Seismic Design of Masonry Buildings - URM Walls Shear Response

1
The University of Adelaide
Department of Civil, Environmental and, Mining engineering
Seismic Design of Masonry Buildings
Semester 2, 2018
Instructor: M.C. Griffith
Homework Assignment #3
“Shear response of URM walls”.
Date: 08-Sep-18
Given the following as in the question outline
Wall length l=4m, height h=2.5m, thickness t=220mm
Length of walletes l= 0.77m, h=0.66m. t=0.220m and a mean failure load of 1016KN
Diagonal compression tests on square panels size l=077m, h=0.77m, t=0.220m and a mean
failure load of115 kN
Unit weight of brick masonry as 19Kn/m3
SOLUTION
f tu
, =0.5 Pu
¿
f tu
, =0.5 × 115
0.77 ×0.220 =339.43 kN
m2
assuming a seismic design category A ; Load combinationA [ 0.6 D+ wL ] will govern
wL=19 × 4 ×2.5 × 0.220=41.8 kN
¿ 0.6 × 4.0+ 41.8=44.2 kN
Maximum moment at h
2 , that is, mid−height of themasonry wall is as follow ;
M max = wL h2
8 → 41.8 × 2.52
8 =32.66 kNm

2
the axial force on the masonry P ;
p=0.6 ×wwall × h
2 =0.6 ×19 × 4
2 =22.8 kN
m2
theflexural tensile stress is calculated as below ;
−P
An
+ 12 M
Sn
,
where , An=te .lw effective cross−sectional areaof masonry
Sn= [ te × lw
2 ]
6
e <0.33 lw for∈− plane bending∨¿
e <0.33 t for ou of plane bending,
An =0.220 ×4=0.88 m2
Sn=0.220× 42
6 =0.587
therefore , flexural tensile stress is calculated by substitution∈the formula as shown ;
−22.8
0.88 + 12× 32.66
0.59 =−25.90+664.27
¿ 638.27 kN /m2
The allowable stress Ft ,is 638 kN
m2 hence , the masorny wallis adequate .
The∈− plane sliding shear resistance V r , along bed jointsbetween masonry courses
V r =0.16 ∅ m √ [ f m
, Auc ] +∅ mμ P1 where :
μ=coefficient of friction
¿ 0.7 for masonry smooth concrete
P1=compressive force∈masonry acting normal ¿ the sliding plane
normallytaken as 0.9×the dead load .
V r =0.16 ×0.85 √ [ 300000 ×0.88 ] +0.85 × 0.5 ×103.5
V r =113.87 kN