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NZS 4230:2004 USER GUIDE

1. Figure 3 0 4 0 02 0 04 0 06 0 08 0 1 0 12 0 14 0 16 2 Flexural Strength of Rectangular Masonry Walls with Uniformly Distributed Reinforcement Confined Wall f 300 MPa Confined Wall f 500 MPa 0 35 0 3 N 0 25 0 2 0 15 0 1 0 05 12 Figure 4 Flexural Strength of Rectangular Masonry Walls with Uniformly Distributed Reinforcement Confined Wall f 500 MPa Table 6 Neutral Axis Depth Ratio c L fj 300 MPa or 500 MPa Unconfined Walls Axial Load Ratio Pit 0 0 05 0 1 0 15 0 2 0 25 0 3 0 35 0 4 0 0 0000 0 0692 0 1384 0 2076 0 2768 0 3460 0 4152 0 4844 0 5536 0 01 0 0135 0 0808 0 1481 0 2155 0 2828 0 3502 0 4175 0 4848 0 5522 0 02 0 0262 0 0918 0 1574 0 2230 0 2885 0 3541 0 4197 0 4852 0 5508 0 04 0 0498 0 1121 0 1745 0 2368 0 2991 0 3614 0 4237 0 4860 0 5483 0 06 0 0712 0 1306 0 1899 0 2493 0 3086 0 3680 0 4273 0 4866 0 5460 0 08 0 0907 0 1473 0 2040 0 2606 0 3173 0 3739 0 4306 0 4873 0 5439 0 1 0 1084 0 1626 0 2168 0 2710 0 3252 0 3794 0 4336 0 4878 0 5420 0 12 0 1247 0 1766 0 2286 0 2805 0 3325 0 3844 0 4364 0 4883 0 5403 0 14 0 1397 0
2. fy where 33p 1 Pw 300 7bars 016 bwd 7x201 440 x 0 8 x 2600 0 0048 and pw 300 gt 33 x 0 0048 x 300 0 16 1 0 since h Ly gt 1 0 Hence Vm 0 16 1 0 pm where vy 0 70 MPa for p 1 and f m 12 MPa Vm 1 16 x 0 70 0 81 MPa Shear stress carried by Vp 0 9 N tana w where 50 kN As illustrated in Figure 10 2 of NZS 4230 2004 it is necessary to calculate the compression depth a in order to establish tana The following illustrates the procedure of establishing compression depth a using Table 6 Tbars x D16 bw xLw 7x201 140 x 2600 0 00387 f pX 9 00387 290 T 12 m 0 0967 _ 58 8 103 fiat 12 2600 140 0 0135 From Table 6 2012 L w 29 0 12 x 2600 312 for unconfined concrete masonry 0 85 0 85 x 312 265 2 mm Therefore Lw a 2600 _ 265 2 14 34 280 262 ana h 2800 0 417 3 ek eG 140 x 0 8 x 2600 0 064 MPa Shear stress to be carried by v Vs Vn Vm Vp 1 37 0 81 0 064 0 50 MPa Vg where 0 8 for a masonry walls S w A 0 50 0 8 Try f 300 MPa and reinforcement spacing 200 mm 140 x 200 A 58 3 mm Therefore use R10 200 crs 78 5 mm per 200 mm spacing It is essential that shear reinforc
3. 13 0 4 0 35 0 3 0 2 0 15 0 1 0 05 Figure 5 0 4 0 35 0 3 025 0 2 0 15 0 1 0 05 Figure 6 Unconfined Wall 77 o3 0 2 M 0 4 0 6 E Neutral Axis Depth of Unconfined Rectangular Masonry Walls with Uniformly Distributed Reinforcement f 300 MPa 500 MPa Confined Wall pfy X N 0 1 0 2 0 3 0 4 0 5 0 6 Lw Neutral Axis Depth of Confined Rectangular Masonry Walls with Uniformly Distributed Reinforcement fy 300 MPa or 500 MPa 14 2 7 2 Curvature Ductility To avoid failure of potential plastic hinge regions of unconfined masonry shear walls the masonry standard limits the extreme fibre compression strain at the full design inelastic response displacement to the unconfined ultimate compression strain of e 0 003 The available ductility at this ultimate compression strain decreases with increasing depth of the compression zone expressed as a fraction of the wall length Section 7 4 6 of NZS 4230 2004 ensures that the available ductility will exceed the structural ductility factor u for walls of aspect ratio less than 3 This section provides justification for the relationship limiting neutral axis depth The most common and desirable sources of inelastic structural deformations are rotations in potential plastic hinges Therefore it
4. Therefore My Np RES ic dau 25x 2 55 2 34 wall 5 Now Ms E M 20 2 34 17 66 kNm m assuming ag vay Mp anadas 08 2 34 18 5 mm M 2 2 3 Therefore gt 22 300x 95 2 45 19 103 72 2 2 707 mm m Try 020 reinforcing bars spaced at 400 mm c c A 785 mm m Check a _ 25 10 785 300 0 851 1 0 0 85 12 103 M Asf 2 5 25x108 00 25 54 190 25 54 2 214 kNm m gt 2 27 3 4 Design of Shear Reinforcement The single storey cantilevered concrete masonry wall of Figure 13 is to resist a shear force while responding elastically to the design earthquake For a wall width of 140 mm f m 12 MPa and N 50 kN design the required amount of shear reinforcement N 50 KN V 300 kN 2800 016 016 016 016 016 016 016 2600 Figure 13 Forces acting on masonry wall Solution N 50 Therefore 2300 58 8 0 85 V 300 kN Require V gt V Therefore V 5 300 0 75 gt 400 Vn 2 Check maximum shear stress V note that d 0 8Ly for walls Ww _ 400 103 140 x 0 8 x 2600 1 37 MPa lt vg Vg 1 50 MPa for f m 12 MPa 28 Now Vn Vm Vp Vs Shear stress carried by Vm Vm
5. fmAg 16 x 5000 x 190 Mn _ 3211 8x109 faL2 t 16 50002 x190 0 0423 From Figure 1 and for f 300 MPa for flexural reinforcement fy 0 028 fm Therefore p DAS IG 0 0015 300 Check Ductility Capacity Using Figure 8 check the available ductility p p4 0 0015 12 0 0011 f 16 m m g 0 0697 From Figure 8 4 5 From Eqn 14 3 3 x 4 5 1 38 3 6 3 98 4 0 1 H3 6 3 6 Hence ductility OK Flexural Reinforcement For p 0 0015 reinforcement per 400 mm will be 114 mm 0 0015 400x190 400 Therefore use 012 400 mm crs 113 mm 400 mm 35 Shear Design To estimate the maximum shear force on the wall the flexural overstrength at the base of the wall Mo needs to be calculated 1 25 provide for Grade 300 reinforcement 16 MPa 0 070 foL ut 13 bars x 113 mm 0 00155 Pprovided 5000 x 190 f and p 0 00155 x 300 0 029 fin 16 From Table 2 M 0 047 Therefore Mhn provided 0 047 16 50007 x 190 3580 kNm The overstrength value is calculated as follow Mo _ 1 25 provided _ 1 25 3580 _ 1 64 Pow M M 2730 Dynamic Shear Magnification Factor For up to 6 storeys 0 9 10 tsis 10 Hence the design shear force at the wall base is Vn GeV 1 5 1 64 x V 2 46 2 46
6. 2 bm fy and C4 33 300 pw 0 00297 for 012 200 crs 0 098 1 0 for simplicity Vm 0 098 1 x 0 2 0 22x 4 16 0 88 MPa Therefore Vm 0 88 x 0 19 x 0 8 x 10 133 kN Hence Va E CT lt ZERO 0 75 Therefore NO horizontal joint steel is required i e Aj 0 The horizontal shear is carried by the horizontal component of the diagonal strut across the joint 56 Vertical Joint Shear x0 9x1 6 ai 2 ses 26 2 kN Vi 0 9 0 8 Vy 2Vg Gravity induced joint shear is considered negligible in this instance z Viv 52 3 Nominal shear stress in the joint Viv _ 52 3 103 Viv 0 17 lt v Therefore OK bchg 190 1600 V 135 mv where Vw 0 see 11 4 6 2 of NZS 4230 2004 Hence Vey 2223 02 69 7 kN 0 75 Therefore the total area of vertical joint shear reinforcement required Vs _ 69 7103 f 300 232 4 mm Take f 300 MPa y Therefore use 4 R10 to give Aj 314 2 Level 3 Joint Shear Design A similar process to that above is required but not tabulated herein see Figure 18 for detailed 57 3 8 Strut and tie Design of Wall with Opening Figure 19 a shows a three storey concrete masonry wall with openings and loading conditions that resemble a design example of a reinforced concrete wall reported by Paulay and Priestley 1992 It is noted that design
7. C2 Vim fy where 33p 1 Pw 300 and pw 0 00297 gt 0 098 48 a 0424 1475 Ww 0 42 4 1 75 x 1200 2 x 1200 1 31 Vm 0 098 1 31 0 15416 where Vom 0 15 for p 2 gt Vm 0 84 MPa Shear stress carried by v 0 9 tana byd Where N 143 5 kN 168 8 0 85 and p 0 00297 Nn i Est 0 046 fy and Pa 0 0557 m From Table 6 c 0 122 w For Pier 3 with Lw 1200 c 0 122 x 1200 146 4 mm Therefore a 0 85 x 146 4 124 4 mm Consequently tana ML 0 90 for pier in double bending 3 vp 0 9x 423x10 0 90 0 64 MPa 190 x 0 8 x 1200 Shear stress to be carried Vn Vm Vp 60 0 84 0 64 Vs Vn Vm Vp 1 0 12 MPa where C 0 8 for masonry walls bws Vs Ay x 300 190 x 200 A 19 0 mm 0 12 0 8 Try fy 300 MPa and reinforcement spacing 200 However this is less than the 0 07 required by clause 7 3 4 3 of the standard Therefore use R6 200 crs 28 2 mm to give 0 074 49 Design of 2 Storey The procedure is the same as for 15 storey and is not repeated here Minimum requirements of D12 200 again govern flexure but shear reinforcement in the outer piers can be reduced to 0 07 of the gross cross sectional area of the wall minimum reinforcement area required by clause 7 3 4 3 Flex
8. 1 03 S Vg From Section 10 3 of NZS 4230 2004 Vn Vm Vp Vs Shear stress carried by Vm C4 here C 33p 27 where 1 Pw 300 and py 0 00297 33 x 0 00297 x 300 300 24 12475 gt C 0 42 4 1 75x1200 2 x 800 1 12 0 098 Vm 0 098 1 12 where Vom 0 15 2 gt Vm 0 73 MPa Shear stress carried by v 0 9 tan a Where N 18 8 kN gt n 22 1 kN 0 00297 LEE 0 0091 fmb wt 47 fy p 0 0557 m From Table 6 c 0 068 Pier 1 with Ly 800 gt c 54 4 mm Therefore a 0 85 x c 46 2 mm 800 46 2 Consequently for pier in double bending tana EU 0 628 3 Vp 0 9x 88x10 0 628 0 087 MPa 190 x 0 8 x 800 Shear stress to be carried v Vm Vp Vs Vn Vm Vp 1 03 0 73 0 087 0 39 MPa Af and Vs where C 0 8 for masonry walls bws gt 0 39 Try f 300 MPa and reinforcement spacing 200 mm 190 x 200 A 61 8 mm Therefore use R10 200 crs 78 5 mm Inner Piers Clearly Pier governs due to lighter compression load V 1 5 2 109 0 219 5 kN 599m kN 0 75 Check shear stress b 190 mm d 0 8 x 1200 960 Vn _ 292 7x10 gt Yn 190 x 960 1 60 MPa Vg Shear stress carried by Vm
9. 1 14 0 15 1 14 Vim 1 29 x 0 50 note that Vp 0 50 MPa 2 0 67 MPa Therefore the shear reinforcement required Vs Vn Vm Vp take v 0 for simplicity gt Vs 1 09 0 67 0 0 42 MPa Af Vg note that C3 0 8 for masonry walls bws A 39402 dey ee try f 300 MPa and 200 mm 190 x 200 A 66 5 Therefore use R10 200 crs 78 5 mm and p EC 0 2 190 200 Shear Design 2 Storey 250 therefore Va 1 1 x 2 31 x 250 635 3 kN Check shear stress b 190 mm d 0 8 x 4600 3680 mm _ 635 3x 10 Vn 0 91 MPa lt v 1 50 MPa 190 x 3680 Shear stress carried by Vm C2 Vim f where C4 33p 1 Pw 300 5bars x DH12 4bars x DH16 500 5bbarsxD12 300 33 x x 33x x bwd 300 bwd 300 0 10 0 03 0 13 and 042x 4 1 75 x 4200 4600 1 01 gt Vm 0 13 1 01 x Vim 1 14 Vom 0 70 MPa since outside plastic hinge region 1 14 x 0 70 0 80 MPa 66 Therefore the shear reinforcement required Vs Vn Vm Vp take v 0 for simplicity gt Vs 0 91 0 80 0 0 11 MPa A Vf Vg where 0 8 for masonry walls bws 0 11 0 8x Apka try fy 300 MPa and 200 mm 190 x 200 17 5 2 28 3 5 Therefore use R6 200 crs 28 3 mm and
10. gt 6 200 zm R10 200 i 4 012 2 020 5 012 4 016 8 016 Figure 22 Reinforcement for Design Example 3 8 Theoretical depth of neutral axis 0 85 x 0 85 x fr x 190 652 1x10 0 85x0 85 x 12 x 190 395 8 mm 0 086L Moment capacity about the centre of the structure M Tas Cm 3 3 Tn 1 9 135 7 652 1 3 3 671 0 x 1 9 3874 6 Therefore 0 75 x 3874 6 2906 kNm gt 62 2 and 3 storey vertical members To avoid the formation of plastic hinges the amount of reinforcement in the 2 and 3 storey vertical members should be sufficient to ensure that yielding does not occur in these members Hence the 2 and 3 storey vertical members are intentionally designed for 5096 higher tension forces than the design level tension forces Consider earthquake Ve as Figure 21 Tie J L 1 5 x 204 8 307 2 kN 3 Therefore T take f 500 MPa 819 2 Try 4 DH16 A 804 2 mm about 2 shortfall note that DH16 is the maximum bar size allowed in Table 1 Tie F H bAcufy 1 5 x 107 3 161 0 kN 3 Therefore a take f 500 MPa 429 3 Try 5 DH12 565 5 mm EN Consider earthquake Vg as in Figure 21 b Tie B D 1 5 x 112 3 168 5 kN 3 Therefore Acp take f 500 MPa 0 75 x 500 449 3 Try 4 0 12 452 4 Tie F H Use 5 DH12 because member force
11. 1900 400 0 165 2 234 8 127 3 473 7 366 2 234 8 Figure 20 Strut and Tie Models for Masonry Wall seismic loading only 58 Figures 20 a and b show the strut and tie models for the squat wall with openings corresponding to the seismic lateral forces being considered For the purpose of limited ductile design particular tension chords should be chosen to ensure yielding can best be accommodated For example members 1 and E F in Figure 20 a represent a good choice for this purpose Corresponding forces in other members should be determined and hence reinforcement provided so as to ensure that no yielding in other ties can occur As these members carry only tension yielding with cyclic displacements may lead to unacceptable cumulative elongations Such elongations would impose significant relative secondary displacement on the small piers adjacent to openings particularly those at l J and The resulting bending moment and shear forces although secondary may eventually reduce the capacity of these vital struts In order to ensure that plastic hinges form inside the 1 storey vertical members the quantity of reinforcement in the 277 and 3 storey vertical members should be sufficient to ensure that yielding does not occur in these members Consequently a simplified procedure is adopted in this example to design the vertical tie members above storey 50 more tension force th
12. 190x200 0 07 Note that p 0 07 is x the minimum reinforcement area required by 7 3 4 3 of NZS 4230 2004 Shear Design 3 Storey V 150 therefore Va 1 1 x 2 31 x 150 381 2 kN Check shear stress b 190 mm d 0 8 x 4600 3680 mm _ 381 2x 103 0 54 v 190 3680 n Shear stress carried by Vm C4 f where 33 1 Pw 300 33x 9bars x DH12 500 5barsxD12 300 byd 300 bud 300 0 08 0 03 0 11 and 0 42 4 1 75 x 3600 4600 1 10 gt Vm 0 1 1 1 10 x V bm Vom 0 70 MPa outside plastic hinge region 1 21 0 70 0 85 MPa gt Since Vm gt Vn the shear reinforcement needed the 3 storey pier is governed by the minimum reinforcement area required by clause 7 3 4 3 i e 0 0796 of the gross cross sectional area Therefore shear reinforcement in the a storey pier can be reduced to R6 200 crs 67 4 PRESTRESSED MASONRY A new addition to NZS 4230 is the inclusion of Appendix A related to the design of prestressed concrete masonry As noted in the commentary this section is primarily for application to wall components but its use for other component types is not precluded Design information for unbonded post tensioning is presented below This form of prestressing is recommended as it minimises structural damage and results in structures that exhibit little or no permanent horizontal deformation follo
