BEHAVIOR OF REINFORCED CONCRETE SHORT RECTANGULAR COLUMNS STRENGTHENED BY STEEL LATTICE FRAMED JACKET
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- Δεσποίνη Βλαβιανός
- 7 χρόνια πριν
- Προβολές:
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1 Composites Behavior of reinforced concrete short rectangular columns strengthened by steel lattice framed jacket XIII International Conference on Computational Plasticity. Fundamentals and Applications COMPLAS XIII E. Oñate, D.R.J. Owen, D. Peric and M. Chiumenti(Eds) BEHAVIOR OF REINFORCED CONCRETE SHORT RECTANGULAR COLUMNS STRENGTHENED BY STEEL LATTICE FRAMED JACKET NAMEER A. ALWASH* AND ALI A. AL-ZAHID *(Professor) College of Engineering Babylon University (Researcher) Key words: Short Columns, Strengthened Column, Composites, Steel Lattice Framed Jacket. Abstract: The complete derivation of the more exact solution that controlled the behaviour of reinforced concrete short rectangular column strengthened by steel lattice frame jacket is presented including the effect of bond slip between the concrete and reinforcement and that between concrete and steel jacket. Bond between these materials is considered well through assuming an existing of distributed frictional springs. An exact displacement field is derived utilizing the principle of minimum strain energy following Euler Cauchy formula. Stress - strain diagram for confined concrete by internal reinforcement and external steel jacket is suggested. An empirical equation that controls the effect of interface spring stiffness between concrete and steel jacket is found through experimental investigation. A computer program is prepared using visual basic language and designated as (ARCC SSJ) 1 including all the above work. The validity and efficiency of the presented finite element is tested through comparison with others experimental results. Present study shows good agreement. 1 INTRODUCTION Sometimes the engineering problem may be solved through expressing it in a mathematical model. The model is often a set of differential equations which have to be solved for the behaviour of the system. A mathematical expression is introduced in the present study that controls the behaviour of reinforced concrete short rectangular columns strengthened by steel lattice framed jacket. The study presents the derivation of tangent stiffness matrix of the case mentioned previously. Several previous studies were presented by others for the same problem following either experimental investigation [1] or presenting an analytical method []. 1 Name of prepared program which is refer to (Analysis of Reinforced Concrete Column Strengthened by Steel Jacket). 471
