URM

1. Module 1:
Unreinforced Masonry
Building Assessment
Session 09: Face loaded walls and out-of-plane response

2. Background mechanics
 Modern assessment methods
are based on dynamic stability
of cracked wall (as opposed to
uncracked strength)
 The response is semi-rigid
rocking
 Semi-rigidity mainly due to
masonry crushing at pivots
 Wall displaced shape
described by two modes (rigid
diaphragm) or four modes
(flexible diaphragm)
Wall displaced shape ;
If diaphragm is rigid then only
Modes 21 and 22 exist

3. Face Loaded Walls
• Critical failure mechanism: most common failure, including parapets
• Assumption: There are sufficient header courses for wall to act as a
whole unit (This will be true for English bond).
…Otherwise consider each wythe acting independently
Four leaf thick wall without
connection between
central two leafs

4. Out-of-plane response is influenced by:
 Vertical span
 Wall thickness
 Overburden
 Diaphragm flexibility (support conditions)
These are the factors we can most
easily influence

5. Modification of NZ technique for Nepal
 NBC 105 Section 12: Secondary elements
 Spectral shape, C(T) has maximum of 0.08
 For walls, KP = 4 (note for ornaments KP= 8)
 Maximum of [CP(T) x KP x P] is 0.32 x P
 P is amplification with height. Value varies 1 ≤ P ≤ 2

6. KP in NBC 105

7. First the theory
 The theory is rather complicated, but tools have been developed
to make things easier
 First the theory
 Then the shortcut

8. Face Loaded Walls
• Step 1: Identify geometry
– Restraint points
– Wall height, h
– Thickness, t
h
t

9. Face Loaded Walls
• Step 2: Establish weight of the wall parts (W)
– Top & Bottom
– Finishes
– Veneer/cavity
• Loads above (P)
– Wall
– Floors, roof
– Determine P/W
P
W

10. Face Loaded Walls
• Step 3: Effective Thickness
– Teff = 0.98 Tnom (lightly loaded)
– Cement plaster (condition)

11. Face Loaded Walls
• Step 4: Assess Boundary Conditions - Walls
We will discuss this in more detail later

12. Face Loaded Walls
• Step 4: Assess Boundary Conditions - Parapets

13. Face Loaded Walls
• Step 5: Assess critical displacement
• Method is based on virtual work
• See Section 10, pages App-14 & 15
• Instability displacement

14. URM Parapet testing
 Note how parapet quickly
becomes unstable once rocking
initiates

16. Face Loaded Walls
• Step 6: Limiting displacement
(wall capacity)
• Experimental modelling and shake table
testing show that the wall becomes
increasingly difficult to reverse as it
approaches critical instability
• Adopt a ‘safety factor’ of 0.6:
∆m = 0.6 ∆i
 Max disp = 60% of instability disp
Displacement capacity

17. What period?
 Loadings standards require a natural period
 Rocking structures have no natural period
 Period is ‘amplitude dependent’

18. What period?
 Instead, we need to select a stiffness (and hence period) that
provides best representation of the actual path taken as rocking
occurs
 Increase stiffness (reduce period) by 60%

19. Face Loaded Walls
• Next establish demand on wall
• Step 7: Period of the wall
• ∆t = 0.6 ∆m = 0.36 ∆i
• Process relies on polar moment of inertia, but
design charts have been developed to avoid
having to do these calculations
• See Section 10 pg 10-94
• J is polar moment of inertia

20. Face Loaded Walls
• Step 7: Period of the wall
• How to tackle veneers (for cavity walls)

21. Face Loaded Walls
• Step 8: Design Response Co-efficient Cp(Tp)
• NZS 1170.5: Parts
– µp=1
– Ci(Tp) = Ch(Tp) Soil C spectra
– Tp ≥ 0.5 s
• CHi the level of input motion
This step modified for Nepal standard

22. Face Loaded Walls
• Step 9: Participation Factor
• This step relates the rocking wall to an
equivalent single degree of freedom oscillator
Text suggests that when in
doubt simply assume as
gamma = 1.5

