u} (kips/ft) 
φV_{n} (kips/ft) 

End Span 
Hand Solution 
spBeam Solution 
Hand Solution 
spBeam Solution 

Interior Negative 
0.69 
0.67 
2.15 
2.08 
Minimum reinforcement requirement governed flexural design in this example. spBeam program enables the user to enter the rotational support springs as boundary conditions for joist supports and evaluate various analysis and design options beyond the limitations of the simplified method. The coefficients traditionally used to determine moments do not address various types of support and geometry.
Typically, in widemodule joist construction, oneway slab is reinforced with single layer reinforcement placed near the top in the primary direction. As seen in this example, this may cause crack control criteria to govern the reinforcement spacing and consequently, it may warrant the use of welded wire fabric reinforcement instead of deformed bar.
The maximum calculated total immediate (instantaneous) deflection (DL + LL) = 0.003 in., this value can be compared with maximum permissible calculated deflection limitation per project criteria in accordance to ACI 31814. ACI 31814 (Table 24.2.2)
In addition to deflection results, parametric studies can be performed in spBeam to optimize design and detailing results. Note in the reinforcement diagram (Figure 2.1.6.8) 2#2 for top reinforcement in the span left and right zones as well as span bottom reinforcement. One layer is suitable by inspection (Figure 2.1.6.5) to meet the required area of steel for top and bottom reinforcement.
The widemodule joists in this floor are considered as beams per ACI 31814 (9.8.1.8). Therefore, the design of the joist shall conform to the requirements of Tbeams per ACI 31814 (9.2.4).
Figure 2.2 – Partial plan view illustrating oneway joist to be design
Joist design involves the following steps:
2.2.1. Determination of span loads
2.2.2. Determination of design moments and shears
2.2.3. Flexural Design
2.2.4. Shear Design
2.2.5. Deflections
2.2.6. Computer Program Solution
2.2.7. Summary and comparison of design results
2.2.8. Conclusions and observations
The following gravity load combinations are considered:
_{} ACI 31814 (Eq. 5.3.1a)
_{} kips/ft
_{} ACI 31814 (Eq. 5.3.1b)
_{}
_{}kips/ft
Span loads are governed by the second load combination.
Note that for Floor Live Load Reduction per ASCE/SEI 710:
_{} ASCE/SEI 710 (Eq. 41)
Where:
Live Load Element Factor, _{}for interior beams ASCE/SEI 710 (Table 42)
Tributary Area _{}ft^{2}
Since_{} ft^{2} _{} ft^{2}, live load reduction is not applicable.
The factored moment and shear can be determined using the simplified method if the requirements are satisfied: ACI 31814 (6.5.1)
ü Members are prismatic.
ü Loads are uniformly distributed.
ü L ≤ 3D (0.48 kips/ft ≤ 3 x 0.62 kips/ft)
ü There are at least two spans.
ü The longer of two adjacent spans does not exceed the shorter by more than 20 percent.
Thus, the approximate coefficients can be used. The factored moments and shears are determined and summarized in the following tables. ACI 31814 (Table 6.5.2 and Table 6.5.3)
Table 2.2.2.1 – OneWay Joist Design Moment Values 

Location 
Design Moment Value 

End Spans 
Exterior Support Negative 
_{}ftkips 
Midspan 
_{}ftkips 

Interior Support Negative 
_{}ftkips 

Interior Spans 
Midspan Positive 
_{}ftkips 
Support Negative 
_{}ftkips 
Table 2.2.2.2 – OneWay Joist Design Shear Values 

