UNIVERSITY OF EASTERN PHILIPPINES

COLLEGE OF ENGINEERING
SCE 104 – PROFESSIONA COURSE 4 (PRESTRESSED CONCRETE DESIGN)
2ND SEM.SY 2024-2025
PROFESSOR: ENGR. RIC L. GONZAGA, MCE

CHAPTER III: ELASTIC DESIGN OF PRESTRESSED

ONCRETE BEAM
The following are the learning outcomes to be discuss in this chapter, to wit:

Learning Outcomes
At the end of this chapter, you will be able to:

1. Determine the Elastic Flexural Analysis in Working Strength Design (WSD)

2. Find the different conditions and limitations of the elastic equation for flexural
stress.
3. Calculate and investigate the design of the section and the stresses of the
beam.

3.1 Introduction

Structural analysis for elastic flexural analysis, you can solve and determine the
different types of problems and approaches utilizing this process shown below:

ELASTIC FLEXURAL ANALYSIS

(WORKING STRENGTH)

Where:
P = Prestressing Force
H = Horizontal Component of P = P cos 𝜃
V = Vertical Component of P = P sin 𝜃 = H tan 𝜃
𝜃 = Angle of Inclination of the tendon centroid at the particular section
Since 𝜃 is normally quite small: cos 𝜃 ≅ 1.0, ∴ 𝐻 = 𝑃

 The magnitude of the prestress force P is not constant for the following reasons.

1. The jacking force (Pj) is immediately reduced to initial prestressing force, Pi because of
elastic shortening of concrete upon transfer, slip of the tendon as the force is transferred
from jacks to beam ends, and loss due to friction between tendon and concrete (post –
tensioning) or between tendon and cable alignment devices (pretensioning).
 UNIVERSITY OF EASTERN PHILIPPINES
COLLEGE OF ENGINEERING
SCE 104 – PROFESSIONA COURSE 4 (PRESTRESSED CONCRETE DESIGN)
2ND SEM.SY 2024-2025
PROFESSOR: ENGR. RIC L. GONZAGA, MCE

2. There is further reduction of force from Pi to the (effective prestress, Pe), occurring over a
long period of time a a gradually decreasing rate, because of concrete creep under the
sustained prestress force, concrete shrinkage and relaxation of stress in the steel.

3.2 ELASTIC EQUATIONS FOR FLEXURAL STRESS

1. DUE TO PRESTRESSING FORCE WHEN PI IS APPLIED WITH AN ECCENTRICITY

E BELOW THE CENTROID OF THE CROSS SECTION AREA AC AND TOP AND
BOTTOM FIBER DISTANCES C1 AND C2 RESPECTIVELY
𝑃 𝑃𝑖 𝑒𝑐1
𝑓𝑖 = − 𝐴 𝑖 + = Top Fiber Stress
𝑐 𝐼𝑐
𝑃 𝑒𝑐1 𝐴𝑐
= − 𝐴 𝑖 (1 − )
𝑐 𝐼𝑐

𝐼
Since 𝑟 = √𝐴𝑐 = radius of gyration of section
𝑐
𝐼𝑐
𝑟2 = 𝐴𝑐

𝑃𝑖 𝑒𝑐1
Then, 𝑓1 = − (1 − 𝐼𝑐 )
𝐴𝑐
𝐴𝑐
𝑃𝑖 𝑒𝑐1
𝑓1 = − (1 − 2 )
𝐴𝑐 𝑟

𝑃𝑖 𝑒𝑐1
Likewise, 𝑓2 = − (1 + ) = 𝐵𝑜𝑡𝑡𝑜𝑚 𝐹𝑖𝑏𝑒𝑟 𝑆𝑡𝑟𝑒𝑠𝑠
𝐴𝑐 𝑟2
 (-) compression; (+) tension
 Normally, as the eccentric prestress force is applied,
the beam deflects upward.