13. 44 _ 0 037 x16 300 0 002 Since the structure is designed as one of limited ductility the requirements of clause 7 4 5 1 of NZS 4230 2004 apply for spacing and bar size Consequently it is required to adopt minimum n x12 bar size of 012 and minimum of 4 bars i e 200 crs With 012 at 200 crs p 4 x 200 x 190 0 00297 This exceeds the p 0 002 required Refer to Figure 18 for details Inner Piers Inner piers are designed for the worst loading conditions of Piers 2 and 3 From Figure 17 it may be determined that Pier 3 governs design due to larger bending moment and lighter axial compressive load Pier 3 143 5 kN M May 1 2 654 66 6 kNm 6 86 IN 1 aN uae 2111 1211112111 1117 1 1 1 He tate HHH 21 12 7 R20 111 111111 012 200 012 200 IIIT T 1 4 121111 11 11111111 11 e Sie ROO Bite 11 0 Ce 4 016 R1 200 H R20 7 114311 121 cil D12 200 D12 200 Figure 18 Reinforcement for Design Example 3 7 45 Therefore N 143 5 0 85 168 8 gt 78 2 Dimensionless Design Parameters 3
14. 210 516 6 Check Maximum Shear Stress _ Va _ 516 6x10 bwd 190 0 8 5000 Vn 0 68 MPa From Table 10 1 of NZS 4230 2004 the maximum allowable shear stress Vg for fm 16 MPa is 1 8 MPa Therefore OK 36 Plastic Hinge Region Within the plastic hinge region Vm 0 Therefore v v 0 68 MPa N bd and Vp 0 9 tana As illustrated in Figure 10 2 of NZS 4230 2004 it is necessary to calculate the compression depth a in order to establish tana To establish compression depth a using Table 6 f p 0 029 and Nn 00697 fin From Table 6 559496 Lw Therefore c 0 126 x 5000 630 mm gt for unconfined concrete masonry 0 85 0 85 x 630 535 5 mm Calculation of tana 6 220954808 3000 5 2500 267 8 _ 0 372 6000 E 2900 2878 0248 9000 N3 3 arc 2500 267 8 0 186 12000 1 en tanag Qo tanas 2500 267 8 2 tana tanas 15000 0 149 Ni 212 tanas tan org 2900 2678 _ 4 454 18000 tana 2 267 8 Figure 15 Contribution of Axial Load 37 3 0 9 18019 0 744 0 112 MPa 190 x 0 8 x 5000 3 20 9 190 10 0 372 0 066 MPa 190 x 0 8 x 5000 3 Vp3 0 9 0X 10 0 248 0 044 MPa 190 x 0 8 x 5000 3 20 9 0X 10 0 86 0 033 MPa 190 x 0 8 5000 3 Vps 0 9 30 10 0 149 0 026 MPa 190 x 0 8 x 5000 3 vpe
15. 3 Consequently _ 0 003 _ 3 18x 4 x 0 0035 Cmax w gt Cmax 0 202 15 3 5 b Neutral axis of ductile masonry wall Find the maximum allowable neutral axis depth for a ductile cantilever wall Aspect ratio of 3 reinforced with grade 500 reinforcement Solution To make allowances in proportions of excess or deficiency of flexural strength ductility demand Eqn 15 can be modified CS 225 x0 202L w Har 31 Substituting Eqn 13 Substituting Eqn 12 Assuming 2 Cmax x0 202L y 2M Cmax Mow ME M 74 0 404L gt Cmax Ha 32 3 6 Ductile Cantilever Shear Wall The 6 storey concrete masonry shear wall of Figure 14 is to be designed for the seismic lateral loads shown which have been based on a ductility factor of 4 0 Design gravity loads of 150 KN including self weight act at each floor and at roof level and the weight of the ground floor and footing are sufficient to provide stability at the foundation level under the overturning moments Wall width should be 190 mm Design flexural and shear reinforcement for the wall 60 m 6 1 150 Ls 150 150 150 150 250 ASS WS Figure 14 Ductile Cantilever Shear Wall Solution Initially fj 12 MPa will be assumed From the lateral loads of Figure 14 the wall base moment is M 3x 60x6 50
16. 50 kN gt 250 05 2 873 o DINE 2 E S E N 3 wie AL 77 300 0 kN 300 0 kN 3300 m 1900 1 470 0 45 2 57 8kN 4737 486 2 304 8 b Figure 21 Strut and Tie Models for Masonry Wall Seismic and Gravity Loads 1 storey vertical members gt Consider earthquake Ve as in Figure 21 a Tie l J bAufy 184 8 184 8 x 10 Theref taking f 300 MP erefore 075 300 taking f 30 a 821 3 Try 4 D16 A 804 2 mm about 296 shortfall Tie E F 85 2 kN Therefore Agr 10 EF 0 75 300 378 7 Clause 7 4 5 1 of the standard requires minimum longitudinal reinforcement of 012 400 crs the potential plastic hinge zone Consequently adopt 5 D12 for Member E F to give 565 5 mm Check moment capacity at wall base Tension forces provided 804 2 x 300 241 3 KN Ter 565 5 x 300 169 6 Therefore total compression force at Node A including gravity load Cm Ty Ter 241 3 169 6 200 0 75 757 6 kN 60 Theoretical depth of neutral axis Cm 0 85 x 0 85 x fm x 190 757 6 x 10 0 85 x 0 85 x12 x 190 459 9 270 100L where L 800 2000 1800 4600 mm C lt 0 2L see clause 7 4 6 1 of NZS 4230 2004 Moment capacity about the centre of the structure M Ty Cm x 3 3 241 3 757 6 x 3 3 3296 4 kNm Therefore p
17. _ 168 8 x10 0 046 frre wt 16 1200 190 and 6 My _ 78 210 0 0179 124 16 12002 x 190 f From Figure1 Fa 0 00 Therefore use 012 200 for the two inner piers to satisfy the requirements of clause 7 4 5 1 Refer to Figure 18 for details Ductility Checks Clause 7 4 6 1 of NZS 4230 2004 requires that for walls with contraflexure point between adjacent heights of lateral support c 0 4512 L where Lw is the wall length and Ln is the unsupported height Note that calculations should be conducted using the amount of reinforcement required Prequirea rather than the amount of reinforcement actually provided as the latter results in a higher moment capacity and hence reduced ductility demand for which a higher value of could be tolerated pee Ls required i rom 1 240 0 006 0 040 40 OK 2 540 0 050 0 000 83 OK 3 540 0 042 0 000 70 OK 4 240 0 082 0 000 91 OK Units mm mm 46 Shear Design 1 Storey From NZS 4230 2004 V gt 2V where 0 75 Outer Piers Pier 1 governs due to the presence of axial tension force V 3 1 2x 48 5 93 9 kN where Vg 3 1kN and Vg 48 5 Vha 125 2 Now for Type A masonry vg 0 451 0 45 x J16 1 8 MPa Check shear stress by 190 mm d 0 8 x 800 640 Mas _125 2 109 bwd 190 640
18. a Part of single storey structure and b Elastic design for all load combinations and Shear stress less than 0 5v ACI 530 02 5 02 TMS 402 Minimum wall thickness of 0 0333L if 02 a Factored axial compression stress less than 0 05f m P 3101 Minimum wall thickness of 0 0333L if gt 0 2 f Ag Otherwise more slender walls permitted see P 3101 for further details 21 3 DESIGN EXAMPLES 3 1 Determine fm From Strengths of Grout and Masonry Units Calculate the characteristic masonry compressive strength f m given that the mean strengths of concrete masonry unit and grout are 17 5 MPa and 22 0 MPa with standard deviations of 3 05 MPa and 2 75 MPa respectively For typical concrete masonry the ratio of the net concrete block area to the gross area of masonry unit is to be taken as 0 45 i e a 0 45 Solution The characteristic masonry compressive strength 5 percentile value fm be calculated from the strengths of the grout and the masonry unit using the equations presented in Appendix B of NZS 4230 2004 Finding the mean masonry compressive strength fin From Eqn B 1 of NZS 4230 2004 fm 0 59afgp 0 90 1 fg 0 59 x 0 45 x 17 5 0 90 x 1 0 45 x 22 0 15 54 MPa Finding the standard deviation of masonry strength Xm From Eqn B 2 of NZS 4230 2004 Xm 0 35a2x3 0 8101 ax 40 350 452 x 3 052 0 81x 1 0 45 x 2 75 1 59 MPa Fi
19. gt amp 0 116 Second iteration using amp 0 116 Eqn 39 0 0115 m Us 0 00387 Eqn 40 AP 9 8 kN 8 3 kN 6 8 kN AP 5 3 APs 3 8 and AP 34 0 kN 0 004 m gt amp 0 116 therefore OK Eqn 32 i 0 972 x18 x 0 19 Eqn 31 M 831 34 x 0 004 Ex 567 x 2272 kNm 10 81 Eqn 33 2 18x10 12 1 0 2 x 10 _227 0 0412 14400 3 6 5 14400x3 6x0 19 1000 d 7 63 0 116 5 40 x 0 116 1 69 x Stress in tendon furthest away from compression end of wall PAP 31 5498 Tm 140 1257 MPa fost Stress in tendon closest to compression end of wall _ 5 5 _ 831 5 3 8 uum 1214 MPa ps5 140 First tendon yield c a B 0 431 0 96 0 449 m 0 96 for confined masonry __ 1517 1257 102 190000 10 15 3 6 2 04 04449 Eqn 45 0 1172 where h hy 10 15 modifies to reflect the actual tendon length Eqn 46 1517 1257 140 3 6 2 0 4 0 449 APy2 32 2 kN 28 1 23 9 kN APyys 19 8 kN Eqn 47 AP 140 4 kN Eqn 49 ay 0 972 x18 0 19 0 463 Eqn 48 36 4 x 3 6 2 0 4 19 8 x 3 6 2 0 4 140 229 kNm 50 831 34 350283 229 2475 m 248 kN Eqn 51 dy 0 041 0 1172 0 158 m 82 O
20. 16 Whenever exceeds the wall possesses reserve strength as higher resistance will be offered by the structure than anticipated when design forces were established The overstrength factors A are taken 1 25 and 1 40 for grade 300 and 500 reinforcement respectively while the strength reduction factor shall be taken as 0 85 It is expected that a corresponding reduction in ductility demand in the design earthquake will result Consequently design criteria primarily affected by ductility capacity may be met for the reduced ductility demand rather than the anticipated ductility Therefore Har HA 13 2 7 3 Ductility Capacity of Cantilevered Concrete Masonry Walls Section 7 4 6 1 of NZS 4230 2004 provides a simplified but conservative method to ensure that adequate ductility can be developed in masonry walls The Standard allows the rational analysis developed by Priestley as an alternative to determine the available ductility of cantilevered concrete masonry walls Figure 8 includes dimensionless design charts for the ductility capacity of unconfined concrete masonry walls whose aspect ratio is A hy Ly 3 For walls of other aspect ratio A the ductility capacity can be found from the us value using Eqn 14 3 3 us EL Pi 1 14 When the ductility capacity found from Figure 8 and Eqn 14 is less than that required redesign is neces
21. 1895 0 2394 0 2893 0 3392 0 3890 0 4389 0 4888 0 5387 0 16 0 1535 0 2014 0 2494 0 2974 0 3453 0 3933 0 4412 0 4892 0 5372 0 18 0 1663 0 2125 0 2587 0 3048 0 3510 0 3972 0 4434 0 4896 0 5358 0 2 0 1782 0 2227 0 2673 0 3118 0 3563 0 4009 0 4454 0 4900 0 5345 Table 7 Neutral Axis Depth Ratio c L f 300 MPa or 500 MPa Confined Walls 2 Axial Load Ratio Kft 0 0 05 0 1 0 15 0 2 0 25 0 3 0 35 0 4 0 0 0000 0 0579 0 1157 0 1736 0 2315 0 2894 0 3472 0 4051 0 4630 0 01 0 0113 0 0679 0 1244 0 1810 0 2376 0 2941 0 3507 0 4072 0 4638 0 02 0 0221 0 0774 0 1327 0 1881 0 2434 0 2987 0 3540 0 4093 0 4646 0 04 0 0424 0 0953 0 1483 0 2013 0 2542 0 3072 0 3602 0 4131 0 4661 0 06 0 0610 0 1118 0 1626 0 2134 0 2642 0 3150 0 3659 0 4167 0 4675 0 08 0 0781 0 1270 0 1758 0 2246 0 2734 0 3223 0 3711 0 4199 0 4688 0 1 0 0940 0 1410 0 1880 0 2350 0 2820 0 3289 0 3759 0 4229 0 4699 0 12 0 1087 0 1540 0 1993 0 2446 0 2899 0 3351 0 3804 0 4257 0 4710 0 14 0 1224 0 1661 0 2098 0 2535 0 2972 0 3409 0 3846 0 4283 0 4720 0 16 0 1351 0 1774 0 2196 0 2618 0 3041 0 3463 0 3885 0 4307 0 4730 0 18 0 1471 0 1879 0 2288 0 2696 0 3105 0 3513 0 3922 0 4330 0 4739 0 2 0 1582 0 1978 0 2373 0 2769 0 3165 0 3560 0 3956 0 4351 0 4747
22. 30 0 35 0 40 0 00 0 000 0 0236 0 0444 0 0625 0 0778 0 0903 0 1000 0 1069 0 1111 0 01 0 0049 0 0280 0 0484 0 0661 0 0809 0 0932 0 1027 0 1094 0 1135 0 02 0 0098 0 0324 0 0523 0 0696 0 0841 0 0961 0 1054 0 1120 0 1159 0 04 0 0191 0 0408 0 0599 0 0765 0 0904 0 1019 0 1107 0 1171 0 1208 0 06 0 0281 0 0489 0 0673 0 0832 0 0967 0 1076 0 1161 0 1221 0 1257 0 08 0 0369 0 0569 0 0746 0 0898 0 1027 0 1133 0 1214 0 1272 0 1306 0 10 0 0454 0 0646 0 0817 0 0962 0 1088 0 1188 0 1267 0 1322 0 1355 0 12 0 0534 0 0720 0 0887 0 1026 0 1146 0 1243 0 1320 0 1372 0 1403 0 14 0 0614 0 0794 0 0956 0 1089 0 1205 0 1298 0 1372 0 1422 0 1452 0 16 0 0692 0 0866 0 1018 0 1151 0 1262 0 1352 0 1424 0 1472 0 1500 0 18 0 0769 0 0938 0 1083 0 1212 0 1319 0 1406 0 1475 0 1522 0 1549 0 20 0 0843 0 1006 0 1148 0 1273 0 1377 0 1460 0 1527 0 1573 0 1598 10 0 4 0 35 0 3 N 0 25 0 2 0 15 0 1 0 05 Figure 1 Unconfined Wall f 300 MPa 0 04 0 06 0 12 0 14 0 16 Flexural Strength of Rectangular Masonry Walls with Uniformly Distributed Reinforcement Unconfined Wall f 300 MPa Unconfined Wall f 500 MPa 0 14 0 16 11 Figure 2 0 4 Flexural Strength of Rectangular Masonry Walls with Uniformly Distributed Reinforcement Unconfined Wall f 500 MPa Confined Wall f 300 MPa wt
23. adopted Consequently the value of Em 15 GPa is specified as a mean value rather than as an upper or a lower characteristic value 2 3 8 Ultimate Compression Strain a NZS 4230 1990 specified an ultimate compression strain for unconfined concrete masonry of amp 0 0025 This value was adopted somewhat arbitrarily in order to be conservatively less that the comparable value of a 0 003 which is specified in NZS 3101 for the design of concrete structures In the period since development of NZS 4230 1990 it has become accepted internationally based upon a wealth of physical test results that there is no evidence to support a value other than that adopted for concrete Consequently when using NZS 4230 2004 the ultimate compression strain of unconfined concrete masonry shall be taken as a 0 003 2 3 4 Strength Reduction Factors Selection of strength reduction factors should be based on comprehensive studies on the measured structural performance of elements when correlated against their predicted strength in order to determine the effect of materials and of construction quality The strategy adopted in NZS 4230 1990 was to consider the values used in NZS 3101 but to then add additional conservatism based on the perception that masonry material strength characteristics and construction practices were less consistent than their reinforced concrete equivalent In NZS 4230 2004 the strength reduction factors have been altered with respect
24. assumption of plane section response and distributed wall cracks results in sufficiently accurate design rules 4 2 1 First Cracking The moment corresponding to first cracking may be evaluated by Eqn 18 The formula is based on the flexural state at which one wall end decompresses and the other end compresses to a stress of twice the average masonry stress fm 69 _ P N 18 n 6 6 TET M Meere 19 e where by is the wall thickness Lw is the wall length Ver is the applied force at the top of the wall corresponding to the 1 cracking moment M and he is the effective wall height The deflection of the top of the wall d at should be based on the concrete masonry wall elastic properties and consists of a component due to shear deformation des and a component due to flexure 212 2 1 v P N 5 20 5 Embw der der der sh where Poisson s ratio may be taken as v 0 2 It should be noted that the shear deformation component desn can be of significant magnitude for squat walls under serviceability loads whereas for the ultimate limit state it becomes increasingly insignificant The curvature at 1 cracking can be calculated as follows 2 P 2 21 der 4 2 2 Maximum Serviceability Moment Typically at this serviceability limit state the applied lateral force has surpassed that necessary to initiate cracking at the base of the wall The