Location 
Design Shear Value 
End Span at Face of First Interior Support 
_{}kips 
At Face of all other Supports 
_{}kips 
* When support beam is wider than the column, the clear span, ln, of the joists is measured from the face of the column. For calculating negative moments, l_{n}, is taken as the average of the adjacent clear spans.
ACI 31814 (6.5.2)
For the oneway joist of a widemodule joist system, the end span moment values govern the design as tabulated in Table 2.2.2.1.
Calculate the required reinforcement to resist the first interior support negative moment:
_{} ftkips
Use #5 reinforcement with 1.5 in. concrete cover. The distance from extreme compression fiber to the centroid of longitudinal tension reinforcement, d, is calculated below:
_{}in.
To determine the area of steel, assumptions have to be made whether the section is tension or compression controlled, and regarding the distance between the resultant compression and tension forces along the slab section (jd). In this example, tensioncontrolled section will be assumed so the reduction factor_{}is equal to 0.9, and jd will be taken equal to 0.9d since we are designing for the negative moment in a Tbeam (narrow compression zone). The assumptions will be verified once the area of steel is finalized.
Assume_{}in.
Joist average width, _{}in.
The required reinforcement at initial trial is calculated as follows:
_{} in.^{2}
Recalculate ‘a’ for the actual A_{s} = 1.79 in.^{2}: _{} in.
_{}in.
_{}
Therefore, the assumption that section is tensioncontrolled is valid.
_{} in.^{2}
The minimum reinforcement shall not be less than
_{} in.^{2} ACI 31814 (9.6.1.2(a))
And not less than
_{} in.^{2} ACI 31814 (9.6.1.2(b))
Part of the negativemoment steel shall be distributed over a width equal to the smaller of the effective flange width (72 in) and _{}in. ACI 31814 (24.3.4)
Where the effective width of the overhanging flange on each side of the beam web is the smallest of the following: ACI 31814 (6.3.2.1)
_{}in., where h is the slab thickness.
_{}in., where s_{w} is the clear distance to the adjacent web.
_{}in
Therefore, the effective flange width is 72 in
Provide 6# 5 bars within 38.4 in width.
_{}in^{2} _{} in^{2} o.k.
Calculate the required reinforcement to resist the positive moment:
_{} ftkips
In the positive moment regions, the beam acts as a Tshaped beam. The effective flange width as was calculated earlier is 72 in.
By assuming #3 bars for joist stirrups and the maximum bar size for joist bottom reinforcement as #7 and following the 1.5 in. concrete cover to reinforcement requirement of beam stirrups per ACI 31814 (20.6.1), the distance from extreme compression fiber to the centroid of longitudinal tension reinforcement, d, is calculated below:
_{}in
Since we are designing for the positive moment in a TBeam (wide compression zone), select a moment arm, jd approximately equal to 0.95d.Assume that _{}in.
_{}in^{2}
Recalculate ‘a’ for the actual A_{s} = 1.23 in.^{2}: _{} in.
_{}in.
_{}
Therefore, the assumption that section is tensioncontrolled is valid.
_{}in.^{2}
Use 2#7 bundled bars with _{}in^{2} > _{} in^{2} o.k.
Figure 2.2.3.1 – Crosssectional view at joist midspan (Section 2/2.2 in Figure 2.2)
Figure 2.2.3.2 – Crosssectional view at joist near support face (Section 3/2.2 in Figure 2.2)
From Table 2.2.2.2 above, the shear value in end span at face of first interior support governs.
_{}kips
The design shear at a distance, d, away from the face of support,
_{}kips / ft
Shear strength provided by concrete
_{} ACI 31814 (Eq. 22.5.5.1)
_{}lb _{}kips
Since_{}, shear reinforcement is required.
Try # 3, Grade 60 doubleleg stirrups with a 90° hook.
The nominal shear strength required to be provided by shear reinforcement is
_{}kips
Check whether V_{s} is less than _{}
If V_{s} is greater than_{}, then the crosssection has to be revised as ACI 31814 limits the shear capacity to be provided by stirrups to_{}. ACI 31814 (22.5.1.2)
_{}lb _{}kips
Since _{}does not exceed_{}. The crosssection is adequate.
Assume # 3 stirrups with two legs (_{} in^{2})
Calculate the required stirrup spacing as
_{}in.
Check whether the required spacing based on the shear demand meets the spacing limits for shear reinforcement per ACI 31814 (9.7.6.2.2).
Check whether V_{s} is less than_{}
_{} lb _{}kips > V_{s} = 12.6 kips
Therefore, maximum stirrup spacing shall be the smallest of d/2 and 24 in. ACI 31814 (Table 9.7.6.2.2)
_{}
This value governs over the required stirrup spacing of 19.7 in which was based on the demand.
Joist minimum shear reinforcement requirements must be checked since widemodule joists do not satisfy ACI 31814 (9.8).
Check the maximum stirrup spacing based on minimum shear reinforcement
_{}in. (does not govern) ACI 31814 (10.6.2.2(a))
_{}in. (does not govern) ACI 31814 (10.6.2.2(b))
Therefore, s_{max} value is governed by the spacing limit per ACI 31814 (9.7.6.2.2), and is equal to 9.35 in.
Use #3 @ 9 in. stirrups
_{} ACI 31814 (22.5.1.1 and 22.5.10.5.3)
_{}kips
_{}kips > _{}kips o.k.
Compute where _{}is equal to_{}, and the stirrups can be stopped
_{}
At interior end of the exterior span, use 16#3 @ 9 in o.c., Place 1^{st} stirrup 2 in. from the face of supporting girder.
Since the preliminary joist depth met minimum depth requirement, the deflection calculations are not required. A lesser depth maybe possible and consequently cost savings can be achieved through deflection computations. Deflection values are calculated and provided for every model created by spBeam Program and can be used by the engineer to make additional optimization decisions.
spBeam Program can be utilized to analysis and design the oneway widemodule joist. A single widemodule joist is modeled as a five span continuousbeam.
The program calculates the internal forces (shear force and bending moment), moment and shear capacities, immediate and longterm deflection results, and required flexural reinforcement. The graphical and text results are provided here for both input and output of the spBeam model.
The ribs are modeled as a rectangular longitudinal beam with an equivalent width of 7.33 in. and 21 in. depth to reflect the sloped sides of the forming pans.
Figure 2.2.6.1 – spBeam Model –OneWay Joist Section
Figure 2.2.6.2 – spBeam Model – Isometric View – OneWay Joist
Figure 2.2.6.3 – spBeam Model – Loads (Including Live Load Patterning)
Figure 2.2.6.4 – spBeam Model – Internal Forces (Shear Force Diagram and Bending Moment Diagram)
Figure 2.2.6.5 – spBeam Model – Moment Capacity Diagram
Figure 2.2.6.6 – spBeam Model – Shear Capacity Diagram
Figure 2.2.6.7 – spBeam Model – Immediate Deflection Diagram
Figure 2.2.6.8 – spBeam Model – Reinforcement Diagram
Table 2.2.7.1 – Comparison of Hand Solution with spBeam Solution 

Flexural Design 

Span Location 
Design Moment (ftkips) 
Reinforcement Required for Flexure (in^{2}) 
Reinforcement Provided for Flexure (in^{2}) 

End Span 
Hand Solution 
spBeam Solution 
Hand Solution 
spBeam Solution 
Hand Solution 
spBeam Solution 
Interior Negative 
136.6 
128.46 
1.78 
1.658 
6#5 
6#5 
Positive 
98.2 
88.98 
1.18 
1.064 
2#7 
2#7 
Shear Design 

Span Location 
V_{u} (kips) 
φV_{n} (kips) 

