Contents Eurocode 2 2004 WALL 002
User Manual: Eurocode 2-2004 WALL-002
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Software Verification PROGRAM NAME: ETABS REVISION NO.: 4 EXAMPLE Eurocode 2-2004 Wall-002 P-M INTERACTION CHECK FOR A WALL EXAMPLE DESCRIPTION The Demand/Capacity ratio for a given axial load and moment are tested in this example. A reinforced concrete wall is subjected to factored axial load Pu = 11605 kN and moments Muy = 15342 kN-m. This wall is reinforced as noted below. The design capacity ratio is checked by hand calculations and the results are compared with ETABS program results. GEOMETRY, PROPERTIES AND LOADING EXAMPLE Eurocode 2-2004 Wall-002 - 1 Software Verification PROGRAM NAME: ETABS REVISION NO.: Material Properties E= ν= 25000 MPa 0.2 4 Design Properties Section Properties tb = 200 mm H = 2500 mm d= 2400 mm s= 460 mm As1= As5 = 4-35M+2-20M (4600 mm^2) As2, As3, As4, As5 = 2-20M (600 mm^2) f ′c = 30 MPa fy = 460 MPa TECHNICAL FEATURES OF ETABS TESTED Concrete wall flexural demand/capacity ratio RESULTS COMPARISON Independent results are hand calculated and compared with ETABS design check. Output Parameter ETABS Independent Percent Difference Wall Demand/Capacity Ratio 1.011 1.00 1.10% COMPUTER FILE: EUROCODE 2-2004 WALL-002 CONCLUSION The ETABS results show an acceptable comparison with the independent results. EXAMPLE Eurocode 2-2004 Wall-002 - 2 Software Verification PROGRAM NAME: ETABS REVISION NO.: 4 HAND CALCULATION Wall Strength Determined as follows: 1) A value of e = 1322 mm was determined using e = M u / Pu where M u and Pu were taken from the ETABS test model PMM interaction diagram for pier P1. Values for M u and Pu were taken near the balanced condition and large enough to produce a flexural D/C ratio very close to or equal to one. The depth to the neutral axis, c, was determined by iteration using an excel spreadsheet so that equations 1 and 2 below were equal. 2) From the equation of equilibrium: Pn = Cc + Cs − T Where Cc =Ccw +Ccf , where Ccw and Ccf are the area of the concrete web and flange in compression α cc fck 0.85 • 30 Ccw= • 200 • ( a − 200 )= • 200 • ( a − 200 )= 3400(a − 200) γm 1.5 α cc fck 0.85(30) = 200 • ( 2500 = − 1000 ) ) − 1000 ) ) 5,100, 000 Ccf ( ( 200 • ( 2500= γm 1.5 α f α f Cs =A1′ fs1 − cc ck + A2′ fs 2 − cc ck γm γm f f f T = As 4 s 4 + As 5 s 4 + As 6 s 4 γs γs γs Pn1= 3400(a − 200) + 5,100, 000 + + As 3 α cc fck fs 3 − γs γm α cc fck + A3′ fs 3 − γm As1 α cc fck fs1 − γs γm As 2 α cc fck + fs 2 − γm γs As 4 A A fs 4 − s 5 fs 5 − s 6 fs 6 − γs γs γs (Eqn. 1) EXAMPLE Eurocode 2-2004 Wall-002 - 3 Software Verification PROGRAM NAME: ETABS REVISION NO.: 4 3) Taking moments about As6: a − tf − tf + Cs1 ( d - d' ) + 1 Ccf ( d - d' ) + Ccw d 2 Pn2 = e′ Cs2 ( 4s ) + Cs3 ( 3s ) − Ts4 ( 2s ) − Ts5 ( s ) = Cs1 where Ts 4 = (Eqn. 2) As1 α cc f ck As 2 As 3 α cc f ck α cc f ck = = f s1 − ; Cs 2 fs3 − fs2 − ; Cs 3 γs γm γs γm γs γm ; As 4 ( f s 4 ) and the bar strains and stresses are determined below. γs The plastic centroid is at the center of the section and d ′′ = 700 mm e′ =e + d ′′ =1322 + 700 =2472 mm. 4) Using c = 1299 mm (from iteration), a= β1c = 0.895 •1299 = 1163 mm 5) Assuming the extreme fiber strain equals 0.0035 and c = 1299 mm, the steel stresses and strains can be calculated. When the bar strain exceeds the yield strain, then fs = f y : c − d′ ε s1 = 0.0035 c c − s − d′ εs 2 = 0.0035 c c − 2s − d ′ εs3 = 0.0035 c d − c − 2s = εs 4 εs6 d −c d −c−s = εs5 εs6 d −c d −c εs6 = 0.0035 c = 0.00323; f s = ε s E ≤ Fy ; f s1 = 460 MPa = 0.00199 f s 2 = 398.2 MPa = 0.00075 f s 3 = 150.3 MPa = 0.00049 f s 4 = 97.5 MPa = 0.00173 f s 5 = 345.4 MPa = 0.00297 f s 6 = 460.00 MPa EXAMPLE Eurocode 2-2004 Wall-002 - 4 Software Verification PROGRAM NAME: ETABS REVISION NO.: 4 Substituting in Eqn. 1 and 2 and iterating the value of the neutral axis depth until the two equations are equal give Pn1 = 11605 kN Pn2 = 11605 kN M= P= 11605(1322) /1000 = 15342 kN-m n ne EXAMPLE Eurocode 2-2004 Wall-002 - 5
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