This article presents a detailed example of designing a one-way post-tensioned car parking structure. It addresses key considerations such as durability and cover requirements, gravity load calculations, prestress estimation, and checks for both stress limits and ultimate bending and shear capacities. The article concludes with images illustrating the design of the same post-tensioned one-way slab using Adapt PT/RC software
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Design Brief
Parking demand in the garage typically peaks around 12 noon, often exceeding available capacity as additional vehicles arrive. Since there is no adjacent land available for expansion, constructing an elevated (suspended) slab has become necessary to meet the growing demand. To maintain unobstructed vehicular circulation on the ground floor, columns supporting the suspended slab must be placed only at predetermined locations.
The layout of the elevated platform, along with these designated column positions, is shown below
Structural proposal
Based on the layout and column positions shown above, the slab spans vary between 7 meters and 12 meters. To accommodate these spans while maintaining a minimal Storey height, a 200 mm thick post-tensioned slab is proposed. The structural layout of the slab is illustrated below
Design Calculations
Materials & Properties
The properties of materials to be used in the design of the slab is as given below:
Properties of tendon
Fpk = 1860MPa (Properties of tendon is from BS 5859 1980)
Fpk0.1 = 0.85fpk = 1581MPa
Fpd = Fpk0.1/1.15 = 1374MPa
Area of tendon = 112mm²
Properties of concrete
Concrete grade (fck) = 35MPa
Material safety factor = 1.5
Design strength of concrete (fcd) = 35/1.5 = 23.3MPa
Slab Cross-section Properties
Area of Slab section = 1 x 0.2 = 0.2m²
Zt = Zb = I/y (where y = h/2 = 0.2/2 = 0.1)
Zt = (b x h³) / (12 x y) = (1000 x 0.2³)/ (12 x 0.1) = 0.0067m³ = 6.7 x 106 mm³
Durability and Cover Requirement of the Slab
The slab would be provided with a roof to keep it dry and the immediate environment is non-aggressive, so from table 4.1, EN 1992-1-1 the exposure class is XC1
Structural class
The default structural class is S4. Based on parameters in Table 4.3N of EN 1992-1-1 we shall workout the design structural class.
Design life of structure: Since the design life of structure is 50years and less than 100years, the structural class remain S4.
Strength class: Since the concrete strength is greater than 30/37 the structural class is reduced to S3
Member with slab geometry: Since the member is a slab, the structural class is reduced to S2
Special Quality control: It is assumed that special quality control would be observed during production, the structural class is reduced to S1
Minimum Cover
Minimum cover with regards to bond is the diameter of the tendons which is 12mm (Table 4.2 EN 1992-1-1)
Minimum cover requirement with regard to durability due to prestressing steel is 15mm (From table 4.5N of EN 1992-1-1, corresponding to structural class S1 and exposure class XC1)
Since Minimum cover requirement with regard to durability due to prestressing steel is greater than the Minimum cover with regards to bond, then the minimum cover is 15mm
Deviation
Although the recommended deviation from EN 1992-1-1 is 10mm, we shall adopt 5mm in this article
Nominal Cover
Cnom = Cmin + CΔdev = 15 + 5 = 20mm
To find the referenced tables above and for background details on durability read, “Durability of Concrete Structures”
Actions
We shall estimate the loads that would be acting on the slab which include: 1) permanent load from self-weight, finishes, prestress force, 2) variable load 3) equivalent load due to effect of prestress and eccentricity of tendons.
Permanent and Variable Loads
Variable load = 2.5KN/m² (Traffic and parking areas of vehicle with gross weight of 30KN, table 6.8 EN 1991-1-1)
Additional permanent load = 1KN/m²
Self-weight = 0.2 x 25 = 5KN/m²
Total permanent load = 5 + 1 = 6KN/m²
Equivalent load
Percentage of dead load to balance = 75% (for further details on load balancing read, Load Balancing approach for post-tensioned concrete design)
Equivalent load in each span = 0.75 x 6 = 4.5KN/m²
NB: It is highly important to ensure the balanced/equivalent load is not beyond the self-weight of the slab to avoid over-balancing as the weight of the super-dead (additional dead) load will not be present on the slab during stressing operations.