25. dimension hp Interior joints 11 4 2 3a of NZS 4230 2004 1600 mm 12 mm Therefore Dus 1600 133 gt 70 12 Exterior joints 11 4 2 5 hy 800 dbc 12 h Therefore P 800 67 This is about 4 shortfall of the requirement therefore bc Minimum horizontal dimension Interior joints 11 4 2 2b 1200 mm dbb 16 mm Therefore fic 21200 75 60 dy 16 Exterior joints 11 4 2 4 required cover Lg 10 100 20d 108 100 20 x 16 10 x 16 580mm lt h provided is 800 mm therefore OK Joint Shear Design The joints should be designed to the provisions of Section 11 of NZS 4230 2004 At level 2 the critical joints are 3 and 4 If there is doubt as to the critical joints then it is prudent to evaluate all joints An estimation of the joint shear force may be found by the appropriate slope of the moment gradient through the joint Paulay and Priestley 1992 Hence the horizontal shear Vj and vertical shear Vj at a joint are approximated by Veg hz M Vin R 2 hb V V hi M Mg Voort Viv hc 53 where M M and Mg are the moments at top bottom left and right of the joint and Vpr are the shears applied to the left and right sides of the joint from the beams and Vcg and Voas the shears applied to the top and bottom of the joint from the columns T
26. is permitted to either adopt capacity design principles or to use a simplified approach 3 7 3 3 In the simplified approach where limits are placed on building height the influence of material overstrength and dynamic magnification are accounted for by amplifying the seismic moments outside potential plastic hinge regions by an additional 5096 Eqn 3 3 and by applying the seismic shear forces throughout the structure by an additional 10096 Eqn 3 4 Consequently the load combinations become 2 Mo 1 5 and V gt Vo Voy 2Ve 2 5 Component Design An important modification to NZS 4230 2004 with respect to its predecessors is the use of a document format that collects the majority of criteria associated with specific components into separate sections This is a departure from earlier versions which were formatted based upon design actions The change was adopted because the new format was believed to be more helpful for users of the document The change also anticipated the next release of NZS 3101 to adopt a similar format and is somewhat more consistent with equivalent Standards from other countries particular AS 3700 2 5 1 Definition of Column Having determined that the design of walls beams and columns would be dealt with in separate sections it was deemed important to clearly establish the distinction between a wall and a column In Section 2 of the standard it is stated that a column is an element having a
27. is useful to relate section rotations per unit length i e curvature to corresponding bending moments As shown in Figure 7 a the maximum curvature ductility is expressed as m 1 1 where 4 is the maximum curvature expected to be attained relied on and is the yield curvature Mr 4 i TE Mah 14 fii gt 1 Cu X us ai E yr m N i 1 l al Lw _ 1 E Curvature 1 aa a Moment Curvature Relationship b First yield Curvature c Ultimate Curvature yl Figure 7 Definition of curvature ductility Yield Curvature For distributed flexural reinforcement as would generally be the case for a masonry wall the curvature associated with tension yielding of the most extreme reinforcing bar will not reflect the effective yielding curvature of all tension reinforcement identified as Similarly may also result from nonlinear compression response at the extreme compression fibre y 5 oo fy 2 w where sy fy Es and is the corresponding neutral axis depth Extrapolating linearly to the nominal moment M as shown in Figure 7 a the yield curvature is given as Mn mM 3 Maximum Curvature The maximum attainable curvature of a section is normally controlled by the maximum compression strain e at the extreme fibre With reference to Figure 7 c this curvature can be expr
28. length not greater than 790 mm and a width not less that 240 mm subject primarily to compressive axial load However the intent of Section 7 3 1 5 was that a wall having a length less than 790 mm and having a compressive axial load less than 0 1 Ag may be designed as either a wall or as a column depending on the intended function of the component within the design strategy recognising that the design criteria for columns are more stringent than those for walls 2 5 2 Moment Capacity of Walls Moment capacity may be calculated from first principles using a linear distribution of strain across the section the appropriate magnitude of ultimate compression strain and the appropriate rectangular stress block Alternatively for Rectangular section masonry components with uniformly distributed flexural reinforcement Tables 2 to 5 overpage may be used These tables list in non dimensional form the nominal capacity of unconfined and confined concrete masonry walls with either Grade 300 or Grade 500 flexural reinforcement for different values of the two salient parameters namely the axial load ratio N PmLwt or and the strength adjusted reinforcement ratio pfy f m or pfy Kf m 5 Masonry Standards Joint Committee 2002 Building Code Requirements for Masonry Structures and Specification for Masonry Structures ACI 530 02 ASCE 5 02 TMS 402 02 USA Charts produced from Tables 2 to 5 are also plotted which enable the user to
29. located closest to the neutral axis Check to see that this bar does indeed yield Es _ Em 100 _ 0 003 195 1 gt g x 95 1 0 00146 0 0015 therefore OK Now taking moment about the neutral axis Cn 2 e m 236 8 195 1 199 33 9 x 195 1 100 33 9 x 500 195 1 33 9 x 900 195 1 33 9 x 1300 195 1 33 9 x 1700 195 1 135 x 290 195 1 247 7 kNm Alternatively use Table 2 to establish flexural strength of the masonry wall As _ 5x113 1 p 0 00224 Lyt 140x1800 f p 0 00224 x 300 0 056 f 12 3 __ 135x10 fhLut 12 1800 140 25 From Table 2 Mn 0 04499 mew M 0 04499 x 12 1800 1 10 245 kNm 26 3 3 Out of Plane Flexure A 190 mm thick fully grouted concrete masonry wall is subjected to N 21 3 kN m and is required to resist an out of plane moment of M 17 kNm m Design the flexural reinforcement using fm 12 MPa and f 300 MPa Solution Axial load wee 25 0 0 85 Fe M Require Mn 2 gt 17 20 kNm m 0 85 is assumed that M M M where M is moment capacity due to axial compression load and M is moment t capacity to be sustained by the flexural reinforcement As shown in Figure 12 moment due to N NatAsfy ases Na 25x 103 82 1 0 851 1 0 0 85 x 12 x 109 2 45 mm I1 P m C 7
30. see Table 10 1 of NZS 4230 2004 Vm C4 where 33py 308 note that 0 00265 0 087 and 1 for beams gt Vm 0 087 1 0 24 16 0 87 Therefore Vs Vn Vm Vp and vp 0 gt Vs 1 06 0 87 0 0 19 MPa note that 1 0 for beams Clause 10 3 2 10 requires spacing of shear reinforcement placed perpendicular to the axis of component not to exceed 0 54 or 600 mm 51 Therefore maximum shear reinforcement spacing Smax 600 mm gt Try 200 mm and f 300 MPa A 300 Vs 190 x 200 2 TM A x 300 190 x 200 gt A 24 1 mm Use 200 crs ie 28 mm per 200 mm This is also the minimum area of reinforcement of 0 07 required by clause 7 3 4 3 of the standard Spandrels 2 3 V 27 04 2x77 2 181 4 kN Vin Aeon 0 75 3 gt pa 28 lt v bwd 190 0 8 1600 Vm C C2 Vbm fy where C4 33py 300 note that py 0 00265 C4 0 087 and 1 0 for beams gt Vm 0 087 1 x 0 2416 0 87 MPa Therefore Vs Vn Vm Vp gt Vs 0 99 0 87 0 note that v 0 Vs 0 12 Try s 200 mm and f 300 MPa 190 x 200 0 12 309 190 x 200 gt A 15 2 mm Therefore use R6 200 crs Design of Level 3 Spandrels The design of level 3 spandrels is similar to above and is not included herein 52 Beam Column Joints Check dimensional limitations Minimum vertical
31. sey where Qu 57 afm wB op In this equation AP should correspond to the actual tendon stress state at the displacement du It is emphasized that Eqn 57 is of idealised nature and simply attempts to relate the lateral displacement to the masonry strain state in the compression toe region at the wall state where initiation of strength degradation due to masonry crushing is anticipated to commence Eqn 56 assumes that the total rotation occurs at a height of h 2 above the wall base This is consistent with the current thinking for plastic hinge zone rotation for reinforced concrete masonry walls For evaluation of d it is acceptable to interpolate between the axial forces calculated at nominal flexural strength first tendon yield and overstrength relative to the displacements d and do as applicable with a maximum of N P The base shear corresponding to d can be based on Eqn 31 using the appropriate axial force or on interpolation between V V and V with a maximum of V 78 5 PRESTRESSED MASONRY SHEAR WALL Consider the wall shown in Figure 30 a It is assumed that the five storey wall is 15 m high 3 6m long 190 mm thick and prestressed with five high strength prestressing strands 140 mm Half height 20 series concrete masonry units 100 mm high are used in the plastic deformation zone regular 20 series masonry units are used elsewhere The wall self weight is calculated to be 225
32. strength according to NZS 4230 2004 is per definition achieved when the concrete masonry fails in compression at the strain equals 0 003 e Overstrength This strength corresponds to the maximum moment strength developed by the wall taking into account stress increase yield and strain hardening of the prestressing tendons At this stage large deformations are expected and the maximum concrete masonry strain is likely to have surpassed 0 003 Past the maximum wall strength the wall resistance gradually degrades until failure All of the above limit states generally need to be evaluated both immediately after prestress transfer and after long term losses Laursen P T 2002 Seismic Analysis and Performance of Post Tensioned Concrete Masonry Walls Doctoral Thesis University of Auckland 281pp 1 Laursen P T and Ingham J M 1999 Design of Prestressed Concrete Masonry Walls Journal of the Structural Engineering Society of New Zealand 12 2 21 39 68 4 2 Flexural Response of Cantilever Walls This section considers the flexural design of prestressed concrete masonry cantilever walls with unbonded prestressing tendons where the lateral force is assumed to be acting at the top of the wall or at some effective height he refer to Figure 23 For other structural shapes and loading configurations the formulae should be modified accordingly Note that the term tendon in the following sections refer to both prest
33. the use of unreinforced concrete masonry is excluded by the Standard The only permitted use of unreinforced masonry in New Zealand is as a veneer tied to a structural element Design of masonry veneers is addressed in Appendix F of NZS 4230 2004 in NZS 4210 2001 in NZS 4229 1999 and also in NZS 3604 1999 Timber Framed Structures Veneer design outside the scope of these standards is the subject of special design though some assistance may be provided by referring to AS 3700 Masonry Structures 2 2 Nature of Commentary Much of the information in NZS 4230 1990 was a significant departure from that contained in both previous New Zealand masonry standards and in the masonry codes and standards of other countries at that time This was primarily due to the adoption of a limit state design approach rather than the previous allowable stress method and because the principle of capacity design had only recently been fully developed Consequently NZS 4230 1990 Part 2 contained comprehensive details on many aspects of structural seismic design that were equally applicable for construction using other structural materials Since release of NZS 4230 1990 much of the commentary details have been assembled within a text by Paulay and Priestley For NZS 4230 2004 it was decided to produce an abbreviated commentary that primarily addressed aspects of performance specific to concrete masonry This permitted the Standard and the commentary
34. to their predecessors because 1 The manufacture of masonry constituent materials and the construction of masonry structures are governed by the same regulatory regimes as those of reinforced concrete 2 There is no measured data to form a basis for adoption of strength reduction values other than those employed in NZS 3101 for concrete structures and the adoption of corresponding values will facilitate designers interchanging between NZS 4230 and NZS 3101 3 The values adopted in NZS 4230 2004 are more conservative than those recently prepared by the Masonry Standards Joint Committee comprised of representatives from the American Concrete Institute the American Society of Civil Engineers and The Masonry Society 2 4 Design Philosophies Table 3 2 of NZS 4230 2004 presents four permitted design philosophies primarily based upon the permitted structural ductility factor Whilst all design philosophies are equally valid general discussion amongst designers of concrete masonry structures tends to suggest that nominally ductile and limited ductile response is most regularly favoured Taking due account for overall structural behaviour in order to avoid brittle failure mechanisms nominally ductile design has the advantage over elastic design of producing reduced seismic without requiring any special seismic detailing 2 4 4 Limited Ductile Design As outlined in section 3 7 3 of NZS 4230 2004 when conducting limited ductile design it
35. would be critical when earthquake force lt acting in Vg direction i e 565 5 mm Refer to Figure 22 for details Design of Tension Reinforcement in Horizontal Members In section 3 7 3 3 of NZS 4230 2004 there are two equations given that permit a simplified capacity design approach to be used However in this example it has been necessary to place a significantly larger quantity of vertical reinforcement than required i e member E F in order to satisfy spacing criteria This has resulted in a concern about relying upon these simplified expressions and instead a full capacity design is conducted below to establish the appropriate horizontal design forces 63 To estimate the maximum tension force in horizontal ties the flexural overstrength at wall base needs to be calculated 1 25Mhn provided lt Consider earthquake Ve as Figure 21 a Mn provided 3296 4 kNm The overstrength value is calculated as follow m 1 25Mn provided M M _ 1 25 x 3296 4 7 2095 1 97 Dynamic magnification factor For up to 6 storeys oy 0 9 40 i35 10 1 1 Hence the design force for Member C G K is calculated as follow 1 1 x 1 97 x 150 325 1 kN Therefore bAcfy 325 1 kN _ 32515103 1 0 500 650 2 mm Ack 1 0 see 3 4 7 and take f 500 MPa Try 4 DH16 804 y Consider earthquake Ve as in Figure 21 b Mh provided 387