End Span 
Hand Solution 
spBeam Solution 
Hand Solution 
spBeam Solution 

Interior Negative 
23.9 
22.96 
35 
35.44 
In this design example, the oneway joist system is modeled as a continuous Tbeam representing single oneway joist. There is a good agreement between the hand solution and computer solution. Note that the coefficients traditionally used to determine moments do not address various types of support and geometry.
The maximum calculated total immediate (instantaneous) deflection (DL + LL) = 0.316 in., this value can be compared with maximum permissible calculated deflection limitation per project criteria in accordance to ACI 31814. ACI 31814 (Table 24.2.2)
In addition to deflection results, parametric studies can be performed in spBeam to optimize design and detailing results. With a minimum spacing of 1 in. between the 2#7, two stirrups, and 1.5 in. cover on each side, a total width of 6.5 in. is required. The rib width at the bar level is 6.385 in. which is slightly less than required. For detailing purposes, one of the following options can be used:
1. Bottom bars can be bundled. This practice is often found in joist construction.
2. Stirrups can be rotated by a small angle to preserve the minimum spacing.
3. Bottom bars can be raised sufficiently to achieve the required width taking into the account the reduction into the moment capacity.
4. Other detailing options provided by the builder/formwork supply.
Figure 2.2.8 – Joist CrossSection
In widemodule joist construction, the supporting beam, sometimes referred to as girder, depth is typically set to match the overall joist depth. Therefore, the beam depth is set to 21 in. This depth need to satisfy the minimum depth requirement of ACI 31814 (Table 9.3.1.1) so that the deflection computations can be waived.
Using the minimum depth for nonprestressed beams in Table 9.3.1.1.
End Span: _{}in (governs) < 21 in. ACI 31814 (Table 9.3.1.1)
Interior Span: _{}in ACI 31814 (Table 9.3.1.1)
Therefore, the preliminary beam depth satisfies the minimum depth requirement.
Figure 2.3 – Partial plan view showing interior beam along grid B
Beam (girder) design involves the following steps:
2.3.1. Determination of span loads
2.3.2. Determination of design moments and shears
2.3.3. Flexural Design
2.3.4. Shear Design
2.3.5. Deflections
2.3.6. Computer Program Solution
2.3.7. Summary and comparison of design results
2.3.8. Conclusions and observations
Dead Load:
Try 36 in width for the beam (slightly larger than the column width that helps facilitate the forming, and reduces the beam longitudinal vs. column vertical bar interference)
Joist & Slab Weight_{ }_{}kips/ft
Beam Weight _{} kips/ft
Superimposed Dead Load, SDL _{} kips/ft
Live Load:
Check for live load reduction per ASCE/SEI 710
_{} ASCE/SEI 710 (Eq. 41)
Where:
_{}reduced design live load per ft^{2} of area supported by the member
_{}unreduced design live load per ft^{2} of area supported by the member = 80 psf
_{}live load element factor = 2 for interior beams ASCE/SEI 710 (Table 42)
_{}tributary area _{}ft^{2}
_{}psf
Which satisfies _{}requirement for members supporting one floor.
ASCE/SEI 710 (4.7.2)
_{}kips/ft
Load Combination:
The following gravity load combinations are considered:
_{} ACI 31814 (Eq. 5.3.1a)
_{} kips/ft
_{} ACI 31814 (Eq. 5.3.1b)
_{}kips/ft
The span loads are governed by the second load combination.
The factored moment and shear can be determined using the simplified method if the requirements are satisfied: ACI 31814 (6.5.1)
ü Members are prismatic.
ü Loads are uniformly distributed.
ü L ≤ 3D (1.52 kips/ft ≤ 3 x 3.83 kips/ft)
ü There are at least two spans.
ü The longer of two adjacent spans does not exceed the shorter by more than 20 percent.
Thus, the approximate coefficients can be used. The factored moment and shear are determined and summarized in the following tables. ACI 31814 (Table 6.5.2 and Table 6.5.3)
Table 2.3.2.1 – Interior Beam Design Moment Values 

Location 
Design Moment Value 

End Spans 
Exterior Support Negative 
_{}ftkips 
Midspan 
_{}ftkips 

Interior Support Negative 
_{}ftkips 

Interior Spans 
Midspan Positive 
_{}ftkips 
Support Negative 
_{}ftkips 
Table 2.3.2.2 – Interior Beam Design Shear Values 

Location 
Design Shear Value 
End Span at Face of First Interior Support 
_{}kips 
At Face of all other Supports 
_{}kips 
For this interior beam, the end span moment values govern the design as tabulated in Table 2.3.2.1.
Calculate the required reinforcement to resist the first interior support negative moment:
_{} ftkips
Use #8 bars with 1.5 in. concrete cover per ACI 31814 (Table 20.6.1.3.1). To avoid interference with joist negative moment reinforcement, the clear cover to the girder top reinforcement is required to be increased by lowering the girder top reinforcement. The distance from extreme compression fiber to the centroid of longitudinal tension reinforcement, d, is calculated below:
_{}in.
To determine the area of steel, assumptions have to be made whether the section is tension or compression controlled, and regarding the distance between the resultant compression and tension forces along the slab section (jd). In this example, tensioncontrolled section will be assumed so the reduction factor_{}is equal to 0.9, and jd will be taken equal to 0.9d since we are designing for the negative moment in a rectangular beam (narrow compression zone). The assumptions will be verified once the area of steel is finalized.
Assume_{}in.
Interior beam width, _{}in.
The required reinforcement at initial trial is calculated as follows:
_{} in.^{2}
Recalculate ‘a’ for the actual A_{s} = 7.59 in.^{2}: _{}
_{}in.
_{}
Therefore, the assumption that section is tensioncontrolled is valid.
_{}
The minimum reinforcement shall not be less than
_{} ACI 31814 (9.6.1.2(a))
And not less than
_{} ACI 31814 (9.6.1.2(b))
Provide 10  # 8 bars:
_{}in^{2} _{} in^{2} o.k.
Maximum spacing allowed:
Check the requirement for distribution of flexural reinforcement to control flexural cracking:
_{} ACI 31814 (Table 24.3.2)
_{}in.
Use _{}ksi ACI 31814 (24.3.2.1)
_{}in. (governs)
_{}in.
Spacing provided for 10# 8 bars_{} _{}in. _{}in. o.k.
Where d_{s} = 2.625 in. for #3 stirrup as shown in the following Figure. CRSI 2002 (Figure 129)
Figure 2.3.3 – Maximum number of bars in beams
Check the spacing, s provided, is greater than the minimum center to center spacing, s_{min} where
_{} CRSI 2002 (Figure 129)
Where maximum aggregate size is ¾”
_{}
Since the spacing provided is greater than 2 in. Therefore, 10#8 bars are o.k.
All the values on Table 2.3.3.1 are calculated based on the procedure outlined above.
Table 2.3.3.1 – Reinforcing Design Summary 


End Span 
Interior Span 


Exterior Negative 
Positive 
Interior Negative 
Positive 
Negative 
Design Moment, M_{u} (ftkips) 
348.2 
397.9 
553.5 
344 
500.3 
Effective depth, d (in.) 
18.0^{*} 
18.625^{**} 
18^{*} 
18.625^{**} 
18^{*} 
A_{s} req’d (in.^{2}) 
4.53 
5.03 
7.45 
4.31 
6.68 
A_{s} min (in.^{2}) 
2.29 
2.37 
2.29 
2.37 
2.29 
Reinforcement 
6#8 
7#8 
10#8 
6#8 
9#8 
^{*} The beam top bars are to be placed below the joist top bars. 