2. INITIAL PRESTRESSING + SELF WEIGHT

When W o = Self weight
Mo = Moment due to W o

Then:,
 UNIVERSITY OF EASTERN PHILIPPINES
COLLEGE OF ENGINEERING
SCE 104 – PROFESSIONA COURSE 4 (PRESTRESSED CONCRETE DESIGN)
2ND SEM.SY 2024-2025
PROFESSOR: ENGR. RIC L. GONZAGA, MCE

3. EFFECTIVE PRESTRESSING AND SELF WEIGHT

4. Pe + SERVICE LOADS IN ADDITION TO SELF WEIGHT OF THE BEAM, PLUS

SERVICE LIVE LOAD

Where:
Mo = Moment due to prestressing
Md = Moment due to Dead Load
Ml = Moment due to Live Load
Ic = Moment of Inertia

KERN POINTS LOCATION

Upper Kern-Point
 UNIVERSITY OF EASTERN PHILIPPINES
COLLEGE OF ENGINEERING
SCE 104 – PROFESSIONA COURSE 4 (PRESTRESSED CONCRETE DESIGN)
2ND SEM.SY 2024-2025
PROFESSOR: ENGR. RIC L. GONZAGA, MCE

𝑟2
Similarly, 𝑘2 = Lower Kern-Point
𝑐1

 For Rectangular Beams:

Ac = bd
𝑑
C2 = C1 = 2
𝑏𝑑 3
𝐼𝑐 =
12
Thus;
𝑑
𝑒𝑐2 𝑒𝑐2 𝐴𝑐 𝑒 ( 2 ) (𝑏𝑑)
= =
𝑟2 𝐼𝑐 𝑏𝑑 3
12
6𝑒
=
𝑑
Likewise:
𝑒𝑐1 6𝑒
=
𝑟2 𝑑
Then:
𝑃𝑒 6𝑒 (𝑀𝑜 + 𝑀𝑑 + 𝑀𝑙 )𝑐1
𝑓1 = − (1 − ) −
𝐴𝑐 𝑑 𝐼𝑐
𝑑
𝑐1 6 6
= 23 = 2
=
𝐼𝑐 𝑏𝑑 𝑏𝑑 𝐴𝑐 𝑑
12

𝑃𝑒 6𝑒 (𝑀𝑜 + 𝑀𝑑 + 𝑀𝑙 )6
𝑓1 = − (1 − ) −
𝐴𝑐 𝑑 𝑏𝑑 2
𝑃𝑒 6𝑒 (𝑀𝑜 + 𝑀𝑑 + 𝑀𝑙 )6
𝑓2 = − (1 + ) +
𝐴𝑐 𝑑 𝑏𝑑 2

𝐼𝑐
𝑟2 𝐴𝑐
𝑘1 = − = −
𝑐1 𝑑
2
𝒅
𝒌𝟏 = −
𝟔
𝒅
𝒌𝟐 =
𝟔
 UNIVERSITY OF EASTERN PHILIPPINES
COLLEGE OF ENGINEERING
SCE 104 – PROFESSIONA COURSE 4 (PRESTRESSED CONCRETE DESIGN)
2ND SEM.SY 2024-2025
PROFESSOR: ENGR. RIC L. GONZAGA, MCE
Activity # 1

INSTRUCTION: Read comprehensively and follow the instruction

carefully.
I. Identification. Identify the following equation/formula pertaining to and
what is the notation of each variable/constant.
𝒅
1. 𝑘𝟐 = 𝟔

𝐼
2. 𝑟 = √ 𝑐
𝐴 𝑐

𝑃𝑖 𝑒𝑐1 𝐴𝑐
3. 𝑓𝑖 = − (1 − )
𝐴𝑐 𝐼𝑐

𝑷𝒆 𝑒𝑐1 𝑴𝒐 𝒄 𝟏
4. 𝑓1 = (1 − )−
𝑨𝒄 𝑟2 𝑰𝒄

5. P = P sin 𝜃 = H tan 𝜃
 UNIVERSITY OF EASTERN PHILIPPINES
COLLEGE OF ENGINEERING
SCE 104 – PROFESSIONA COURSE 4 (PRESTRESSED CONCRETE DESIGN)
2ND SEM.SY 2024-2025
PROFESSOR: ENGR. RIC L. GONZAGA, MCE

3.3 ELASTIC DESIGN OF PRESTRESS BEAMS (WSD)

In Elastic Design of Prestress Beam, Preliminary design procedure based on the

assumption that there is a little or no tension in the concrete.