36. 0 2004 Design of Reinforced Concrete Masonry Structures The purpose of this user guide is to provide additional information explaining the rationale for new or altered clauses within the new Standard and to demonstrate the procedure in which it is intended that the new Standard be used 1 1 Background The New Zealand masonry design standard was first introduced in 1985 as a provisional Standard NZS 4230P 1985 This document superseded NZS 1900 Chapter 9 2 and closely followed the format of NZS 3101 Code of practice for the design of concrete structures The document was formally introduced in 1990 as NZS 4230 1990 Since 1985 the Standard has been subject to significant amendment as a result of the publication of the revised loadings standard NZS 4203 1992 This latter document contained significant revisions to the formatting of seismic loadings typically dominating design for most New Zealand structures and is itself currently subject to replacement by the joint loadings standard AS NZS 1170 1 2 Related Standards Whilst a variety of Standards are referred to within NZS 4230 2004 several documents merit special attention e As noted above NZS 4230 2004 is the material design standard for reinforced concrete masonry and is to be used in conjunction with the appropriate loadings standard defining the magnitude of design actions and loading combinations to be used in design This has proven somewhat problematic as the former
37. 0 9 18019 0 124 0 022 MPa 190 x 0 8 x 5000 Vp Vp1 Vp2 Vp3 Vps 0 30 MPa gt Therefore the required shear reinforcement where 0 8 for a wall and the maximum spacing of transverse reinforcement 200 mm since the wall height exceeds 3 storeys Try f 300 MPa A x 300 0 38 0 8 190 x 200 A 60 2 mm 200 mm vertical spacing Therefore use R10 200 crs within plastic hinge region 78 5 200 mm spacing Outside Plastic Hinge Region For example immediately above level 2 1 5 1 64 x 60 50 40 30 443 kN Therefore 3 44310 _ 0 58 MPa bd 38 From 10 3 2 6 of NZS 4230 2004 Vm C4 f where 33p 15 13bars 113 y byd 300 13x113 300 190x0 8x5000 300 0 064 and 1 0 since hJL gt 1 0 Therefore Vm C41 C2 Vbm 0 064 1 0 24 16 0 85 MPa gt v Since Vm gt Vn only minimum shear reinforcement of 0 07 is required Take s 400 mm A 0 07 x 400 x190 53 2 Therefore use R10 400 crs outside plastic hinge region 39 3 7 Limited Ductile Wall with Openings The seismic lateral loads for the 2 storey masonry wall of Figure 16 are based on the limited ductile approach corresponding to 2 Design gravity loads both dead and live including self weight are 20 at the roof and 30 at levels 0 and 1 It is required to design the
38. 0618 0 0765 0 0882 0 0971 0 1029 0 1059 0 01 0 0049 0 0279 0 0480 0 0652 0 0795 0 0909 0 0995 0 1052 0 1079 0 02 0 0097 0 0322 0 0518 0 0686 0 0826 0 0937 0 1020 0 1075 0 1102 0 04 0 0190 0 0406 0 0593 0 0753 0 0886 0 0992 0 1070 0 1122 0 1146 0 06 0 0280 0 0487 0 0665 0 0818 0 0945 0 1045 0 1120 0 1168 0 1190 0 08 0 0367 0 0566 0 0735 0 0881 0 1002 0 1099 0 1169 0 1215 0 1235 0 10 0 0451 0 0641 0 0804 0 0944 0 1059 0 1152 0 1218 0 1261 0 1279 0 12 0 0534 0 0713 0 0871 0 1005 0 1116 0 1204 0 1267 0 1307 0 1324 0 14 0 0613 0 0783 0 0936 0 1064 0 1171 0 1255 0 1315 0 1353 0 1369 0 16 0 0690 0 0853 0 0999 0 1123 0 1225 0 1306 0 1363 0 1399 0 1414 0 18 0 0762 0 0922 0 1062 0 1181 0 1279 0 1357 0 1411 0 1445 0 1459 0 20 0 0832 0 0989 0 1124 0 1238 0 1332 0 1406 0 1459 0 1491 0 1503 Table 3 M for unconfined wall with f 500 MPa f Lt 2 Axial Load Ratio ELA 0 0 05 0 10 0 15 0 20 0 25 0 30 0 35 0 40 0 00 0 000 0 0235 0 0441 0 0618 0 0765 0 0882 0 0971 0 1029 0 1059 0 01 0 0049 0 0279 0 0480 0 0652 0 0794 0 0908 0 0993 0 1049 0 1076 0 02 0 0097 0 0322 0 0517 0 0685 0 0824 0 0934 0 1015 0 1068 0 1093 0 04 0 0190 0 0405 0 0591 0 0750 0 0881 0 0984 0 1059 0 1107 0 1128 0 06 0 0280 0 0484 0 0662 0 0813 0 0937 0 1033 0 1103 0 1147 0 1163 0 08 0 0365 0 0561 0 0731 0 0874 0 0992 0 1081 0 1147 0 1186 0 1199 0 10 0 0448 0 0635 0 0797 0 0934 0 1043 0 1129 0 1190 0 1225 0 1234 0 12 0 0528 0 0707 0 0862 0 0992 0 1096 0 1176 0 1233 0 1264 0 1271 0 14 0 0605 0 0777 0 09
39. 1990 While it is expected that the notes provided here will not address all potential queries it is hoped that they may provide significant benefit in explaining the most significant changes presented in the latest release of the document 2 1 Change of Title and Scope The previous version of this document was titled NZS 4230 1990 Code of Practice for the Design of Masonry Structures The new document has three separate changes within the title The word Code has ceased to be used in conjunction with Standards documents to more clearly delineate the distinction between the New Zealand Building Code NZBC and the Standards that are cited within the Code NZS 4230 2004 is intended for citation in Verification Method B1 VM1 of the Approved Documents for NZBC Clause B1 Structure The previous document was effectively intended to be used primarily for the design of reinforced concrete masonry structures but did not preclude its use in the design of other masonry materials such as clay or stone As the majority of structural masonry constructed in New Zealand uses hollow concrete masonry units and because the research used to underpin the details within the Standard almost exclusively pertain to the use of concrete masonry the title was altered to reflect this Use of the word reinforced is intentional Primarily because the majority of structural concrete masonry in New Zealand is critically designed to support seismic loads
40. 2 Storey 1 0 8 0 64 0 154 48 5 27 7 2 1 2 1 44 0 346 109 0 62 3 3 1 2 1 44 0 346 109 0 62 3 4 0 8 0 64 0 154 48 5 27 7 x 4 16 1 0 315 180 Table 10 Pier Shear Forces and Moments Parameter Units Pier 1 Pier 2 Pier 3 Pier 4 gt First Storey Vie kN 48 5 109 0 109 0 48 5 315 07 kNm 29 1 65 4 65 4 29 1 Ma kNm 67 9 152 6 152 6 67 9 Second Store Vie kN 27 7 62 3 62 3 27 7 180 7 kNm 16 6 37 4 37 4 16 6 Ma top Z kNm 27 7 62 3 62 3 27 7 bottom kNm 38 8 87 2 87 2 38 8 1 a Moments at critical pier i section Moments at spandrel centrelines pier i Note that in Table 10 the pier shear forces are used to establish the pier bending moments For instance the first storey bending moments of pier 1 are found from Ve x 5 48 5 29 1 kNm Spandrel moments and shears are found by extrapolating the pier moments to the pier spandrel intersection points then imposing moment equilibrium of all moments at a joint At interior joints the moments in the spandrels on either side of the joint are estimated considering equilibrium requirements by the assumption that the spandrel moment on one side of a joint centreline is equal to the ratio of the lengths of the adjacent span times the spandrel moment on the other side of the joint For example with regard to Figure 17b at joint 2 the beam moment to the left of the centreline may be expressed
41. 25 0 1047 0 1147 0 1223 0 1275 0 1303 0 1307 0 16 0 0680 0 0844 0 0986 0 1103 0 1198 0 1269 0 1318 0 1342 0 1344 0 18 0 0752 0 0910 0 1045 0 1157 0 1247 0 1315 0 1359 0 1381 0 1380 0 20 0 0823 0 0974 0 1104 0 1211 0 1297 0 1359 0 1400 0 1420 0 1417 Table 4 EM for confined wall with f 300 MPa Axial Load Ratio Ra 0 0 05 0 10 0 15 0 20 0 25 0 30 0 35 0 40 0 00 0 000 0 0236 0 0444 0 0625 0 0778 0 0903 0 1000 0 1069 0 1111 0 01 0 0049 0 0280 0 0484 0 0661 0 0810 0 0933 0 1027 0 1095 0 1136 0 02 0 0098 0 0324 0 0523 0 0696 0 0842 0 0962 0 1055 0 1121 0 1161 0 04 0 0191 0 0409 0 0599 0 0766 0 0905 0 1020 0 1108 0 1173 0 1211 0 06 0 0281 0 0491 0 0673 0 0833 0 0967 0 1078 0 1163 0 1224 0 1261 0 08 0 0369 0 0569 0 0746 0 0899 0 1029 0 1135 0 1217 0 1275 0 1311 0 10 0 0454 0 0645 0 0818 0 0964 0 1089 0 1191 0 1271 0 1326 0 1360 0 12 0 0537 0 0720 0 0888 0 1027 0 1149 0 1246 0 1323 0 1377 0 1410 0 14 0 0616 0 0794 0 0956 0 1090 0 1209 0 1302 0 1376 0 1428 0 1459 0 16 0 0692 0 0867 0 1021 0 1152 0 1267 0 1357 0 1428 0 1479 0 1509 0 18 0 0767 0 0939 0 1085 0 1214 0 1324 0 1412 0 1480 0 1530 0 1558 0 20 0 0841 0 1009 0 1149 0 1275 0 1381 0 1466 0 1532 0 1581 0 1608 Table 5 i for confined wall with f 500 MPa Ply Axial Load Ratio Kf wt 0 0 05 0 10 0 15 0 20 0 25 0
42. 4 6 kNm The overstrength value is calculated as follow 1 25 provided M _ 1 25 x 3874 6 2095 2 31 gt dow 1 97 when considering Ve 64 Dynamic magnification factor For up to 6 storeys 1 1 Hence the design force for Member N F J is calculated as follow 1 1 x 2 31 x 250 635 3 kN Therefore bAnufy 635 3 kN 3 apon take f 500 MPa 1 0 x 500 1270 6 mm Try 7 DH16 1407 4 Design of Shear Reinforcement It is assumed that shear forces are to be resisted by the bigger wall elements adjacent to openings such that only these elements require design of shear reinforcement For other part of the wall structure it is only required to satisfy 0 07 i e use R6 200 crs As Vau are typically negligible therefore dV 2 VE where 1 0 3 4 7 of NZS 4230 2004 Shear Design 1 Storey 300 kN _ 1 1x2 31x 300 1 0 762 3 Therefore Vi Check shear stress b 190 mm d 0 8 x 4600 3680 mm 3 ieee 1 09 MPa lt v 1 50 MPa for fm 12 MPa 190 x 3680 From Section 10 3 of NZS 4230 2004 Vn Vm Vp Vs Shear stress carried by Vm C2 Vom fy where 33p 1 Pw 300 9bars x D12 8bars x D16 2bars 020 bwd note that pw 3254 7 190 x 0 8 x 4600 0 0046 65 Therefore 0 15 and 0 42 4 1 75 3400 4600
43. 6 x 10 m 1 5 m and 1 08 is assumed The The use of he 2 3hy is an approximate presentation of moment and shear characteristic in a multi storey wall with a triangular distribution of lateral loads For specified lateral loads and storey heights the relationship may be accurately evaluated from h X h Vj Y V 79 confinement plate length is taken 2 L or about 800 mm It is assumed that the height of the plastic hinge zone is 0 076 x h 0 76 m the value of 0 076 was found experimentally by Laursen and the ultimate flexural strain is 0 008 taken from section 7 4 6 4 or Figure 7 1 of NZS 4230 2004 Solution Summary First tendon yield Overstrength 250 4 hy 15m 10 Nominal strength ly 3 6 by 0 19m 200 4 fm 18 MPa capacity aximum 2 Aps 700 mm PN Ji 0 ps z serviceability moment Drift ratio y 1 24 E 190 GPa X 1504 fj 1517 MPa 2 N 567 kN o P 831 kN 0 114 d fps 1187 MPa 100 4 fm 2 04 MPa First cracking h 0 76 m Emy 0 008 1 08 0 9K 0 972 ees 0 96 9 Prediction A Ultimate Displacement Capacity 0 50 100 150 200 250 300 Displacement mm Figure 31 Predicted in plane response Table 12 Predicted force and displacement First Maximum Nominal Ultimate First ten Wall over serviceability displacement cracking strength don yield strength moment capacity V
44. 7 214C 16 6 3 4 135 29127 4 654 7 29 1 4 1 485 109 0 109 0 a 485 103 87 6 5C I 6 51 103 8C 29 1 654 654 29 1 28 NE 30 EM 28 J b Seismic Loading E Figure 17 Forces and Moments for the 2 Storey Masonry Wall Forces Shears in kN Moment in kNm Axial Forces in parentheses Design of 1 Storey Piers Flexural Design Outer piers Outer piers are designed for the worst of Pier 1 and Pier 4 loading Since the piers have been chosen as the ductile elements the moments in Figure 17 are the design moments i e M Mq ME Pier 1 18 8 M May 2 0 29 1 27 1 kNm Note that Mg 2 0 kNm Therefore 2188 0 85 22 1 kN 43 27 1 gt 0 85 gt 31 9 kNm Dimensionless Design Parameters 3 _ 22 1 10 0 0091 fl wt 16 800 190 and 6 My _ 31 9x10 0 0164 16x800 x190 f From Figure 1 18 0 037 4 188 8 2 0 29 1 31 1 kNm Therefore _ 188 8 0 85 222 1 Mn gt M 25 31 1 0 85 gt 36 6 kNm Dimensionless Design Parameters Na 222 1 103 0 091 16 800x190 and _ 36 6x109 n_ 2 0 0188 124 16x 8002 x190 f From Figure 1 p lt 0 00 Pier 1 governs Now f Pan 0 037 for f 300 MPa and f m 16 MPa m
45. 83 9 182 227 242 248 253 kN d 2 9 10 8 41 2 124 158 310 mm AP 0 0 34 109 140 231 kN Table 12 and Figure 31 present the predicted wall in plane response with the base shear V lateral displacement d and tendon force increase AP related to the equivalent structure shown in Figure 30 b Material properties and wall dimensions are specified in Figure 31 Specific details on the calculation example may be found over page It is seen in Figure 31 that wall softening initiates between the maximum serviceability moment and the nominal strength limit states The wall ultimate displacement capacity is reached 83 mm after the nominal strength limit state The displacement at first tendon yield and wall overstrength is in this case only of theoretical interest 80 Solution calculations First cracking limit state _ 567 831 3 6 Eqn 18 Mer 6 839 kNm Eqn 19 Ver 2999 10 2 Eqn 20 em 2 56783110 2 1 0 2 567 831 _ 0 0029 m 3 14400x3 6 x0 19 5 14400 x 0 19 Maximum serviceability moment Eqn 24 2 04 x os 1 21 x 3 6 x 0 19 1820 kNm 21550 182 kN 2 Equos 2548010 tO tom 18 14400 3 6 5 14400x3 6x0 19 1000 Nominal strength First iteration using amp 0 114 Eqn 39 Ue 0 0117 m and Us 0 00384 m Eqn 40 AP 10 1 kN 8 5 kN 7 0 kN 5 5 kN 3 9 and AP 35 0 kN ei 0 004 m
46. Generally the tendon ultimate strain is of the order of 5 which would result in unrealistically high displacement Consequently concrete masonry failure is expected Confinement by the foundation is likely to increase the failure masonry strain beyond 0 003 As the extreme concrete masonry fibres fail there is a tendency for the compression zone to migrate towards the centre of the wall reducing the wall strength gradually Experiments at the University of Auckland have shown drift ratio capacities of 1 2 for prestressed grouted concrete masonry walls of various aspect ratios suggesting high displacement capacity It is noted that this limit state may occur before tendon yielding depending on the wall aspect ratio the prestressing steel area and the initial tendon stress fse The drift ratio or the drift angle is defined as the ultimate displacement d divided by the effective height ga 55 77 Figure 29 Vertical Strain Evaluation at Ultimate Displacement Capacity Evaluation of the extreme masonry strain at displacements beyond nominal flexural strength necessitates definition of a plastic hinging zone at the bottom of the wall Assuming that all lateral displacement at the top of the wall is due to rotation of the plastic hinge as shown in Figure 29 the masonry extreme fibre strain e can be related to the wall lateral displacement du y h d 3 and 0 h 56 C L
