The University of Adelaide - URM Walls Shear Response Analysis

This document presents a solved homework assignment from the University of Adelaide's Seismic Design of Masonry Buildings course, focusing on the shear response of unreinforced masonry (URM) walls. The assignment involves analyzing a URM wall with given dimensions and material properties to...Read More

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 designcategory 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=44.2kN
Maximum moment at h
2 , that is , mid−height of the masonry 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 stressis 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 planebending ,
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

3
Thus ,∈this case :P1= 90
100 ×115=103.5 kN
Auc=uncracked portion of the effective cross sectional area of the wall that provides shear bond capacity
We have two cases¿ consider :
Case 1
wall behaves as cantilever ,¿ atathe base with free rotation at thetop .
: Auc=te × dv
where dv isthe effective wall depthequal ¿ 0.8 lw
Auc =0.220 × 0.8× 4=0.704 m2
Case 2
Top∧bottom bases remain∥(¿−¿ position).
Auc=0.220 × 4=0.88 m2
the sliding shear resistance at the base of the wall V r=∅ mμC
Where:
C=compression fprce∈masonryacting normal ¿ the sliding plane∧tension∈the y
direction T y is zero , that is , C=Pd +T y But the latter is 0.
Eccemtricity e= M f
Pf
= 32.66
22.8 =1.43
for ∈plane bending e ≤ 0.33l w=0.33 × 4=1.32 hence , okay that is for case one

4
for outof plane bending e ≤ 0.33 te=0.33× 0.220=0.726 , hence okay ∈case 2
ˇfor flexural strengthof masonry wall :
maximum stress musnot exceed ∅ m f m for tension∧¿
∅ m f t for compression
maximum stress at wall ends is at follows ;
Maximum f c= Pf
Ae
+ M f
Sn
= 22.8
0.88 + 32.66
0.59 ≤ ∅ m f m
, =0.85 ×300000 →{0.3M/mm2 into N/M2}
Maximum f c=81.27 ≤255000. therefore okay .
the compression capaciy of the masonry wall Pr = [ 0.85 ẓ ∅ m f m
, ] te ( lw
2 −e ).2
Pr =[0.85 ×0.5 × 0.66× 300000]0.220 ( 4
2 −1.43 ) .2
The Compression capacity of te masonry wall Pr =21.10 kN /m2
f sp= 2 P
π an
=2× 103.5
π ×0.77 × 0.66 =129.65 kN /m2
V ¿=V test
Ab
−PD +L(ASTM C 496)
V ¿= 1016
4 × 2.5 −22.8=78.8 kN
V m =0.56 V t + 0.75 PD
A ( Section A 106.3.3 .5 )
V m =0.56 ×115+0.75 × 10.88
4 ×0.220 =73.67 kN
V = 0.75 SDS W
R ( Equation A 1−7 )
V =0.75× 3.426 × 44.2
1.5 ( Valueof R as ∈table ASTM 496 )
V =75.72 kN

5