Contents ACI 530 11 Masonry WALL 002
User Manual: ACI 530-11 Masonry-WALL-002
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Software Verification PROGRAM NAME: REVISION NO.: ETABS 0 EXAMPLE ACI 530-11 Masonry Wall-002 P-M INTERACTION CHECK FOR WALL EXAMPLE DESCRIPTION The Demand/Capacity ratio for a given axial loading and moment is tested in this example. The wall is reinforced as shown below. The concrete core wall is loaded with a factored axial load Pu = 1496 k and moments Mu3 = 7387 k-ft. The design capacity ratio is checked by hand calculations and the results are compared with ETABS program results. GEOMETRY, PROPERTIES AND LOADING EXAMPLE ACI 530-11 Masonry Wall-002 - 1 Software Verification PROGRAM NAME: REVISION NO.: Material Properties E= ν= G= 3600 k/in2 0.2 1500 k/in2 ETABS 0 Design Properties Section Properties tb = 8 in h = 98 in As1= As6 = 2-#10,2#6 (5.96 in^2) As2, As3, As4 and As5 = 2-#6 (0.88 in^2) f ′c = 4 k/in2 fy = 60 k/in2 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 Column Demand/Capacity Ratio 0.998 1.00 -0.20% COMPUTER FILE: ACI 530-11 MASONRY WALL-002 CONCLUSION The ETABS results show an acceptable comparison with the independent results. EXAMPLE ACI 530-11 Masonry Wall-002 - 2 Software Verification PROGRAM NAME: REVISION NO.: ETABS 0 HAND CALCULATION Wall Strength under compression and bending 1) A value of e = 59.24 inches was determined using e = M u / Pu where M u and Pu were taken from the ETABS test model interaction diagram. The values of M u and Pu were 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: Pn1 = Cc + Cs − T where Cc = β1 fm′ ab = 0.8 • 2.5 •12a = 24.0a Cs = A1′ ( fs1 − 0.8 fm′ ) + A2′ ( fs 2 − 0.8 fm′ ) + A3′ ( fs 3 − 0.8 fm′ ) T = As4 f s4 + As5 f s5 + As6 f s6 Pn1 = 24a + A1′ ( fs1 − 0.8 fm′ ) + A2′ ( fs 2 − 0.8 fm′ ) + (Eqn. 1) A3′ ( fs 3 − 0.8 fm′ ) − As 4 fs 4 − As 5 fs 5 − As 6 fs 6 3) Taking moments about As6: a −tf 1 Ccf ( d − d ') + Ccw d − 2 Pn 2 = e′ Cs 3 ( 3s ) − Ts 4 ( 2s ) − Ts 5 ( s ) + Cs1 ( d − d ') + Cs 2 ( 4s ) + (Eqn. 2) Cs1 A1′ ( f s1 − 0.8 f m′ )= where= ; Csn An′ ( f sn − 0.8 f m′ ) ; Tsn = f sn Asn ; and the bar strains are determined below. The plastic centroid is at the center of the section and d ′′ = 45 inch e′ =e + d ′′ =59.24 + 45 =104.24 inch. EXAMPLE ACI 530-11 Masonry Wall-002 - 3 Software Verification PROGRAM NAME: REVISION NO.: ETABS 0 4) Iterating on a value of c until equations 1 and 2 are equal c is found to be c = 41.15 inches. • c 0.8 • 41.15 a 0.8 = = = 32.92 inches 5) Assuming the extreme fiber strain equals 0.0025 and c = 41.15 inches, the steel stresses and strains can be calculated. When the bar strain exceeds the yield strain, then f s = f y : c−d ' 0.0025 c c−s−d ' ε s2 = 0.0025 c c − 2s − d ' ε s3 = 0.0025 c d − c − 2s ε s4 = ε s6 d −c d −c−s ε s5 = ε s6 d −c d −c ε s6 = 0.0025 c ε s1 = = 0.00226;= f s ε s E ≤ Fy ; f s1 = 60.00 ksi = 0.00116 f s 2 = 33.74 ksi = 0.00007 f s 3 = 2.03 ksi = 0.00102 f s 4 = 29.7 ksi = 0.00212 f s 5 = 60.00 ksi = 0.00321 f s 6 = 60.00 ksi Substituting the above values of the compression block depth, a, and the rebar stresses into equations Eqn. 1 and Eqn. 2 give Pn1 = 1662 k Pn2 = 1662 k M= P= 1662(41.15) /12 = 8208 k-ft n ne 6) Calculate the capacity, φPn = 0.9 (1622 ) = 1496 kips φM n = 0.9 ( 8208 ) = 7387 k-ft. EXAMPLE ACI 530-11 Masonry Wall-002 - 4
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