Under working load, z = 0.65d (on the average)

Where: d = overall depth

b = overall width
t = f = effective prestressing force
M = Tz = T (0.65d)
Or
M = Fz = F (0.65d)

 𝑴
F = 𝟎.𝟔𝟓 𝒅 (effective prestressing force)
But F = Asfs T=C
𝑀
Asfs = 0.65 𝑑
 Average Unit Stress on Concrete, 𝛿c =𝐴
𝐶
𝑐
𝑪 𝑨𝒔 𝒇𝒔
𝟎. 𝟓𝟎 𝒇𝒄 = =
𝑨𝒄 𝑨𝒄

Note: for preliminary design, the ave. stress is assumed to be ½ fc or 0.5 fc

 Approximate proportion of depth of beam and moment

d = k √𝑀 = depth of beam in centimeters

k = coefficient which ranges from 3.3 to 4.4 – from experience and
experiment
M = Max. Bending Moment in KN-m
Mg = girder or beam moment
Mt = Total Moment
Ml = Mt - Mg
𝑀𝑔
When 𝑀𝑡
is small, use the following approximate relation:

𝑴
F = 𝟎.𝟓𝟎𝒍 𝒅
 UNIVERSITY OF EASTERN PHILIPPINES
COLLEGE OF ENGINEERING
SCE 104 – PROFESSIONA COURSE 4 (PRESTRESSED CONCRETE DESIGN)
2ND SEM.SY 2024-2025
PROFESSOR: ENGR. RIC L. GONZAGA, MCE
Example 1: Problem: Design a rectangular prestress concrete beam to carry a liveload of 30
KN/m on a span of 9.15m, using a straight post tensioned wires using stress relived
solid wires 6mm in a diameter with a final effective stress of 0.60 f y after loss of
stress due to elastic shortening of concrete, shrinkage and creep. Assume the solid
wires to be placed 10 cm above the bottom of the beam to give ample protection
or covering. Determine the initial prestressing force that must be applied and its
location assuming 15% loss of stress. Allowable compressive stress of concrete is
0.45 f’c and allowable tensile stress of concrete is 0.54 √𝑓′𝑐. Assume concrete
wont crack in tension. Ultimate stress of stress – relieved solid wires is f y = 1655
MPa, f’c = 20.7 MPa.

Given: W = 30 KN/m
L = 9.15 m
fy = 1655 MPa,
f’c = 20.7 MPa
Final effective stress of 0.60 f y after loss of stress due to elastic shortening of
concrete, shrinkage and creep
Allowable compressive stress of concrete is 0.45 f’c
Allowable tensile stress of concrete is 0.54 √𝑓′𝑐
Assume concrete wont crack in tension

Required: Design a rectangular prestress concrete beam

Solution:

I. PRELIMINARY DESIGN

𝑊𝐿2 30 (9.15)2
1. MLL = = = 314 𝐾𝑁 − 𝑚 = 314 x 102 KN – cm
8 8

2. Trial Value of “d”

d = k √𝑀 ; using k = 3.3 – if possible to economize
d = 3.3 √314 = 58.48 𝑐𝑚 = 584.76 𝑐𝑚
say d = 60 cm
𝑀
3. F = 0.65 𝑑M
314 𝑥 102 𝐾𝑁−𝑐𝑚
= 0.65 (60𝑐𝑚)
= 805.13 KN