2 THEORETICAL MODEL.1 Displacement field Starting from concept of strain energy that is defined as the energy stored in a body due to axial deformation (i.e. for axial member) is equal to: 1 UUUU = llll σσσσ(xxxx) εεεε(xxxx) AAAA(xxxx) ddddxxxx (1) εεεε(xxxx) = uuuu (xxxx) () σσσσ EEEE = (3) εεεε σσσσ(xxxx) = EEEE uuuu (xxxx) (4) π p = 1 llll uuuu (xxxx)eeee AAAA(xxxx)ddddxxxx (5) Utilizing the process suggested by Alwash [3],the strain energy of axial element composed of three materials including the effect of bond slip can be written as: 1 UUUU = σσσσ ccccεεεε cccc llll AAAA cccc + 1 llll σσσσ rrrrrrrrεεεε rrrrrrrr AAAA rrrrrrrr + 1 llll σσσσ rrrrssss εεεε rrrrssss The above equation can be re-written as : UUUU = 1 1 llll uuuu EEEE cccc AAAA cccc llll kkkk rrrrrrrrssss (uuuu(xxxx) rrrrssss uuuu(xxxx) cccc ) + 1 llll uuuu EEEE rrrrrrrr AAAA rrrrrrrr AAAA rrrrssss + 1 llll kkkk rrrrrrrrrrrr (uuuu(xxxx) rrrrrrrr uuuu(xxxx) cccc ) + 1 llll uuuu EEEE rrrrssss AAAA rrrrssss + ddddxxxx (6) + 1 llll kkkk rrrrrrrrrrrr (uuuu(xxxx) rrrrrrrr uuuu(xxxx) cccc ) ddddxxxx (7) + 1 llll kkkk rrrrrrrrssss (uuuu(xxxx) rrrrssss uuuu(xxxx) cccc ) where: AAAA cccc, AAAA rrrrrrrr, AAAA rrrrssss = cross section area of concrete, reinforcement and steel jacket respectively. kkkk rrrrrrrrrrrr =interface spring stiffness between concrete and reinforcement kkkk rrrrrrrrssss =interface spring stiffness between concrete and steel jacket Then, for minimum U, utilizing Euler Lagrange formula: dddd cccc ddddxxxx dddd rrrrrrrr ddddxxxx dddd rrrrssss ddddxxxx = (8) cccc = (9) rrrrrrrr = (1) rrrrssss For EEEE cccc AAAA cccc = ccccccccccccrrrrcccccccccccccccc, EEEE rrrrrrrr AAAA rrrrrrrr = ccccccccccccrrrrcccccccccccccccc, EEEE rrrrssss AAAA rrrrssss = ccccccccccccrrrrcccccccccccccccc, kkkk rrrrrrrrrrrr = ccccccccccccrrrrcccccccccccccccc, kkkk rrrrrrrrssss = ccccccccccccrrrrcccccccccccccccc and from equations (7), (8),(9), (1), the following differential equations for the problem are obtained as : uuuu cccc + kkkk rrrrrrrrrrrr EEEE cccc AAAA cccc (uuuu rrrrrrrr uuuu cccc ) + kkkk rrrrrrrrssss EEEE cccc AAAA cccc uuuu rrrrssss uuuu cccc = (11) 47
3 kkkk rrrrrrrrrrrr uuuu rrrrrrrr (uuuu EEEE rrrrrrrr AAAA rrrrrrrr uuuu cccc ) = rrrrrrrr (1) kkkk rrrrrrrrssss uuuu rrrrssss uuuu EEEE rrrrssss AAAA rrrrssss uuuu cccc = rrrrssss (13) Solving the differential equations (11), (1) and (13) to find the displacement field for considered case: where: uuuu cccc = AAAA1 eeee λλλλ 1 xxxx + AAAA eeee λλλλ1xxxx + AAAA3 eeee λλλλxxxx + AAAA4 eeee λλλλxxxx + AAAA5 xxxx + AAAA6 (14) uuuu rrrrrrrr = BBBB1 eeee λλλλ1xxxx + BBBB eeee λλλλ1xxxx + BBBB3 eeee λλλλxxxx + BBBB4 eeee λλλλxxxx + BBBB5 xxxx + BBBB6 (15) uuuu rrrrssss = CCCC1 eeee λλλλ1xxxx + CCCC eeee λλλλ1xxxx + CCCC3 eeee λλλλxxxx + CCCC4 eeee λλλλxxxx + CCCC5 xxxx + CCCC6 (16) kkkk rrrrrrrrrrrr EEEE cccc