^{**} The beam bottom bars are to be placed at the bottommost layer. The joist bottom bars, then, shall be spliced at joistbeam intersection. 
From Table 2.3.2.2 above, the shear value in end span at face of first interior support governs.
_{}kips
The design shear at a distance, d, away from the face of support,
_{}kips
Shear strength provided by concrete
_{} ACI 31814 (Eq. 22.5.5.1)
_{}lb _{}kips
Since_{}, shear reinforcement is required.
Try # 3, Grade 60 fourleg stirrups (A_{v} = 0.44 in.^{2}) with a 90° hook.
The nominal shear strength required to be provided by shear reinforcement is
_{}kips
Check whether V_{s} is less than _{}
If V_{s} is greater than_{}, then the crosssection has to be revised as ACI 31814 limits the shear capacity to be provided by stirrups to_{}. ACI 31814 (22.5.1.2)
_{}lb _{}kips
Since _{}does not exceed_{}. The crosssection is adequate.
Calculate the required stirrup spacing as
_{}in. ACI 31814 (22.5.10.5.3)
Check whether the required spacing based on the shear demand meets the spacing limits for shear reinforcement per ACI 31814 (9.7.6.2.2).
Check whether V_{s} is less than_{}
_{} lb _{}kips > V_{s} = 46 kips
Therefore, maximum stirrup spacing shall be the smallest of d/2 and 24 in. ACI 31814 (Table 9.7.6.2.2)
_{}
This value governs over the required stirrup spacing of 19.7 in which was based on the demand. Note that since the stirrup spacing is governed by s_{max}, the size of the stirrup can be kept as # 3. Selecting # 4 stirrup size will produce capacity more than what is required and therefore, be uneconomical.
Check the maximum stirrup spacing based on minimum shear reinforcement
_{}in. (does not govern) ACI 31814 (10.6.2.2(a))
_{}in. (does not govern) ACI 31814 (10.6.2.2(b))
Therefore, s_{max} value is governed by the spacing limit per ACI 31814 (9.7.6.2.2), and is equal to 9 in.
Use # 3 @ 8 in. stirrups
_{} ACI 31814 (22.5.1.1 and 22.5.10.5.3)
_{}kips
_{}kips > _{}kips o.k.
Compute where _{}is equal to_{}, and the stirrups can be stopped
_{}
At interior end of the exterior span, use 16# 3 @ 8 in o.c., Place 1^{st} stirrup 2 in. from the face of the column.
Since the preliminary beam depth met minimum depth requirement, the deflection calculations are not required. A lesser depth maybe possible and consequently cost savings can be achieved through deflection computations. Deflection values are calculated and provided for every model created by spBeam Program and can be used by the engineer to make additional optimization decisions.
spBeam Program can be utilized to analyze and design the interior continuous beam along grid B. The beam is modeled as a three span continuous rectangular beam.
The program calculates the internal forces (shear force and bending moment), moment and shear capacities, immediate and longterm deflection results, and required flexural reinforcement. The graphical and text results are provided here for both input and output of the spBeam model.
The beam is modeled as a 36 in. by 21 in. deep rectangular longitudinal beam with column supports. The supports can be modeled as pinned, fixed, or using actual geometric properties of the beamcolumn joint. A value of 100 is used in this model for column stiffness share, indicating the actual column stiffness. When the percentage lies between zero and 100%, the joint stiffness contribution by the column is multiplied by that percentage. The default value is 100%.
Figure 2.3.6.1 – spBeam Model – Support Data
Figure 2.3.6.2 – spBeam Model – Isometric View – Interior Continuous Beam along Grid B
Figure 2.3.6.3 – spBeam Model – Loads (Including Live Load Patterning)
Figure 2.3.6.4 – spBeam Model – Internal Forces (Shear Force Diagram and Bending Moment Diagram)
Figure 2.3.6.5 – spBeam Model – Moment Capacity Diagram
Figure 2.3.6.6 – spBeam Model – Shear Capacity Diagram
Figure 2.3.6.7 – spBeam Model – Immediate Deflection Diagram
Figure 2.3.6.8 – spBeam Model – Reinforcement Diagram
Table 2.3.7.1 – Comparison of Hand Solution with spBeam Solution 

Flexural Design 

Span Location 
Design Moment (ftkips) 
Reinforcement Required for Flexure (in^{2}) 
Reinforcement Provided for Flexure (in^{2}) 

End Span 
Hand Solution 
spBeam Solution 
Hand Solution 
spBeam Solution 
Hand Solution 
spBeam Solution 
Interior Negative 
553.5 
486.01 
7.51 
6.45 
10#8 
9#8 
Positive 
344 
316.4 
4.22 
3.94 
6#8 
5#8 
Shear Design 

Span Location 
V_{u} (kips) 
φV_{n} (kips) 