4. T = F = ASfy
85.13x103 = As(0.60)f y
8.05x103 = As(0.60)(1655 MPa)
As = 810.8 mm2 say 811 mm2
 UNIVERSITY OF EASTERN PHILIPPINES
COLLEGE OF ENGINEERING
SCE 104 – PROFESSIONA COURSE 4 (PRESTRESSED CONCRETE DESIGN)
2ND SEM.SY 2024-2025
PROFESSOR: ENGR. RIC L. GONZAGA, MCE

Using 6mmØ wire:

𝑝𝑖
Ab = 4
(6mm)2 = 28.27 mm2
As 811
No. of wires req’d = N = Ab = 28.27 = 28.69 say 29 wires
Actual As = 28.27 mm2 (29) = 819.83 mm2 > 811 mm2

& C=T
1/2 fc Ac = Asfs
fc = 0.45 fc’ = 0.45(20.7)
= 9.315 MPa
1/2 (9.315) Ac = 811(0.6)(1655)
Ac = 172909 mm2
Ac = bd = b (600 mm)

172909 = b (600)
b = 288.19 mm say 300mm

INCREASE SECTION BY
Adding 50 mm to b
& 150 mm to d
Why?...This is to compensate for the self-weight of the beam

Then, try 350 mm x 750 mm

II. INVESTIGATE
1. DEAD LOAD
W DL = 2400(9.81)(0.35)(0.75) = 6180.30 N/m
WDL L2 6180.30(9.15)2
MDL = 8
= 8
= 64.68 KN-m
6MDL 6(64.68𝑥106 )
fDL = = = 1.971 MPa
bd2 (350)(750)2
2. LIVE LOAD
MLL = 314 x 106 N-mm
6MLL 6(314𝑥106 )
FLL = bd2
= (350)(750)2
= 9.570 MPa
3. DUE TO PRE-STRESS
P = Actual AS (fs)
fs = 0.60 (1655) = 993 MPa
P = 819.83 (993)
= 814091.19 N
750
e = 2 - 100 = 275 mm
 UNIVERSITY OF EASTERN PHILIPPINES
COLLEGE OF ENGINEERING
SCE 104 – PROFESSIONA COURSE 4 (PRESTRESSED CONCRETE DESIGN)
2ND SEM.SY 2024-2025
PROFESSOR: ENGR. RIC L. GONZAGA, MCE

𝑃 6𝑒 814091.19 6(275)
f1 = bd [1 - d
]= 350(750)
[1 – 750
] = -3.10 (1 - 2.2) MPa = + 3.72 (TENSION)
𝑃 6𝑒 814091.19 6(275)
f2 = bd [1 + d
]= 350(750)
[1 + 750
] = -3.10 (1 + 2.2) MPa = - 9.92 (COMPRESSION)
STRESS DIAGRAMS:

Answer:
fC = 7.821 MPa
ft = 1.621 MPa

Allowable stresses:
fC = 0.45fc’ MPa = 0.45(20.7) = 9.315 MPa
ft = 0.54 √fc’ = 0.54√20.7 =2.46 MPa

since, actual fC < Allow. fC

actual ft < Allow. ft ∴ safe!

Example 2: Problem: A 500 mm by 760 mm concrete beam of 7.5m simple span is loaded with
a uniform load of 4,470 kg/m including its own weight. The center of gravity of the
prestressing tendon is located 220mm above the bottom of the beam and produces
an effective prestress of 16400 kg. Compute the fiber stress in the concrete at the
midspan section and show by diagrams the resulting stress distribution.

Given: w = 4,470 kg/m

e = 220 mm
Pe = 16,400 kg
L = 7.5 m

Required: Compute the fiber stress in the concrete at the midspan section and show by diagrams
the resulting stress distribution.

