# Lecture 4 5 Urm Shear Walls

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- Classnotes for ROSE School Course in: Masonry Structures Notes Prepared by: Daniel P. Abrams Willett Professor of Civil Engineering University of Illinois at Urbana-Champaign October 7, 2004 Lessons 4 and 5: Lateral Strength and Behavior of URM Shear Walls flexural strength, shear strength, stiffness, perforated shear walls
- Existing URM Buildings
- Damage to Parapets 1994 Northridge Earthquake, Filmore 1996 Urbana Summer
- Damage Can Be Selective 1886 Charleston, South Carolina
- Damage to Corners 1994 Northridge Earthquake, LA
- Damage to In-Plane Walls 1994 Northridge Earthquake, Hollywood URM cracked pier, Hollywood
- Damage to Out-of-Plane Walls 1886 Charleston, South Carolina 1996 Yunnan Province Earthquake, Lijiang
- Likely Consequences St. Louis Firehouse 1999 Armenia, Colombia Earthquake
- 2001 Bhuj Earthquake
- Lateral Strength of URM Shear Walls
- URM Shear Walls Ref: BIA Tech. Note 24C The Contemporary Bearing Wall - Introduction to Shear Wall Design NCMA TEK 14-7 Concrete Masonry Shear Walls P 3 P b h i H 3 H i H 1 P i P 1 flexural tension crack flexural compression cracks V b M b diagonal tension crack
- URM Shear Walls Design Criteria (a) allowable flexural tensile stress: -f a + f b < F t F t given in UBC 2107.3.5 (Table 21 - I); F t = 0 per MSJC Sec. 2.2.3.2 pg. cc-35 of MSJC Commentary reads: Note, no values for allowable tensile stress are given in the Code for in-plane bending because flexural tension in walls should be carried by reinforcement for in-plane bending. where: F a = allowable axial compressive stress (UBC 2107.3.2 or MSJC 2.2.3) F b = allowable flexural compressive stress = 0.33 f´ m (UBC 2107.3.3 or MSJC 2.2.3) (b) allowable axial and flexural compressive stress: MSJC Sec. 2.2.3.1 and UBC 2107.3.4 unity formula:
- Allowable Tensile Stresses, F t MSJC Table 2.2.3.2 and UBC Table 21-I 40 25 68* 80 50 80* 30 19 58* 60 38 60* 24 15 41* 48 30 48* 15 9 26* 30 19 29* * grouted masonry is addressed only by MSJC all units are (psi) Direction of Tension and Type of Masonry Mortar Type Portland Cement/Lime or Mortar Cement Masonry Cement/Lime M or S M or S N N tension normal to bed joints solid units hollow units fully grouted units tension parallel to bed joints solid units hollow units fully grouted units
- URM Shear Walls Design Criteria (c) allowable shear stresses: UBC Sec. 2107.3.7 shear stress, unreinforced masonry: clay units: F v = 0.3 (f’ m ) 1/2 < 80 psi (7-44) concrete units: with M or S mortar F v = 34 psi with N mortar F v = 23 psi allowable shear stress may be increased by 0.2 f md where f md is compressive stress due to dead load Per UBC Sec. 2107.3.12 shear stress is average shear stress,
- URM Shear Walls Design Criteria (c) allowable shear stresses: MSJC Sec. 2.2.5.2: shear stress, unreinforced masonry: F v shall not exceed the lesser of: (a) 1.5 (f’ m ) 1/2 (b) 120 psi (c) v + 0.45 N v /A n where v = 37 psi for running bond, w/o solid grout 37 psi for stack bond and solid grout 60 psi for running bond and solid grout (d) 15 psi for masonry in other than running bond Note: Per MSJC Sec. 2.2.5.1, shear stress is maximum stress,
- URM Shear Walls Design Criteria (c) allowable shear stresses: f vmax f vavg for rectangular section
- URM Shear Walls Possible shear cracking modes. strong mortar weak units through masonry units Associated NCMA TEK Note #66A: Design for Shear Resistance of Concrete Masonry Walls (1982) low vertical compressive stress sliding along bed joints weak mortar strong units stair step through bed and head joints
- Example: URM Shear Walls Determine the maximum base shear per UBC and MSJC. 5000 lb. DL H H 9’- 4” 9’- 4” 6’ - 8” 8” CMU’s with face shell bedding block strength = 2800 psi Type N Portland cement lime mortar special inspection provided during construction Net section with face shell bedding: 80” 1.25”
- Example Forces and Stresses: Maximum base shear capacity per UBC shear stress flexural tensile stress
- flexural compressive stress Example Maximum base shear capacity per UBC Maximum base shear capacity per MSJC
- In lieu of prism tests, a lower bound compressive strength of 1861 psi will be used based on the 2800 psi unit strength and Type N mortar per MSJC Spec. Table 2.

