2. Boundary Conditions
Boundary conditions control out-of-plane failure modes
Typically, the 1-way bending assumption is a lower bound estimate of
actual strength
But, it may be excessively conservative
Therefore 2-way bending needs to be considered
But first – why so many boundary condition cases in Section 10
appendices?
5. 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
BPR
=
1.05
6. What if we used the wrong chart?
Check that you have correct wall thickness
h/t = 4100/350
= 11.7
P/W≈1.25
h/t = 4100/350
= 11.7
BPR
=
0.3
7. What is the effect of the correct
vertical boundary conditions?
The correct boundary conditions are:
eb = +t/2 and ep = +t/2, giving BPR = 1.05 (Top of page App-31)
The worst possible mistake gives:
eb = 0 and ep = -t/2, giving BPR = 0.3 (Top of page App-31)
If you used the wrong boundary conditions you could mistakenly
assume the wall is weaker than it is:
0.3/1.05 = 30% or in reverse, 1.05/0.3 = 350% stronger than it is
11. Boundary Condition Examples
Rocks on outside
of lower wall
Rocks on centre. Roof and parapet unable to provide
rigid restraint. Parapet and roof loads combine
roughly at centre, or evaluate a combined eccentricity
13. Boundary Condition Examples
Rocks on inside
of lower wall
Rocks on centre of parapet
Empirical evidence
from Chch says
this failure is less
likely than previous
slide
Possible
influence
of wall
linings?
18. Boundary Condition Examples
Single leaf cavity has no rigid support at top.
Inner leaf load‐bearing so outer leaf
probably critical
Rocks on rigid diaphragm
20. Boundary Condition Examples
Full length cavity with
Proper ties and mid‐height
Wall‐diaphragm anchorage (not drawn)
Should achieve composite action
Rocks on outside of lower wall
24. Boundary Condition Examples
Need to evaluate
weight of tributary roof
compared to weight of
parapet to establish
which governs
Also need to check
cantilever overturning
mode
26. Boundary Condition Examples
Could be either. Need to satisfy that
previous strengthening is sound
plus consider overburden versus
thrust force of parapet. Check both?
28. Boundary Condition Examples
Overburden
on inner edge
for both
rocking directions
Would need to consider
response of far‐side wall
at same time, assuming
roof support is moving
31. Beware possible cantilever action
Note that gable ended walls have failed mid-way up building
height
Perhaps check multiple possible failure scenarios
36. Effect of inter-storey drift
Differences most detectable for low P and high h/t
0.08
0.13
This is still a topic being discussed currently
37. FEMA, ASCE 41-06
Procedure is popular because of its simplicity
Sharif et al. (2007) from UBC: Using principles of rocking
mechanics; suggests that ASCE (2007) is conservative
Pass
Versus
No pass
39. What is wrong with our vertical one-
way bending assumption?
1. We don’t actually have diaphragm anchorages in most cases
2. We have overlooked the end walls
Are these end walls reliable to stabilise the face loaded wall?
50. Kinematics of rigid bodies
The Italian code has a method for
‘macro-elements’ where rigid body
motion can be conceived
The method is based on the
principles of Virtual Work
For a rocking cantilever, this
method reduced to an equation of
moment equilibrium
51. Consider a rational design process
English bond overlap = 240 / 4 = 60 mm
Reduce for mortar head joint, and poor workmanship
Say effective average overlap = 60 – 10 – 10 = 40 mm
Page 10-79
52. Consider a rational design process
Average vertical wall stress
= 0.5 x height x density
Assume density = 18 kN/m3
Assume height is 16.0 m,
thickness = 0.98 m
Assume 0.9 DL + 1.25 E
Average axial stress =
0.9 x 18 x 16/2 = 130 kPa
Vertical stress distribution
on wall
53. Nepali baked brick, mud mortar
Property Range Recommended
f’m Masonry compressive strength 3-13 MPa 4.2 MPa
f’j Mud mortar compressive strength 0.2 MPa 0.2 MPa
f’b Baked brick compressive strength 5-15 MPa 10 MPa
Em Elastic (Young’s) Modulus of
masonry
10-20 GPa 10 GPa (small ε)
1.3 GPa (large ε)
c Mud mortar cohesion 0.0 - 0.2 MPa 0.05 MPa
μf Mud mortar coefficient of friction 0 - 0.6 0.3
