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Software Verification PROGRAM NAME: REVISION NO.:
ETABS 0
EXAMPLE ACI 318-14 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 = 2384 k and moments Mu3 = 9293k-ft. The design capacity ratio is checked by hand calculations and results are compared with ETABS program results. GEOMETRY, PROPERTIES AND LOADING
EXAMPLE ACI 318-14 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
Wall Demand/Capacity Ratio
0.999
1.00
0.10%
COMPUTER FILE: ACI 318-14 WALL-002 CONCLUSION The ETABS results show an acceptable comparison with the independent results.
EXAMPLE ACI 318-14 Wall-002 - 2
Software Verification PROGRAM NAME: REVISION NO.:
ETABS 0
HAND CALCULATION Wall Strength under compression and bending 1) A value of e = 46.78 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 =Ccw +Ccf , where Ccw and Ccf are the area of the concrete web and flange in compression = Ccw 0.85 fc′ • 8 • ( a − 8 ) = Ccf 0.85 fc′ ( 8 • ( 98 − 40 ) )
Cs =A1′ ( fs1 − 0.85 fc′) + A2′ ( fs 2 − 0.85 fc′) + A3′ ( fs 3 − 0.85 fc′) T = As4 f s4 + As5 f s5 + As6 f s6
= Pn1 0.85 fc′ • 8 • ( a − 8 ) + 0.85 fc′ ( 8 • ( 98 − 40 ) ) + A1′ ( fs1 − 0.85 fc′) + A2′ ( fs 2 − 0.85 fc′) + A3′ ( fs 3 − 0.85 fc′) + As 4 fs 4 + As 5 fs 5 + As 6 fs 6 3) Taking moments about As6: a −tf − t f + Cs1 ( d − d ′ ) + 1 Ccf ( d − d ′ ) + Ccw d − 2 Pn 2 = e′ Cs 2 ( 4s ) + Cs 3 ( 3s ) − Ts 4 ( 2s ) − Ts 5 ( s )
(Eqn. 1)
(Eqn. 2)
Cs1 A1′ ( fs1 − 0.85 fc′)= where= , Csn An′ ( f sn − 0.85 f c′) , Tsn = f sn Asn , and the bar strains are determined below. The plastic centroid is at the center of the section, d ′′ 98 − 8 = 45 inches = 2 e′ =e + d ′′ =46.78 + 45 =91.78 inches
EXAMPLE ACI 318-14 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 = 44.58 inches. a = 0.85 • c=0.85 • 44.58 = 37.89 inches
5) Assuming the extreme fiber strain equals 0.003 and c = 44.58 inches, the steel stresses and strains can be calculated. When the bar strain exceeds the yield strain then, f s = f y : c−d ' 0.003 c c−s−d ' ε s2 = 0.003 c c − 2s − d ' ε s3 = 0.003 c d − c − 2s ε s4 = ε s6 d −c d −c−s ε s5 = ε s6 d −c d −c ε s6 = 0.003 c
ε s1 =
= 0.00273;= f s ε s E ≤ Fy ; f s1 = 60.00 ksi = 0.00152
f s 2 = 44.07 ksi
= 0.00310
f s 3 = 8.94 ksi
= 0.00090
f s 4 = 26.2 ksi
= 0.00211
f s 5 = 60.00 ksi
= 0.00333
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 = 3148 k Pn2 = 3148 k M= P= 3148(46.78) /12 = 12,273 k-ft n ne
6) Determine if φ is tension controlled or compression controlled. εt = 0.00332 , ε y = 0.0021 for ε y < εt < 0.005 ; φ =
0.005 − εt = 0.757 − ε 0.005 y
( φt − φc )
EXAMPLE ACI 318-14 Wall-002 - 4
Software Verification PROGRAM NAME: REVISION NO.:
ETABS 0
7) Calculate the capacity,
φPn = 0.757 ( 3148 ) = 2384 kips 9, 293 k-ft. φM n = 0.757 (12, 273) =
EXAMPLE ACI 318-14 Wall-002 - 5
ETABS 0
EXAMPLE ACI 318-14 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 = 2384 k and moments Mu3 = 9293k-ft. The design capacity ratio is checked by hand calculations and results are compared with ETABS program results. GEOMETRY, PROPERTIES AND LOADING
EXAMPLE ACI 318-14 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
Wall Demand/Capacity Ratio
0.999
1.00
0.10%
COMPUTER FILE: ACI 318-14 WALL-002 CONCLUSION The ETABS results show an acceptable comparison with the independent results.