Solution: W = 4470 kg/m = 43,850.70 N/m

 UNIVERSITY OF EASTERN PHILIPPINES
COLLEGE OF ENGINEERING
SCE 104 – PROFESSIONA COURSE 4 (PRESTRESSED CONCRETE DESIGN)
2ND SEM.SY 2024-2025
PROFESSOR: ENGR. RIC L. GONZAGA, MCE
𝑊𝑜 𝐿2
M= = 308.33 x 10^3 N-m
8
760𝑚𝑚
e = 2 – 220mm = 160mm
𝑃𝑒 = 16,400 kg = 1608.84 KN

FOR RECTANGULAR BEAMS

6𝑀 6 ( 308.33 𝑥 103 ) (1000)

𝑓𝑡 = = = 6.41 MPa
𝑏𝑑2 500𝑚𝑚 ( 760𝑚𝑚) 2

𝑃𝑒 6𝑒 −1608.84 𝑥 103 𝑁 6 (160𝑚𝑚)

𝑓1 = − (1 − ) = (1 − )
𝑏𝑑 𝑑 500𝑚𝑚 ( 760𝑚𝑚) 760𝑚𝑚

𝒇𝟏 = +𝟏. 𝟏𝟏 𝑴𝑷𝒂 (𝑻𝒆𝒏𝒔𝒊𝒐𝒏)

𝑃𝑒 6𝑒 −1608.84 𝑥 103 𝑁 6 (160𝑚𝑚)

𝑓2 = − (1 + ) = (1 + )
𝑏𝑑 𝑑 500𝑚𝑚 ( 760𝑚𝑚) 760𝑚𝑚

𝒇𝟐 = −𝟗. 𝟓𝟖 𝑴𝑷𝒂 (𝑪𝒐𝒎𝒑𝒓𝒆𝒔𝒔𝒊𝒐𝒏)

Answer:

Example 3: Problem: A simply supported prestressed concrete beam, 9.0m long carries a uniform
load of 58 KN/m including its own weight and two equal concentrated loads of 110
KN at its third points as shown. The size of the beam is 460mm x 920mm. If an
effective prestress of 1250 KN is produced on a tendon located 0.15m from the
bottom of the beam, calculate the fiber stress of the beam at the point of maximum
moment. Draw the stress distribution diagrams and the resultant stress diagram.
Given:
 UNIVERSITY OF EASTERN PHILIPPINES
COLLEGE OF ENGINEERING
SCE 104 – PROFESSIONA COURSE 4 (PRESTRESSED CONCRETE DESIGN)
2ND SEM.SY 2024-2025
PROFESSOR: ENGR. RIC L. GONZAGA, MCE
Required: Calculate the fiber stress of the beam at the point of maximum moment. Draw the
stress distribution diagrams and the resultant stress diagram.

Solution:

1. Moment due to Uniform Load,

𝑊𝐿2 (58)(9)2
𝑀𝐷𝐿 = 8 = 8 = 587.25 𝐾𝑁 − 𝑚
2. Stress due to Uniform Load,

6𝑀𝐷𝐿 (6) 587.25 𝑥 106

𝑓𝐷𝐿 = = = 9.05 MPa
𝑏𝑑2 460 (920)2

3. Moment due to Concentrated Loads at middle thirds

𝑃𝐿 (110) (9)
𝑀𝐿𝐿 = = = 330.0 KN-m
3 3

4. Stress due to Concentrated Loads

6𝑀𝐿𝐿 (6)330 𝑥 106

𝑓𝐿𝐿 = = = ± 5.09 𝑀𝑃𝑎
𝑏𝑑2 460 (920) 2

5. Stress due to effective Prestress

𝑃 6𝑒
𝑓1 = − 𝑏𝑑𝑒 (1 − ) Top fiber (tension)
𝑑
𝑃 6𝑒
𝑓2 = − 𝑏𝑑𝑒 (1 + )
𝑑
𝑑 920𝑚𝑚
e = 2 – 150mm = – 150mm = 310 mm
2
𝑃𝑒 = 1250 KN
1250 𝑥 103 𝑁 6(310)
𝑓1 = − 460 (920) 𝑚𝑚2 (1 − ) = -2.95 MPa (1 – 2.02)
920
Answer: 𝒇𝟏 = −𝟐. 𝟗𝟓 (−𝟏. 𝟎𝟐) 𝑴𝑷𝒂 = +𝟑. 𝟎𝟎𝟗 𝑴𝑷𝒂
1250 𝑥 103 𝑁 6(310)
𝑓2 = − 460 (920) 𝑚𝑚2 (1 + ) = -2.95 MPa (1 + 2.02)
920
𝒇𝟐 = −𝟖. 𝟗𝟎𝟗 𝑴𝑷𝒂