23. Face Loaded Walls
• Step 10: Displacement Response
Double integration of accelerations
Angular frequency: ߱ ൌ
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‫݈݁ܿܿܣ‬ ൌ ߛ‫ܥ‬௣ ܶ௣ ܴ௣݃
‫ܦ‬௣௛ ൌ ߛ
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ଶగ
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‫ܥ‬௣ ܶ௣ ܴ௣݃ Displacement demand

24. NBC 105 Load combination
 Earthquake loads are multiplied by 1.25
 Demand increases by extra 25%

25. Face Loaded Walls
• Step 11: Express as percentage of the capacity
required to match the loadings standard

26. Face Loaded Walls – short cut
 Shortcut for Regular Walls – Appendix 10 C

27. What is the vertical axes plotting?
 The ‘Building Performance Ratio’ is the result of the virtual work
calculation for this particular:
 wall geometry
 vertical loading
 Boundary conditions
 Unfortunately, the method has the NZ spectra ‘built into’ the
charts

28. Correction factor
 To map the NZ spectra onto the Nepal spectra we need a
Correction Factor (CF):
 For 0 < Tp < 0.4, CF = 6.6
 For 0.4 ≤ Tp ≤ 1.0, CF = 7 T2 – 15 T + 11.5
 For 1.0 < Tp, CF = 3.7
 Corrected BPR = Chart BPR * CF
 Note: The 1.25E load combination has been built into the
correction

29. Correction factor

30. Correction Factor
For 0 < Tp < 0.4, CF = 6.6
For 0.4 ≤ Tp ≤ 1.0, CF = 7 T2 – 15 T + 11.5
For 1.0 < Tp, CF = 3.7

31. Check resonance

32. Example 1: Full code
 For your wall:
 Geometry
 Vertical loading
 Boundary conditions
 You obtain Chart BPR = 0.5
 You calculate Tp
 You calculate CF = 5.0
 Correct BPR = 0.5 x 5.0 =2.5
 If on ground in Kathmandu:
 Z=1, I=1, P=1
 Then wall is 250% of code

33. Example 2: Fraction of code
 For your wall:
 Geometry
 Vertical loading
 Boundary conditions
 You obtain BPR = 0.5
 You calculate Tp
 You calculate CF = 3.7
 Corrected BPR = 0.5 x 3.7 =1.85
 If Z=1.1 (highest seismic zone),
P=2.0 (highest part of building),
 Then wall is 1.85/1.1/2.0
= 84% of Nepal Code

34. Example 2: Fraction of code
 For your wall:
 Geometry
 Vertical loading
 Boundary conditions
 You obtain BPR = 0.5
 You calculate Tp
 You calculate CF = 3.7
 Corrected BPR = 0.5 x 3.7 =1.85
 If Z=1.1 (highest seismic zone),
P=2.0 (highest part of building),
 K=8 (double the demand because
of parapet)
 Then wall is 1.85/1.1/2.0/2.0
= 42% of Nepal Code

35. Example 3: Compare to NZ code
 To compare with NZ code, simply ignore correction
factor
 BPR = 0.5, Ch(0)=1.12, Z=0.38, P=1, therefore
115% of NZ Code (compare to 250%)
 NZ gets about half the score of Nepal
 BPR = 0.5, Ch(0)=1.12, Z=0.45, P=2, therefore
50% of NZ code (compare to 84%)
 Difference in score reduces for long periods

36. Boundary Condition Examples
Rocks about
hard edge

37. See top plot page App 31
Check that you have correct wall thickness
h/t = 4100/350
= 11.7
P/W≈1.25
h/t = 4100/350
= 11.7

38. Face Loaded Walls – short cut
 Basic Performance Ratio = 1.1
 Now work out:
 Wall period: See Section 10, page 10-99, Eq. 10.23
 h = 4.1, (P/W) = 1.25, Tp = (0.28x4.1/(1+2x1.25))^0.5 = 0.57 seconds

39. Corrected BPR
 BPR from chart = 1.1
 Period Tp = 0.57 sec
 Correction Factor = 7 x (0.57)2 – 15 (0.57) + 11.5 = 5.2
 Corrected BPR = 1.1 x 5.2 = 5.7