EXAMPLE ACI 318-14 Wall-002 - 2
Software Verification PROGRAM NAME: REVISION NO.:
ETABS 0
HAND CALCULATION Wall Strength under compression and bending 1) A value of e = 46.78 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 =Ccw +Ccf , where Ccw and Ccf are the area of the concrete web and flange in compression = Ccw 0.85 fc′ • 8 • ( a − 8 ) = Ccf 0.85 fc′ ( 8 • ( 98 − 40 ) )
Cs =A1′ ( fs1 − 0.85 fc′) + A2′ ( fs 2 − 0.85 fc′) + A3′ ( fs 3 − 0.85 fc′) T = As4 f s4 + As5 f s5 + As6 f s6
= Pn1 0.85 fc′ • 8 • ( a − 8 ) + 0.85 fc′ ( 8 • ( 98 − 40 ) ) + A1′ ( fs1 − 0.85 fc′) + A2′ ( fs 2 − 0.85 fc′) + A3′ ( fs 3 − 0.85 fc′) + As 4 fs 4 + As 5 fs 5 + As 6 fs 6 3) Taking moments about As6: a −tf − t f + Cs1 ( d − d ′ ) + 1 Ccf ( d − d ′ ) + Ccw d − 2 Pn 2 = e′ Cs 2 ( 4s ) + Cs 3 ( 3s ) − Ts 4 ( 2s ) − Ts 5 ( s )
(Eqn. 1)
(Eqn. 2)
Cs1 A1′ ( fs1 − 0.85 fc′)= where= , Csn An′ ( f sn − 0.85 f c′) , Tsn = f sn Asn , and the bar strains are determined below. The plastic centroid is at the center of the section, d ′′ 98 − 8 = 45 inches = 2 e′ =e + d ′′ =46.78 + 45 =91.78 inches
EXAMPLE ACI 318-14 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 = 44.58 inches. a = 0.85 • c=0.85 • 44.58 = 37.89 inches
5) Assuming the extreme fiber strain equals 0.003 and c = 44.58 inches, the steel stresses and strains can be calculated. When the bar strain exceeds the yield strain then, f s = f y : c−d ' 0.003 c c−s−d ' ε s2 = 0.003 c c − 2s − d ' ε s3 = 0.003 c d − c − 2s ε s4 = ε s6 d −c d −c−s ε s5 = ε s6 d −c d −c ε s6 = 0.003 c
ε s1 =
= 0.00273;= f s ε s E ≤ Fy ; f s1 = 60.00 ksi = 0.00152
f s 2 = 44.07 ksi
= 0.00310
f s 3 = 8.94 ksi
= 0.00090
f s 4 = 26.2 ksi
= 0.00211
f s 5 = 60.00 ksi
= 0.00333
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 = 3148 k Pn2 = 3148 k M= P= 3148(46.78) /12 = 12,273 k-ft n ne
6) Determine if φ is tension controlled or compression controlled. εt = 0.00332 , ε y = 0.0021 for ε y < εt < 0.005 ; φ =
0.005 − εt = 0.757 − ε 0.005 y
( φt − φc )
EXAMPLE ACI 318-14 Wall-002 - 4
Software Verification PROGRAM NAME: REVISION NO.:
ETABS 0
7) Calculate the capacity,
φPn = 0.757 ( 3148 ) = 2384 kips 9, 293 k-ft. φM n = 0.757 (12, 273) =
EXAMPLE ACI 318-14 Wall-002 - 5
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