6. Stress Diagrams
 UNIVERSITY OF EASTERN PHILIPPINES
COLLEGE OF ENGINEERING
SCE 104 – PROFESSIONA COURSE 4 (PRESTRESSED CONCRETE DESIGN)
2ND SEM.SY 2024-2025
PROFESSOR: ENGR. RIC L. GONZAGA, MCE

Activity # 2

INSTRUCTION, Read the problem comprehensively and follow the

instruction carefully. Write your solutions and answers on a
separate sheet of paper.

I. A concrete beam has dimensions of 300mm width and 570mm total

depth. The simple span is 9m. fc’ = 21 MPa. If a live load of 14 KN/m
is imposed on the beam, investigate if the section is adequate under
the
a.) Conventional reinforced concrete beam design
n = 9; fc = 0.45 fc’ = 9.45 MPa fs = 138 MPa
Protective covering = 65mm

b.) Prestressed concrete beam design

Allowable concrete stress loss in prestress are:
fc = 9.45 MPa ft = 0
Loss in prestress is 15%
Show computations and calculations
 UNIVERSITY OF EASTERN PHILIPPINES
COLLEGE OF ENGINEERING
SCE 104 – PROFESSIONA COURSE 4 (PRESTRESSED CONCRETE DESIGN)
2ND SEM.SY 2024-2025
PROFESSOR: ENGR. RIC L. GONZAGA, MCE

INSTRUCTION, Read the problem comprehensively and follow the

instruction carefully. Write your solutions and answers on the
space provided.

II. Determine the safe liveload (uniform load) that 300mm x 450mm
concrete beam could carry if it has a simple span of 9.0m.
Assuming that the maximum final (prestress – stress) for top and
bottom are 12.0 kg/cm2 and 140.6 kg/cm2 respectively, determine
also the prestressing force that could be applied to the beam and
the corresponding eccentricity, without tension resulting at the
bottom. What is the stress at the top of the beam?
 UNIVERSITY OF EASTERN PHILIPPINES
COLLEGE OF ENGINEERING
SCE 104 – PROFESSIONA COURSE 4 (PRESTRESSED CONCRETE DESIGN)
2ND SEM.SY 2024-2025
PROFESSOR: ENGR. RIC L. GONZAGA, MCE
SUMMARY

 Elastic flexural analysis can be solved and determined the different types of problems
and approaches utilizing these process shown below:
ELASTIC FLEXURAL ANALYSIS
(WORKING STRENGTH)

Where:P = Prestressing Force

H = Horizontal Component of P = P cos 𝜃
V = Vertical Component of P = P sin 𝜃 = H tan 𝜃
𝜃 = Angle of Inclination of the tendon centroid at the particular section
Since 𝜃 is normally quite small: cos 𝜃 ≅ 1.0, ∴ 𝐻 = 𝑃

 The magnitude of the prestress force P is not constant for the following
reasons.

1. The jacking force (Pj) is immediately reduced to initial

prestressing force, Pi because of elastic shortening of concrete
upon transfer, slip of the tendon as the force is transferred from
jacks to beam ends, and loss due to friction between tendon and
concrete (post – tensioning) or between tendon and cable
alignment devices (pretensioning).

2. There is further reduction of force from P i to the (effective

prestress, Pe), occurring over a long period of time a a gradually
decreasing rate, because of concrete creep under the sustained
prestress force, concrete shrinkage and relaxation of stress in the
steel.

