ACI 318 14 PT SL Example 001

User Manual: ACI 318-14 PT-SL Example 001

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Software Verification
PROGRAM NAME:
REVISION NO.:

ETABS
0

ACI 318-14 PT-SL EXAMPLE 001
Design Verification of Post-Tensioned Slab using the ACI 318-14 code
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 10
inches thick by 36 inches wide and spans 32 feet, as shown in shown in Figure 1.
A 36-inch-wide design strip was centered along the length of the slab and was
defined as an A-Strip. B-strips were placed at each end of the span perpendicular
to the Strip-A (the B-Strips are necessary to define the tendon profile). A tendon,
with two strands having an area of 0.153 square inches each, was added to the AStrip. The self weight and live loads were added to the slab. The loads and posttensioning forces are shown below. 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 were compared with the ETABS results
and summarized for verification and validation of the ETABS results.
Loads: Dead = self weight, Live = 100psf

ACI 318-14 PT-SL EXAMPLE 001 - 1

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Figure 1 One-Way Slab
GEOMETRY, PROPERTIES AND LOADING
Thickness,
Effective depth,
Clear span,
Concrete strength,
Yield strength of steel,
Prestressing, ultimate

T, h
d
L
f 'C
fy

=
=
=
=
=

f pu =

Prestressing, effective
fe
Area of Prestress (single strand), AP
Concrete unit weight,
wc
Modulus of elasticity,
Ec
Modulus of elasticity,
Es
Poisson’s ratio,

Dead load,
wd
Live load,
wl

10
9
384
4,000
60,000

in
in
in
psi
psi

270,000 psi

=

175,500 psi

=
=
=
=
=
=
=

0.153
0.150
3,600
29,000
0
self
100

sq in
pcf
ksi
ksi
psf
psf

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.

ACI 318-14 PT-SL EXAMPLE 001 - 2

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RESULTS COMPARISON
The ETABS total factored moments, required mild steel reinforcing and slab
stresses are compared to the independent hand calculations in Table 1.
Table 1 Comparison of Results
FEATURE TESTED

INDEPENDENT
RESULTS

ETABS
RESULTS

DIFFERENCE

1429.0

1428.3

-0.05%

2.21

2.21

0.00%

0.734

0.735

0.14%

0.414

0.414

0.00%

1.518

1.519

0.07%

1.220

1.221

0.08%

1.134

1.135

0.09%

0.836

0.837

0.12%

Factored moment,
Mu (Ultimate) (k-in)
Area of Mild Steel req’d,
As (sq-in)
Transfer Conc. Stress, top
(D+PTI), ksi
Transfer Conc. Stress, bot
(D+PTI), ksi
Normal Conc. Stress, top
(D+L+PTF), ksi
Normal Conc. Stress, bot
(D+L+PTF), ksi
Long-Term Conc. Stress,
top (D+0.5L+PTF(L)), ksi
Long-Term Conc. Stress,
bot (D+0.5L+PTF(L)), ksi

COMPUTER FILE: ACI 318-14 PT-SL EX001.EDB
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.

ACI 318-14 PT-SL EXAMPLE 001 - 3

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CALCULATIONS:

Design Parameters:
 =0.9
Mild Steel Reinforcing
fc = 4000 psi
fy = 60,000 psi

Post-Tensioning
fj
Stressing Loss
Long-Term Loss
fi
fe

= 216.0 ksi
= 27.0 ksi
= 13.5 ksi
= 189.0 ksi
= 175.5 ksi

Loads:
Dead, self-wt = 10 / 12 ft  0.150 kcf = 0.125 ksf (D)  1.2 = 0.150 ksf (Du)
Live,
0.100 ksf (L)  1.6 = 0.160 ksf (Lu)
Total =0.225 ksf (D+L)
0.310 ksf (D+L)ult

 =0.225 ksf  3 ft = 0.675 klf,

Ultimate Moment, M U 

 u = 0.310 ksf  3ft = 0.930 klf

wl12
= 0.310 klf  322/8 = 119.0 k-ft = 1429.0 k-in
8

ACI 318-14 PT-SL EXAMPLE 001 - 4

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f'c
(span-to-depth ratio > 35)
300  P
4, 000
 175,500  10, 000 
300  0.000944 
 199, 624 psi  205,500 psi

