• • •
Pre-stressed concrete = Pre-compression concrete Pre-compression stresses is applied applied at the place when when tensile stress occur Concrete weak in tension but strong in compression c ompression
• • •
Pre-stressed concrete = Pre-compression concrete Pre-compression stresses is applied applied at the place when when tensile stress occur Concrete weak in tension but strong in compression c ompression
Pretension Posttension
• • •
• •
Steel tendon is first stressed Concrete is then poured After harden and reaching the required strength, the steel tendon is cut
Concrete is first poured in the mould After harden, steel tendon is then stressed and anchored at both ends
Phase 1
Phase 2
Phase 3
Phase 1
Phase 2
Phase 3
Steel Tendon
Concrete Between C30 and C60
7 or 19 nos of strand
High workability during wet and high strength when hardened
High strength steel
Concrete strength at transfer, f ci 0.6f ck(t) N/mm2
Steel strength usually 1650 N/mm2
Section 5.10.2.2(3): MS EN 1992-1-1: 2004
Stresses
Loading Stage Transfer Symbol Value or Equation
Service Symbol Value or Equation
Compressive
f ct
0.6 f ck (t )
f cs
0.6 f ck
Tensile
f tt
f ctm
f ts
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
b Top fibre
y t
h
N.A Ungrouted duct
y b
e
Bottom fibre Prestressing tendon
Reinforcement / bonded tendon
w = 10 kN/m b = 200 mm
8m
m m 0 0 5 = h
f c +
y t y b
_ f t
Mmax = wL2/8 = 80 kNm
Tensile stress, f t = f c = M max /Z = M maxy /I where I = b h 3 /12 and y t = y b = h /2 Therefore; = (w L 2 /8)(h /2) f t = f c = 9.65 N/mm2
(b h 3 /12)
Prestress force, P is then applied to eliminate tensile stress P
P
N.A
f c = 9.65 N/mm2
14 N/mm2
+
9.65
+
+
=
_ f t = 9.64 N/mm2
/A P
Stress due to loading
Stress due to force P
0 N/mm2 Total stress
+ P/A = 0 Prestress force, P required to eliminate tensile stress = 9.65A = 965 kN
If the prestress force, P applied is not at the centroid N.A e
P
P
/Z t Pe
f c = 9.65 N/mm2
14 N/mm2
+
+
+
+
=
_
+
f t = 9.65 N/mm2
Stress due to loading
Pe /Z b
/A P
Stress due to force P 9.65
0 N/mm2 Total stress
+ P/A + Pe/Z = 0 Prestress force, P required to eliminate tensile stress = 9.65 / [(1/A) + (e/Z)] = 283 kN
(a) At Transfer M i = Moment due to self weight
= Coefficient of short term losses
e
P
P M /i Z t
f 1t
/Z t Pe
/A P
f tt
+
+
+
_ /i Z b M Stress due to self weight
/A P
=
+ + /Z b Pe
Stress due to force P
+ f 2t
f ct
Total stress
(a) At Service w s kN/m (Service Load)
e
= Coefficient of long term losses
P
M /i Z t
M s /Z t
+
+
+ _ /i Z b M Stress due to self weight
+
_
Stress due to w s kN/m
/Z t Pe
/A P
f 1s
f cs
+
+
M s /Z b
P
+
= +
P /A
/Z b Pe
Stress due to force P
f 2s
f ts
Total stress
From both Figures: At Transfer M i/Z t + P / A Pe/Z t f tt M i/Z b + P / A + Pe/Z b f ct –
–
–
-------------------
(1) (2)
-------------------
(3) (4)
At Service M i/Z t + M s/Z t + P / A Pe/Z t f cs M i/Z b M s/Z b + P / A + Pe/Z b f ts –
–
–
–
Rearranging Eqs. (1) – (4): P/ A(eA/Z t
----------
(5)
P/ A(eA/Z b + 1) M i/Z b f ct –
----------
(6)
P/ A(1 eA/Z t) + M i/Z t + M s/Z t f cs
----------
(7)
----------
(8)
–
1) M i/Z t f tt –
–
P/ A(1 + eA/Z b) + M i/Z b + M s/Z b f ts
–
w s = 20 kN/m
e
P
P
15 m
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 f ck = 40 N/mm2 draw the stress distribution diagram at midspan during transfer and service. Also check that these stresses are within the allowable limits.