 ELASTIC EQUATIONS FOR FLEXURAL STRESS can bs classified according to:

 DUE TO PRESTRESSING FORCE WHEN PI IS APPLIED WITH AN

ECCENTRICITY E BELOW THE CENTROID OF THE CROSS SECTION
AREA AC AND TOP AND BOTTOM FIBER DISTANCES C1 AND C2
RESPECTIVELY
 UNIVERSITY OF EASTERN PHILIPPINES
COLLEGE OF ENGINEERING
SCE 104 – PROFESSIONA COURSE 4 (PRESTRESSED CONCRETE DESIGN)
2ND SEM.SY 2024-2025
PROFESSOR: ENGR. RIC L. GONZAGA, MCE

𝑃 𝑃𝑖 𝑒𝑐1
𝑓𝑖 = − 𝐴 𝑖 + = Top Fiber Stress
𝑐 𝐼𝑐
𝑃𝑖 𝑒𝑐 𝐴
= − 𝐴 (1 − 𝐼1 𝑐 )
𝑐 𝑐
𝑃𝑖 𝑒𝑐1
𝑓2 = − (1 + 2 ) = 𝐵𝑜𝑡𝑡𝑜𝑚 𝐹𝑖𝑏𝑒𝑟 𝑆𝑡𝑟𝑒𝑠𝑠
𝐴𝑐 𝑟
 (-) compression; (+) tension
 Normally, as the eccentric prestress force is applied, the beam
deflects upward.

 INITIAL PRESTRESSING + SELF WEIGHT

When W o = Self weight
Mo = Moment due to W o

 EFFECTIVE PRESTRESSING AND SELF WEIGHT

 Pe + SERVICE LOADS IN ADDITION TO SELF WEIGHT OF THE BEAM,

PLUS SERVICE LIVE LOAD

Where:
Mo = Moment due to prestressing
Md = Moment due to Dead Load
Ml = Moment due to Live Load
Ic = Moment of Inertia

𝑑
For Rectangular Beams:𝑘1 = − 6
𝑑
𝑘2 =
6
 UNIVERSITY OF EASTERN PHILIPPINES
COLLEGE OF ENGINEERING
SCE 104 – PROFESSIONA COURSE 4 (PRESTRESSED CONCRETE DESIGN)
2ND SEM.SY 2024-2025
PROFESSOR: ENGR. RIC L. GONZAGA, MCE

 ELASTIC DESIGN OF PRESTRESS BEAMS (WSD)

In Elastic Design of Prestress Beam, Preliminary design procedure based on the
assumption that there is a little or no tension in the concrete.

Under working load, z = 0.65d (on the average)

Where: d = overall depth
b = overall width
t = f = effective prestressing force
M = Tz = T (0.65d)
or
M = Fz = F (0.65d)
 F = 0.65𝑀 𝑑 (effective prestressing force)
But F = Asfs T=C
𝑀
Asfs = 0.65 𝑑
 Average Unit Stress on Concrete,
𝐶
𝛿 c =𝐴
𝑐
𝐶 𝐴𝑠 𝑓𝑠
0.50 𝑓𝑐 = =
𝐴𝑐 𝐴𝑐

Note: for preliminary design, the ave. stress is assumed to be ½ fc or 0.5 fc

 Approximate proportion of depth of beam and moment

d = k √𝑀 = depth of beam in centimeters
k = coefficient which ranges from 3.3 to 4.4 – from experience and experiment
M = Max. Bending Moment in KN-m
Mg = girder or beam moment
Mt = Total Moment
Ml = Mt - Mg
𝑀𝑔
When 𝑀 is small, use the following approximate relation:
𝑡
𝑴
F = 𝟎.𝟓𝟎𝒍 𝒅
 UNIVERSITY OF EASTERN PHILIPPINES
COLLEGE OF ENGINEERING
SCE 104 – PROFESSIONA COURSE 4 (PRESTRESSED CONCRETE DESIGN)
2ND SEM.SY 2024-2025
PROFESSOR: ENGR. RIC L. GONZAGA, MCE

HOW MUCH HAVE YOU LEARNED?