End Span 
Hand Solution 
spBeam Solution 
Hand Solution 
spBeam Solution 

Interior Negative 
103.2 
95.14 
112.9 
109.45 
In this design example, the interior beam is modeled as a continuous rectangular longitudinal beam. There is a good agreement between the hand solution and computer solution. Note that the coefficients traditionally used to determine moments do not address various types of support and geometry.
The maximum calculated total immediate (instantaneous) deflection (DL + LL) = 0.529 in., this value can be compared with maximum permissible calculated deflection limitation per project criteria in accordance to ACI 31814. ACI 31814 (Table 24.2.2)
In addition to deflection results, parametric studies can be performed in spBeam to optimize design and detailing results.
The reinforcement diagram (Figure 2.3.6.8) shows the minimum length required (including the development length) for flexural design. The bars can be extended and detailed to provide the required support for shear stirrups.
Figure 2.3.8 – Interior Beam CrossSection (Near the First Interior Support)
In the widemodule joist construction, the supporting beam depths shall be same as the overall joist depth. Therefore, the beam depth is set to 21 in. This beam depth need to satisfy the minimum depth requirement of ACI 31814 (Table 9.3.1.1) so that the deflection computations can be waived. The beams of the exterior frame shall be designed and detailed for the combined effects of flexure, shear, and torsion according to ACI 318.
Using the minimum depth for nonprestressed beams in Table 9.3.1.1.
End Span: _{}in (governs) < 21 in. ACI 31814 (Table 9.3.1.1)
Interior Span: _{}in ACI 31814 (Table 9.3.1.1)
Therefore, the preliminary beam depth satisfies the minimum depth requirement.
Figure 2.4 – Partial plan view showing interior beam along grid B
Beam (girder) design involves the following steps:
2.4.1. Determination of span loads
2.4.2. Determination of design moment, shear, and torsion
2.4.3. Flexural and torsion design
2.4.4. Shear and torsion design
2.4.5. Deflections
2.4.6. Computer program solution
2.4.7. Summary and comparison of design results
2.4.8. Conclusions and observations
Dead Load:
Try 24 in width for the beam (slightly larger than the column width that helps facilitate the forming, and reduces the beam longitudinal vs. column vertical bar interference)
Joist & Slab Weight_{ }_{}kips/ft
Beam Weight_{}kips/ft
Superimposed Dead Load, SDL _{}kips/ft
Live Load:
Check for live load reduction per ASCE/SEI 710
_{} ASCE/SEI 710 (Eq. 41)
Where:
_{}reduced design live load per ft^{2} of area supported by the member
_{}unreduced design live load per ft^{2} of area supported by the member = 80 psf
_{}2 (edge beams without cantilever slabs) ASCE/SEI 710 (Table 42)
_{}tributary area _{}ft^{2}
_{}psf
Which satisfies _{}requirement for members supporting one floor. ASCE/SEI 710 (4.7.2)
_{}kips/ft
Load Combination:
The following gravity load combinations are considered:
_{} ACI 31814 (Eq. 5.3.1a)
_{} kips/ft
_{} ACI 31814 (Eq. 5.3.1b)
_{}kips/ft
The span loads are governed by the second load combination.
For factored torsional moment calculations, the beam selfweight is not included since it is applied along the beam section centerline. And the moment arm is the distance from the midspan to the centerline of the exterior beam section = 16/2 – (24/2  20/2)/12 = 7.83 ft.
Thus, the following load combinations are used for the calculation of the factored torsional moment:
_{} ACI 31814 (Eq. 5.3.1a)
_{} kips.ft/ft
_{} ACI 31814 (Eq. 5.3.1b)
_{}kips.ft/ft
The span factored torsional moments are governed by the second load combination.
The factored moment and shear can be determined using the simplified method if the requirements are satisfied: ACI 31814 (6.5.1)
ü Members are prismatic.
ü Loads are uniformly distributed.
ü L ≤ 3D (0.97 kips/ft ≤ 3 x 2.10 kips/ft)
ü There are at least two spans.
ü The longer of two adjacent spans does not exceed the shorter by more than 20 percent.
Thus, the approximate coefficients can be used. The factored moments and shears are determined and summarized in the following tables. ACI 31814 (Table 6.5.2 and Table 6.5.3)
Table 2.4.2.1 – Exterior Beam Design Flexural Moment Values 

Location 
Design Flexural Moment Value 

End Spans 
Exterior Support Negative 
_{}ftkips 
Midspan 
_{}ftkips 

Interior Support Negative 
_{}ftkips 

Interior Spans 
Midspan Positive 
_{}ftkips 
Support Negative 
_{}ftkips 
Table 2.4.2.2 – Exterior Design Shear Values 

Location 
Design Shear Value 
End Span at Face of First Interior Support 
_{}kips 
At Face of all other Supports 
_{}kips 
Any structural analysis method can be used to calculate the torsional. The following table shows the torsional moment values at the centerline of the supports:
Table 2.4.2.3 – Exterior Design Torsional Moment Values 

Location 
Design Torsional Moment Value 
At the centerline of all Supports 
_{}ftkips 
For this exterior beam, the end span moment values govern the design as tabulated in Table 2.4.2.1.
Calculate the required reinforcement to resist the first interior support negative moment:
_{} ftkips
Use #8 bars with 1.5 in. concrete cover per ACI 31814 (Table 20.6.1.3.1). To avoid interference with joist negative moment reinforcement, the clear cover to the girder top reinforcement is required to be increased by lowering the girder top reinforcement. The distance from extreme compression fiber to the centroid of longitudinal tension reinforcement, d, is calculated below:
_{}
To determine the area of steel, assumptions have to be made whether the section is tension or compression controlled, and regarding the distance between the resultant compression and tension forces along the slab section (jd). In this example, tensioncontrolled section will be assumed so the reduction factor_{}is equal to 0.9, and jd will be taken equal to 0.9d since we are designing for the negative moment in a rectangular beam (narrow compression zone). The assumptions will be verified once the area of steel is finalized.
Assume_{}in.
Interior beam width, _{}in.
The required reinforcement at initial trial is calculated as follows:
_{} in.^{2}
Recalculate ‘a’ for the actual A_{s} = 4.49 in.^{2}: _{}
_{}in.
_{}
Therefore, the assumption that section is tensioncontrolled is valid.
_{}
The minimum reinforcement shall not be less than
_{} ACI 31814 (9.6.1.2(a))
And not less than
_{} ACI 31814 (9.6.1.2(b))
All the values on the following table are calculated based on the procedure outlined above.
Table 2.4.3.1 – Reinforcing Design Summary (Flexure only) 


End Span 
Interior Span 


Exterior Negative 
Positive 
Interior Negative 
Positive 
Negative 
Design Moment, M_{u} (ftkips) 
204.7 
233.9 
327.5 
204.7 
297.7 
Effective depth, d (in.) 
18.0^{*} 
18.625^{**} 
18.0^{*} 
18.625^{**} 
18.0^{*} 
A_{s, req’d} (in.^{2}) 
2.65 
2.93 
4.36 
2.55 
3.94 
A_{s, min} (in.^{2}) 
1.53 
1.58 
1.53 
1.58 
1.53 
^{*} The beam top bars are to be placed below the joist top bars. 

