📖 Explanation
This is a question of calculation of elastic shortening loss in post tensioned member. Given data, length = 15 m b x D = 450 mm x 450 mm Numberof tendons = 3 Cross-section area of each tendon (As)=200mm2 . Prestress = 1500 MPa Modular ratio (m) = 6 From the given data, eccentricity (e) = 450/2-125 = 100 mm Force in each cable (P)=1500×200×10−3=300kN The tendons are tensioned one after another, and hence when tendon (1) is pulled no loss in tenson (1) When tendon (2) is pulled, loss in tendon (1) but no loss in tendon (2). When tendon (3) is pulled, loss in tendon (1) and (2), but no loss in tendon (3). Hence, there will be 2 times losses in tendon (1), time loss in tendon (1) and no loss in tendon (3). While calculating elastic shortening loss, self weight of the structure is neglected to be on the conservative side. Consider tensioning of tendon-1 No loss in tendon (1) Consider tensioning of tendon-2 Stress in concrete at the level of prestressing tendon
fce=AP+IPe=450×450300×103+12450×4503300×103×100×100=2.36MPa
I = moment of inertial of the section about the centroidal axis. As the tendons are horizontal and at the same level fc,avg=fc . Loss due to elastic deformation =mfc=6×2.36=14.16MPa. Considering tensioning of tendon 3 Loss due to elastic deformation in (1) =mfc=6×2.36=14.16MPa. Loss due to elastic deformation in (2) =mfc=6×2.36=14.16MPa. Total loss in tendon (1) = 2 x 14.16 = 28.32 MPa Total loss in tendon (1) = 2 x 14.16 = 28.32 MPa In tendon (2) = 14.16 MPa In tendon (3) = 0 Average loss of pre-stress, considering all three tendons is =328.32+14.16+0=14.16MPa