Pre-stressed concrete = Pre-compression concrete Pre-compression stresses is applied at the place when tensile stress occur Concrete weak in tension

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2 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

3 Steel tendon is first stressed Concrete is then poured After harden and reaching the required strength, the steel tendon is cut Pretension Posttension Concrete is first poured in the mould After harden, steel tendon is then stressed and anchored at both ends

4 Phase 1 Phase 2 Phase 3

6 Phase 1 Phase 2 Phase 3

9 Concrete Between C30 and C60 Steel Tendon 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/mm 2 Steel strength usually 1650 N/mm 2

11 Section (3): MS EN : 2004

16 Stresses Loading Stage Transfer Service Symbol Value or Equation Symbol Value or Equation Compressive f ct 0.6f ck (t) f cs 0.6f ck Tensile f tt f ctm f ts 0

17 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

18 b Top fibre y t h y b e N.A Ungrouted duct Prestressing tendon Bottom fibre Reinforcement / bonded tendon

19 h = 500 mm w = 10 kn/m b = 200 mm y t f c 8 m y b _ f t M max = wl 2 /8 = 80 knm Tensile stress, f t = f c = M max /Z = M max y/i where I = bh 3 /12 and y t = y b = h/2 Therefore; f t = f c = (wl 2 /8)(h/2) (bh 3 /12) = 9.65 N/mm 2

20 Prestress force, P is then applied to eliminate tensile stress P N.A P f c = 9.65 N/mm 2 14 N/mm 2 _ f t = 9.64 N/mm 2 Stress due to loading P/A Stress due to force P = 0 N/mm 2 Total stress 9.65 P/A = 0 Prestress force, P required to eliminate tensile stress = 9.65A = 965 kn

21 If the prestress force, P applied is not at the centroid P e N.A P f c = 9.65 N/mm 2 _ f t = 9.65 N/mm 2 Stress due to loading P/A Pe/Z t Stress due to force P Pe/Z b = 14 N/mm 2 0 N/mm 2 Total stress 9.65 P/A Pe/Z = 0 Prestress force, P required to eliminate tensile stress = 9.65 / [(1/A) (e/z)] = 283 kn

22 (a) At Transfer M i = Moment due to self weight P e = Coefficient of short term losses P M i /Z t P/A Pe/Z t f 1t f tt _ = M i /Z b Stress due to self weight P/A Pe/Z b Stress due to force P f 2t f ct Total stress

23 (a) At Service w s kn/m (Service Load) P e = Coefficient of long term losses P M i /Z t M s /Z t P/A Pe/Z t = f 1s f cs M i /Z b M s /Z b P/A Pe/Z b f 2s f ts Stress due to self weight Stress due to w s kn/m Stress due to force P Total stress

24 From both Figures: At Transfer M i /Z t P/A Pe/Z t f tt (1) M i /Z b P/A Pe/Z b f ct (2) At Service M i /Z t M s /Z t P/A Pe/Z t f cs (3) M i /Z b M s /Z b P/A Pe/Z b f ts (4)

25 Rearranging Eqs. (1) (4): P/A(eA/Z t 1) M i /Z t f tt (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) P/A(1 ea/z b ) M i /Z b M s /Z b f ts (8)

26 w s = 20 kn/m P e 15 m P Rectangular beam = b h = 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/mm 2 draw the stress distribution diagram at midspan during transfer and service. Also check that these stresses are within the allowable limits.

27 Cross sectional area, A c = b h = = mm 2 Moment of inertia, I xx = bh 3 /12 = /12 = mm 4 Modulus of section, Z t = Z b = I xx /y = / 475 = mm 3

28 Stress limit for f ck = 40 N/mm 2 and f ci = 28 N/mm 2 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/mm 2 At Service f cs = 0.6f ck = 0.6(40) = 24.0 N/mm 2 f ts = 0.0 N/mm 2

29 At Transfer Stress Distribution Self weight, w i = (A c )(25) = ( )(25) 10-6 = 7.12 kn/m M i at mid-span = w i L 2 /8 = (7.12)(15 2 )/8 = knm Short-term losses, = (1 0.10) = 0.90 Stress at top fibre f 1t = M i /Z t P/A Pe/Z t = 5.20 N/mm 2 Stress at bottom fibre f 2t = M i /Z b P/A Pe/Z b = N/mm 2 f 1t ( 5.20) f tt ( 2.77) _ f 2t (17.84) f ct (16.8) PASS FAIL

30 At Service Stress Distribution Service load, w s = 20 kn/m M s at mid-span = w s L 2 /8 = (20)(15 2 )/8 = 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 Pe/Z t = 8.33 N/mm 2 Stress at bottom fibre f 1s (8.33) f cs (24.0) PASS f 2s = M i /Z b M s /Z b P/A Pe/Z b = 3.23 N/mm 2 _ f 2k ( 3.23) f ts (0) PASS

31 Eq. (5) Eq. (7) M i /Z t M i /Z t M s /Z t ( f cs f tt ) Z t ( )M i M s ( f cs f tt ) (9) Eq. (6) Eq. (8) M i /Z b M i /Z b M s /Z b ( f ts f ct ) Z b ( )M i M s ( f ts f ct ) (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.

32 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/mm 2. 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.

33 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/mm 2

34 M s = (20) 20 2 /8 = 1000 knm (not including the self weight) From Eq. (9): ( f cs f tt ) = 0.9(30) 0.8(2.90) = N/mm 2 Z t ( )M i M s ( f cs f tt ) Z t (0.1M i 900) 10 6 / From Eq. (10): ( f ts f ct ) = 0.9(0) 0.8(18.0) = 14.4 N/mm 2 Z b ( )M i M s ( f ts f ct ) Z b (0.1M i 900) 10 6 / 14.4 Since Z b Z t, only Z b is used & checked to find the suitable cross section of the beam

35 For rectangular cross section Z t = Z b = I xx /y = bh 3 /12 (h/2) = bh 2 /6 Try 300 mm 1000 mm Self weight, w i = = 7.5 kn/m Moment due to self weight, M i = 375 knm Z b ( ) 10 6 / mm 3 Z = /6 = mm 3 Z b Increase size

36 Try 300 mm 1400 mm Self weight, w i = = 10.5 kn/m Moment due to self weight, M i = 525 knm Z b ( ) 10 6 / mm 3 Z = /6 = mm 3 Z b OK

37 TARMAC TOPFLOOR

38 TARMAC TOPFLOOR

PURE BENDING. If a simply supported beam carries two point loads of 10 kn as shown in the following figure, pure bending occurs at segment BC.

PURE BENDING. If a simply supported beam carries two point loads of 10 kn as shown in the following figure, pure bending occurs at segment BC. BENDING STRESS The effect of a bending moment applied to a cross-section of a beam is to induce a state of stress across that section. These stresses are known as bending stresses and they act normally