Prestressing force
As preliminary estimate the prestressing force shall be kept constant while the tendon drapes vary. The largest middle span shall be taken as the controlling span
Mid span
Required prestressing force = Ps = (wunb x l²)/8 x e
Balanced/equivalent load, wb = 4.5KN/m²
Length, l = 12m
How much drape that can be achieved is dictated by physical constraints such as allowance for concrete cover or links, when necessary. Since the nominal cover have been estimated to be 20mm hence the drape equals:
Eccentricity at mid span = 200/2 – 20 = 80mm
Drape of tendon = [e2 – e2] = [80 – 0] = 80mm
Required prestressing force, Ps = (4.5 x 12²) / (8 x 0.08) = 1012.5KN
Estimation of tendon drape in End Spans and Interior Supports
Since the prestress force of 1012.5KN has been determined and is going to be kept constant across the spans then we shall calculate the drape that will achieve a balanced load of 4.5kN/m2 on the end spans
Drape on End spans
e = (wb x l²)/8 x Ps
e = (4.5 x 12²)/8 x 1012.5 = 0.027m = 27mm
Drape over Interior Supports
Over the interior support, the tendons would be made to achieve the maximum drape possible.
Eccentricity over support = 200/2 – 20 = 80mm
Drape of tendon = [e2 – e1] = [80 – 0] = 80mm
Drape over Interior Supports
Over the interior support, the tendons would be made to achieve the maximum drape possible.
Eccentricity over support = 200/2 – 20 = 80mm
Drape of tendon = [e2 – e1] = [80 – 0] = 80mm
Analysis for Serviceability Verifications
Separate analyses shall be performed for permanent loads, variable (live) loads, and equivalent loads. The variable load will be applied across all spans, as well as in a patterned configuration to capture the most critical loading effects. While hand calculations using the moment distribution method are feasible, this article employs a computer-based approach. The slab is analyzed for each load case using STAAD Pro software. The analysis results are presented below.”
Estimate the stress at critical sections
The stresses at critical section shall be evaluated. Due to symmetry of the structure, the stress shall be checked for one interior support, one end span, and the middle span.
Interior Supports
Check the stress at top of the section
σt = $\left(\frac{-P}{A}+\frac{M_{\r }}{Z_t}\right) $
Max Moment due to variable load = 27.860KNm
Moment due to permanent load = 62.12KNm
Moment due to equivalent load = 46.590KNm
Resultant Moment = 27.860 + 62.12 – 46.590 = 43.39
Stress at top fiber $$
\frac{-\text{1012.5}\times \,\,10^3}{\text{0.2}\times \,\,10^6}\,\,+\,\,\frac{\text{32.73}\times \,\,10^6}{\text{6.7}\times \,\,10^6}
$$
Stress at top fiber σt = -5.0 + 6.48
σt = 1.48MPa
Check the stress at bottom of the section
Stress at bottom fiber σb = $\left(\frac{-P}{A}+\frac{M_{\r }}{Z_b}\right) $
Resultant Moment = -27.860 – 62.12 + 46.590 = -43.39KNm
Stress at top fiber σb = $$
\frac{-\text{1012.5}\times \,\,10^3}{\text{0.2}\times \,\,10^6}\,\,+\,\,\frac{-\text{32.73}\times \,\,10^6}{\text{6.7}\times \,\,10^6}
$$
σb = – 5.0 – 6.48
σb = -11.48MPa
End Spans
Check the stress at top of the section
σt = $\left(\frac{-P}{A}+\frac{M_{\r }}{Z_t}\right) $