AAAA cccc = cccc1 kkkk rrrrrrrrssss EEEE cccc AAAA cccc = cccc kkkk rrrrrrrrrrrr EEEE rrrrrrrr AAAA rrrrrrrr = kkkk rrrrrrrrssss EEEE rrrrssss AAAA rrrrssss = λλλλ 1 = NNNN cccc + NNNN cccc 4( +cccc1 +cccc ) λλλλ = NNNN cccc NNNN cccc 4( +cccc1+cccc ).. Stiffness matrix To find the relation between the constants (A1, A, A3, ) and(b1, B, B3, ) use equation number (1) while the relation between the constants (A1, A, A3, ) and(c1, C, C3, ) will be obtained through using equation (13) Re-writing the displacement field in terms of constants (A) as: uuuu cccc = AAAA1 eeee λλλλ 1 xxxx + AAAA eeee λλλλ1xxxx + AAAA3 eeee λλλλxxxx + AAAA4 eeee λλλλxxxx + AAAA5 xxxx + AAAA6 (17) AAAA1 uuuu rrrrrrrr = eeee λλλλ1xxxx AAAA + eeee λλλλ1xxxx AAAA3 + eeee λλλλxxxx AAAA4 + eeee λλλλxxxx + AAAA5 xxxx + AAAA6 λλλλ 1 λλλλ 1 λλλλ λλλλ (18) AAAA1 uuuu rrrrssss = eeee λλλλ1xxxx AAAA + eeee λλλλ1xxxx AAAA3 + eeee λλλλxxxx AAAA4 + eeee λλλλxxxx + AAAA5 xxxx + AAAA6 λλλλ 1 λλλλ 1 λλλλ λλλλ (19) The displacement field should be re-written in terms of nodal displacements instead of the constants (A1, A, A3,.. ) Applying the following boundary xxxx = uuuu cccc = uuuu cccc1, uuuu rrrrrrrr = uuuu rrrrrrrr1 cccccccccccc uuuu rrrrssss = uuuu rrrrssss xxxx = LLLL uuuu cccc = uuuu cccc, uuuu rrrrrrrr = uuuu rrrrrrrr cccccccccccc uuuu rrrrssss = uuuu rrrrssss 473
4 uuuu cccc1 = AAAA1 + AAAA+AAAA3 + AAAA4 + AAAA6 () uuuu cccc = AAAA1 eeee λλλλ 1 llll + AAAA eeee λλλλ1llll + AAAA3 eeee λλλλllll + AAAA4 eeee λλλλllll + AAAA5 llll + AAAA6 (1) uuuu rrrrrrrr1 = AAAA1+ AAAA+ AAAA3+ AAAA4+AAAA6 λλλλ 1 λλλλ 1 λλλλ λλλλ () AAAA1 uuuu rrrrrrrr = eeee λλλλ1llll AAAA + eeee λλλλ1llll AAAA3 + eeee λλλλllll AAAA4 + eeee λλλλllll + AAAA5 llll + AAAA6 λλλλ 1 λλλλ 1 λλλλ λλλλ (3) uuuu rrrrssss 1 = AAAA1+ AAAA+ AAAA3+ AAAA4+AAAA6 λλλλ 1 λλλλ 1 λλλλ λλλλ (4) AAAA1 uuuu rrrrssss = eeee λλλλ1llll AAAA + eeee λλλλ1llll AAAA3 + eeee λλλλllll AAAA4 + eeee λλλλllll + AAAA5 llll + AAAA6 λλλλ 1 λλλλ 1 λλλλ λλλλ (5) Putting equations (), (1),(), (3), (4) and (5) in a matrix form: uuuu cccc1 uuuu cccc uuuu rrrrrrrr1 uuuu = rrrrrrrr uuuu rrrrssss 1 uuuu rrrrssss eeee λλλλ 1 llll eeee λλλλ 1llll eeee λλλλ llll eeee λλλλ llll llll 1 1 λλλλ 1 λλλλ λλλλ AAAA1 AAAA eeee λλλλ 1llll eeee λλλλ llll eeee λλλλ llll AAAA3 llll 1 λλλλ 1 λλλλ λλλλ AAAA4 1 AAAA5 λλλλ 1 λλλλ λλλλ eeee λλλλ 1llll eeee λλλλ llll eeee λλλλ AAAA6 llll llll 1 λλλλ 1 λλλλ λλλλ λλλλ 1 eeee λλλλ 1llll λλλλ 1 λλλλ 1 eeee λλλλ 1llll λλλλ 1 (6) In a symbolic form {eeee} dddd.cccc.ffff. = [RRRR] 6 6 {AAAA} 6 1 (7) {AAAA} 6 1 = [RRRR] 1 {eeee} = [GGGG]{eeee} (8) Use the Gauss-Jordan elimination method to find the inverse of matrix[rrrr]: IIII1 IIII IIII3 IIII4 IIII5 IIII6 IIII7 IIII8 IIII9 IIII1 IIII11 IIII1 [RRRR] 1 = IIII13 IIII14 