Dan Abrams + Magenes Course on Masonry

1. Seismic design and assessment of
Seismic design and assessment of
Masonry Structures
Masonry Structures
Lesson 14, continued
October 2004
Masonry Structures, lesson 14 part 2 slide 1
Example of assessment with rigid-body mechanism analysis
Out-of-plane failure of gable walls
1.5 m
6.0 m
9.0 m
6.0 m
Masonry Structures, lesson 14 part 2 slide 2

2. 1.50 cm
o
1.5 m
6.0 m
6.00 m
0.40 m
9.0 m
6.0 m
Masonry Structures, lesson 14 part 2 slide 3
P
1.50 cm
αP' ∆x1
o
1.50 m
∆x0
αW
6.00 m
0.40 m
W
.50 m
∆y0 O
.40 m
Masonry Structures, lesson 14 part 2 slide 4

3. P
weight/m2 of roof (including truss structures):
αP' ∆x1
1.4 kN/ m2
weight of masonry: 20 kN/m3
assume each gable wall carries 1/3 of vertical
weight carried by top ridge beam, P = 26 kN
P’=1/2 of weight = 1.5 P = 39 kN
1.50 m
∆x0 centre of mass of gable wall: 1/3 of height
αW
W assume P applied at centreline of wall
.50 m
Direct moment equilibrium at incipient rocking:
∆y0 O α 0 (1.5 ⋅ P'+0.5 ⋅W ) − 0.2 ⋅ P − 0.2 ⋅W = 0
0.2 ⋅ P + 0.2 ⋅W
.40 m α0 = = 0.162
1.5 ⋅ P'+0.5 ⋅W
Masonry Structures, lesson 14 part 2 slide 5
P or, through application of PVW:
αP' ∆x1
α 0 P '⋅∆ x1 + α 0W ⋅ ∆ x 0 − P ⋅ ∆ y 0 − W ⋅ ∆ y 0 = 0
∆ x 0 = θ ⋅ 0.5; ∆ x1 = θ ⋅1.5; ∆ y 0 = θ ⋅ 0.2
α 0 P'⋅1.5θ + α 0W ⋅ 0.5θ − P ⋅ 0.2θ − W ⋅ 0.2θ = 0
1.50 m
0.2 ⋅ P + 0.2 ⋅ W
αW
∆x0 α0 = = 0.162
1.5 ⋅ P'+0.5 ⋅ W
W
Evaluation of effective mass M*:
.50 m
θ
use ∆x1 as control displacement, ∆x0 = ∆x1 /3
2
⎛ n +m ⎞
⎜ ∑ Piδ x,i ⎟
∆y0 O
M * = ⎝ i =n1+ m ⎠ = (P'⋅1 + W ⋅ 0.333)2 = 6.162
.40 m g ∑ Piδ x,i
2 (
9.81 ⋅ P'⋅12 + W ⋅ 0.3332 )
i =1
Masonry Structures, lesson 14 part 2 slide 6

4. P effective mass ratio e* :
αP' ∆x1
n+m
9.81⋅ 6.162
e* = gM * / ∑ Pi = = 0.806
i =1 W + P'
Evaluation of effective static acceleration threshold a0*:
n+ m
α 0 ∑ Pi
1.50 m
αW
∆x0 α0 g 0.162 ⋅ 9.81
a0 =
* i =1
*
= *
= = 1.972 m/s 2
W
M e 0.806
.50 m
θ
“Linear” static safety check (ultimate limit state):
∆y0 O a gS ⎛ Z⎞
a* ≥
0 ⎜1 + 1.5 ⎟ with q = 2.0
q ⎝ H⎠
.40 m
Masonry Structures, lesson 14 part 2 slide 7
1.50 cm
“Linear” static safety check (ultimate limit state):
o a gS ⎛ Z⎞
a* ≥
0 ⎜1 + 1.5 ⎟ with q = 2.0
q ⎝ H⎠
Z where Z= 7.02 m is height of
= 0.78
H centroid of weights P’ and W with
6.00 m
0.40 m
respect to ground and Z = 7.5 m
ag S
a0 ≥
*
(1 + 1.5 ⋅ 0.936) = 1.202ag S
2.0
mechanism is verified if agS ≤ a0*/1.202 = 1.641 m/s2
= 0.167g
Masonry Structures, lesson 14 part 2 slide 8

5. P Noninear static safety check (ultimate limit state)
αP' ∆x1
Evaluate static α-∆x1 (α-dk ) relationship for
finite displacement. All forces are proportional
to weigths, therefore α-dk relationship is
linear:
α = α 0 (1 − d k / d k ,0 )
1.50 m
∆x0
αW
Evaluate displacement dk,0 at zero horizontal
W
force (i.e. zero restoring moment):
.50 m
θ
W ⋅ (0.2 − ∆ x 0 ) − P ⋅ (0.2 − ∆ x1 ) = 0
∆y0 O W ⋅ (0.2 − ∆ x1 / 3) − P ⋅ (0.2 − ∆ x1 ) = 0
W ⋅ 0.2 + P ⋅ 0.2
.40 m ∆ x1, 0 = d k , 0 = = 0.326 m
W /3+ P
Masonry Structures, lesson 14 part 2 slide 9
P
αP' ∆x1
Evaluate effective displacement of equivalent
sdof system:
n+m
∑ Pi δ x,i P '⋅1 + W ⋅ 0.333
i =1
d = dk
*
n+m
= dk = 0.68 ⋅ d k
1⋅ ( P'+W )
δ x,k ∑ Pi
1.50 m
∆x0 i =1
αW
W d * = 0.68 ⋅ d k,0 = 0.222 m
.50 m
0
θ
d * = 0.4 ⋅ d * = 0.089 m
u 0
∆y0 O
d * = 0.4 ⋅ d * = 0.035 m
s u
.40 m
Masonry Structures, lesson 14 part 2 slide 10

6. d * = 0.68 ⋅ d k,0 = 0.222 m
0
a* d * = 0.4 ⋅ d * = 0.089 m
u 0
d s* d * = 0.4 ⋅ d * = 0.035 m
a0 * Ts* = 2π *
s u
as
a s* Ts* = 0.92 sec
(2π/T s*)2
ds*=0.4du* du*=0.4d0* d0* d*
Masonry Structures, lesson 14 part 2 slide 11
Assume fundamental period of building had been determined previously as
T1=0.2 sec
Then Ts*=0.92 >1.5 T1=0.3 sec
therefore, assuming, e.g. agS = 0.2g = 1.962 m/s2 , the effective displacement
demand is:
T1Ts* ⎛ Z⎞
∆ (Ts* ) = a g S ⋅1.5 ⋅ 2 ⎜
1.9 + 2.4 ⎟ = 0.071 m < 0.089 m = d *
4π ⎝
u
H⎠
Masonry Structures, lesson 14 part 2 slide 12