47. Mn 0 75 x 3395 7 2472 3 kNm gt M et Consider earthquake Vg as in Figure 21 b Tie A B bAasfy 77 3 Therefore A apfy 77 3 10 taking f 300 MPa 0 75 x 300 25 341 6 Try 4 D12 A 452 4 mm Tie M N 473 7 kN 473 7 x 10 Theref A MI me m taking f MP erefore MN 7 575 300 taking f 300 MPa 2105 3 mm Try 8 D16 and 2 020 A 2236 8 mm note that D20 is the maximum bar size allowed for 190 mm wide masonry wall Tie E F Use 5 D12 because member force would be critical when earthquake force acting in Ve direction i e 565 mm Refer to Figure 22 for details Check moment capacity at wall base Tension forces provided 452 4 x 300 135 7 kN Twn 2236 8 x 300 671 0 Ter 565 5 300 169 6 kN Therefore total compression force at Node including gravity load Tag Tun Tun Ter 135 7 671 0 671 0 169 6 o 0 75 652 1 kN Note that in the above calculation it is recognised that the vertical component of strut E N matches the force in tie M N 61 T R6 200 R6 200 1 Reg200 ft ft tT Li I I DIL LI L l LIII LLL 4 DH16 1 44 R6 200 4 DH12 1 R6 200 i I 4 DH16 5 DH12 f See 6 3 9 ofthe Standard for H minimum length of lap splices 7 DH16 11 FII FT H 5 D12
48. USER S GUIDE TO NZS 4230 2004 DESIGN OF REINFORCED CONCRETE MASONRY STRUCTURES New Zealand Concrete Masonry September 2004 Association Inc ACKNOWLEDGEMENT This document was written by Jason Ingham and Kok Choon Voon of the Department of Civil and Environmental Engineering University of Auckland The authors wish to acknowledge the role of Standards New Zealand and of the committee members responsible for drafting NZS 4230 2004 The authors wish to thank David Barnard and Mike Cathie for their assistance in formulating the design notes and in development of the design examples included in this guide Peter Laursen and Gavin Wight are thanked for their significant contributions pertaining to the design of unbonded post tensioned masonry walls It is acknowledged that the contents of this user guide and in particular the design examples are derived or adapted from earlier versions and the efforts of Nigel Priestley in formulating those design examples is recognised It is acknowledged that the strut and tie model in section 3 8 is an adoption of that reported in Paulay and Priestley 1992 DISCLAIMER This document is not intended as a substitute for professional engineering consulting services and it needs to be read in conjunction with NZS 4230 2004 Users of this document are advised that the information and design examples are provided in good faith and while all care has been exercised to ensure that the contents of the docum
49. an design levels From the given lateral forces the total overturning moment at 300 mm below the wall base is M 150 x 8 6 0 3 100 x 5 8 0 3 50 x 2 7 0 3 2095 kNm Whilst the use of strut and tie analysis is specifically endorsed in section 7 4 8 1 of NZS 4230 2004 no advice is given in section 3 4 7 for an appropriate value to be used in conjunction with the analysis At the time of preparing this guide the draft version of the next NZS 3101 has adopted the factor recommended in ACI 318 of 0 75 This corresponds to the factor used for shear and torsion which is consistent with the strut and tie procedure Consequently 0 75 is adopted here for use in strut and tie analysis of concrete masonry structure Design of Tension Reinforcement in Vertical Members The area of tension reinforcement required in vertical ties after considering the effect of axial loads can be evaluated as follows Nn T Ast x Tj Therefore QA gif Ti Ni 8 Figure 21 on page 60 shows the strut and tie model for the squat wall when both seismic and gravity loads are considered 8 Paulay and Priestley 1992 adopted the procedure of QA gif as this would result in a more conservative design 59 o w TEAM T 77 3 07 3 REM 112 3 150 Tie 15 xum TELE 15 100 kN gt lt 30 157 3
50. as length of spandrel 2 3 16 gel length of spandrel 1 2 PARES 16 Meo length of spandrel 2 3 pier centreline 17 length of spandrel 1 2 length of spandrel 2 3 moments at joint 2 More sophisticated analyses are probably inappropriate because of the deep members large joints and influence of cracking and shear deformations The resulting pier and spandrel moments and shears are plotted in Figure 17b Axial forces in the piers are found from the resultant of beam shear vertical equilibrium and these are presented in Table 11 Table 11 Revised Total Axial Load Pier Nova Ne 15 Storey 2 Storey 1 85 103 8 18 8 34 21 4 12 6 2 156 5 61 3 3 143 5 58 7 4 188 8 55 4 42 0 2 08 6 02 d2 D 08 02 8 5 sel 74 asd va CUT pea 616 8 5 0 9 8 2 0 9 8 5 6 6 62 o a 62 34 0C 60 0 7 60 0C 34 0C 0 8 0 1 0 1 0 8 1 3 0 42 8 14 1 4 B 42 3 0 4 T DSL b 4 1 Wh NU FB Cs 062 27 0 14 0 282 278 2 0 12 12 i 2 0 31 15 4 15 31 7 8500 7 150 0C 150 0C 85 0C 18 07 07 18 L 28 3 0 2 8 a Gravity Loading Qu 180 185 1801 1180 D 194 m 18 Q 20 1 214 ido 4 1 27
51. ck is assumed with a stress of 0 85f a 0 85 and an extreme fibre strain of 0 003 corresponding to the definition of nominal strength in NZS 4230 2004 for unconfined concrete masonry For confined masonry NZS 4230 2004 recommends using an average stress of 0 9 a 0 9K with fm based on unconfined prism strength and e 0 008 The corresponding moment M and lateral force V can be evaluated by simple equilibrium as shown in Figure 26 with the following equation et ce 2 31 2 2 2 where is the length of the equivalent ultimate compression block given by P AP N 32 In these equations AP accounts for the increase in tendon force that arises from the flexural deformation and accounts for the associated tendon force eccentricity Both AP and may initially be assumed to equal zero for simple use This approach is similar to the method used in NZS 3101 1995 A better estimate of the nominal strength may be obtained from Eqn 31 when taking into account the tendon force increase AP and the associated tendon force eccentricity is observed from Figure 26 that there is moment reversal near the top of the wall due to which results in reversal of curvature This effect is not taken into account below when calculating wall deformations because it has a negligible effect on the predicted wall behaviour at nominal flexural strength 72 5 hi
52. ed in New Zealand and abroad have successfully shown shear strength of reinforced masonry walls significantly in excess of that allowed by NZS 4230 1990 Consequently new shear strength provisions are provided in section 10 3 2 of NZS 4230 2004 As outlined in clause 10 3 2 2 Eqn 10 5 masonry shear strength shall be evaluated as the sum of contributions from individual components namely masonry Vm shear reinforcement vs and applied axial compression load vp Masonry Component v It has been successfully demonstrated through experimental studies that masonry shear strength Vm increases with fm However the increase is not linear in all ranges of fm but the rate becomes gradually lower as fm increases Consequently it is acceptable that Vm increases approximately in proportion to Eqn 10 6 of NZS 4230 2004 is a shear expression recently developed by Voon and Ingham for concrete masonry walls taking into account the beneficial influence of the dowel action of tension longitudinal reinforcement and the detrimental influence of wall aspect ratio These conditions are represented by the and C terms included in Eqn 10 6 of NZS 4230 2004 The Vim specified in table 10 1 was established for a concrete masonry wall that has the worst case aspect ratio of hJL gt 1 0 and reinforced longitudinally using grade 300 reinforcing steel with the minimum specified py of 0 07 7 3 4 3 For masonry walls that have aspect ratios o
53. ement be adequately anchored at both ends to be fully effective on either side of any potentially inclined crack This generally required a hook or bend at the end of the reinforcement Although hooking the bar round the end vertical reinforcement in walls is the best solution for anchorage it may induce excessive congestion at end flues and result in incomplete grouting of the flue Consequently bending the shear reinforcement up or down into the flue is acceptable particularly for walls of small width 30 3 5 Concrete Masonry Wall Ductility Considerations 3 5 a Neutral axis of limited ductile masonry wall Find the maximum allowable neutral axis depth for a limited ductile cantilever wall with aspect ratio of 3 The wall is reinforced with grade 500 reinforcement Solution 500 gt SUA i 200 103 y y 0 0025 Cy Assume E amp w 0000 c For the purpose of an approximation that will generally overestimate the yield curvature it may be assumed that me 0 001 This value would necessitate a rather large quantity of uniformly distributed vertical reinforcement in a rectangular wall in excess of 1 596 With this estimate the extrapolated yield curvature can be evaluated using Eqn 2 _ 0 0025 0 001 0 0035 Using Eqn 2 4 Lw Lw 4 4 0 0035 Using Eqn 9 gt o z g Eq 3 Using Eqn 11 3 18 for p 2 h Ly
54. ent are correct it is the sole responsibility of the user to determine that the engineering solutions offered are fit for the intended purpose The New Zealand Concrete Masonry Association and the Cement and Concrete Association of New Zealand accepts no liability arising from its use COPYRIGHT New Zealand Concrete Masonry Association Inc 2004 Except where the Copyright Act allows no part of this publication may be reproduced stored in any retrieval system in any form or transmitted by any means without prior permission in writing of the New Zealand Concrete Masonry Association Inc New Zealand Concrete Masonry Association Inc Level 6 142 Featherston Street Wellington P O Box 448 Wellington Telephone 0 4 499 8820 Facsimile 0 4 499 7760 Email admin cca org nz CONTENTS 1 INTRODUGC TIGON 2 erem eit nd eden RP 4 1 1 Backgrounder eim recede e pede 4 1 2 Related Standards oni Hu et ed die Re redes 4 2 DESIGN INOTES teinte dett atem es 4 2 1 Change of Title and Scope cedes dete e Ree n ce dee Feed dts 5 2 2 Nature ot Commentaly s a ce br iae eds 5 2 9 Material Strengtlis heit cen ERU De EHE Dod eae 5 2 9 1 Compression Strength Fioriere nemen 6 2 3 2 Modulus of Elasticity of Masonry 6 2 3 3 Ultimate Compression Strain cccccceeeeeeeseseseeeseseeeseeeeseeseseseseeeeeseeees 6 2 3 4 Strength Reduction Factors ssssssssseeeeneeemee
55. ers may elect to consider a more sophisticated loading pattern with horizontal loads apportioned within the wall based upon tributary areas rather than the simple lumped horizontal forces shown in Figure 19 a The concrete masonry wall shown in Figure 19 a is to be designed for the seismic lateral forces corresponding with an assumed ductility of 2 The relatively small gravity loads are approximated by a number of forces at node points given in Figure 19 a and the strut and tie model for the gravity loads is represented in Figure 19 b Wall width should be 190 mm and f m 12 MPa It is required to design the flexural and shear reinforcement for the wall e 7400 800 2000 1800 2000 800 15 20 15 150 kN Y io 1 z 7 E e i 2 amp 2 2 N 15 20 15 100 kN 2 1 5 4 k E 9 o o e e CD 20 40 20 50kN Y 1 5 F J g o o o 2 E E e e Be E E 40 20 20 Y NNN ae o 77778 7477777 7777 77877777 1 2 a E e 70 kN 120 kN 70 kN 70 kN 120 kN 70 kN a Wall Geometry and Loading Condition b Strut and Tie Model for Gravity Loading Figure 19 Limited Ductile 3 storey Masonry Wall with Openings Solution Level 3 150 P K 4 100 kN 250 Ou 1273 M 1213 27 kd El e p a 7y 7 THEA ZA VE 477 7 A M E I 300 0 KN 300 0 KN
56. essed as 4 15 Displacement and Curvature Ductility The displacement ductility for a cantilever concrete masonry wall can be expressed as A Ay A DOCE MEE LE 5 y y consequently HA 1 Ay Yield Displacement The yield displacement for a cantilever wall of height hy may be estimated as Ay 2 3 6 Plastic Displacement The plastic rotation occurring in the equivalent plastic hinge length L is given by 0 pbp y Lp 7 Assuming the plastic rotation to be concentrated at mid height of the plastic hinge the plastic displacement at the top of the cantilever wall is A 0 h 0 56 lh 0 5L 8 Substituting Eqns 6 and 8 into Eqn 5 gives bn by Lp hw 0 5L Ha 1 by h 3 L L 1 4 9 Aw 2hw Rearranging Eqn 9 1 10 3 Lp hw 2 Paulay and Priestley 1992 indicated that typical values of the plastic hinge length is 0 3 lt L Ly lt 0 8 For simplicity the plastic hinge length L may be taken as half the wall length Ly and Eqn 10 may be simplified to sis X 3 11 X1 L 4hw hi 2 2A where A is the wall aspect ratio hy Ly Reduced Ductility The flexural overstrength factor is used to measure the extent of any over or undersign 0 W flexural overstrength _ 12 moment resulting from loading S tan dard forces 12 bow
57. f 0 25 lt hJL lt 1 0 and or pw greater then 0 07 the v may be amplified by the and terms to give vm In order to guard against premature shear failure within the potential plastic hinge region of a component the masonry standard assumes that little strength degradation occurs up to a component ductility ratio of 1 25 followed by a gradual decrease to higher ductility This behaviour is represented by table 10 1 of NZS 4230 2004 Axial Load Component vp Unlike NZS 4230 1990 the shear strength provided by axial load is evaluated independently of Vm in NZS 4230 2004 Section 10 3 2 7 of NZS 4230 2004 outlined the formulation which considers the axial compression force to enhance the shear strength by arch action forming an inclined strut Limitations of lt 0 1f and lt 0 are included to prevent excessive dependence on v in a relatively squat masonry component and to avoid the possibility of brittle shear failure of a masonry component In addition the use of N when calculating v is to ensure a more conservative design than would arise using Np Shear Reinforcement Component v The shear strength contributed by the shear reinforcement is evaluated using the method incorporated in NZS 3101 but is modified for the design of masonry walls to add conservatism based on the perception that bar anchorage effects result in reduced efficiency of shear reinforcement in masonry walls when compared with the use of e
