SS CP 65 1999 PT SL Example 001

User Manual: SS CP 65-1999 PT-SL Example 001

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SS CP 65-99 PT-SL EXAMPLE 001
Post-Tensioned Slab Design
PROBLEM DESCRIPTION
The purpose of this example is to verify the slab stresses and the required area of
mild steel strength reinforcing for a post-tensioned slab.
A one-way, simply supported slab is modeled in ETABS. The modeled slab is 254
mm thick by 914 mm wide and spans 9754 mm, as shown in shown in Figure 1.
Prestressing tendon, Ap
Mild Steel, As

229 mm
254 mm
25 mm

Length, L = 9754 mm

914 mm

Section

Elevation

Figure 1 One-Way Slab

SS CP 65-99 PT-SL EXAMPLE 001 - 1

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A 254-mm-wide design strip is centered along the length of the slab and has been
defined as an A-Strip. B-strips have been placed at each end of the span,
perpendicular to Strip-A (the B-Strips are necessary to define the tendon profile).
A tendon with two strands, each having an area of 99 mm2, has been added to the
A-Strip. The self weight and live loads have been added to the slab. The loads and
post-tensioning forces are as follows.
Loads:

Live = 4.788 kN/m2

Dead = self weight,

The total factored strip moments, required area of mild steel reinforcement, and
slab stresses are reported at the mid-span of the slab. Independent hand
calculations are compared with the ETABS results and summarized for verification
and validation of the ETABS results.
GEOMETRY, PROPERTIES AND LOADING
Thickness
Effective depth
Clear span

T, h =
d =
L =

Concrete strength
Yield strength of steel
Prestressing, ultimate
Prestressing, effective
Area of Prestress (single strand)
Concrete unit weight
Modulus of elasticity
Modulus of elasticity
Poisson’s ratio

f 'c
fy
fpu
fe
Ap
wc
Ec
Es


Dead load
Live load

wd =
wl =

254
229
9754

=
30
=
400
=
1862
=
1210
=
198
=
23.56
= 25000
= 200,000
=
0
self
4.788

mm
mm
mm
MPa
MPa
MPa
MPa
mm2
kN/m3
N/mm3
N/mm3
kN/m2
kN/m2

TECHNICAL FEATURES OF ETABS TESTED
 Calculation of the required flexural reinforcement
 Check of slab stresses due to the application of dead, live, and post-tensioning
loads
RESULTS COMPARISON
Table 1 shows the comparison of the ETABS total factored moments, required
mild steel reinforcing, and slab stresses with the independent hand calculations.

SS CP 65-99 PT-SL EXAMPLE 001 - 2

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Table 1 Comparison of Results
FEATURE TESTED
Factored moment,
Mu (Ultimate) (kN-m)
Area of Mild Steel req’d,
As (sq-cm)
Transfer Conc. Stress, top
(D+PTI), MPa
Transfer Conc. Stress, bot
(D+PTI), MPa
Normal Conc. Stress, top
(D+L+PTF), MPa
Normal Conc. Stress, bot
(D+L+PTF), MPa

INDEPENDENT
RESULTS

ETABS
RESULTS

DIFFERENCE

174.4

174.4

0.00%

19.65

19.80

0.76%

5.058

5.057

-0.02%

2.839

2.839

0.00%

10.460

10.467

0.07%

8.402

8.409

0.08%

COMPUTER FILE: SS CP 65-1999 PT-SL EX001.EDB
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.

SS CP 65-99 PT-SL EXAMPLE 001 - 3

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HAND CALCULATIONS:
Design Parameters:
Mild Steel Reinforcing
f’c = 30MPa
fy = 400MPa

m, steel = 1.15

Post-Tensioning
fpu = 1862 MPa
fpy = 1675 MPa
Stressing Loss = 186 MPa
Long-Term Loss = 94 MPa
fi = 1490 MPa
fe = 1210 MPa

m, concrete = 1.50
Prestressing tendon, Ap
Mild Steel, As

229 mm
254 mm
25 mm

Length, L = 9754 mm
Elevation

914 mm

Section

Loads:
Dead, self-wt = 0.254 m  23.56 kN/m3 = 5.984 kN/m2 (D)  1.4 = 8.378 kN/m2 (Du)
Live,
= 4.788 kN/m2 (L)  1.6 = 7.661 kN/m2 (Lu)
Total = 10.772 kN/m2 (D+L) = 16.039 kN/m2 (D+L)ult

 =10.772 kN/m2  0.914 m = 9.846 kN/m, u = 16.039 kN/m2  0.914 m = 14.659 kN/m
wl12
Ultimate Moment, M U 
= 14.659  (9.754)2/8 = 174.4 kN-m
8

SS CP 65-99 PT-SL EXAMPLE 001 - 4

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f pu Ap 
7000 
 1  1.7

l/d 
f cu bd 
7000 
1862(198) 
 1210 
1  1.7

9754 / 229 
30(914)(229) 

Ultimate Stress in strand, f pb  f pe 

 1358 MPa  0.7 f pu  1303 MPa
K factor used to determine the effective depth is given as:
174.4
M
K
 0.1213 < 0.156
=
2
f cu bd
30000(0.914)(0.229) 2


K 
  0.95d = 192.2 mm
z  d  0.5  0.25 
0.9 

Ultimate force in PT, Fult , PT  AP ( f PS )  2  99 1303 1000  258.0 kN
Ultimate moment due to PT,
M ult , PT  Fult , PT ( z ) /   258.0  0.192  1.15  43.12 kN-m
Net Moment to be resisted by As,
M NET  MU  M PT
 174.4  43.12  131.28 kN-m
The area of tensile steel reinforcement is then given by:
M NET
131.28
1e6   1965 mm 2
As 
=
0.87 f y z X
0.87  400 192 
Check of Concrete Stresses at Midspan:
Initial Condition (Transfer), load combination (D+PTi) = 1.0D+0.0L+1.0PTI

Tendon stress at transfer = jacking stress  stressing losses = 1490  186 = 1304 MPa
The force in the tendon at transfer, = 1304 197.4  1000  257.4 kN
Moment due to dead load, M D  5.984  0.914  9.754  8  65.04 kN-m
2

Moment due to PT,

Stress in concrete,

M PT  FPTI (sag)  257.4 102 mm  1000  26.25 kN-m
F
M  M PT
257.4
65.04  26.23
f  PTI  D


A
S
0.254  0.914 
0.00983
where S = 0.00983m3
f  1.109  3.948 MPa
f  5.058(Comp) max, 2.839(Tension) max
SS CP 65-99 PT-SL EXAMPLE 001 - 5

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Normal Condition, load combinations: (D+L+PTF) = 1.0D+1.0L+1.0PTF

Tendon stress at normal = jacking  stressing  long-term = 1490  186  94 = 1210 MPa
The force in tendon at normal, = 1210 197.4  1000  238.9 kN
Moment due to dead load, M D  5.984  0.914  9.754  8  65.04 kN-m
2

Moment due to live load, M L  4.788  0.914  9.754  8  52.04 kN-m
2

Moment due to PT,

M PT  FPTI (sag)  238.9 102 mm  1000  24.37 kN-m

Stress in concrete for (D+L+PTF),
F
M
 M PT
238.8
117.08  24.37
f  PTI  D  L


A
S
0.254  0.914 
0.00983
f  1.029  9.431
f  10.460(Comp) max, 8.402(Tension) max

SS CP 65-99 PT-SL EXAMPLE 001 - 6



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