Maximum moment due to variable load = 13.169KNm
Moment due to permanent load = 5.691KNm
Moment due to equivalent load = 4.268KNm
Resultant Moment = -13.169 – 5.691 + 4.268 = -14.592KNm
Stress at top fiber σt = $$
\frac{-\text{1012.5}\times \,\,10^3}{\text{0.2}\times \,\,10^6}\,\,+\,\,\frac{-\text{14.592}\times \,\,10^6}{\text{6.7}\times \,\,10^6}
$$
σt = – 5.0 – 2.17
σt = – 7.17MPa
Check the stress at bottom of the section
Stress at bottom fiber
σb = $\left(\frac{-P}{A}+\frac{M_{\r }}{Z_b}\right) $
Resultant Moment = 13.169 + 5.691 – 4.268 = 14.592
σb = $$
\frac{-\text{1012.5}\times \,\,10^3}{\text{0.2}\times \,\,10^6}\,\,+\,\,\frac{\text{14.592}\times \,\,10^6}{\text{6.7}\times \,\,10^6}
$$
σb = – 5.0 + 2.17
σb = -2.83MPa
Mid Spans
Check the stress at top of the section
Stress at top fiber
σt = $\left(\frac{-P}{A}+\frac{M_{\r }}{Z_t}\right) $
Max moment due to variable load = 21.260KNm
Moment due to permanent load = 45.880KNm
Moment due to equivalent load = 34.41KNm
Resultant Moment = -21.260 – 45.880 + 34.41 = -32.73
Stress at top fiber σt = $$
\frac{-\text{1012.5}\times \,\,10^3}{\text{0.2}\times \,\,10^6}\,\,+\,\,\frac{-\text{32.73}\times \,\,10^6}{\text{6.7}\times \,\,10^6}
$$
σt = – 5.0 – 4.89
σt = -9.89MPa
Check the stress at bottom of the section
Stress at bottom fiber
σb = $\left(\frac{-P}{A}+\frac{M_{\r }}{Z_b}\right) $
Resultant Moment = 21.260 + 45.880 – 34.41 = 32.73
tress at top fiber σb = $$
\frac{-\text{1012.5}\times \,\,10^3}{\text{0.2}\times \,\,10^6}\,\,+\,\,\frac{\text{32.73}\times \,\,10^6}{\text{6.7}\times \,\,10^6}
$$
σb = – 5.0 + 4.89
σb = 0.11MPa
Compare the Stresses with Permissible Stress
Permissible tensile stress = 0.3 x fck2/3 = 0.3 x 352/3 = 3.2N/mm²
Permissible compressive stress = 0.3fck = 0.3 x 35 = 10.5N/mm²
The stress not within the permissible stress is compressive stress of 11.48MPa. This would not cause immediate crack as it is not a tensile stress violation but can lead to excessive creep in the long run. For further details on creep effect read, “Creep Deformation of Concrete Structures”
Detailing of tendons
From properties of tendon
Capacity of tendon = fpd x As = 1374 x 112 x 10-3 = 154KN
Number of Tendons = 1012.5/154 = 7 tendons
Flexural Strength
The flexural strength of the slab shall also be verified at critical sections such as internal support and mid spans
Analysis for flexural Strength Verification
The analysis for bending moment for ultimate limit state verifications entails analyzing the slab for the combination of factored permanent and variable load, and secondary moment due to prestress.
The bending moment diagram below each represent a combination of a distinct case of factored pattern variable load and factored permanent load. Moment due to secondary effect of prestress is computed by hand and then added to the factored moment for mid-span moment or subtracted from factored moment for moment at interior support.
Computation of Secondary Moment and Net Moment at critical Sections
To estimate the total ultimate moment, the secondary moment shall be calculated and then added to the critical moment due to factored permanent and variable load.