^{**} The beam bottom bars are to be placed at the bottommost layer. The joist bottom bars, then, shall be spliced at joistbeam intersection. 
Torsion requirements for longitudinal steel have to be determined and combined with reinforcement area required for flexure.
Calculate the required reinforcement to resist torsion:
Check if torsional effects can be neglected:
If T_{u} < φT_{th}, it shall be permitted to neglect torsional effects. ACI 31814 (22.7.1.1)
Where:
_{}ftkips
_{}ftkips = Threshold torsion (the calculation of φT_{th} is shown in the next section)
Since T_{u} > φT_{th}, the torsional effects must be considered.
Check if the factored design torsion can be reduced:
It is permitted to reduce T_{u} to φT_{cr}; due to redistribution of internal forces after torsional cracking; if the exterior continuous beam meet the following requirements: ACI 31814 (22.7.3.2)
1. The beam is statically indeterminate (continuous beam).
2. T_{u} ≥ φT_{cr}.
To check the second condition, φT_{cr} need to be calculated as follows:
Since the beams are cast monolithically with slab and joists, A_{cp} (area enclosed by outside perimeter of concrete cross section) and p_{cp} (outside perimeter of concrete cross section) for the beam can include a portion of the adjoining slab. The effective width, b_{e}, of the overhanging flange must conform to ACI 31814 (8.4.1.8):
_{}_{ }in. (governs)
_{}_{ }in.
_{}in.^{2}
_{} in.
_{} in.^{3}
The torsional properties of the beam ignoring the overhanging flange are the following:
_{}in.^{2}
_{}in.
_{} in.^{3} > 2796 in.^{3}
Therefore, consider the rectangular section. ACI 31814 (9.2.4.4(b))
_{} ACI 31814 (Table 22.7.5.1(a))
_{} in.lbs = 49.9 ftkips
_{} ftkips
Checking the second condition of ACI 31814 (22.7.3.2):
_{}ftkips > _{}ftkips
Thus, ACI 31814 permits to reduce T_{u} to φT_{cr}.
_{}ftkips ACI 31814 (22.7.3.2)
It is assumed that the torsional loading on the beam is uniformly distributed along the span.
Determine the adequacy of crosssectional dimensions for the torsion:
For solid sections, the limit on shear and torsion is given by:
_{} ACI 31814 (22.7.7.1)
Where:
A_{oh} = area enclosed by centerline of outermost closed transverse torsional reinforcement.
p_{h} = perimeter of centerline of outermost closed transverse torsional reinforcement.
Using d = 18.0 in., the factored shear force at the critical section located at a distance d from the face of the support is:
_{} kips.
Also, the nominal shear strength provided by the concrete is:
_{} ACI 31814 (Eq. 22.5.5.1)
Using a 1.5in. clear cover to # 3 closed stirrups at bottom and 2.125 in clear cover to # 3 closed stirrups at top.
_{}in^{2}
_{}in
_{}psi_{}psi
Therefore, the section is adequate.
Determine the transverse reinforcement required for torsion:
_{} ACI 31814 (Eq. 22.7.6.1a)
Where
_{}in^{2} ACI 31814 (22.7.6.1.1)
_{} ACI 31814 (22.7.6.1.2(a))
Therefore,
_{}in^{2} / in per leg
Determine the transverse reinforcement required for shear:
From Table 2.3.2.2 above, the maximum shear value occurs at the face of the first interior support in the end span.
_{}kips
The design shear at a distance, d, away from the face of support,
_{}kips
Shear strength provided by concrete
_{} ACI 31814 (Eq. 22.5.5.1)
_{}lb _{}kips
Since_{}, shear reinforcement is required.
The nominal shear strength required to be provided by shear reinforcement is
_{}kips
Check whether V_{s} is less than _{}
If V_{s} is greater than_{}, then the crosssection has to be revised as ACI 31814 limits the shear capacity to be provided by stirrups to_{}. ACI 31814 (22.5.1.2)
_{}lb _{}kips
Since V_{s} does not exceed_{}. The crosssection is adequate.
Calculate the required transverse reinforcement for shear as
_{}in.^{2}/in. ACI 31814 (22.5.10.5.3)
Calculate total required transverse reinforcement for combined shear and torsion:
_{}
Minimum transverse reinforcement for shear and torsion is calculated as follows:
_{} ACI 31814 (9.6.4.2)
_{} in.^{2}/in. < 0.0626 in.^{2}/in.
Then, provide _{}
Calculate the required spacing:
Maximum spacing of transverse torsion reinforcement:
_{} ACI 31814 (9.7.6.3.3)
_{}
Maximum spacing of transverse shear reinforcement:
Check whether the required spacing based on the shear demand meets the spacing limits for shear reinforcement per ACI 31814 (9.7.6.2.2).
Check whether V_{s} is less than_{}
_{} lb _{}kips > V_{s} = 19.43 kips
Therefore, maximum stirrup spacing shall be the smallest of d/2 and 24 in.
_{} ACI 31814 (Table 9.7.6.2.2)
_{}(governs)
Using a bundle of 2#3 closed stirrups with 2 legs (area per leg _{}in^{2}), the required spacing, s, at the critical section is:
_{}in _{}in
Provide a bundle of 2#3 closed stirrups with 2 legs spaced at 7 in. on center. #3 bars are selected for consistency with the transverse reinforcement size used for the joist and interior beam. The stirrups are bundled at outer legs to maintain A_{0} value in calculation of A_{t}/s for torsion. The bundle of 2#3 bars are defined as userdefined reinforcement of size #2 in spBeam Program.
The designer may choose to utilize #4 closed stirrups with 2 legs at 6 in. on center alternatively.
In view of the shear and torsion distribution along the span length, this same reinforcement and spacing can be provided throughout the span length.
Calculate the additional required longitudinal reinforcement for torsion:
_{} ACI 31814 (Eq. 22.7.6.1b)
Where
_{}in.^{2} ACI 31814 (22.7.6.1.1)
_{} ACI 31814 (22.7.6.1.2(a))
Therefore,
_{}in.^{2}
The minimum total area of longitudinal torsional reinforcement:
_{} ACI 31814 (9.6.4.3)
_{} in.^{2}
Since A_{l} > A_{l,min}, use A_{l} = 1.68 in.^{2}
The longitudinal reinforcement is to be distributed around the perimeter of the stirrups, with a maximum spacing of 12 in. There shall be at least one longitudinal bar in each corner of the stirrups.
ACI31814 (9.7.5.1)
Longitudinal bars shall have a diameter at least 0.042 times the stirrup spacing, but not less than 3/8 in.
ACI31814 (9.7.5.2)
To meet the maximum spacing requirement, a bar has to be provided between corner bars at all four sides. This configuration leads to eightbars; three at top, three at bottom, and one at each side. Therefore, the reinforcement area per bar is A_{s} = 1.68/8 = 0.21 in.^{2}
Then, use #5 bars for longitudinal bars which also meets minimum bar diameter requirement of 3/8 in. A_{l }shall be provided in addition to the required flexural reinforcement at the negative moment regions (supporttop) and positive moment region (midspanbottom). At midspantop region where flexural reinforcement is not required for flexure, 3#5 bars shall be provided. Class B lap splice is to be provided.
Table 2.4.3.2 – Reinforcing Design Summary (Flexure + Torsion) 