58. f the wall h which is significantly longer than h for multi storey building Aps is the area of the jth tendon and E is the elastic modulus of the prestressing steel It must be ensured that P AP does not exceed the tendon yield strength Iteration process for calculation of and calculate amp using Eqn 37 using AP 0 calculate and using Eqns 38 or 39 calculate AP X AP using Eqn 40 calculate amp using Eqn 37 using AP from 3 repeat steps 2 to 4 until convergence of amp n calculate M using Eqn 31 and d using Eqn 33 Oak WN gt 74 The masonry design codes BS 5628 1995 and AS 3700 1998 present formulae for calculating the tendon stress increase but are not applicable for in plane wall bending because they were developed for out of plane response NZS 3101 1995 recognises that the design tendon force for unbonded tendons will exceed the tendon force following losses Using the notation presented here the increase in tendon force is given by AP 70 MPa 41 100 i ae fos lt fpy and fps lt fse 400 MPa 42 ps where Aps is the total prestressing tendon area fy is the resulting average tendon stress corresponding to P AP fy is the tendon yield stress and fse is the tendon stress corresponding to P This equation seems to provide reasonable results but has not been validated for all wall configurations It would be prudent to assume a total tend
59. gation Ue tension end and shortening compression end us assuming a linear variation of vertical deformation across the wall top as shown in Figure 27 The following equations were established for unconfined and confined concrete masonry fh Unconfined 4 0152 2 375 0 835 38 3 3652 2 126 0 073 ate S 5 n Em fh Confined Ug 22 52 10 4 1 83 39 m us 1 6722 1 646 0 142 In these equations elongation is positive and shortening is negative It is clear that the tendon force increase due to vertical deformation will increase the axial load ratio Iteration using Eqns 38 or 39 is therefore needed to find AP ZAP such that the calculated axial force ratio at nominal flexural strength injected in the equations on the right hand side in fact corresponds to the calculated tendon force increase on the left hand side of the equations The effective total tendon force eccentricity relative to the wall centre line can be evaluated by where 40 and AP are the initial tendon force and tendon force increase of the j th tendon and y is the horizontal location of the j th tendon with respect to the wall centre line taken as positive towards the tension end of the wall The tendon vertical extension uj is defined in Figure 27 and L is the tendon length approximately the height o
60. he hy and h are the beam and column depths respectively where hj 0 9h and 0 9h The h and hj are approximate distance between the lines of action of the flexural compression found in the beams and columns on opposite sides of the joints Level 2 Joint Shear Design Joint 3 Horizontal Joint Shear Gravity induced joint shear 0 1 1 2 1 26 2 x 0 9 1 2 2 0 60 kN As illustrated here joint shear resulted from gravity loads is small Consequently gravity induced joint shear could be considered negligible in this instance Earthquake induced joint shear 37 4 65 4 572 82 4 x0 9x1 2 VE in 11 5 kN 0 9 1 6 Limited ductility design requires Vin Va oujn 2VE jh gt Vi 0 2 11 5 Gravity induced joint shear is considered negligible 23 0 kN Nominal shear stress in the joint Vin _ 23 0x10 boho 1901200 Vin 0 10 lt vg 0 45416 1 8 MPa Therefore OK From section 11 4 5 2 since beams remain elastic i e no hinging Vsh Vinh where Vinh 0 5 11 5 kN but need not be taken less than Vinh Vmbch where Vm bm fy and C4 33py 300 pw 0 00297 for 012 200 crs gt 0 098 1 0 for simplicity 54 Vm 0 098 1 x 0 2 0 22 x 416 0 88 MPa Therefore Vm 0 88 x 0 19 x 1 2 x 10 200 6 kN Hence Vsh 2302 200 6 lt ZERO 0 75 Therefore NO horizontal join
61. ily at the design of slender precast reinforced concrete walls but it would seem appropriate that any adopted criteria for reinforced concrete walls be applied in a suitably adjusted manner to reinforced concrete masonry walls Recognising that at the current time there is considerable engineering judgement associated with the design of slender walls the position taken by the committee tasked with authoring NZS 4230 2004 was to permit a maximum wall thickness of 0 05L where L is the smaller of the clear vertical height between effective line of horizontal support or the clear horizontal length between line of vertical support For free standing walls an effective height of twice that of the actual cantilever height should be adopted This 0 05L minimum wall thickness criteria without permitting relaxation to 0 03L in special low stress situations is more stringent than that provided previously in NZS 4230 1990 more stringent than that permitted in the US document ACI 530 02 ASCE 5 02 TMS 402 02 and more stringent than the criteria in the draft version of P 3101 currently in development Consequently designers may elect to use engineering judgement to design outside the scope of NZS 4230 2004 at their discretion The appropriate criteria from these other documents is reported in Table 8 below Table 8 Wall slenderness limits in other design standards Standard Limits NZS 4230 1990 Minimum wall thickness of 0 03L if
62. ionship limiting the neutral axis depth is presented in sections 2 7 2 and 3 4 An outline of the procedure for conducting a special study to determine the available ductility of cantilevered concrete masonry walls is presented in section 2 7 3 2 7 1 Neutral Axis Depth Neutral axis depth may be calculated from first principles using a linear distribution of strain across the section the appropriate level of ultimate compression strain and the appropriate rectangular stress block Alternatively for Rectangular section structural walls Tables 6 and 7 may be used These list in non dimensional form the neutral axis depth of unconfined and confined walls with either Grade 300 or Grade 500 flexural reinforcement for different values of axial load ratio N f mLwt or N Kf mLyt and reinforcement ratio pfy f pfy Kf m where p is the ratio of uniformly distributed vertical reinforcement Charts produced from Tables 6 and 7 are also plotted which enable the user to quickly obtain a value for c Ly given the axial load ratio Lut or and different value of or These charts are shown as Figures 5 and 6 Table 2 n for unconfined wall with f 300 MPa ftw pfy Axial Load Ratio d a fr 0 0 05 0 10 0 15 0 20 0 25 0 30 0 35 0 40 0 00 0 000 0 0235 0 0441 0
63. isplacement is inversely proportional to wall length This means that the basic presumption of the traditional approach to allocate lateral load to walls in proportion to their stiffness as a means to obtain simultaneous yielding of the walls and hence uniform ductility demand is impossible to achieve It was also shown by Paulay that the yield curvature of a structural wall is insensitive to axial load ratio As a consequence it is possible to define as a function of wall length alone The moments and shears in the piers can be found from the method suggested by Paulay This design approach assigns lateral force between piers in proportion to the product of element area An bwLw and element length Lw rather than the second moment of area of the section as would result from a stiffness approach i e the pier strength should be allocated in proportion to L2 rather than p Consequently the pier shear forces and moments are as summarised in Tables 9 and 10 9 Priestley M J N and Kowalsky M J 1998 Aspects of Drift and Ductility Capacity of Rectangular Cantilever Structural Walls Bulletin of NZNSEE Vol 31 No 2 pp 73 85 Paulay T 1997 Review of Code Provision for Torsional Seismic Effects in Buildings Bulletin of NZNSEE Vol 30 No 3 pp 252 263 41 Table 9 Pier Shear Forces Pier Length Lw L2 12 Ve KN WI m m si 1 Storey
64. kN and the additional dead load of the floors and roof amounts to 0 5 MPa at the base of the wall unbonded post tensioning i 5 4 5 3 E 2 x 2 0 1 LLLA ILL ULE 7 3 6 2 3 6 s a Prototype b Equivalent structure Figure 30 Post tensioned concrete masonry cantilever wall Solution Gravity load N Wall self weight additional dead load 225 kN 0 5 x 3600 mm x 190 mm 225 kN 342 kN 567 kN Calculations are performed on the equivalent single degree of freedom structure shown in Figure 30 b with an assumed effective height he 2 3 x hw 10m The tendons are placed symmetrically about the wall centre line at zero 200 mm and 400 mm eccentricities from the wall centre line the five strands are represented with one line in Figure 30 In the calculation the tendon elastic elongation capacity is based on the actual tendon length approximated as hy using an effective tendon elastic modulus of x h hy An initial tendon stress of 0 67fp is selected based on an estimated first tendon yield at a lateral drift of about 1 596 assuming that the wall rocks as a rigid body around the lower corners A total prestressing force of Aps x fps 700 x 1187 831 kN is found resulting in an initial axial load ratio of 0 114 fm 18 MPa Confinement plates are imagined embedded in the horizontal bed joints in the wall corners by the base over a height of 2 x h 2 x 0 07
65. lding given by 1 and Py Apsfpy 53 At this state it is assumed that the tendon closest to the flexural compression zone has reached its yield stress The resulting displacement can then be evaluated using the following equation which is similar to Eqn 45 Z 2 fpy fps h _ 54 Eps In this equation ey is the distance from the compression end of the wall to the closest tendon and 1 is the tendon stress in the same tendon at nominal strength It is noted that Eqn 54 is not appropriate if the closest tendon is located within the flexural compression zone i e lt and that if the tendon closest to the compression zone is near to the location of the flexural neutral axis unrealistically large values of d are calculated When all tendons are located near the wall centreline the wall yield strength coincides with the wall overstrength It can be argued for conservatism that the tendon yield stress f in Eqn 53 should be replaced with the tendon ultimate strength fpu in order to establish the maximum credible wall flexural strength It is however unnecessary to modify Eqn 54 accordingly because the tendon strain at ultimate strength is of the order of 596 and therefore not attainable in reality for walls of any geometry 4 2 6 Ultimate Displacement Capacity The ultimate displacement is limited by the strain capacity of the tendons as well as the crushing strain of the masonry
66. loading standard NZS 4203 1992 is currently being superseded by AS NZS 1170 with the seismic design criteria for New Zealand presented in part 5 or NZS 1170 5 Unfortunately release of NZS 1170 5 has encountered significant delay such that NZS 4230 2004 has been released before NZS 1170 5 The potential therefore exists for this matter to result in minor amendments to NZS 4230 2004 The issue is briefly addressed in the Foreword to NZS 4230 2004 NZS 4230 2004 is to be used in the design of concrete masonry structures The relevant document stipulating appropriate masonry materials and construction practice is NZS 4210 2001 Masonry construction Materials and workmanship NZS 4230 2004 is a specific design standard Where the structural form falls within the scope of NZS 4229 1999 Concrete Masonry Buildings Not Requiring Specific Engineering Design this latter document may be used as a substitute for NZS 4230 2004 NZS 4230 2004 is to be used in the design of concrete masonry structures general form is intended to facilitate consultation with NZS 3101 The design of concrete structures standard particularly for situations that are not satisfactorily considered in NZS 4230 but where engineering judgement may permit the content of NZS 3101 to indicate an appropriate solution 2 DESIGN NOTES The purpose of this chapter is to record and detail aspects of the Standard that differ from the previous version NZS 4230
67. nclosed stirrups in beams and columns As the shear strength provisions of NZS 4230 2004 originated from experimental data of masonry walls and because the new shear strength provisions generated significantly reduce shear reinforcement requirements sections 8 3 11 and 9 3 6 and Eqn 10 9 of NZS 4230 2004 must be considered to establish the quantity and detailing of minimum shear reinforcement required in beams and columns Voon K C and Ingham J M 2003 Shear Strength of Concrete Masonry Walls School of Engineering Report No 611 University of Auckland 20 2 9 Design of Slender Wall Slender concrete masonry walls are often designed as free standing vertical cantilevers in applications such as boundary walls and fire walls and also as simply supported elements with low stress demands such as exterior walls of single storey factory buildings In such circumstances these walls are typically subjected to low levels of axial and shear stress and NZS 4230 1990 permitted relaxation of the criteria associated with maximum wall slenderness in such situations Recently there has been considerable debate within the New Zealand structural design fraternity regarding both an appropriate rational procedure for determining suitable slenderness criteria and appropriate prescribed limits for maximum wall slenderness alternatively expressed as a minimum wall thickness for a prescribed wall height This debate has been directed primar