Middle Span
eccentricity = 200/2 – 20 = 80mm
secondary moment = Total moment – primary moment (P x e)
secondary moment = 34.41 – 1012.5 x 0.08 = -46.59KNm
Net Ultimate moment at middle span = 97.520KNm + 46.59 = 144.11KNm
End Span
eccentricity = 27mm
secondary moment = Total moment – primary moment (P x e)
secondary moment = 4.268 – 1012.5 x 0.027 = -23.0KNm
Net Ultimate moment at middle span = 26.315KNm + 23.0 = 49.315KNm
Interior Support
eccentricity = 200/2 – 20 = 80mm
secondary moment = Total moment – primary moment (P x e)
secondary moment = 46.59 – 1012.5 x 0.08 = -34.41KNm
Net Ultimate moment at interior support = 130.792KNm – 46.59 = 84.202KN
Verify the Flexural Strength at Mid Span
STEP 1: Evaluate the total tensile force produced by the tendons.
Here we will assume all the tendons have yielded, hence the tension in the tendons equals;
Tensile force = fpd x As = 7 x 1374 x 112 x 10-3 = 1077.83KN
STEP 2: Assume the depth of stress block
Assume the depth of neutral axis and calculate the depth of the compressive stress block
Let assume a depth of neutral axis (x) = 56.25mm
Depth of compressive stress block (s) = λx = 0.8 x 56.25 = 45mm
STEP 3: Calculate the compressive force.
Comp force = fcd x area of rectangular stress block
= (23.3 x 45 x 1000)/1000 = 1050KN
The difference between the compressive and tensile force = 1077.8 – 1050 = 27.8KN.
This is small and can be ignored, so the compressive and tensile force can be assumed to be in equilibrium.
STEP 4: Calculate the Ultimate Moment of the beam.
Calculate the ultimate moment by taking moment of forces about the soffit of the beam. Take tensile force as positive.
Compressive Moment
Compressive force = 1050KN
Lever arm of compressive force to the soffit (z) = 200 – 45/2 = 177.5mm
Compressive moment = (1050 x 177.5)/1000 = 186.4KNm
Tensile Moment
Tensile fore = fpd x As x no of tendons
Tensile fore = 1374 x 112 x 10-3 x 7 = 1077.83KN
Lever arm (z) = 20mm
Moment = (1077.83 x 20)/1000 = 21.6KNm
Ultimate Bending Capacity at mid span = 21.6 – 186.4 = – 164.81KNm
The ultimate bending capacity of the beam (164.81KNm) is greater than the ultimate bending moment (144.1KNm), the ultimate bending strength at mid-span is adequate.
Also, since the bending strength at mid-span is adequate then the same applies to end span as the mid-span is more critical.
Verify the Flexural Strength over Internal Supports
STEP 1: Evaluate the total tensile force produced by the tendons.
Here we will assume all the tendons have yielded, hence the tension in the tendons equals;
Tensile force = fpd x As = 7 x 1374 x 112 x 10-3= 1077.83KN
STEP 2: Assume the depth of stress block
Assume the depth of neutral axis and calculate the depth of the compressive stress block
Let assume a depth of neutral axis (x) = 62.5mm
Depth of compressive stress block (s) = λx = 0.8 x 62.5 = 45mm
STEP 3: Calculate the compressive force.
Comp force = fcd x area of rectangular stress block
= (23.3 x 45 x 1000)/1000 = 1050KN
The difference between the compressive and tensile force = 1077.8 – 1050 = 27.8KN.
This is small and can be ignored, so the compressive and tensile force can be assumed to be in equilibrium.
STEP 4: Calculate the Ultimate Moment of the beam.
Calculate the ultimate moment by taking moment of forces about the soffit of the beam. Take tensile force as positive.