40. Face Loaded Walls – short cut
 Corrected BPR = 5.7
 Curve already accounts for C(Ti) and K
 Need to account for Z and I (if not equal to 1)

41. Face Loaded Walls – short cut
 Now just need to account for P:

42. Face Loaded Walls – short cut
 H is building height
 h is height to centre of wall
 For building sitting on ground, P= 1
 Kathmandu, Z=1
 Importance I=1
 Then: % of Nepal Standard is Corrected BPR / [ P x Z x I ]
 = 5.7 > 100% of NBC 105

43. NBC 105:1994 vs NZS 1170.5:2004

44. NBC 105:1994 vs NZS 1170.5:2004
Napier

45. Using NZS 1170.5:2004
 BPR = 1.1
 Ch(0) = 1.12
 Z = 0.38
 CHi = 2.125
 % = 1.1/1.12/0.38/1
= 2.6
 (about half the score given
by Nepal code)

46. Exercise 6: Exterior wall
 Establish the face load capacity of the top storey exterior wall
 Building located in Kathmandu

47. Exercise 6: Exterior wall
 Assumptions:
 Assume for now that all walls are appropriately tied to the diaphragms
 Ignore the presence of the concrete band beams
CF = 7T2 – 15 T + 11.5
% of code = BPR x CF / Z / I / P

48. Exercise 6: Exterior wall
 Analyse per meter length of wall
 Tributary weight of slab = (1.6 + 0.6) x 0.1 x 1 x 24 = 5.3 kN
 Weight of wall = 2.7 x 0.35 x 1 x 18 = 17.0 kN
 P/W = 0.3
 h = 2.7, t = 0.35, h/t = 2.7/0.35 = 7.7
Roof slab
Cross-section
view of wall

49. Exercise 6 – Page App-31, Fig a)
BPR=0.8

50. Exercise 6: Exterior wall
 Wall period: See Section 10, page 10-99, Eq. 10.23
 h = 2.7, (P/W) = 0.3, Tp = (0.28x2.7/(1+2x0.3))^0.5 = 0.7 seconds

51. Exercise 6: Exterior wall
 BPR=0.8
 Correction Factor = 7 * 0.7^2 – 15 * 0.7 +11.5 = 4.5
 Corrected BPR = 0.8 * 4.5 = 3.1
 Importance = 1, Zone = 1, P = 1+5/6 = 1.83
 Percent of Nepal Code = 3.1 / 1.83 = 168% of code

52. Parts period
Resonance!!!
Storey P (kN) w (kN) h (m) Tp (sec) T1/Tp
3 5.2 17 2.7 0.68 0.35
2 25 17 2.7 0.44 0.55
1 44.8 17 2.7 0.35 0.69

53. Exercise 6: Check resonance
 Need to check Part period versus main period
 In session 2 we calculated:
 Period in long direction: 0.24 sec
 Period in short direction: 0.28 sec
 Ratio = 0.28/0.70 = 0.4
 Hence no amplification

54. Exercise 6: Exterior wall (page 10-99)
 Check answer using the long method

55. Exercise 6: Exterior wall
 Maximum usable displacement = 0.6 t = 0.6 * 0.35 = 0.21 m
 Wall period: See Section 10, page 10-99, Eq. 10.23
 h = 2.7, (P/W) = 0.3, Tp = (0.28x2.7/(1+2x0.3))^0.5 = 0.7 seconds

56. Exercise 6: Exterior wall
 Spectral acceleration = 0.4g

57. Exercise 6: Exterior wall
 Participation factor: Assume maximum of 1.5
 Displ. demand = 1.5 * (0.7/2/pi)^2 * 0.4 * 1.83 * g = 0.128 m
 % of code = 0.21 / 0.13 = 160% of code
(approx. same answer as before)
 Check NZ code:
 BPR = 0.8, Z = 0.38, P=1+(5/6*3*2.7/6)=2.125, Ch(0)=1.12
 % = 0.8/1.12/0.38/2.125 = 88% (again, about half of Nepal code)

58. Exercise 7: Interior wall
 Establish the face load capacity of the top storey interior wall
 Building located in Kathmandu
CF = 7T2 – 15 T + 11.5
% of code = BPR x CF / Z / I / P