End Span 
Interior Span 


Top Reinforcing for Exterior Negative Moment 
Bottom Reinforcing for Positive Moment 
Top Reinforcing for Interior Negative Moment 
Bottom Reinforcing for Positive Moment 
Top Reinforcing for Interior Negative Moment 
Required Longitudinal Reinforcement (in.^{2}) 
2.65 
2.93 
4.36 
2.55 
3.94 
Required Torsional Longitudinal Reinforcement (in.^{2}) 
0.63 
0.63 
0.63 
0.63 
0.63 
Required Total Longitudinal Reinforcement (in.^{2}) 
3.28 
3.56 
4.99 
3.18 
4.57 
Reinforcement 
5#8 
5#8 
7–#8 
4#8 
6#8 
Maximum spacing allowed:
Check the requirement for distribution of flexural reinforcement to control flexural cracking:
_{} ACI 31814 (Table 24.3.2)
_{}in.
Use _{}ksi ACI 31814 (24.3.2.1)
_{}in. (governs)
_{}in.
Spacing provided for 4# 8 bars_{} _{}in. _{}in. o.k.
Where d_{s} = 2.625 in. for #3 stirrup. CRSI 2002 (Figure 129)
Check the spacing, s provided, is greater than the minimum center to center spacing, s_{min} where
_{} CRSI 2002 (Figure 129)
Where maximum aggregate size is ¾”
_{}
Spacing provided for 7# 8 bars_{} _{}in. _{}in. o.k.
Therefore, the reinforcement selections in the previous table meet the spacing requirements.
Since the preliminary beam depth met minimum depth requirement, the deflection calculations are not required. A lesser depth maybe possible and consequently cost savings can be achieved through deflection computations. Deflection values are calculated and provided for every model created by spBeam Program and can be used by the engineer to make additional optimization decisions.
spBeam Program can be utilized to analyze and design the exterior continuous beam along grid A. The beam is modeled as a three span continuous rectangular beam.
The program calculates the internal forces (shear force and bending moment), moment and shear capacities, immediate and longterm deflection results, and required flexural reinforcement. The graphical and text results are provided here for both input and output of the spBeam model.
The beam is modeled as a 24 in. by 21 in. deep rectangular longitudinal beam with column supports at 100% stiffness share.
The reinforcement database is selected as Userdefined in order to define a bundle of 2#3 bars as #2 with crosssectional area of 0.22 in^{2}. (Different than #2 defined earlier for welded wire).
Torsion analysis was engaged using the torsion analysis and design check box located on the solve options tab in the input dialog box. The design for torsion is based on a thinwalled tube, space truss analogy. spBeam allows both equilibrium and compatibility torsion conditions. In the equilibrium mode, which is assumed by default, unreduced total value of the torsional design moment is used in the design. In the compatibility mode, factored torsional moments that exceed cracking moment T_{cr} are reduced to the value of T_{cr}. However, it is user’s responsibility to determine which mode is appropriate and the program does not perform any redistribution of internal forces if compatibility torsion is selected. In this model, the following solve options were used.
Figure 2.4.6.1 – spBeam Model – Isometric View – Exterior Continuous Beam along Grid A
Figure 2.4.6.2 – spBeam Model – Loads (Including Live Load Patterning)
Figure 2.4.6.3 – spBeam Model – Internal Forces (Shear Force, Torsion, and Bending Moment Diagrams)
Figure 2.4.6.4 – spBeam Model – Moment Capacity Diagram
Figure 2.4.6.5 – spBeam Model – Shear Capacity Diagram
Figure 2.4.6.6 – spBeam Model – Immediate Deflection Diagram
Figure 2.4.6.7 – spBeam Model – Reinforcement Diagram
Table 2.4.6.1 – Comparison of Hand Solution with spBeam Solution 

Span 
M_{u }(ftkips) 
T_{u} (ftkips) 
A_{t}/s (in^{2}/in per leg) 

