• Pre-stressed concrete = Pre-compression concrete

• Pre-compression stresses is applied at the place when tensile stress occur

• Concrete weak in tension but strong in compression

• Steel tendon is first stressed

• Concrete is then poured

• After harden and reaching the required strength, the steel tendon is cut

Pre-tension

• Concrete is first poured in the mould

• After harden, steel tendon is then stressed and anchored at both ends

Post-tension

Phase 1

Phase 2

Phase 3

Phase 1

Phase 2

Phase 3

Between C30 and C60

High workability during wet and high strength when hardened

Concrete strength at transfer, fci 0.6fck(t) N/mm2

Concrete

7 or 19 nos of strand

High strength steel

Steel strength usually 1650 N/mm2

Steel Tendon

Section 5.10.2.2(3): MS EN 1992-1-1: 2004

Stresses Loading StageTransfer Service

Symbol Value or Equation

Symbol Value or Equation

Compressive fct 0.6fck (t) fcs 0.6fck

Tensile ftt fctm fts 0

Some losses are immediate affecting the prestress force as soon as it

transferred. Other losses occur gradually with time. Known as short-term

and long-term losses, they include:

Short-term

(a)Elastic shortening of the concrete

(b)Slip or movement of tendons at anchorage

(c) Relaxation of prestressing steel

(d)Friction at the bend due to curvature of tendons

Long-term

(a)Creep and shrinkage of the concrete under sustained compression

(b)Relaxation of the prestressing steel under sustained tension

Top fibre

Bottom fibre

e

N.A

yb

yt

h

b

Prestressing

tendon

Ungrouted duct

Reinforcement /

bonded tendon

Mmax = wL2/8 = 80 kNm

8 m

w = 10 kN/m

Tensile stress, ft = fc = Mmax/Z = Mmaxy/I

where I = bh3/12 and yt = yb = h/2

Therefore;

ft = fc = (wL2/8)(h/2) (bh3/12)

= 9.65 N/mm2

yt

ft

fcb = 200 mm

h=

500 m

m

yb_

+

P P

Prestress force, P is then applied to eliminate

tensile stress

N.A

ft = 9.64 N/mm2

fc = 9.65 N/mm2

_

+

Stress due to

loading

=

0 N/mm2

14 N/mm2

Total stress

P/A

+

Stress due to

force P

+

9.65 + P/A = 0

Prestress force, P

required to eliminate

tensile stress

= 9.65A

= 965 kN

P P

If the prestress force, P applied is not at the centroid

eN.A

ft = 9.65 N/mm2

fc = 9.65 N/mm2

_

+

Stress due to

loading

=

0 N/mm2

14 N/mm2

Total stress

P/A

+

Stress due to force P

Pe/Zb

++

+

9.65 + P/A + Pe/Z = 0

Prestress force, P required to eliminate tensile stress

= 9.65 / [(1/A) + (e/Z)] = 283 kN

Pe/Zt

(a) At Transfer

P P

Mi/Zb

Mi/Zt

_

+

Stress due to

self weight

e

P/A

+

Stress due to force P

Pe/Zb

++

+

Pe/ZtP/A

=

f1t ftt

Total stress

f2t fct

+

Mi = Moment due to self weight

= Coefficient of short term losses

(a) At Service

P P

Mi/Zb

Mi/Zt

_

+

Stress due to

self weight

e

P/A

+

Stress due to force P

Pe/Zb

++

+

Pe/ZtP/A

ws kN/m (Service Load)

Stress due to

ws kN/m

Ms/Zb

Ms/Zt

_

+

+ =

f1s fcs

Total stress

f2s fts

+

= Coefficient of long term losses

From both Figures:

At Transfer

Mi/Zt + P/A – Pe/Zt –ftt ---------- (1)

–Mi/Zb + P/A + Pe/Zb fct ---------- (2)

At Service

Mi/Zt + Ms/Zt + P/A – Pe/Zt fcs ---------- (3)

–Mi/Zb – Ms/Zb + P/A + Pe/Zb –fts ---------- (4)

Rearranging Eqs. (1) – (4):

P/A(eA/Zt – 1) – Mi/Zt ftt ---------- (5)

P/A(eA/Zb + 1) – Mi/Zb fct ---------- (6)

P/A(1 – eA/Zt) + Mi/Zt + Ms/Zt fcs ---------- (7)

–P/A(1 + eA/Zb) + Mi/Zb + Ms/Zb fts ---------- (8)

Rectangular beam = b h = 300 950 mm. Prestress force, P =

2000 kN acted at e = 200 mm below centroid. The short-term

and long-term losses of the prestress are 10% and 20%,

respectively.