ρ Density 1700-2100 kg/m3 18 kN/m3
ν Poisson's ratio 0.15-0.35 0.25
NBC 2002 : 1994
Houben and Guillard, 1994
Niker Research project - Deliverable 4.5, Vertical elements
Brick sector in Nepal - Overview and policy issues – Pandit 2013
54. Cohesion and friction
We assume that:
Fh = (0.5c + mu * vert stress)
Fcohesion = 0.5 x 0.05 x A =
0.025A
A is area of 1 shear plane x
number of shear planes
Consider moment capacity
Cohesion
Friction
Assume soft ground and no restoring moment due to self weight
56. NBC 109 reports standard brick size
Masonry is is a modular material
Dimensions are scales of the basic module size
57. Moment due to cohesion
Height = 16,000 mm
Height of brick approx 70 mm (including joint)
Shear plane = 0.98 m x 0.04 m = 0.039 m2 ( = 393 cm2)
Number of shear planes = 16000 / 70 (brick height) = 229 planes
Total cohesion force = 0.025 x 0.039 x 229 = 224 kN
Moment due to cohesion F = 224 x (16/2) m = 1796 kNm
58. Moment due to friction
Average axial stress = 0.9 x 18 x 16/2 = 130 kPa
Shear plane = 0.98 m x 0.04 m = 0.039 m2 ( = 393 cm2)
Number of shear planes = 16000 / 70 (brick height) = 229 planes
Total friction force = (0.3 x 130) x 0.039 x 229 = 349 kN
Moment due to friction F = 349 x (16/3) m = 1862 kNm
Total moment capacity = 1796 + 1862 = 3657 kNm
59. What about vertical acceleration?
Assume 20% reduction
in resistance due to
loss of gravity
60. Moment due to friction
Average axial stress = 0.9 x 18 x 6/2 = 48.6 kPa
Shear plane = 0.98 m x 0.04 m = 0.039 m2 ( = 393 cm2)
Number of shear planes = 16000 / 70 (brick height) = 229 planes
Total friction force = (0.3 x 130) x 0.039 x 229 = 349 kN
Moment due to friction F = 349 x (16/3) m = 1862 kNm
Total moment capacity = 1796 + 1862 = 3657 kNm
Reduced moment capacity = 80% x (1796 + 1862) = 2926 kNm
61. Determine period of cantilever
h = 16.0 m, t = 0.98 m
Cantilever period = 3.25 seconds
66. Capacity / demand
Capacity = 2926 kNm
Demand = 2352 kNm
Ratio is that wall has 119% of required strength.
So this explains why most did not fail
But consider:
20% drop in cohesion: Ratio drops to 107%
20% drop in cohesion (poor mud mortar), 20% drop in friction, 25%
increase in shaking at site
Ratio drops to only 76% of code
67. Capacity / demand
Capacity = 2926 kNm
Demand = 2352 kNm
Ratio is that wall has 119% of required strength.
So this explains why most did not fail
But consider:
20% drop in cohesion: Ratio drops to 107%
20% drop in cohesion (poor mud mortar), 20% drop in friction, 25%
increase in shaking at site
Ratio drops to only 76% of code
However, do recall that I also ignored
selfweight as a restoring force.
If rocking about an edge then the wall
would have a much greater restoring force
68. Capacity / demand
Capacity = 2926 kNm
Demand = 2352 kNm
Ratio is that wall has 119% of required strength.
So this explains why most did not fail
But consider:
20% drop in cohesion: Ratio drops to 107%
20% drop in cohesion (poor mud mortar), 20% drop in friction, 25%
increase in shaking at site
Ratio drops to only 76% of code
However, do recall that I also ignored
selfweight as a restoring force.
If rocking about an edge then the wall
would have a much greater restoring force
Note that the Italian method typically assumes zero tensile
strength of masonry, so self weight restoring is a factor
that is important to consider, even if then discounted
70. Collapse Mechanisms
for churches
Much work done in Europe on
this ‘macro element’ concept
These procedures should be
directly relevant to Rana
Palaces and other similar
structures
Italians recommend a 1.35
‘safety factor’ associated with
imperfect knowledge
76. Learning lessons
Boundary conditions significantly influence results
If needed, you may need to revisit the site to check assumptions
1-way bending is a lower bound value, but only applies when
diaphragm anchorages exist
2-way bending generates demand on end walls
Macro-element analysis is widely used in Italy for modelling
palaces and other monumental buildings
English bond particularly vulnerable to end wall separation
Sensitivity study is an very useful exercise