- Example Maximum base shear capacity per MSJC flexural tensile stress flexural compressive stress shear stress tension compression axial and flexural stress UBC MSJC 7890 11,857 1194 V b max Summary: 10,629 11,934 794
- URM Shear Walls Post-Cracked Behavior h toe f m < F a e L/2 width = b heel H P [1] [2] [3]
- URM Shear Walls Note: shear strength should be checked considering effects of flexural cracking Post-Cracked Behavior Lateral Load, H Lateral Deflection at Top of Wall first flexural cracking resultant load, P, shifts toward toe toe crushing 2 to 3 times cracking load MSJC/UBC assumed behavior
- Perforated URM Shear Walls
- Lateral Stiffness of Shear Walls Cantilevered shear wall H L h
- Lateral Stiffness of Shear Walls Pier between openings H H h L
- Lateral Stiffness of Shear Walls 0.2 0 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 1.6 1.4 1.2 1.0 0.8 0.6 0.4 0.2 cantilever fixed pier
- References Associated NCMA TEK Note: 61A Concrete Masonry Load Bearing Walls - Lateral Load Distribution (1981) Associated BIA Technical Note: 24C The Contemporary Bearing Wall - Introduction to Shear Wall Design 24D The Contemporary Bearing Wall - Example of Shear Wall Design 24I Earthquake Analysis of Engineered Brick Masonry Structures
- Example: Lateral-Force Distribution Determine the distribution of the lateral force, H, to walls A, B and C. ? k i = 0.2776 bE m *based on cantilever action type of masonry and wall thickness is the same for each wall A 10’ 1.50 0.0556 bE m 0.20 10’-0” h=15’ A H 18’-0” B B 18’ 0.83 0.2077 bE m 0.75 C 6’-0” C 6’ 2.50 0.0143 bE m 0.05
- Lateral-Force Distribution to Piers Perforated Shear Walls h 3 L 1 L 2 H V 1 V 2 L 3 L 2 V 3 h 1 h 2 equilibrium: shear force attracted to single pier: overall story stiffness:
- Example: Lateral Force Distribution to Piers Determine the distribution of story shear, H, to each pier. A H 56” a 40” 112” 24” 64” 24” 7.63” V a Section A-A Elevation b 40” 32” V b A 8”grouted concrete block c V c
- Example: Lateral Force Distribution to Piers piers a and c 40” 7.63” 48” 7.63” 671 7501 pier b 64” 7.63”
- Perforated Shear Walls Axial Force due to Overturning f max f ai = ave. axial stress across pier “i” c y 1 y 2 y 3 p 1 p 2 p 3 y 1 y 2 y 3 y M [1] equilibrium of pier axial forces: [5] equilibrium of moments: [6] from similar triangles: substituting in [5]: [7] [8] [2] [3] [4]
- Perforated Shear Walls Axial Force due to Overturning [10] solving for f max : substituting in [6]: [11] [12] [13] distribution factor for overturning moment
- Perforated Shear Walls Design Criteria for Piers between Openings P P = P dead + P live + P lateral V V h M P M=V i h/2 flexure: reinforced piers flexure: unreinforced piers
- Perforated Shear Walls Design Criteria for Piers between Openings P V V h M P D+L D+L P max for small lateral load M=V i h/2 0.75(D+L+W/E) D+L+W/E P max and M max for large lateral 0.9D-0.75E 0.9D+E P min for smallest moment capacity D+W shear: unreinforced piers shear: reinforced piers UBC MSJC Sec. 2.1.1 Effect Loading Combinations
- Example: Perforated Shear Wall Check stress per the UBC for the structure shown below. Design pier reinforcement if necessary. Gravity Loads Level Dead Live 3 50 kip 80 kip Special inspection is provided f’ m = 2500 psi fully grouted but unreinforced Grade 60 reinforcement Type N mortar with Portland Cement 2 60 kip 80 kip 1 60 kip 80 kip total 170 kip 240 kip Earthquake Loads 14.9 kip 7.4 kip 10’-0” 10’-0” 9’-8” 14.9 kip 14.9 kip 18’-8”
- Example: Perforated Shear Wall 18’-8” Pier Dimensions 9’-4” 8” grouted concrete block 3’-4” 40” 32” 3’-4” 5’-4” 3’-4” 3’-4” 3’-4” 2’-8” 4’-0” 2’-8” 7.63” a b c
- Example: Perforated Shear Wall Stiffness of Pier “a” 7.63” 32” 7.63” 40” y a
- Example: Perforated Shear Wall Stiffness of Pier “b” b 7.63” 40”
- Example: Perforated Shear Wall Stiffness of Pier “c” c 32” 7.63” 40” (same as Pier a) b 1.43 E m 0.409 15.2 c 0.38 E m 0.109 4.0 ? k = 3.50 E m 1.000 37.2 k pier k i DF i V i a 1.69 E m 0.483 18.0 Distribution of Story Shear to Piers