IIII15 IIII16 IIII17 IIII18 IIII19 IIII IIII1 IIII IIII3 IIII4 IIII5 IIII6 IIII7 IIII8 IIII9 IIII3 IIII31 IIII3 IIII33 IIII34 IIII35 IIII36 (9) Thus, the displacement field became: uuuu cccc = IIII1 eeee λλλλ 1 xxxx + IIII7 eeee λλλλ1xxxx + IIII13 eeee λλλλxxxx + IIII19 eeee λλλλxxxx + IIII5 xxxx + IIII31 uuuu cccc1 + IIII eeee λλλλ 1 xxxx + IIII8 eeee λλλλ1xxxx + IIII14 eeee λλλλ xxxx + IIII eeee λλλλxxxx + IIII6 xxxx + IIII3 uuuu cccc + IIII3 eeee λλλλ 1 xxxx + IIII9 eeee λλλλ1xxxx + IIII15 eeee λλλλxxxx + IIII1 eeee λλλλxxxx + IIII7 xxxx + IIII33 uuuu rrrrrrrr1 + IIII4 eeee λλλλ 1 xxxx + IIII1 eeee λλλλ1xxxx + IIII16 eeee λλλλxxxx + IIII eeee λλλλxxxx + IIII8 xxxx + IIII34 uuuu rrrrrrrr + IIII5 eeee λλλλ 1 xxxx + IIII11 eeee λλλλ1xxxx + IIII17 eeee λλλλxxxx + IIII3 eeee λλλλxxxx + IIII9 xxxx + IIII35 uuuu rrrrssss 1 + IIII6 eeee λλλλ 1 xxxx + IIII1 eeee λλλλ1xxxx + IIII18 eeee λλλλxxxx + IIII4 eeee λλλλxxxx + IIII3 xxxx + IIII36 uuuu rrrrssss (3) 474
5 uuuu rrrrrrrr = IIII1 eeee λλλλ 1 xxxx + IIII7 eeee λλλλ1xxxx + IIII13 eeee λλλλxxxx + IIII19 eeee λλλλxxxx + IIII5 xxxx + IIII31 uuuu λλλλ 1 λλλλ 1 λλλλ λλλλ cccc1 + IIII eeee λλλλ 1 xxxx + IIII8 eeee λλλλ1xxxx + IIII14 eeee λλλλxxxx + IIII eeee λλλλxxxx + IIII6 xxxx + IIII3 uuuu λλλλ 1 λλλλ 1 λλλλ λλλλ cccc + IIII3 eeee λλλλ 1 xxxx + IIII9 eeee λλλλ1xxxx + IIII15 eeee λλλλxxxx + IIII1 eeee λλλλxxxx + IIII7 xxxx + IIII33 uuuu λλλλ 1 λλλλ 1 λλλλ λλλλ rrrrrrrr1 + IIII4 eeee λλλλ 1 xxxx + IIII1 eeee λλλλ1xxxx + IIII16 eeee λλλλxxxx + IIII eeee λλλλxxxx + IIII8 xxxx + IIII34 uuuu λλλλ 1 λλλλ 1 λλλλ λλλλ rrrrrrrr + IIII5 eeee λλλλ 1 xxxx + IIII11 eeee λλλλ1xxxx + IIII17 eeee λλλλxxxx + IIII3 eeee λλλλxxxx + IIII9 xxxx + IIII35 uuuu λλλλ 1 λλλλ 1 λλλλ λλλλ rrrrssss 1 + IIII6 eeee λλλλ 1 xxxx + IIII1 eeee λλλλ1xxxx + IIII18 eeee λλλλxxxx + IIII4 eeee λλλλxxxx + IIII3 xxxxiiii36 uuuu λλλλ 1 λλλλ 1 λλλλ λλλλ rrrrssss (31) uuuu rrrrssss = IIII1 eeee λλλλ 1 xxxx + IIII7 eeee λλλλ1xxxx + IIII13 eeee λλλλxxxx + IIII19 eeee λλλλxxxx + IIII5 xxxx + IIII31 uuuu λλλλ 1 λλλλ 1 λλλλ λλλλ cccc1 + IIII eeee λλλλ 1 xxxx + IIII8 eeee λλλλ1xxxx + IIII14 eeee λλλλxxxx + IIII eeee λλλλxxxx + IIII6 xxxx + IIII3 uuuu λλλλ 1 λλλλ 1 λλλλ λλλλ cccc + IIII3 eeee λλλλ 1 xxxx + IIII9 eeee λλλλ1xxxx + IIII15 eeee λλλλxxxx + IIII1 eeee λλλλxxxx + IIII7 xxxx + IIII33 uuuu λλλλ 1 λλλλ 1 λλλλ λλλλ rrrrrrrr1 + IIII4 eeee λλλλ 1 xxxx + IIII1 eeee λλλλ1xxxx + IIII16 eeee λλλλxxxx + IIII eeee λλλλxxxx + IIII8 xxxx + IIII34 uuuu λλλλ 1 λλλλ 1 λλλλ λλλλ rrrrrrrr + IIII5 eeee λλλλ 1 xxxx + IIII11 eeee λλλλ1xxxx + IIII17 eeee λλλλxxxx + IIII3 eeee λλλλxxxx + IIII9 xxxx + IIII35 uuuu λλλλ 1 λλλλ 1 λλλλ λλλλ rrrrssss 1 + IIII6 eeee λλλλ 1 xxxx + IIII1 eeee λλλλ1xxxx + IIII18 eeee λλλλxxxx + IIII4 eeee λλλλxxxx + IIII3 