68. nding the characteristic masonry compressive strength fm From Eqn B 3 of NZS 4230 2004 fh fm 1 65 15 54 1 65x 1 59 12 9 MPa Note that the values for mean and standard deviation of strength used here for masonry units and for grout correspond to the lowest characteristic values permitted by NZS 4210 with a resultant fm in excess of that specified in table 3 1 of NZS 4230 2004 for observation types B and A Note also that these calculations have established a mean strength of approximately 15 MPa supporting the use of Em 15 GPa as discussed here in section 2 3 2 22 3 2 In plane Flexure 3 2 a Establishing Flexural Strength of Masonry Beam Calculate the nominal flexural strength of the concrete masonry beam shown in Figure 10 Assume the beam is unconfined fm 12 MPa and f 300 MPa 0 003 m 1 y D12 S EN 8 v N 012 140 a Cross section b Strain profile Figure 10 Concrete Masonry beam Solution Assume that both D12 bars yield in tension Therefore tension force due to reinforcement is 122 4 113 1 mm gt 2 113 1 x 300 67 85 kN Now consider Force Equilibrium Gin where Cm 0 85f mab gt 0 85f mab 67 85 kN 3 2187 8510 0 85f x 140 55 9 0 85 Check to see if the upper reinforcing bar indeed yields g _ Em 100 c _ 0 003 _ 55 9 gt g x 44 1 0 00237 gt 0 0015
69. nns 6 2 4 4 Design Philosophies n eed nOD PR P I 7 2 4 4 Limited Ductile KRURA ERRI 7 2 5 Component Design eo oret oe P Pie PR ER e i eyes 7 2 5 4 Definition of Column 7 2 5 2 Moment Capacity of Walls aseisti rri riia 7 2 6 Maximum Bar Diameters 8 2 Ductility Considerations eee 8 2 7 1 Neutral Axis 8 2 1 2 Curvature 5 2 4528 Ue 15 2 73 Ductility Capacity of Cantilevered Concrete Masonry Walls 17 2 7 4 Walls with Openings 19 2 8 Masonry In plane Shear 19 2 9 Design of Slender Wall tics een eee 21 3 DESIGN EXAMPLES 2 nette anes ah anes ee es 22 3 1 Determine f From Strengths of Grout and Masonry Units 22 3 2 Ineplane Plexure ede hiv eee od ee ited dian etd eae 23 3 2 a Establishing Flexural Strength of Masonry 23 3 2 b Establishing flexural strength of mason
70. on force increase of 75 times the result calculated by Eqn 41 when the prestressing tendons are approximately evenly distributed along the length of the wall Eqn 43 evaluates the resulting tendon eccentricity ei due to the total tendon force increase assuming that the tendon force increase AP acts at an eccentricity of L 6 and that the tendons are evenly distributed across the wall AP e _ 43 6 P AP Having calculated and e the nominal flexural strength M and corresponding displacement d can then be evaluated using Eqns 31 and 33 4 2 4 Yield strength Contrary to reinforced concrete walls the yield strength for unbonded prestressed walls is typically found at displacements beyond the displacement at nominal flexural strength Structural testing has consistently shown that the behaviour of unbonded prestressed walls loaded beyond the nominal strength is dominated by rocking as illustrated in Figure 28 Even for walls without specially placed confinement plates experimental observations consistently demonstrate that the wall is able to support compression strains far beyond 0 003 In Figure 28 the wall has rocked over by a displacement d corresponding to a rotation At this state it is assumed that the extreme tendon at the tension side of the wall yields resulting in a tendon strain increase of fps 44 ps where Eps the modulus of elasticity for the tendon steel and f
71. ps is taken as the tendon stress in the extreme tendon at nominal strength If a wall is displaced laterally beyond dy some reduction of prestress should be anticipated upon unloading Notably this does not mean that wall strength is permanently reduced because the tendons can be fully activated by subsequent loading excursions The wall rotation 6 can be related to the wall displacement increase at first tendon yield dj and the tendon strain increase in the following way 2 2 _ fps dy 0hg 45 ete Cte Ens 11 BS 5628 1995 Part 2 Code of Practice for use of Masonry Structural Use of Reinforced and Prestressed Masonry British Standards Institution London 2 AS 3700 1998 Masonry Structures Standard Association of Australia Homebush NSW Australia 75 Figure 28 Rocking Response where and it is assumed 0 85 for unconfined masonry and 0 96 for confined masonry In this equation e is the eccentricity of the extreme tendon at the wall tension side with respect to the compressive end of the wall The length of the compression zone c is calculated at the nominal strength based on Eqn 32 thus assuming that the wall rocks about an axis at the distance c from the extreme compression fibre in the wall As dy is considered as the displacement increment beyond d the stress state in the extreme tendon should rigorou
72. quickly obtain a value for pfy f m or pf Kfn given the axial load ratio or and the moment ratio t or M KfimLwt These charts are shown as Figures 1 to 4 On the charts each curve represents a different value for pfy fm or pf Kfm For points which fall between the curves values can be established using linear interpolation 2 6 Maximum Bar Diameters Whilst not changed from the values given in NZS 4230 1990 it is emphasised here that there are limits to the permitted bar diameter that may be used for different component types as specified in 7 3 4 5 8 3 6 1 and 9 3 5 1 Furthermore as detailed in C7 3 4 5 there are limits to the size of bar that may be lapped which makes a more restrictive requirement when using grade 500 MPa reinforcement Consequently the resulting maximum bar sizes are presented below Table 1 Maximum bar diameter for different block sizes Block size Walls and beams Columns mm 300 MPa f 500 MPa f 300 MPa f 500 MPa 140 D16 DH12 5 D10 3 DH10 190 020 DH16 3 D16 DH16 240 D25 DH20 2 D20 DH20 390 032 DH32 2 7 Ductility Considerations The Standard notes in section 7 4 6 that unless confirmed by a special study adequate ductility may be assumed when the neutral axis depth of a component is less than an appropriate fraction of the section depth Section 2 7 1 below lists the ratios c Lw for masonry walls while justification for the relat
73. re along the height of the wall at the maximum serviceability moment assuming plane section response The curvature varies from at the base to at the height her at which the 1 cracking occurs Between the heights he and he the curvature varies linearly between and zero It can be shown that the curvature varies linearly with the non dimensional crack length y as defined in Figure 24 Eqn 26 defines the non dimensional crack length at the base of the wall at the maximum serviceability moment again assuming 0 55 2124 gm 26 ra and Eqn 27 defines the resulting cracked wall height Me M her 27 71 The total displacement d of the top of the wall can then be calculated by integration along the wall height with the following result de desh 28 21 y h y deg he ha ral Ser he 1288 29 EmLwYe 1 Ye Ye 1 Ye 3 which may be approximated assuming 0 55 as fm foh defi T me IB and 12 1 v he V 30 5 30 d esh In Eqns 29 and 30 and de sn represent the flexural and shear deformations respectively At this flexural state it is assumed that the relatively small deformations of the wall do not result in significant tendon force increase or migration of the tendon force eccentricity 4 2 3 Nominal Strength At the ultimate limit state an equivalent rectangular stress blo
74. reinforcement for the wall based on the limited ductility provisions of NZS 4230 2004 using fim 16 MPa and f 300 MPa The wall thickness is 190 mm a 1 8 1 2 1 8 1 2 1 8 4 55 20 kN m 180 mq os 1 12 28 30 kN 135 mm pp ee ee ee 1 1 2 28 1 2 3 4 30 kN m o 08 IT EJ oup qtd 1 3 7 3231 4 9 91 SOR ORO ACRE RE RO RN ANO RX RN RX AE AE RX RX RX RX RX RX RO ANO RX RX ANC KOA RX ANO ANC RX RN ANO 237 0 5 0 5 10 4 Figure 16 Limited Ductile 2 Storey Masonry Wall with Openings Solution As the structure is 2 storeys high it may be designed for pier hinging or spandrel hinging as outlined in section 4 4 5 10 of NZS 3101 1995 Because of the relative proportions it is expected that pier hinging will initiate first and this behaviour is assumed below Consequently the piers are identified as potential hinging areas In accordance with section 3 7 3 3 of the standard the spandrels are required to be designed for 50 higher moments than design level moments with shear strength enhanced by 100 in spandrels and piers Axial load Assume each pier is loaded by the appropriate tributary area Axial load 1 storey Piers 1 and 4 Ne au 20 30 x 0 8 0 9 85 Piers 2 and 3 Noa 50 x 1 2 1 8 150 kN Axial load 2 storey Piers 1 and 4 Ne qu 20 x 0 8 0 9 34 kN Pie
75. ressing strands and bars P N V Wall Thickness he bw Lw SANAN 4 9 ENS M OON Figure 23 Definition of Wall The applied forces and loads represented by the symbols V M N and P used in the following equations are all factored loads calculated according to the applicable limit state as defined in the New Zealand loading standard NZS 4203 1992 The axial force N is due to dead and live loads P is the prestressing force initial force after anchor lock off or force after all long term losses and V is the applied lateral force due to lateral actions It is assumed that moment M only arises from lateral forces V i e permanent loads and prestressing do not introduce permanent moment in the wall Figure 23 shows the various definitions of wall dimensions and forces It is assumed for the flexural calculations that plane sections remain plane i e a linear strain distribution across the wall length This assumption enables analytical calculation of strength stiffness and displacement and implies distributed cracking up the wall height From laboratory wall tests it was observed that PCM wall flexural response was primarily due to rocking where a crack opened at the base and that distributed flexural cracking did not develop This type of rocking behaviour is a feature of prestressing with unbonded tendons Despite this discrepancy between theory and observation it appears that the
76. rs 2 and 3 Ne qu 20 x 1 2 1 8 60 Within this user guide pier refers to the part of a wall or column between two openings and spandrel refers to the deep beam above an opening 40 Dimensional Limitations Minimum thickness of piers by 190 mm L 1200 mm bw 190 _g 45 La 1200 This is more than the general seismic requirement of bw gt 0 075L cited by the standard 7 4 4 1 of NZS 4230 2004 Dimensional limitations of spandrels Spandrels at level 1 are more critical due to deeper beam depth Therefore bw 190 mm h 1600 mm and L 1800 mm Ln _ 1800 95 lt 20 b 190 and L h _ 1800 x 1600 zn 79 8 80 b2 190 The spandrels are within the dimensional limitations required by the standard clause 8 4 2 3 Determination of Seismic Lateral Forces 1 Storey Piers It is assumed that the spandrels are sufficiently stiff to force mid height contraflexure points in the piers The traditional approach of allocating lateral force to inelastically responding members in proportion to their assumed stiffness has been reported to commonly lead to significant errors regardless of whether gross stiffness or some fraction of gross stiffness is assumed This is because walls of different length in the same direction will not have the same yield displacement This can be illustrated by substituting Eqns 2 and 3 into Eqn 6 to give Mp Ey 4 ae y Mi x E 3 which indicates that the yield d
77. ry 24 3 3 JQutofPlane FleXUre eee 27 3 4 Design of Shear ener 28 3 5 Concrete Masonry Wall Ductility Considerations sse 31 3 5 a Neutral axis of limited ductile masonry 31 3 5 b Neutral axis of ductile masonry 31 3 6 Ductile Cantilever Shear Wall sse 33 3 7 Limited Ductile Wall with Openings ss mmm 40 3 8 Strut and tie Design of Wall with Opening ssseee mm 58 4 PRESTRESSED MASONRY nere tete teet ict es 68 4 1 Limit states 6 oa enne te eed a tee ua 68 4 2 Flexural Response of Cantilever Walls sse 69 4 2 1 First Grackirig ie eie ent ett deste der 69 4 2 Maximum Serviceability Moment seem 70 4 2 3 Nominal Strength anode fU e ete fe e dd 72 4 2 4 Yield strength s ni feine tp ee debe Io dee dta 75 4 2 5 Flexural Overstrength nedum o ee eere tee d 77 4 2 6 Ultimate Displacement Capacity 4 77 5 PRESTRESSED MASONRY SHEAR WALL 79 1 INTRODUCTION NZS 4230 is the materials standard specifying design and detailing requirements for masonry structures The latest version of this document has the full title NZS 423
78. s edge elongates a This edge shortens Figure 26 Wall Equilibrium at Nominal Flexural Strength The total lateral displacement dn is given by the sum of the flexural displacement and shear displacement dnsn corresponding to M and may be evaluated using Eqn 33 dn dag dag where 33 2 fatis Unconfined dg 2 3065 1 386 0 856 34 EmLw 2 fmh Confined dag 7 6385 5 4084 1 69 35 EmLw Y 1 F i i i VIDA DNI NAONNANA Figure 27 Wall Deformation at Nominal Flexural Strength 12 1 v h 36 SE ml wow 73 En 37 fm wbw Eqns 34 and 35 were developed using numerical integration and curve fitting and are thus of an approximate nature and are valid for axial load ratios of 0 05 to 0 25 The extreme fibre strain was taken as 0 003 for unconfined concrete masonry and 0 008 for confined concrete masonry Detailed information on derivation of these equations may be found in Laursen The total tendon force increase AP at e of 0 003 or 0 008 is difficult to evaluate for pre stressed walls with unbonded tendons because the tendon stress increase depends on the deformation of the entire wall between points of anchorage However the force increase or decrease in each tendon in the wall cross section may be evaluated based on the estimated wall end elon
79. sary to increase ductility The most convenient and effective way to increase ductility is to use a higher design value of fm for Type A masonry This will reduce the axial load ratio Nf mAg where and the adjusted reinforcement ratio p p12 f m proportionally From Figure 8 the ductility will therefore increase Where the required increase in cannot be provided a second alternative is to confine the masonry within critical regions of the wall The substantial increase in ductility capacity resulting from confinement is presented in Figure 9 A third practical solution is to increase the thickness of the wall In Figures 8 and 9 the reinforcement ratio is expressed in the dimensionless form p where for unconfined walls p 1 fm for confined walls Kfm f and K 1 p 25 fm Priestley M J 1981 Ductility of Unconfined Masonry Shear Walls NZNSEE Vol 14 No 1 pp 3 11 Priestley M J N 1982 Ductility of Confined Masonry Shear Walls NZNSEE Vol 5 No 1 pp 22 26 17 10 0 600 0 r 80070 F 20070 900 0 900 0 700 0 80070 20070 F 10070 10 0 6000 800 0 2000 900 0 9000 7000 00 0 c00 0 1000 12 fm p p 12 of Unconfined Concrete Masonry Walls for Aspect Ra