SELF ASSESSMENT EXAMINATION:

INSTRUCTION: Read the problem comprehensively and follow the instruction carefully.
Write your solutions and answers on the space provided..

1. Determine the prestressing steel area required in the precast concrete T-Beam,
given the following data:
MTOTAL = 434 KN-m
Effective Prestress for steel = 862 N/mm2
Allowable Stress for concrete under working load is 11.03 N/mm2
Consider zero stress at the bottom of the beam
What is the concrete stress at the top of the beam?
 UNIVERSITY OF EASTERN PHILIPPINES
COLLEGE OF ENGINEERING
SCE 104 – PROFESSIONA COURSE 4 (PRESTRESSED CONCRETE DESIGN)
2ND SEM.SY 2024-2025
PROFESSOR: ENGR. RIC L. GONZAGA, MCE

2. A 9.00 m span of a post-tensioned beam is to be designed to carry a uniformly

distributed load. In addition to its weight, it must carry a dead load of 5.50 KN/m and
a service live load of 12.50 KN/m. Concrete strength of 35 MPa at 28 days, at time
of transfer of prestress force the strength will be 20.7 MPa. Prestress loss may be
assumed at 20% of initial prestress. On the basis that about 25% of the liveload will
be sustained over a substantial time period, kb = 0.25 in determining the balanced
load.
 UNIVERSITY OF EASTERN PHILIPPINES
COLLEGE OF ENGINEERING
SCE 104 – PROFESSIONA COURSE 4 (PRESTRESSED CONCRETE DESIGN)
2ND SEM.SY 2024-2025
PROFESSOR: ENGR. RIC L. GONZAGA, MCE

3. A 350 mm x 800 mm concrete beam is prestressed with a prestressing force of

3000 KN at an eccentricity of 150 mm. the beam carrying a superimposed load of
300 KN/m. Determine the maximum span that the beam should have in order that
no tension will result in the bottom fiber if all the loads are acting. What would be
the resulting stress at the top of the beam?
 UNIVERSITY OF EASTERN PHILIPPINES
COLLEGE OF ENGINEERING
SCE 104 – PROFESSIONA COURSE 4 (PRESTRESSED CONCRETE DESIGN)
2ND SEM.SY 2024-2025
PROFESSOR: ENGR. RIC L. GONZAGA, MCE

4. A 300 mm x 600 mm concrete beam having a span of 6.5 m was applied with a
post-tensioning force of 550 KN. By what amount was the form at the midspan
was lifted after the application of the force applied 150 mm from the bottom of the
beam. Assume concrete won’t crack in tension. Modulus of elasticity of concrete
is 13734 MPa.
 UNIVERSITY OF EASTERN PHILIPPINES
COLLEGE OF ENGINEERING
SCE 104 – PROFESSIONA COURSE 4 (PRESTRESSED CONCRETE DESIGN)
2ND SEM.SY 2024-2025
PROFESSOR: ENGR. RIC L. GONZAGA, MCE

5. Find the safe live load that the prestressed T-beam shown which is
reinforced with a bonded tendon having an area Aps = 1,500 mm 2,
fc’ = 34.50 MPa, fpu = 1862 MPa, and an effective stress after losses
fse = 1102 MPa. fpy = 1713 MPa, span of beam is 6.00 m.
What is the implication to the stress?
 UNIVERSITY OF EASTERN PHILIPPINES
COLLEGE OF ENGINEERING
SCE 104 – PROFESSIONA COURSE 4 (PRESTRESSED CONCRETE DESIGN)
2ND SEM.SY 2024-2025
PROFESSOR: ENGR. RIC L. GONZAGA, MCE

ASSIGNED READINGS AND/OR ENRICHMENT

https://www.aceec.ac.in/wp-content/uploads/2019/01/IV-Prestressed-Concrete

https://www.slideshare.net/PriodeepChowdhury/prestressed-concrete-design-
56871214

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