End Span 
Hand Solution 
spBeam Solution 
Hand Solution 
spBeam Solution 
Hand Solution 
spBeam Solution 
Interior Negative 
327.5 
281.9 
49.9 
49.9 
0.0223 
0.0223 
Span 
A_{v}/s (in^{2}/in) 
(A_{v}+2A_{t})/s (in^{2}/in) 
A_{l }(in^{2}) 

End Span 
Hand Solution 
spBeam Solution 
Hand Solution 
spBeam Solution 
Hand Solution 
spBeam Solution 
Interior Negative 
0.018 
0.011 
0.0626 
0.0555 
1.68 
1.68 
Table 2.4.6.2 – Comparison of Hand Solution with spBeam Solution (Reinforcement) 

Span Location 
Required Reinforcement Area for Flexure + Torsion (in^{2}) 
Reinforcement Provided for Flexure + Torsion 

End Span 
Hand Solution 
spSlab Solution 
Hand Solution 
spSlab Solution 
_{Interior } _{Negative} 
4.91 
3.716+1.680*(3/8)=4.346 
Top Bar: 7#8 
Top Bar: 5#8 + 3#8 = 8#8 
_{Positive} 
3.53 
2.203+1.680*(3/8)=2.833 
Bottom Bar: 5#8 
Bottom Bar: 3#8 + 3#8 = 6#8 
In this design example, the exterior beam is modeled as a continuous rectangular longitudinal beam. There is a good agreement between the hand solution and computer solution. Note that the coefficients traditionally used to determine moments do not address various types of support and geometry.
The maximum calculated total immediate (instantaneous) deflection (DL + LL) = 0.369 in., this value can be compared with maximum permissible calculated deflection limitation per project criteria in accordance to ACI 31814. ACI 31814 (Table 24.2.2)
In addition to deflection results, parametric studies can be performed in spBeam to optimize design and detailing results.
The following observations can be made regarding the reinforcement diagram (Figure 2.4.6.7):
· The longitudinal reinforcement for the flexural design only and does not include (A_{l}).
· A_{l} is shown in the text output report (Design Results) because it has to be detailed and distributed to meet the torsional spacing requirements:
o At least one longitudinal bar must be present at each corner of the stirrups.
o And a bar has to be provided between corner bars at all four sides to meet the maximum spacing requirement (max 12 in.).
o This configuration leads to eightbars in this example: three at top, three at bottom, and one at each side.
o Then, for the top bar for the interior negative section for the end span we need:
1) The longitudinal reinforcement for the flexural design 5#8 is required for flexure.
2) The longitudinal reinforcement for the torsional design 3#8 (3 top bars of the 8 bars distributed around the perimeter of the stirrups). See Table 2.4.6.2.
o Also, for the bottom bar for the positive section for the end span we need:
3) The longitudinal reinforcement for the flexural design 3#8 is required for flexure.
4) The longitudinal reinforcement for the torsional design 3#8 (3 bottom bars of the 8 bars distributed around the perimeter of the stirrups). See Table 2.4.6.2.
· Transverse reinforcement shown reflects the total size and quantity of stirrups to resist the combined effects of shear and torsion.
· Top bars minimum length required (including the development length) for flexural design is shown. The bars can be extended and detailed to provide the required support for shear stirrups.
Figure 2.4.7 – Exterior Beam CrossSection
This section includes the design of interior, edge, and corner columns using spColumn software. The preliminary dimensions for these columns were calculated previously in section 1.2.
Interior Column:
Total Factored Load on 1^{st} story interior column (@ 1^{st} interior support) are reorganized based on the calculations on section 1.2 as follows:
P_{D} = Total service dead load _{} kips
P_{L} = Total service live load_{}kips
_{}
M_{u,x} = unbalance flexural moment at the Interior support for the interior beam (see Figure 2.3.6.3)
= 595.20  545.96 = 49.24 ftkips
M_{u,y} = unbalance flexural moment at the interior support for the joist (see Figure 2.2.6.3)
=154.53 – 141.78 = 12.75 ftkips
The factored loads are then input into spColumn to construct the axial load – moment interaction diagram as shown in the sample input below.
Edge (Exterior) Column:
Total Factored Load on 1^{st} story edge column (@ 1^{st} interior support) are reorganized based on the calculations on section 1.2 as follows:
P_{D} = Total service dead load _{}kips
P_{L} = Total service live load_{} kips
P_{u} = _{}
M_{u,x} = unbalance flexural moment at the Interior support (see Figure 2.4.6.3)
= 333.89 – 317.85 = 16.04 ftkips
M_{u,y} = torsional moment at the Interior support for the exterior beam (see Section 2.4.3) + unbalance flexural moment at exterior support for the joist (see Figure 2.2.6.3)
= 2T_{u}_{ }+ 73.96 = 2 x 49.9 + 73.96 = 173.76 ftkips
Corner Column:
Total Factored Load on 1^{st} story corner column (@ exterior support) are reorganized based on the calculations on section 1.2 as follows:
P_{D} = Total service dead load _{}kips
P_{L} = Total service live load_{} kips
P_{u} = _{}
M_{u,x} = unbalance flexural moment at the Exterior support (see Figure 2.4.6.3) = 250.25 ftkips
M_{u,y} = torsional moment at the Exterior support for the exterior beam (see Section 2.4.3) + unbalance flexural moment at exterior support for the joist (see Figure 2.2.6.3)
= T_{u}_{ }+ 73.96 = 49.9 + 73.96 = 123.86 ftkips
The axial force and moment interaction diagram is constructed using spColumn based on geometry and load input shown below:
Interior Column:
Figure 2.5.2.1 – spColumn Model – Biaxial Moment Interaction
Figure 2.5.2.2 – spColumn Model –Axial Moment Interaction
Figure 2.5.2.3 – spColumn Model – Nominal and Factored Failure Surfaces
Edge (Exterior) Column:
Figure 2.5.2.4 – spColumn Model –Axial Moment Interaction
Corner Column:
Figure 2.5.2.5 – spColumn Model –Axial Moment Interaction