(a) If fck = 40 N/mm2 draw the stress distribution diagram at mid-

span during transfer and service. Also check that these

stresses are within the allowable limits.

P Pe

ws = 20 kN/m

15 m

Cross sectional area, Ac = b h

= 300 950

= 2.85 105 mm2

Moment of inertia, Ixx = bh3/12

= 300 9503/12

= 2.14 1010 mm4

Modulus of section, Zt = Zb = Ixx/y

= 2.14 1010 / 475

= 4.51 107 mm3

Stress limit for fck = 40 N/mm2 and fci = 28 N/mm2

At Transfer

fct = 0.6fck(t) = 0.6(28) = 16.8 N/mm2

ftt = fctm = 0.30 (28)2/3 = 2.77 N/mm2

At Service

fcs = 0.6fck = 0.6(40) = 24.0 N/mm2

fts = 0.0 N/mm2

Stress Distribution

At Transfer

Self weight, wi = (Ac)(25) = (2.85 105)(25) 10-6 = 7.12 kN/m

Mi at mid-span = wiL2/8 = (7.12)(152)/8 = 200.2 kNm

Short-term losses, = (1 – 0.10) = 0.90

Stress at top fibre

f1t = Mi/Zt + P/A – Pe/Zt

= 5.20 N/mm2

Stress at bottom fibre

f2t = – Mi/Zb + P/A + Pe/Zb

= 17.84 N/mm2

PASS

FAILf2t (17.84) fct (16.8)

f1t (5.20) ftt ( 2.77)

+

_

Stress Distribution

At Service

Service load, ws = 20 kN/m

Ms at mid-span = wsL2/8 = (20)(152)/8 = 562.5 kNm

Long-term losses, = (1 – 0.20) = 0.80

Stress at top fibre

f1s = Mi/Zt + Ms/Zt + P/A – Pe/Zt

= 8.33 N/mm2

Stress at bottom fibre

f2s = –Mi/Zb – Ms/Zb + P/A + Pe/Zb

= 3.23 N/mm2

PASS

PASSf1s (8.33) fcs (24.0)

f2k (3.23) fts (0)

+

_

Eq. (5) + Eq. (7)

Mi/Zt – Mi/Zt + Ms/Zt (fcs + ftt)

Zt ( – )Mi + Ms (fcs + ftt) ---------- (9)

Eq. (6) + Eq. (8)

Mi/Zb – Mi/Zb + Ms/Zb (fts + fct)

Zb ( – )Mi + Ms (fts + fct) ---------- (10)

A prestressed concrete beam with an effective length of 20 m is

simply supported at both ends. During service, a characteristics

load 20 kN/m is applied to the beam apart from its self weight. The

concrete strength is 50 N/mm2 and the transfer is done when the

concrete achieve the strength of 30 N/mm2. The prestressing force

applied is 2000 kN at the eccentricity of 500 mm at mid-span. The

short-term and long-term losses is 10% and 20%, respectively.

Design the suitable beam cross section if a rectangular section is

used.

Stress limit;

At Transfer

fct = 0.6fck(t) = 0.6(30) = 18.0 N/mm2

ftt = fctm = 0.30 (30)2/3 = 2.90 N/mm2

At Service

fcs = 0.6fck = 0.6(50) = 30.0 N/mm2

fts = 0.0 N/mm2

Ms = (20) 202/8 = 1000 kNm (not including the self weight)

From Eq. (9):

(fcs + ftt) = 0.9(30) + 0.8(2.90) = 29.32 N/mm2

Zt ( – )Mi + Ms (fcs + ftt)

Zt (0.1Mi + 900) 106 / 29.32

From Eq. (10):

(fts + fct) = 0.9(0) + 0.8(18.0) = 14.4 N/mm2

Zb ( – )Mi + Ms (fts + fct)

Zb (0.1Mi + 900) 106 / 14.4

Since Zb Zt, only Zb is used & checked to find

the suitable cross section of the beam

For rectangular cross section

Zt = Zb = Ixx/y = bh3/12 (h/2) = bh2/6

Try 300 mm 1000 mm

Self weight, wi = 0.3 1.0 25 = 7.5 kN/m

Moment due to self weight, Mi = 375 kNm

Zb (0.1 375 + 900) 106 / 14.4

65.1 106 mm3

Z = 300 10002/6 = 50 106 mm3 Zb Increase size

Try 300 mm 1400 mm

Self weight, wi = 0.3 1.4 25 = 10.5 kN/m

Moment due to self weight, Mi = 525 kNm

Zb (0.1 525 + 900) 106 / 14.4

66.1 106 mm3

Z = 300 14002/6 = 98.0 106 mm3 Zb OK

TA

RM

AC

TO

PF

LO

OR

TA

RM

AC

TO

PF

LO

OR