CIV 6120: Residential Masonry and Timber Design - Detailed Report

This report provides a comprehensive analysis of residential masonry and timber design, addressing key aspects of unreinforced masonry and timber frame structures. It includes calculations for compressive capacity, seismic force analysis, and out-of-plane capacity checks. The report also details the design of a reinforced masonry retaining wall, including bending moment and shear force diagrams, and flexural and shear reinforcement design based on AS 3700 and AS 1170.4 standards. Timber frame design considerations are also addressed according to the assignment brief provided.

CIV 6120 RESIDENTIAL MASONRY STRUCTURAL DESIGN
PART 1: UNREINFORCED MASONRY DESIGN
Q2: COMPRESSIVE CAPACITY OF UNREINFORCED MASONRY
(a) Ultimate axial force on wall W07 at all levels of the building
This is given by: 1.2G+ 1.5Q
G= dead load and Q= live load
Now, assuming uniform axial load distribution such that at all levels the loading is equal
hence from the question the Q is given as: 3.0kPa (from the drawing attached) and G=
1.0kPa
Axial force Fa= 1.2x1.0+1.5x3.0 = 5.7kN applied at all levels
Q G
1.2G+1.5
Q
7557.9
2
1578.9
6 11437.95
15115.
8
1403.5
2 20244.29
22673.
8
526.32
1 27997.99
(b) Checking capacity of wall W07 “design by simple rules” in section 7.33 AS 3700
From section 7.33, we can use the following criterion: ‘
Fd<kFo
Fd= design compressive force
Fo= basic compressive strength capacity
K= a reduction factor for slenderness and eccentricity
The design compressive force Fd is attained:

Fd= Fuc xAd
Ad= 0.29 x2 = 0.58m2 , note that the size of the cross section of the wall is obtained from
the drawing attached and the thickness is derived as per claus 4.5.1
Fuc= 25MPa hence Fd= 25 x 0.58 = 14.5MN or 14500kN
The basic compressive capacity is derived as follows as per section 7.3.2
Now, since there are no testing done, hence:
Fo= φfmAd…..(1)
The bedded thickness (as per clause 4.5.1) = 230+110-50= 0.29m
Ad= 0.29 x 2= 0.58m2
Φ= capacity reduction factor ( obtained as per clause 4.4)
Since the member is under compression and is hollow between the leaves we select φ=
0.5
Fm= is obtained as per clause 3.3.2
F’m= km(fuc)0.5
Km= compressive strength factor (is obtained from table 3.1) hence 1.4 and fuc’= 25MPa
(provided in the question):
F’m= 1.4(25)0.5= 1.4 x5= 7.0
Ad= 0.58m2
Hence Fo= 0.5 x 7.0 x 0.58 = 2.03MN
Hence it is established that the member capacity is inadequate
(c) Alternate floor framing arrangement was used for the building with a 300mm wide
reinforced concrete beam ( bearing area = 300mm x 230mm) transmitting an uktimate
load of 250kN to W07. Checking the capacity of the wall under similar conditions :
The bedded area Ab= 300 x230= 0.069m2 and Fd = 250kN
The wall capacity is still checked using similar criterion

Now, member is solid and the φ can therefore be obtained from section 3.3.2 which is
actually 0.75
Again the member is a concrete beam in compression hence from section 3.3.2 km= 1.4
and Fuc’ = 25MPa ( as provided earlier);
Fm= 7.0
K= reduction factor for slenderness and eccentricity =1.0
Hence Fo = 0.75 x7.0 x 0.069= 0.36225= 362.25kN
Therefore from the criterion above it can be established that the design is safe.
Q3: In-plane analysis of unreinforced masonry
(a) Wall W34 determine the dead load, live load for all levels of the building hence
determine 0.9Wt for W34 at ground floor of the building
From table 3.1 of AS 1170.1 we determine the imposed floor actions Q, we shall obtain
the appropriate values of the loads then multiply by reduction factor as per clause 3.4.2
(assume loading is uniform on the floors for all levels)
For level 3, we have 1.5+1.5+2.0+0.5 = 5.5kN
The reduction factor is 0.5
Q3= 0.5x 5.5= 2.75kN
For level 2, Q2= 2.75 x2 = 5.5kN
And Level 3, Q3= 3x2.75= 8.25kN
For G values we still use AS 1170.1 as follows:
(b) Calculate relative proportion of seismic forces hence shear and moment to each of the
walls in the North-south direction (Y-direction)
In handling this part, assume the following:
 Centre of mass and centre of stiffness coincide
 No accidental eccentricity needs is to be considered
 Seismic loading is only applied in the y direction (N-S direction)