Cross sectional area, Ac
= b h = 300 950 = 2.85 105 mm2
Moment of inertia, I xx
= bh3/12 = 300 9503/12 = 2.14 1010 mm4
Modulus of section, Z t = Z b
= I xx/y = 2.14 1010 / 475 = 4.51 107 mm3
Stress limit for f ck = 40 N/mm2 and f ci = 28 N/mm2 At Transfer f ct = 0.6f ck(t) = 0.6(28) = 16.8 N/mm 2 f tt = f ctm = 0.30 (28)2/3 = 2.77 N/mm2 At Service f cs = 0.6f ck = 0.6(40) = 24.0 N/mm 2 f ts = 0.0 N/mm2
Stress Distribution At Transfer Self weight, w i = ( Ac)(25) = (2.85 105)(25) 10-6 = 7.12 kN/m M i at mid-span = w iL2/8 = (7.12)(152)/8 = 200.2 kNm Short-term losses, = (1 0.10) = 0.90 –
Stress at top fibre f 1t = M i/Z t + P / A = 5.20 N/mm2
–
Pe/Z t
Stress at bottom fibre f 2t = M i/Z b + P / A + Pe/Z b = 17.84 N/mm 2 –
f 1t ( 5.20)
_
f tt ( 2.77)
PASS
+ f 2t (17.84)
f ct (16.8)
FAIL
Stress Distribution At Service Service load, w s = 20 kN/m M s at mid-span = w sL2/8 = (20)(152)/8 = 562.5 kNm Long-term losses, = (1 0.20) = 0.80 –
Stress at top fibre f 1s = M i/Z t + M s/Z t + P / A = 8.33 N/mm 2
–
Pe/Z t
f 1s (8.33)
PASS
+
Stress at bottom fibre f 2s = M i/Z b M s/Z b + P / A + Pe/Z b = 3.23 N/mm2 –
f cs (24.0)
–
_ f 2k ( 3.23)
f ts (0)
PASS
Eq. (5) + Eq. (7) M i/Z t Mi /Z t + Ms /Z t ( fcs + ftt ) –
Z t
+ f tt) ( )M i + M s ( fcs –
---------- (9)
Eq. (6) + Eq. (8) ) M i/Z b M i/Z b + Ms /Z b ( f ts + fct –
Z b
( )M i + Ms ( f ts + 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/mm 2 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 f ct = 0.6f ck(t) = 0.6(30) = 18.0 N/mm 2 f tt = f ctm = 0.30 (30)2/3 = 2.90 N/mm 2 At Service f cs = 0.6f ck = 0.6(50) = 30.0 N/mm 2 f ts = 0.0 N/mm2
M s = (20) 202/8 = 1000 kNm (not including the self weight ) From Eq. (9): ( f cs + f tt) = 0.9(30) + 0.8(2.90) = 29.32 N/mm2 Z t ( )M i + M s ( f cs + f tt) Zt (0.1M i + 900) 106 / 29.32 –
From Eq. (10): ( f ts + f ct) = 0.9(0) + 0.8(18.0) = 14.4 N/mm 2 Z b ( )M i + Ms ( f ts + fct ) Z b (0.1M i + 900) 106 / 14.4 –
Since Z b Z t, only Z b is used & checked to find the suitable cross section of the beam
For rectangular cross section Z t = Z b = I xx/y = bh3/12 (h/2) = bh2/6 Try 300 mm
1000 mm
Self weight, w i = 0.3 1.0 25 = 7.5 kN/m Moment due to self weight, M i = 375 kNm Z b
(0.1 375 + 900) 106 / 14.4 65.1 106 mm3
Z = 300 10002/6 = 50
106 mm3 Z b
Increase size
Try 300 mm
1400 mm
Self weight, w i = 0.3 1.4 25 = 10.5 kN/m Moment due to self weight, M i = 525 kNm Z b
(0.1 525 + 900) 106 / 14.4 66.1 106 mm3
Z = 300 14002/6 = 98.0
106 mm3 Z b
OK
R O O L F P O T C A M R A T
R O O L F P O T C A M R A T
