Solved Example: Design of a load-bearing brick (unreinforced masonry) wall (BS 5628)

Articles > Solved Example: Design of a load-bearing brick (unreinforced masonry) wall (BS 5628)

Question:

The internal load-bearing brick wall shown below, supports an ultimate axial load of 140 kN per metre run including self-weight of the wall. The wall is 102.5 mm thick and 4 m long. Assuming the masonry units conform to Category II and the construction control category is ‘normal’, design the wall.

Solution:

LOADING
Ultimate design load, N = 140 kN m−1 = 140 N mm−1

DESIGN VERTICAL LOAD RESISTANCE OF WALL
â–ºCharacteristic compressive strength
Basic value = f_k

â–ºCheck modification factor
Small plan area – modification factor does not apply since horizontal cross-sectional area of wall, A = 0.1025 × 4.0 = 0.41 m² > 0.2 m².
Narrow brick wall – modification factor is 1.15 since wall is one brick thick.
Hence modified characteristic compressive strength is 1.15f_k

â–ºSafety factor for materials (γ_m)
Manufacture and construction controls categories are, respectively, ‘II’ and ‘normal’. Hence from Table 5.10, γ_m for compression = 3.5

â–ºCapacity reduction factor (β)
Eccentricity
Since wall is axially loaded assume eccentricity of loading, ex < 0.05t
Slenderness ratio (SR)
Concrete slab provides ‘enhanced’ resistance to wall:

$h_{ef}=0.75*height=0.75*2800=2100;mm$

$t_{ef}=actual;thickness(single;leaf)=102.5mm$

$SR=frac{h_{ef}}{t_{ef}}=frac{2100}{102.5}=10.05<(permissible=27)$

Hence, from Table below, β = 0.68.

â–ºDesign vertical load resistance of wall (NR)

$N_R=frac{eta * modified;characteristic;strength*t}{gamma_m} =frac{0.68*1.15f_k*102.5}{3.5}$

DETERMINATION OF f_k:

$N_R geq N$

$frac{0.68*1.115f_k*102.5}{3.5}geq 140$

$f_k geq frac{140}{22.9}=6.1N mm^{-2}$

SELECTION OF BRICK AND MORTAR TYPE:

From Table above (Table 2 of BS 5628), any of the following brick/mortar combinations would be appropriate:

The actual brick type that will be specified on the working drawings will depend not only upon the structural requirements but also durability, acoustics, fire resistance, buildability and cost, amongst others. In this particular case the designer may specify the minimum requirements as HD clay units, compressive strength 30 N mm−2, F0 (i.e. passive environment) and S0 (i.e. no limit on soluble salt content) in mortar designation (iii). Assuming the wall will be plastered on both sides, the appearance of the bricks is not an issue.

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