 The floor slabs act as a rigid diaphragm in-plane ( with no flexural coupling aout
of plane . that is wall act as 12m tall cantilever )
First of all we establish the earthquake design category (EDC) is found based on
the given information. We know that the building seats on a rocky place (class B), IL=2, 1000
year return period.
AS 1170.4 (2007) and using the parameters: IL=2, class B, structure height- 15m, table 2.1
provides the EDC for this case as EDC II
Base shear is determined from the formula :
V= [KpZCh(T1)Sp/u]Wt ….(1) (obtained from section 6 of the standard)
Now, Wt is the seismic weight given by the sum of level weights and this is obtained from clause
6.2.2 as shown below:
Wi= Gi+ϕQi where ϕ= 0.3 since this is a nonstorage
Substituting in equation: = 8+0.3x 4+1 = 10.2kPa per level
Hence total Wt= 5x 10.2= 51kPa
Sp/u= Structural performance factor per structural ductility factor
his is obtained from clause 6.5, since structure is RC with ordinary moment resisting frames, we
select the 3rd entry hence , Sp/U= 0.38
Ch(T1) is the spectral shape factor and this is obtained from clause 6.4) using the following
parameters: 1000 year return period, hence P (secs) = 1.0, subsoil classification is Be (from table
2.1) hence Ch(T1) = 0.88
KpZ is obtained from table 3.5, since we know that: return period is 1000 years hence
probability of exceedance is 1/1000, KpZ= 0.10
Therefore, the base shear is obtained by substituting these parameters in the formula (eqn 1
above)
V= [0.1 x 0.88x 0.38]x51= 1.70544kPa

Lateral force distribution
For this case, we use clause 5.4.2.3 by first checking if criteria is met: remember L= 15m hence
minimum requirements to invoke this clause is met
We therefore use the formula: Fi= Ks [KpZSp/u] Wi at each level
For individual Ks we deduce the following information:
Floor type Ks
5th 3.1
4th 2.5
3rd 1.8
2nd 1.2
1st 0.6
For 1st floor with Ks=0.6,
F1= 0.6[0.1x0.38] x51= 1.1628kPa
And other floors same:
F2= 1.2(0.1x0.38) x51= 2.3256kPa
F3= 1.8(0.1x0.38) x51= 3.4884kPa
F4= 2.5(0.1x0.38) x51= 4.845kPa
F5= 3.1 (0.1x0.38)x 51= 6.0078kPa
 Distribution of Bending Moment
This is given by the formula:
BM= Vx where V is the shear force and X is the height between floor levels.
Therefore BM for each floor is determined:
1st level: Mx= 1.70544 x 3 = 5.116kN-m
2nd floor Mx= 6x 1.79544= 10.232kN-m
3rd floor Mx= 9x1.70544= 15.348kN
4th floor = 12 x 1.70544= 20.464kN-m
5th floor Mx= 15 x 1.70544 = 25.58kN-m

 Lateral force
F= afloor [Icx ac/Rc]Wc<0.5Wc
Wc= 3kN, the minimum criteria to be met: 0.5x3= 1.5kN ( answer obtained must not be
more than this)
From section 8.2, we obtaine the following:
Ic= 1.0 (for all other components), ac= 1.0 and Rc= 2.5 (since it is noncritical ductile)
and af= KpZCh= 0.38 x 0.88 = 0.3344
Hence F= 0.3344 (1x1/2.5)x 3= 0.40128kN
(c) Check wall W34 for the following
(i) In-plane shear
(ii) Heel tension and toe compression
QN4): Out of plane analysis
(a) Check the out of plane capacity of W26 under seismic loading
Assume: inner (230mm) skin of bricks is simply supported at the roof
Second floor levels a one vertical edge is supported laterally on wall W36
Assume outer skin of bricks is connected to the inner skin using heavy duty tie
Assume inertia forces are shared between the two leaves of the wall
QN 5) Reinforced Masonry retaining wall
(a) Draw the bending moment diagram and shear force diagram for the retaining wall
The shear force= V= 2x18/2= 18kN
The bending moment (max): WL2/8 = 18x2x2/8= 9kN/m

Hence the equation: wx2/8= 18/8x X2= 9/4x2
BMD
1 2 3 4 5 6
0
1
2
3
4
5
6
7
8
9
10
Bm
Bm
The Shear force diagram
1 2 3 4 5 6
0
2
4
6
8
10
12
14
16
18
20
SF
SF
(b) Using a block size of 290x190x290 units, design flexural reinforcement and shear
reinforcement for the retaining wall
Assume: M4 mortar to be used units are faceshell bedded
Grout compressive strength : f’cg= 20MPa and Block compression strength f’uc=
15MPa
(c) Neat sketch of the design
PART 2: Timber frame design