### Solved Example: Analysis of a one-way spanning unreinforced masonry wall panel (BS 5628)

Articles > Solved Example: Analysis of a one-way spanning unreinforced masonry wall panel (BS 5628)

### Question:

Estimate the characteristic wind pressure that the cladding panel shown below can resist assuming that it is constructed using bricks having a water absorption of < 7% and mortar designation (iii). Assume γf = 1.2 and γm = 3.0.

### Solution:

ULTIMATE DESIGN MOMENT (M)
Since the vertical edges are unsupported, the panel must span vertically and, therefore, following equations

$M_{prep}=\alpha W_k \gamma_f L^2$

$M_{par}=\mu \alpha W_k \gamma_f L^2$

cannot be used to determine the design moment here. The critical plane of bending will be parallel to the bed joint and the ultimate design moment at mid-height of the panel, M, is given by (clause 32.4.2 of BS 5628)

$M=ultimate\;load*height/8$

Ultimate load on the panel = wind pressure * are

$=(\gamma_fW_k)(height*length\;of\;panel)$$=1.2W_k*3000*1000=3.6W_k10^6Nm^{-1}$

Hence

$M=\frac{3.6W_k*10^6*3000}{8}=1.35*10^9W_k\;N \; mm \; m^{-1}$

MOMENT OF RESISTANCE (Md)
Section modulus (Z):

$Z=\frac{bd^2}{6}=\frac{10^3*102.5^2}{6}=1.75*10^6mm^3\;m^{-1}$

Moment of resistance:
The design moment of resistance, Md, is equal to the moment of resistance when the plane of bending is parallel to the bed joint, Mk par. Hence

$M_d=M_{k\;par}=\frac{f_{kx\;par}Z}{\gamma_m}=\frac{05*1.75*10^6}{3.0}=0.292*10^6N\;mm\;m^{-1}$

where fkx par = 0.5 N mm−2 from table below (Table 3 BS 5628), since water absorption of bricks < 7% and the mortar is designation (iii).

DETERMINATION OF CHARACTERISTIC WIND PRESSURE (Wk)

For structural stability:

$M\leq M_d$

$1.35*10^9W_k \leq0.292*10^6$

$W_k \leq 0.216*10^{-3}N\;mm^{-2}$

Hence the characteristic wind pressure that the panel can resist is 0.216 × 10−3 N mm−2 or 0.216 kN m−2.

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