Aci 530 11 Masonry Wall 002

Software Verification PROGRAM NAME: REVISION NO.:

ETABS 2013 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 2013 program results. GEOMETRY, PROPERTIES AND LOADING

EXAMPLE ACI 530-11 Masonry Wall-002 - 1


Material Properties E= ν= G=

3600 k/in2 0.2 1500 k/in2

ETABS 2013 0

Design Properties

Section Properties

f ′c = 4 k/in2 fy = 60 k/in2

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)

TECHNICAL FEATURES OF ETABS 2013 TESTED  Concrete wall flexural Demand/Capacity ratio RESULTS COMPARISON Independent results are hand calculated and compared with ETABS 2013 design check.

Output Parameter

ETABS 2013

Independent

Percent Difference

Column Demand/Capacity Ratio

0.999

1.00

0.10%

COMPUTER FILE: ACI 530-11 MASONRY WALL-002 CONCLUSION The ETABS 2013 results show an acceptable comparison with the independent results.



ETABS 2013 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 2013 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  1 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 Pn2    e  Cs3  3s   Ts 4  2s   Ts5  s 

    Cs1  d  d '  Cs 2  4s      

(Eqn. 2)

where Cs1  A1  f s1  0.8 f m  ; 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.



ETABS 2013 0

4) Iterating on a value of c until equations 1 and 2 are equal c is found to be c = 41.15 inches. a  0.8 • c  0.8 • 41.15  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 '  s2    0.0025 c    c  2s  d '   s3    0.0025 c    d  c  2s   s4     s6  d c   d cs   s5     s6  d c   d c   s6    0.0025  c 

 s1  

= 0.00226; f s   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 n  Pn e  1662(41.15) /12 = 8208 k-ft 6) Calculate the capacity, Pn = 0.9 1622   1496 kips M n = 0.9  8208   7387 k-ft.


Aci 530 11 Masonry Wall 002

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