80. serviceability moment is limited by Me which occurs when the stress in the extreme compression fibre at the base of the wall has reached kfm as shown in Figure 24 For prestressed concrete symbol adopted in this manual is set out in Table A 1 of NZS 4230 2004 which is reproduced from Table 16 1 of NZS 3101 1995 with k typically ranging between 0 4 and 0 6 dependent on load category Lw a Masonry Wall yLw C Stress Distribution and Crack Length Figure 24 Maximum Serviceability Moment It is noted that Eqn 22 must be satisfied before use of the equations relating to the maximum serviceability moment can be applied though this requirement is generally fulfilled gt 2fm 22 70 The masonry is assumed to remain linearly elastic hence the masonry strain 4 corresponding to can be found from Ems ES 23 m By adopting 0 55 from load category IV infrequent transient loads it may be shown that the maximum serviceability moment can be calculated as Me m 3 Atm LA by fm 05 12 LA by Vehe 24 6 Kfz fh where V is the corresponding lateral force The corresponding curvature at the wall base is 2 f 2 pe m 25 Me 0 Moment Curvature Figure 25 Curvature Distribution at Maximum Serviceability Moment Ws DTS ASA SANSA Figure 25 shows the variation of moment and curvatu
81. sly be taken as fps however using fse initial tendon stress in unloaded state instead of fps in Eqn 45 generally results in little error Given 6 the force increase in the individual tendons can be calculated as 0 eg ej AP EsApsi foy fps Apsj PRX 46 he AP DAP yy 47 where is the location of the j th tendon with respect to the compression end of the wall A is the area of the j th tendon and AP is the total tendon force increase above that at M Note that Eqn 46 assumes linear variation of the tendon force increase with respect to the lateral location of the tendons The resulting moment increase My is then given by n ay n ay 2 gt 6 g me AB 48 j 1 2 j 1 2 where n is the total number of tendons along the length of the wall and the compression zone length at first yield may be calculated as N ay Sa 49 w Finally the yield moment M and displacement d can be evaluated as L a My 50 dy 2 dy dy 51 76 4 2 5 Flexural Overstrength The maximum credible strength of an unbonded prestressed wall may be evaluated by assuming that all tendons have reached their yield strength Consequently the flexural overstrength Mo may be evaluated as L Nep n 2 Vohe 52 where a is the length of the equivalent ultimate compression block and P is the total tendon force when all tendons are yie
82. t steel is required i e Aj 0 The horizontal shear is carried by the horizontal component of the diagonal strut across the joint Vertical Joint Shear Earthquake induced joint shear 69 5 4 74 6 22341080 hoa 6 ER 19 2 kN 0 9x1 2 Vn Vi 2Ve Gravity induced joint shear is considered negligible in this instance gt 38 4 Nominal shear stress in the joint _ _ 384x10 jv 0 13 v Therefore OK bchg 190x1600 V 27 where Vm 0 since potential plastic hinge regions are expected to form in the pier above and below the joint see 11 4 6 2 of NZS 4230 2004 Hence 7394 _0 51 2 0 75 and the total area of vertical joint shear reinforcement required RA 51 2x 10 f 300 170 7 mm Take f 300 MPa y Therefore use 6 R6 to give 169 6 mm 55 Joint 4 Horizontal Joint Shear Earthquake induced joint shear 16 6 29 1 1 x82 4x0 9x0 8 0 9 1 6 VE in 11 1 Limited ductility design requires Vn Vin 2VeEjn Gravity induced joint shear is considered negligible in this instance gt Vin 2x11 1 22 2 KN Nominal shear stress in the joint _ Vh _ 222x10 beh 190 800 0 15 MPa lt w From section 11 4 5 2 since beams remain elastic i e no hinging Vin Vsh n Vmh where V 0 5Vi 11 1 but need not be taken less than Vinh Vmbch where vq
83. therefore bar yielded Now taking moment about the neutral axis Mn Cm x c a 2 T x di c 67 85 x 55 9 47 5 2 33 9 x 100 55 9 33 9 x 290 55 9 11 6 kNm 23 Alternatively use Table 2 to establish flexural strength of the masonry beam p 2 2202 A 140 390 f pb 2000 222 ft 12 0 103 0 gt From Table 2 Ai 0 0451 mh t 0 0451 x 12 x 390 1x10 M 11 5 kNm 3 2 b Establishing flexural strength of masonry wall Calculate the nominal flexural strength of the 140 mm wide concrete masonry wall shown in Figure 11 Assume the wall is unconfined 12 MPa f 300 MPa and 115 kN 115 kN D12 D12 D12 D12 D12 _ 100 400 1800 l Figure 11 Solution Axial load at Base Na mu 115 135 gt 0 85 T T 0 85f 0 85 _ Concrete Masonry wall 135 kN 24 Assume 4 D12 yield in tension and 1 D12 yields in compression 2 Area of 1 D12 113 1 mm Therefore total tension force from longitudinal reinforcement gt T 4 113 1 x 300 135 1 kN and 113 1 x 300 33 9 Now consider Force Equilibrium Cn C T N where C 0 85f ab 0 85f pab 135 1 135 33 9 gt 0 85f mab 236 8 kN 3 0 85f x 140 1598 195 1 0 85 The reinforcing bar in compression is
84. tic strength for these constituent materials it follows that a default value of 12 MPa is appropriate for Observation This is supported by a large volume of masonry prism test results and an example of the calculation conducted to establish this value is presented here in section 3 1 2 3 2 Modulus of Elasticity of Masonry Em As detailed in section 3 4 2 of NZS 4230 2004 the modulus of elasticity of masonry is to be taken as Em 15 GPa This is only 60 of the value adopted previously Discussion with committee members responsible for development of NZS 4230P 1985 has indicated that the previously prescribed value of Em 25 GPa was adopted so that it would result in conservatively large stiffness resulting in reduced periods and therefore larger and more conservative seismic loads However this value is inconsistent with both measured behaviour and with a widely recommended relationship of 1000f representing a secant stiffness passing through the point f Em 0 001 on the stress strain curve Note also that application of this equation to 3 4 2 captures the notion that 12 MPa is the lower 5 characteristic strength but that 15 GPa is the mean modulus of elasticity This is quantitatively demonstrated here in section 3 1 It is argued that whilst period calculation may warrant a conservatively high value of Em serviceability design for deformations merits a correspondingly low value of Em to be
85. tio A Figure 8 18 0 004 0 005 0 006 0 007 0 008 0 009 0 01 0 004 0 005 0 006 0 007 0 008 0 009 0 01 0 001 0 002 0 003 0 001 0 002 0 003 Figure 9 Ductility of Confined Concrete Masonry Walls for Aspect Ratio A 3 2 7 4 Walls with Openings Section 7 4 8 1 requires that for ductile cantilever walls with irregular openings appropriate analyses such as based on strut and tie models shall be used to establish rational paths for the internal forces Significant guidance on the procedure for conducting such an analysis is contained within NZS 3101 and an example is presented here in section 3 8 2 8 Masonry In plane Shear Strength At the time NZS 4230 1990 was released it was recognised that the shear strength provisions it contained were excessively conservative However the absence at that time of experimental 19 data related to the shear strength of masonry walls when subjected to seismic forces prevented the preparation of more accurate criteria The shear resistance of reinforced concrete masonry components is the result of complex mechanisms such as tension of shear reinforcement dowel action of longitudinal reinforcement as well as aggregate interlocking between the parts of the masonry components separated by diagonal cracks and the transmission of forces by diagonal struts forming parallel to shear cracks More recent experimental studies conduct
86. to be produced as a single document which was perceived to be preferable to providing the document in two parts Consequently designers may wish to consult the aforementioned text or NZS 4230 1990 Part 2 if they wish to refresh themselves on aspects of general structural seismic design such as the influence of structural form and geometry on seismic response or the treatment of dynamic magnification to account for higher mode effects In addition care has been taken to avoid unnecessarily replicating information contained within NZS 3101 such that that Standard is in several places referred to in NZS 4230 2004 2 3 Material Strengths In the interval between release of NZS 4230 1990 and NZS 4230 2004 a significant volume of data has been collected pertaining to the material characteristics of concrete masonry This has prompted the changes detailed below Paulay T and Priestley M J 1992 Seismic Design of Reinforced Concrete and Masonry Buildings John Wiley and Sons New York 768 pp 2 3 1 Compression Strength f m The most significant change in material properties is that the previously recommended compressive strength value for Observation Type B masonry was found to be unduly conservative As identified in NZS 4210 the production of both concrete masonry units and of block fill grout is governed by material standards Accounting for the statistical relationship between the mean strength and the lower 5 characteris
87. ural Design Level 2 Spandrels Section 3 7 3 of NZS 4230 2004 requires gt 1 5 1 2 and 3 4 Design for the maximum moments adjacent to Joint 3 Mqy 42 kNm 74 6 kNm Therefore 4 2 1 5 74 6 116 1 kNm Note that beam depth 1 6 and N 0 M 116 1 136 6 kNm 0 85 Dimensionless Design Parameter 6 Mn _ 0 0176 finLt 16 16005 x 190 From Table 2 f p 0 037 fin re 0 037 x 16 0 00197 300 Therefore use D16 400 crs average p 0 00265 i e cells 1 3 6 and 8 from top See Figure 18 for details Spandrel 2 3 Design for the maximum moment of M 1 4 1 5 69 5 105 7 kNm adjacent to Joint 2 Therefore M To s 124 4 kNm 0 85 Dimensionless Design Parameter 6 Mn __ 1244x105 ooe fhL2 t 16 16002 190 50 From Table 2 fy m 0 034 x 16 gt p 300 0 0018 Therefore continue D16 400 crs right through Spandrel 2 3 Shear Design Level 2 Spandrels Design requirement gt V Ve 0 75 for shear Spandrels 1 2 and 3 4 V 27 8 2 82 4 192 6 kN adjacent to Joint 4 228 956 8 0 75 3 gt Vp 777 lt vg bud 190x0 8x1600 Since beams are assumed not to be hinging pier flexural demand was met therefore flexural capacity of spandrels has an additional reserve strength of 1 5Mp Consequently 0 24
88. verstrength Eqn 53 Py 5x140 x 1517 1062 kN 10624567 0 972 18 0 19 Eqn 52 1062 567 x 3 0490 2533 kNm 2953 253 kN 10 2 54 0 0412 10 0 310 190000 10 15 3 6 2 0 4 0 490 0 96 Ultimate displacement capacity First iteration 1 1 a ay 0 431 0 463 2 2 0 466 0 96 0 76 10 278 Eqn 57 dy 0 008 0 126 m 0 466 Second iteration Using d found in Eqn 57 interpolate between a and a to find c 0 463 0 431 ata 9 04314 x 0 126 0 041 0 96 0 76 x 10 Eqn 57 dy 0 008 0 124 gt OK 0 473 The wall strength at d is found by interpolation between nominal strength and first tendon yield limit states with respect to displacement _ Vu Ve L du dp 227 248 227_ 0 124 0 041 242 kN dy dn 0 158 0 041 83
89. wing earthquake excitation It is noted that the provided information is more comprehensive than will be required for most conventional designs and is included as background for the following example For additional information refer to research conducted by Laursen and Ingham at the University of Auckland 4 1 Limit states The flexural design procedure presented here is based on Limit State Design as outlined by NZS 4203 1992 which identifies two limit states namely the Serviceability limit state and the Ultimate limit state The flexural serviceability limit state for prestressed masonry is concerned with flexural strength stiffness and deflections The following flexural states represent the limiting flexural moments for a wall to remain elastic for uncracked and cracked sections First Cracking This limit state corresponds to the state when the extreme fibre of the wall decompresses the tensile strength of concrete masonry is disregarded e Maximum Serviceability moment At this cracked section state the compressive stress in the extreme compression fibre has reached its elastic limit set out by the standard as a stress limitation Reinforcement and concrete masonry remain elastic in this state The flexural ultimate limit state for prestressed masonry is primarily concerned with flexural strength Additionally for ductility purposes overstrength stiffness and deflections should be considered Nominal strength The nominal
90. x5 40x4 30x3 20x2 10 2730 kNm Require gt M Therefore M 2 M NS 2730 0 85 23211 kNm Axial load at Base N 6 x 150 900 kN N 5 gt 900 0 85 1058 8 33 Check Dimensional Limitations Assuming a 200 mm floor slab the unsupported interstorey height 2 8 m Pw _ 190 0068 0 075 L 2800 This is less than the general seismic requirement cited by the standard clause 7 4 4 1 However from Table 6 c lt 0 3L see Page 37 Hence the less stringent demand of bw gt 0 05 L n applies here clause 7 3 3 and this is satisfied by the geometry of the wall Flexure and Shear Design Dimensionless Design Parameters 6 7 0 0563 fubit 12x 5000 x 190 3 ana 1058 8 x 10 0 0929 12 5000x190 From Figure 1 and assuming f 300 MPa for flexural reinforcement P 0 04 m Therefore 0 0016 Check Ductility Capacity Check this using the ductility chart Figure 8 p12 0 0016 and Nn 0 0929 fh fA m mg Figure 6 gives 3 3 Actual aspect ratio A NI 3 6 Therefore from Eqn 14 3 3 3 3 1 x 3 6 1 3 0 lt u 4 assumed 3 6 Thus ductility is inadequate and redesign is necessary 34 Redesign for fm 16 MPa Note that this will require verification of strength using the procedures reported in Appendix B of NZS 4230 2004 Now new Dimensionless Design Parameters 3 1058 8x10

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