Ultimate Stress in strand, f PS  f SE  10000 

Ultimate force in PT, Fult , PT  AP  f PS   2  0.153199.62   61.08 kips
Ultimate force in RC, Fult , RC  As  f  y  2.00(assumed)  60.0   120.0 kips
Total Ultimate force, Fult ,Total  61.08  120.0  181.08 kips
Stress block depth, a 

Fult ,Total
0.85 f ' cb



181.08
 1.48 in
0.85  4   36 

a

 1.48   
Ultimate moment due to PT, M ult , PT  Fult , PT  d     61.08  9 
 0.9  454.1 k-in
2
2 


Net ultimate moment, M net  M U  M ult , PT  1429.0  454.1  974.9 k-in

Required area of mild steel reinforcing, AS 

M net
974.9

 2.18 in 2
a
1.48



 f y  d   0.9  60   9 

2
2 



Note: The required area of mild steel reinforcing was calculated from an assumed amount of
steel. Since the assumed value and the calculated value are not the same a second iteration can be
performed. The second iteration changes the depth of the stress block and the calculated area of
steel value. Upon completion of the second iteration the area of steel was found to be 2.21in2

ACI 318-14 PT-SL EXAMPLE 001 - 5

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Check of Concrete Stresses at Mid-Span:
Initial Condition (Transfer), load combination (D + L + PTi) = 1.0D + 1.0PTI

The stress in the tendon at transfer = jacking stress  stressing losses = 216.0  27.0
= 189.0 ksi


The force in the tendon at transfer, = 189.0 2  0.153  57.83 kips
2
Moment due to dead load, M D  0.125  3 32  8  48.0 k-ft  576 k-in

Moment due to PT, M PT  FPTI (sag)  57.83  4 in   231.3 k-in
F
M  M PT 57.83 576.0  231.3
Stress in concrete, f  PTI  D
, where S = 600 in3


A
S
10  36 
600
f  0.161  0.5745
f = -0.735(Comp)max, 0.414(Tension)max
Normal Condition, load combinations: (D + L + PTF) = 1.0D + 1.0L + 1.0PTF

Tendon stress at normal = jacking  stressing  long-term = 216.0  27.0  13.5 = 175.5 ksi
The force in tendon at Normal, = 175.5  2   0.153  53.70 kips
2
Moment due to dead load, M D  0.125  3 32  8  48.0 k-ft  576 k-in
Moment due to dead load, M L  0.100  3 32  8  38.4 k-ft  461 k-in
2

Moment due to PT,

M PT  FPTI (sag)  53.70  4 in   214.8 k-in

FPTI M D  L  M PT 53.70 1037.0  214.8



10  36 
600
A
S
f  0.149  1.727  0.358
f  1.518(Comp) max,1.220(Tension) max

Stress in concrete for (D + L+ PTF), f 

Long-Term Condition, load combinations: (D + 0.5L + PTF(L)) = 1.0D + 0.5L + 1.0PTF

Tendon stress at normal = jacking  stressing  long-term = 216.0  27.0  13.5 = 175.5 ksi
The force in tendon at Normal, = 175.5  2   0.153  53.70 kips
2
Moment due to dead load, M D  0.125  3 32  8  48.0 k-ft  576 k-in
Moment due to dead load, M L  0.100  3 32  8  38.4 k-ft  460 k-in
2

Moment due to PT,

ACI 318-14 PT-SL EXAMPLE 001 - 6

M PT  FPTI (sag)  53.70  4 in   214.8 k-in

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Stress in concrete for (D + 0.5L + PTF(L)),
M
 M PT 53.70 806.0  214.8
F
f  PTI  D  0.5 L


A
S
10  36 
600
f  0.149  0.985
f  1.134(Comp) max, 0.836(Tension) max

ACI 318-14 PT-SL EXAMPLE 001 - 7



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