Compressive Moment
Compressive force = 1050
Lever arm of compressive force to the soffit (z) = 45/2 = 22.5mm
Compressive moment = (1050 x 22.5)/1000 = 23.625KNm
Tensile Moment
Tensile fore = fpd x As x no of tendons
Tensile fore = 1374 x 112 x 10^ (-3) x 7 = 1077.83KN
Lever arm (z) = 200 – 20 = 180mm
Moment = (1077.83 x 180)/1000 = 194KNm
Ultimate Bending Capacity at mid span = 194 – 23.625 = 170.375KNm
The ultimate bending capacity of the beam (170.375KNm) is greater than the ultimate bending moment (84.202KNm) at interior support, the ultimate strength at support is adequate.
Provision of Minimum Reinforcement
Since the tendons would always be along a single span, minimum reinforcement should be provided in the orthogonal direction to minimize crack due to thermal stress and shrinkage.
Verification of Shear Strength
The shear strength of the slab shall be verified at the most critical section of the slab for combination of factored permanent and variable loads. The secondary shear is often neglected for beam and slab system as it is inconsequential most times.
Analysis for shear Strength Verification
The shear force diagram for each combination is provided below
From the shear force diagram, the most critical shear VEd to be designed for = 75.117KN
Shear resistance without reinforcement of Uncracked Section in Bending
$ V_{Rdc\,\,}=\,\,\frac{I.b_w}{S}\,\,\sqrt{\left( f_{ctd} \right) ^2+\,\,\propto _1\sigma _{cp}f_{ctd}} $
σcp = NEd/Ac
σcp = 1012.5 x 103/(2.0 x 105) ≤ 0.2fcd
σcp = 5.06 ≤ 4.67MPa
σcp = 4.67MPa
α1 = 1
S = b x (h/2 x h/2)/2 = 5 x 106
$ V_{Rdc\,\,}=\,\,\frac{2.0 x 10^5 x 1000}{5 x 10^6}\,\,\sqrt{\left( 1.5 \right) ^2+\,\,1 x 4.67 x 1.5} $ = 405.17KN
Since the shear resistance (405.17KN) is greater than the shear force (75.117KN), the section is adequate in shear
Shear Resistance of Cracked Section without Shear Reinforcement
vRdc x bwd = {CRdc K(100ρLfck)1/3 + K1σcp} bwd ≥ (vmin + K1σcp)bwd
σcp = NEd/Ac σcp = 1012.5 x 103/(2.0 x 105) ≤ 0.2fcd
σcp = 4.67MPa
CRdc = 0.18/ϒ
CRdc = 0.18/1.5
CRdc = 0.12
K = (1 +√200/d) ≤ 2.0
d = e + yt = 90 + 100 = 190 (The actual effective depth of the tendon is 183 but 190 is adopted in this calculation for convenience)
K = (1 +√200/190) = 1.29
ρL = Asl/bwd ≤ 0.02
Asl = 112 x 7 = 784mm²
ρl = (784) / (1000 × 190) = 0.004
K1 = 0.15
When the value of the parameters into the equation
vRdc = {0.12 x 0.15 (100 x 0.004 x 35)1/3 + 0.5 x 4.67} x 1000 x 190
VRdc = 352.52KN
Verifying whether the minimum permitted resistance using VRdc = {0.035K3/2fck1/2} shows that the requirement of minimum resistance is satisfied.
The shear resistance of uncracked section (405.17KN) is greater than the shear resistance of cracked section (352.52KN), hence the resistance of cracked section shall be considered as critical. The shear resistance of the slab without reinforcement is adequate as the minimum resistance (352.526KN) is greater than shear force (75.117KN)
NB: There are lot of important details omitted in the design, such as stress check at transfer, due to the constraint of hand computation. Using software can help optimize design. Adapt PT/RC shall be used to design the same slab in the section below.
Appendices
Design of One-way PT Slab using Adapt PT/RC
Below are images that shows the modelling, input parameters, analysis and design result of one-way post-tensioned slab using Adapt PT/RC.
The diagram above shows the model of 1m width strip of the slab together with the variable load and additional permanent load.