xxxx + IIII36 uuuu λλλλ 1 λλλλ 1 λλλλ λλλλ rrrrssss (3) The strain displacement relation is given in the following expressions: Then, in a matrix form: εεεε cccc = dddddddd cccc ddddxxxx εεεε rrrrrrrr = dddddddd rrrrrrrr ddddxxxx εεεε rrrrssss = dddddddd rrrrssss ddddxxxx (33) (34) (35) Or, in a symbolic form where uuuu cccc1 uuuu εεεε cccc cccc QQQQ1 QQQQ QQQQ3 QQQQ4 QQQQ5 QQQQ6 uuuu εεεε rrrrrrrr1 rrrrrrrr = QQQQ7 QQQQ8 QQQQ9 QQQQ1 QQQQ11 QQQQ1 εεεε rrrrssss QQQQ13 QQQQ14 QQQQ15 QQQQ16 QQQQ17 QQQQ18 uuuu rrrrrrrr uuuu rrrrssss 1 uuuu rrrrssss (36) {εεεε} = [BBBB]{uuuu} (37) 475
6 QQQQ1 QQQQ QQQQ3 QQQQ4 QQQQ5 QQQQ6 [BBBB] = QQQQ7 QQQQ8 QQQQ9 QQQQ1 QQQQ11 QQQQ1 QQQQ13 QQQQ14 QQQQ15 QQQQ16 QQQQ17 QQQQ18 QQQQ1 = λλλλ 1 IIII1 eeee λλλλ 1xxxx λλλλ 1 IIII7 eeee λλλλ 1xxxx + λλλλ IIII13 eeee λλλλ xxxx λλλλ IIII19eeee λλλλ xxxx + IIII5 QQQQ = λλλλ 1 IIII eeee λλλλ 1 xxxx λλλλ 1 IIII8 eeee λλλλ 1xxxx + λλλλ IIII14 eeee λλλλ xxxx λλλλ IIII eeee λλλλ xxxx + IIII6 QQQQ3 =λλλλ 1 IIII3 eeee λλλλ 1 xxxx λλλλ 1 IIII9 eeee λλλλ 1xxxx + λλλλ IIII15 eeee λλλλ xxxx λλλλ IIII1 eeee λλλλ xxxx + IIII7 QQQQ4 = λλλλ 1 IIII4 eeee λλλλ 1 xxxx λλλλ 1 IIII1 eeee λλλλ 1xxxx + λλλλ IIII16 eeee λλλλ xxxx λλλλ IIII eeee λλλλ xxxx + IIII8 QQQQ5 = λλλλ 1 IIII5 eeee λλλλ 1 xxxx λλλλ 1 IIII11 eeee λλλλ 1xxxx + λλλλ IIII17 eeee λλλλ xxxx λλλλ IIII3 eeee λλλλ xxxx + IIII9 QQQQ6 = λλλλ 1 IIII6 eeee λλλλ 1 xxxx λλλλ 1 IIII1 eeee λλλλ 1xxxx + λλλλ IIII18 eeee λλλλ xxxx λλλλ IIII4 eeee λλλλ xxxx + IIII3 QQQQ7 = QQQQ8 = QQQQ9 = λλλλ 1 λλλλ 1 IIII1 eeeeλλλλ 1 xxxx λλλλ 1 λλλλ 1 IIII7 eeee λλλλ 1xxxx + λλλλ λλλλ IIII13 eeeeλλλλ xxxx λλλλ λλλλ IIII19 eeee λλλλ xxxx + IIII5 λλλλ 1 λλλλ 1 IIII eeeeλλλλ 1 xxxx λλλλ 1 λλλλ 1 IIII8 eeee λλλλ 1xxxx + λλλλ λλλλ IIII14 eeeeλλλλ xxxx λλλλ λλλλ IIII eeee λλλλ xxxx + IIII6 λλλλ 1 λλλλ 1 IIII3 eeeeλλλλ 1 xxxx λλλλ 1 λλλλ 1 IIII9 eeee λλλλ 1xxxx + λλλλ λλλλ IIII15 eeeeλλλλ xxxx QQQQ1 = λλλλ 1 λλλλ 1 IIII4 eeeeλλλλ 1 xxxx QQQQ11 = λλλλ λλλλ IIII1 eeee λλλλ xxxx + IIII7 λλλλ 1 λλλλ 1 IIII1 eeee λλλλ 1xxxx + λλλλ λλλλ IIII16 eeeeλλλλ xxxx λλλλ λλλλ IIII eeee λλλλ xxxx + IIII8 λλλλ 1 λλλλ 1 IIII5 eeeeλλλλ 1 xxxx λλλλ 1 λλλλ 1 IIII11 eeee λλλλ 1xxxx + λλλλ λλλλ IIII17 eeeeλλλλ xxxx λλλλ λλλλ IIII3 eeee λλλλ xxxx + IIII9 QQQQ1 = λλλλ 1 λλλλ 1 IIII6 eeeeλλλλ 1 xxxx λλλλ 1 λλλλ 1 IIII1 eeee λλλλ 1xxxx + λλλλ λλλλ IIII18 eeeeλλλλ xxxx λλλλ λλλλ IIII4 eeee λλλλ xxxx + IIII3 QQQQ13 = λλλλ 1 λλλλ 1 IIII1 eeeeλλλλ 1 xxxx λλλλ 1 λλλλ 1 IIII7 eeee λλλλ 1xxxx + λλλλ λλλλ IIII13 eeeeλλλλ xxxx λλλλ λλλλ IIII19 eeee λλλλ xxxx + IIII5 QQQQ14 = λλλλ 1 λλλλ 1 IIII eeeeλλλλ 1 xxxx QQQQ15 = λλλλ 1 λλλλ 1 IIII3 eeeeλλλλ 1 xxxx QQQQ16 = λλλλ 1 λλλλ 1 IIII4 eeeeλλλλ 1 xxxx QQQQ17 = λλλλ 1 λλλλ 1 IIII8 eeee λλλλ 1xxxx + λλλλ λλλλ IIII14 eeeeλλλλ xxxx λλλλ λλλλ IIII eeee λλλλ xxxx + IIII6 λλλλ 1 λλλλ 1 IIII9 eeee λλλλ 1xxxx + λλλλ λλλλ IIII15 eeeeλλλλ xxxx λλλλ λλλλ IIII1 eeee λλλλ xxxx + IIII7 λλλλ 1 λλλλ 1 IIII1 eeee λλλλ 1xxxx + λλλλ λλλλ IIII16 eeeeλλλλ xxxx λλλλ λλλλ IIII eeee λλλλ xxxx + IIII8 λλλλ 1 λλλλ 1 IIII5 eeeeλλλλ 1 xxxx λλλλ 1 λλλλ 1 IIII11 eeee λλλλ 1xxxx + λλλλ λλλλ IIII17 eeeeλλλλ xxxx λλλλ λλλλ IIII3 eeee λλλλ xxxx + IIII9 QQQQ18 = λλλλ 1 λλλλ 1 IIII6 eeeeλλλλ 1 xxxx λλλλ 1 λλλλ 1 IIII1 eeee λλλλ 1xxxx + λλλλ λλλλ IIII18 eeeeλλλλ xxxx λλλλ λλλλ IIII4 eeee λλλλ xxxx + IIII3 The stiffness matrix will be in an integration form as: where LLLL [KKKK] 6 6 = [BBBB] cccc 6 3 [EEEE] 3 3 [BBBB] 3 6 ddddxxxx (38) 476
7 EEEE cccc [EEEE] 3 3 = EEEE rrrrrrrr EEEE rrrrssss A numerical integration was carried out, using 3- point Gaussian Quadrature rule, to evaluate the stiffness matrix..3 Proposed stress strain diagram for confined concrete: The proposed stress strain diagram for confined concrete will be achieved by modifying a stress strain diagram for unconfined concrete. At a pointed value of strain, the corresponding stress value will be larger in confined concrete, the ultimate strength of confined concrete will be taken form Eurocode 8 provision and the effect of confining pressure will based on Campione [] equations..3.1 Stress strain diagram for unconfined concrete The proposed compressive uniaxial stress-strain relationship for the concrete model was obtained using the following equations to complete the multilinear isotropic stress-strain curve for the concrete. σσσσ = EEEE ccccεεεε 1+( εεεε εεεεcccc ) εεεε 1 < εεεε < εεεε cccc (39) σσσσ = ffff cccc εεεεε εεεε cccc (4) εεεε cccc = ffff cccc (41) EEEE cccc σσσσ = EEEE cccc. εεεε εεεε < εεεε 1 (4) εεεε 1 =.3 ffff cccc (43) EEEE cccc where σ =stress at any strain, ε = strain at stress σ ε o = strain at the ultimate compressive strength (f c ) The multilinear isotropic stress-strain relationship requires the first point of the curve to be satisfying Hooke's law. The simplified stress-strain curve for concrete model is constructed from six points connected by straight lines. The curve starts at zero stress and strain. Point number 1, at.3ffff cccc, is calculated for the stress-strain relationship of the concrete in the linear range from equation (4). Point of numbers, 3, and 4 are obtained from equation (39), in which (εεεε cccc ) is calculated from equation (41). Point of number 5 is at (εεεε cccc ) and (ffff cccc ). In this study, an assumption was made of perfectly plastic behaviour after point number