- Example: Perforated Shear Wall 7.63” 40.0” 12.82” Distribute Overturning Moments to Piers pier A i y i A i y i a 549 12.8” 7038 a ? A i =1403 ? A i y i =160,807 124.0” b b 305 124.0” 37,820 c 211.2” c 549 211.2” 115,949
- Example: Perforated Shear Wall total story moment = M 1 (@top of window opening, first story) = 14.9k x 23.0’ + 14.9k x 13.0’ + 7.4k x 3.0’ = 558k-ft Distribute Overturning Moments to Piers b 305 -9.38 27 41 68 -2.9 -1.8 c 549 -96.58” 5120 76 5196 -53.0 -32.1 pier (in 2 ) A i (in) (1000 in 4 ) (1000 in 4 ) (1000 in 4 ) (kips) (1000 in 3 ) I a 549 101.8” 5689 76 5765 55.9 33.9
- Example: Perforated Shear Wall * based on tributary wall length: pier a: (32” + 40” + 32”)/288 = 0.361 (assuming that floor loads are pier b: (32” + 40” + 20”)/288 = 0.319 applied uniformly to all walls) pier c: (20” + 40” + 32”)/288 = 0.319 Summary of Pier Forces pier % gravity * P d P l P eq V eq M eq =V eq (h/2) (kips) (kips) (kips) (kips) (kip-in) a 0.361 61.4 86.6 33.9 18.0 432 b 0.319 54.2 76.6 -1.8 15.2 365 c 0.319 54.2 76.6 -32.1 4.0 224
- Example: Perforated Shear Wall Loading Combinations * UBC 2107.1.7 for Seismic Zones 3 and 4 axial compressive force, P moment, M shear, V d case 1 case 2 case 3 pier D+L 0.75(D+L+E) 0.9D-0.75E 0.75M eq 0.75V eq x1.5 * (kips) (kips) (kips) (kip-in) (kips) (in.) a 148.0 136.4 29.8 327 20.3 36 b 130.8 99.5 47.4 274 17.1 36 c 130.8 122.2 24.7 168 4.5 36
- Example: Perforated Shear Wall Axial and Flexural Stresses, Load Case 1 = D + L pier P D+L f a F a * f a /F a (kips) (psi) (psi) * F a = 0.25f’ m [1-(h/140r) 2 ] Note that conservative assumption is used for F a calculation, r is the lowest and h is the full height. a y a 148.0 270 543 0.497 < 1.0 ok b b 130.8 430 543 0.792< 1.0 ok y c c 130.8 239 543 0.440< 1.0 ok
- Example: Perforated Shear Wall Axial and Flexural Stresses, Load Case 2: 0.75 (D + L + E) * minimum S g is taken to give maximum f b for either direction of building sway ** F b = 0.33f’ m = 833 psi pier 0.75(P D+L+EQ ) f a =P/A F a f a /F a 0.75M e S g f b f b /F b ** f a /F a +f b /F b (kips) (psi) (psi) (kip-in) (in 3 ) (psi) a y a 136.4 249 543 0.459 327 2813 * 116 0.139 0.598 < 1.0 ok b b 99.5 326 543 0.600 274 2035 135 0.162 0.762 < 1.0 ok y c c 122.2 223 543 0.411 168 2813 * 60 0.072 0.483 < 1.0 ok
- Example: Perforated Shear Wall minimum axial compression: check tensile stress with F t = 30 UBC Sec 2107.3.5 Axial and Flexural Stresses, Load Case 3: 0.9D - 0.75P eq * minimum S g is taken to give maximum f b for either direction of building sway ** tensile stresses pier (0.9P D -0.75P EQ ) f a =P/A 0.75M eq S g f b - f a +f b (kips) (psi) (kip-in) (in 3 ) (psi) (psi) ** a y a 29.8 54 327 2813 * 116 62 > 30 psi provide reinf. b b 47.4 155 274 2035 135 -20 < 30 psi ok y c c 24.7 45 168 2813 * 60 15 < 30 psi ok
- Example: Perforated Shear Wall Pier Shear Stress, Load Case 4 : 0.75E * from Case 3 0.9P d -0.75P eq ** UBC 2107.3.7 pier V=0.75V eq x 1.5 f v = V/A web f ao = P/A * F v = 23 + 0.2f ao ** (kips) (psi) (psi) (psi) a y a 20.3 67 54 34 < 67 provide shear reinf. b b 17.1 56 155 54 < 56 provide shear reinf. y c c 4.5 15 45 32 > 15 ok
- Case Study: Large-Scale Test
- Georgia Tech Large-Scale Test photo from Roberto Leon 24’
- Final Crack Pattern slide from Roberto Leon Load Direction
- Final Crack Pattern Load Direction slide from Roberto Leon
- Results- Global Behavior Wall 1 Force-Displacement Response
- Overturning Effect (Vertical Stress) Base strains recorded during loading in the push and pull direction slide from Roberto Leon
- USA CERL Shaking Table Tests photos from S. Sweeney 12’
- Damage on North Wall Permanent offsets of 0.25” – 0.35” due to rocking of pier. Final Cracking Pattern slide from S. Sweeney
- Peak Force vs. Deflection slide from S. Sweeney
- End of Lessons 4 & 5