59. Exercise 7: Interior wall
 Tributary weight of slab = 3.2 x 0.1 x 1 x 24 = 15.4 kN
 Weight of wall = 2.7 x 0.23 x 1 x 18 = 11.2 kN
 P/W = 0.7
 h = 2.7, t = 0.23, h/t = 2.7/0.23 = 11.7
Roof slab
Cross-section
view of wall

60. Interior wall failure
 Example from Christchurch

61. Exercise 7 – Page App-28, Fig a)
BPR=0.7

62. Exercise 7: Interior wall
 Basic Performance Ratio = 0.7
 Period = sqrt [(0.28 x 2.7) / (1 + 2 x 0.7)] = 0.56 sec
 Check (0.28 sec / 0.56 sec) = 0.5 No Resonance of primary and local periods
 Correction factor = 5.2
 Corrected BPR = 5.2 x 0.7 = 3.64
 % of code = 3.64 / 1.83 = 200% of Nepal code

63. Exercise 8: Reanalyse exterior wall
assuming no intermediate wall ties
 Establish the face load capacity of the exterior wall
 Building located in Kathmandu

64. Exercise 8: Full-height exterior wall
 Assumptions:
 Assume for now that all walls are appropriately tied to the diaphragms
 Ignore the presence of the concrete band beams
CF = 7T2 – 15 T + 11.5
% of code = BPR x CF / Z / I / P

65. Exercise 8: Full-height exterior wall
 Analyse per meter length of wall
 Tributary weight of slab = (1.6 + 0.6) x 0.1 x 1 x 24 = 5.3 kN
 Weight of wall = 2.7 x 0.35 x 1 x 18 = 17.0 kN
 P/W = 0.3 = 0.3/3 = 0.1
 h = 2.7, t = 0.35, h/t = 2.7/0.35 = 7.7x3 = 23.1
Roof slab
Cross-section
view of wall

66. Exercise 8 – Page App-31, Fig a)
BPR=0.32

67. Exercise 8: Full-height exterior wall
 Wall period: See Section 10, page 10-99, Eq. 10.23
 h = 2.7 x 3 = 8.1 m
 (P/W) = 0.1, Tp = (0.28x2.7x3/(1+2x0.1))^0.5 = 1.37 seconds

68. Exercise 8: Full-height exterior wall
 BPR=0.32
 Correction Factor = 3.7
 Corrected BPR = 0.32 * 3.7 = 1.18
 Importance = 1, Zone = 1, P = 1
 Percent of Nepal Code = 1.18 / 1 = 118% of code

69. Exercise 9: Durber High School
 Should this wall have failed?

70. Exercise 9: Durber High School
 5 leaf thick:
 110 x 5 + 10 x 4 = 590 mm thick wall
 Wall is too thick for charts. Now need to
use the theory
 No overburden:
 P/W = 0
 Wall is a gabled vertical cantilever
extending from mid-height diaphragm

71. Exercise 9: Durbar High School
 Wall height about 6000 mm including gable, so h/t = 6000/590 = 10.2
 Assume P = 1 + (top storey mid height) / total H = 1.8

72. Exercise 9: Durbar High School
 Wall thickness too great for charts
 T= sqrt (0.65 x 6.0 x(1+ 0.59/6.0)^2) = 1.98 sec
 (t/h)2 = 0.01

73. Exercise 9 (see page 10-100)
 C(T=1.98) = 0.32
 Participation factor = 1.5 / [1 + 0.010] = 1.49
Cp(Tp) = Cd(Tp) x Z x I x P x (K?)

74. Exercise 9 (page 10-96 & 10-100)
 Gamma (Participation factor) = 1.49
 Period = 1.98 sec
 Cp(Tp) = 0.2 from spectra x P (Height factor = 1.8) = 0.36
 Z = 1, I = 1.5 for school
 Dph = 1.49 x (1.98/ 2pi)2 x [0.2 x 1.8] x 1.5 x 9.81 = 980 mm
 % = 30 X 590 / 980= 18% Wall should have failed!!!
 (see eq 10.26, page 10-100)

75. Exercise 10: In your own time, check this wall