8 .3. Eurocode provisions for compressive strength in confined concrete The strength of confined concrete may be evaluated from: ffff cccccccc = ffff cccc CCCCCCCC.87 ffff (44) cccc Equation (44) will be adopted to express the ultimate strength of confined concrete. The strain εεεε cccccccc is evaluated according to Mander et al. [4] as: εεεε cccccccc = εεεε cccccccc [ ffff cccccccc ffff cccc 1 ] (45) The simplified stress-strain curve for confined concrete model is also constructed from six points connected by straight lines with an assumption was made of perfectly plastic behaviour after point number (5) as shown in figure (1)..3.3 Confining pressure Figure (1): Proposed Stress Strain Diagram for Confined Concrete Confining pressure (Cp) due to steel jacket is calculated adopting the model of Campione [] : where ffff yyyyyyyy = of battens, Cp steel jacket = 1.33 ffff yyyyyyyy eeee 1.5rrrr yyyy rrrr 1 ( LLLLrrrr + LLLL ) 1 (46) rrrr cccc cccc rrrr cccc yyyy 478
9 SSSS= spacing between two successive battens, L s = angle side length, cccc cccc =thickness of steel angles, cccc yyyy =thickness of steel battens, rrrr =width of battens, b = column width, L = ( b - L s ) /, Confining pressures due to internal longitudinal bars and stirrups (by Campione [] ) will be: Cp reinforcement = ππππππππ yyyy φφφφ rrrrrrrr + cccc ππππππππ 4 rrrr rrrrrrrr yyyy yyyy 3 rrrr4 EEEE llll yyyy (vvvvεεεε rrrr rrrrrrrr 64 rrrrssss εεεε yyyyrrrrrrrr ) (47) where φφφφ rrrrrrrr = diameter of transverse stirrups, φφφφ llll = diameter of longitudinal reinforcement, rrrr rrrrrrrr =spacing between two successive stirrups, cccc yyyy =the number of side bars EEEE rrrrrrrr =modulus of elasticity for reinforcement, vvvv = Poisson's ratio εεεε rrrrssss =the strain corresponding to concrete cover spalling εεεε yyyyrrrrrrrr =yield strain of the stirrups The pressure that has obtained from equation (46) in addition to that determined from equation (47) are presenting the total confining for the system: Cp =Cp steel jacket + Cp reinforcement (48) 3 INTERFACE SPRING BETWEEN STEEL JACKET AND CONCRETE The bond strength between a rebar and the surrounding concrete is generally due to three components: 1. Chemical adhesion;. Friction; 3. Mechanical interlock between reinforcement and concrete. The interface spring stiffness between the steel jacket and concrete do not include the chemical adhesion nor mechanical interlock also there is no available data to estimate it so this was found experimentally. Another reason that leads to do this experimentally is that the parameters of any formula which is used to evaluate the stiffness of the interface spring between the concrete and steel jacket do not take the additional bond which is from the effect of transverse expansion of the axially loaded column due to Poisson's ratio. The suggested empirical equation of the interface spring stiffness between concrete and steel jacket becomes KKKKrrrrrrrr = 1.83 LLLL rrrrssss (EEEE rrrr + EEEE cccc ) LLLL rrrr Cp steel jacket (49) 479
