chapter 7

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

Phase 1

Cl. 4.3Cl. 4.1.3 4.1.8Cl. 4.3.4Cl. 4.12Cl. 4.8Cl. 4.3.4 4.3.5Cl. 4.3.6Cl. 4.11Cl. 4.3.7Cl. 4.3.8Cl. 4.11.3

COMPRESSIONTENSION

AT TRANSFER

AT SERVICEABILITY

RELAXATION OF STEEL

ELASTIC DEFORMATION OF CONCRETE

CONCRETE SHRINKAGE

CONCRETE CREEP

SLIP AND MOVEMENT DURING ANCHORING

LOSS OF PRESTRESS DUE TO FRICTION

Tensile stress, ft = fc = Mmax/Z = Mmaxy/Iwhere I = bh3/12 and y1 = y2 = h/2

Therefore;ft = fc = (wL2/8)(h/2) (bh3/12)= 0.75wL2/bh2= Let say 7 N/mm2

PPPrestress force, P is then applied to eliminate tensile stress 7 + P/A = 0Prestress force, P required to eliminate tensile stress = 7A = 7bd kN

PPIf the prestress force, P applied is not at centroid7 + P/A + Pe/Z = 0Prestress force, P required to eliminate tensile stress = 7 / [(1/A) + (e/Z)] kNe

AT TRANSFERPPeMa = Moment due to self weight

AT SERVICEPPeStress due to wk kN/m

At Transfer

Ma/Z1 + P/A Pe/Z1 ftpMa/Z2 + P/A + Pe/Z2 fmp

At Service

Ma/Z1 + Mk/Z1 + P/A Pe/Z1 fmk Ma/Z1 Mk/Z1 + P/A + Pe/Z2 ftk

Rectangular beam = b h = 300 950 mm. Prestress force, P = 2000 kN acted at e = 200 mm below centroid. Prestress losses at transfer and service is 10% and 20%, respectively.

If fcu = 40 N/mm2 and beam is class as Class 2 structure, determine the allowable stress limit according to BS 8110.Draw the stress distribution diagram at mid-span during transfer and service. Then, compare the value with the allowable stress limit in (a).PPe15 m

Stress limit for Class 2, fcu = 40 N/mm2, fci = 28 N/mm2

At Transfer (Cl. 4.3.5)fmp = 0.5fci = 0.5(28) = 14 N/mm2ftp = 0.45fci = 0.4528 = 2.38 N/mm2

At Service (Cl. 4.3.4)fmk = 0.33fcu = 0.5(40) = 13.2 N/mm2ftk = 0.45fcu = 0.4540 = 2.90 N/mm2Cross sectional area, Ac = 300 950 = 2.85 105 mm2Moment of inertia, Ixx = (300)(950)3/12 = 2.14 1010 mm4Modulus of section, Z1 = Z2 = Ixx/y = 2.14 1010 / 475 = 4.51 107 mm3

Stress DistributionAt Transfer

Self weight, wa = (Ac)(24) = (2.85 105)(24) 10-6 = 6.84 kN/mMa maximum at mid-span = waL2/8 = (6.84)(152)/8 = 192.4 kNm

Stress at top fibref1p= Ma/Z1 + P/A Pe/Z1 = 5.37 N/mm2

Stress at bottom fibref2p= Ma/Z2 + P/A + Pe/Z2 = 18.01 N/mm2

Stress DistributionAt Service

Live load, wk = 20 kN/mMk maximum at mid-span = wkL2/8 = (20)(152)/8 = 562.5 kNm

Stress at top fibref1k= Ma/Z1 + Mk/Z1 + P/A Pe/Z1= 8.16 N/mm2

Stress at bottom fibref2k= Ma/Z2 Mk/Z2 + P/A + Pe/Z2 = 3.06 N/mm2

At TransferMa/Z1 + P/A Pe/Z1 ftp---------- (1)Ma/Z2 + P/A + Pe/Z2 fmp---------- (2)

At ServiceMa/Z1 + Mk/Z1 + P/A Pe/Z1 fmk ----------(3) Ma/Z2 Mk/Z2 + P/A + Pe/Z2 ftk----------(4)

Rearranging Eqs. (1) (4):P/A(eA/Z1 1) Ma/Z1 ftp ----------(5)P/A(eA/Z2 + 1) Ma/Z2 fmp ----------(6)P/A(1 eA/Z1) + Ma/Z1 + Mk/Z1 fmk ----------(7) P/A(1 + eA/Z2) + Ma/Z2 + Mk/Z2 ftk ----------(8)

Eq. (5) + Eq. (7)Ma/Z1 Ma/Z1 + Mk/Z1 (fmk + ftp) Z1 ( )Ma + Mk (fmk + ftp)----------(9)

Eq. (6) + Eq. (8)Ma/Z2 Ma/Z2 + Mk/Z2 (ftk + fmp) Z2 ( )Ma + Mk (ftk + fmp)----------(10)

A Class 1 prestressed concrete beam with an effective length of 20 m is simply supported at both ends. A dead and live service load of12 kN/m and 10 kN/m, respectively is applied to the beam apart from its self weight. The concrete strength is 40 N/mm2 and the transfer is made on day 7. The short-term and long-term losses is 10% and 20%, respectively. Design the suitable beam cross section if a rectangular section is used.

From Table 7.1: BS 8110: Part 2: 1985, for fcu = 40 N/mm2, then fci = 28 N/mm2

Stress limit for Class 1;

At Transfer (Cl. 4.3.5)fmp = 0.5fci = 0.5(28) = 14 N/mm2ftp = 1.0 N/mm2

At Service (Cl. 4.3.4)fmk = 0.33fcu = 0.5(40) = 13.2 N/mm2ftk = 0 N/mm2

Mk = (10 + 12) 202/8 = 1100 kNm

From Eq. (9):(fmk + ftp) = 0.9(13.2) + 0.8(1.0) = 12.68 N/mm2Z1 ( )Ma + Mk (fmk + ftp) Z1 (0.1Ma + 990) 106 / 12.68

From Eq. (10):(ftk + fmp) = 0.9(0) + 0.8(14.0) = 11.2 N/mm2Z2 ( )Ma + Mk (ftk + fmp)Z2 (0.1Ma + 990) 106 / 11.2

Since Z2 Z1, only Z2 is checked to find the suitable cross section of the beam

For rectangular cross sectionZ1 = Z2 = Ixx/y = bh3/12 (h/2) = bh2/6

Try 400 mm 1000 mmSelf weight, wa = 0.4 1.0 24 = 9.6 kN/mMoment due to self weight, Ma = 480 kNm

Z2 (0.1 480 + 990) 106 / 11.2 92.6 106 mm3

Z = 400 10002/6 = 66.7 106 mm3 Z2 Increase size

Try 400 mm 1200 mmSelf weight, wa = 0.4 1.2 24 = 11.52 kN/mMoment due to self weight, Ma = 576 kNm

Z2 (0.1 576 + 990) 106 / 11.2 93.5 106 mm3

Z = 400 12002/6 = 96.0 106 mm3 Z2 OK

TARMAC TOPFLOOR

chapter 7

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