10 4 APPLICATIONS Case (1): Comparison between Campione experimental results with those of the present study. The dimensions and properties of column specimens tested by Campoione are suarized in Table (1). Cross section dimension length Column Fc' 1 Table (1): Details of Campione Specimen Longitudinal bar Stirrups Angles Strips 15*15 diam. 1 diam. 6 Size 3 Size 3 1 No. of bar 4 yield stress 461 Spacing 4 Thickness 3 39 Thickness Spacing 3 39 The results of Campione experimental work and present study are shown in figure (): Experimental by Campione Present study Applied load (kn) Concrete strain Figure (): Comparison Between The Experimental Results (Campoione ) and The Present Study (Loaded Angles) 48
11 Case (): Load carrying capacity with corresponding failure state when eccentric load is applied, three cases has been taken into consideration. Those were tested by G. Roca et. al. (1) which have the same details and different eccentricities as shown in Table (). Table (): Details of Garzon Roca et al. (1) [] Specimen Column Longitudinal bar Stirrups Angles Strips Cross section dimension 6*6 diam. 1 diam. 6 Size 6 Size 14 length Fc' 1 1 No. of bar 4 5 Spacing 5 Thickness 6 75 Thickness Spacing The results of experimental test by G. Roca and present study were briefed in Table (3). Table (3): Experimental Results (Garzon-Roca et al. 1) [] and Present Study (Loaded Angles) Eccentricity Experimental Axial load capacity kn Present study Axial load capacity kn Experimental Moment capacity kn.m Present study Moment capacity kn.m CONCLUSIONS The present study introduced a new simple and efficient element which can understand the behaviour of reinforced concrete short rectangular columns strengthened by steel lattice 481
12 framed jacket. New model that controls the interface spring stiffness between concrete and steel jacket including the effect of transverse expansion of the axially loaded column due to Poisson's ratio.the model can be used to study the interface nonlinearity between concrete and steel jacket which is not considered before. REFERENCES [1] A.M. Tarabia and Albakry H.F. "Strengthening of R.C. columns by steel angles and strips ", Alexandria Engineering Journal,53, (14). [] G. Campione "Strength and ductility of R.C. columns strengthened with steel angles and battens", Science Direct - Construction and building materials,35, 8-87 (1). [3] N.A. Alwash "One dimensional composite axial finite element including bond slip" (Published by Journal of Babylon University) (6). [4] J. B. Mander, M. J. N. Priestley and R. Park " Theoretical stress strain model for confined concrete",journal of structural engineering, ASCE, 114, 8, (1988). 48
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