PRIVATE

PRIVATE

SOLUTIONS

LATERAL LOADING

Problem 1

Determine the required thickness of a single leaf wall supported as shown above using the following criteria:

Characteristic wind load

= 0.45 kN/m2Height of wall to free edge

= 4.5 m

Length of wall between restraints

= 4.5 m

Concrete blocks (solid) strength

= 7.3 N/mm2Normal category construction control)

)

m = 3.0

Category I or I I masonry units )

Panel not providing stability to structure

f = 1.2

Flexural DesignThe design bending moment per unit height of the wall is given by the following expression for a 2 way panel assumed to crack vertically

The bending moment coefficient depends on

(i)

orthogonal ratio (but including self weight)

(ii)

aspect ratio h/L

(iii)

support conditions

(i)

Try thickness of 190 mm, 7.3 N/mm2 block

Determine fkx

for 7.3 N/mm2 at 100 (fkb)

= 0.25

for 7.3 N/mm2 at 250

= 0.15

Therefore by linear interpolation at 190 mm (fkb) = 0.19

for 7.3 N/mm2 at 100(fkp)

= 0.6

for 7.3 N/mm2 at 250

= 0.35

Therefore at 190 mm (fkp)

= 0.45

. Orthogonal ratio to include allowance for self weight. According to clause 32.4.2,

fkb = fkx + mgd, where gd is determined half way down the wall

gd = 0.5 x 4.5 x 1200 x 9.81 x 10-6 x 0.9 = 0.024

nb - f for max criteria = 0.9

(ii)Aspect ratio

(iii)Support conditions-simple

From BS 5628 : Part 1, Table 8A

forh = 1.0

L

= 0.083 with = 0.5

= 0.080 with = 0.6Therefore by linear interpolation

= 0.080 + (0.06 0.058)/(0.06 0.05)[0.083 0.080] 0.0806 when = 0.58Since Wk = 0.45 kN/m2 ,f = 1.2, L=4.5

then applied design moment per unit height

m= 0.0806 x 0.45 x 1.2 x 4.52

= 0.881 kNm/mThe design moment of resistance is given as (Clause 32.4.3)

For a panel bending in two directions

Now, since fkp = 0.45 and t = 190

thenM=0.45 x 1902 x 10-3

3.0 x 6

=0.903 kNm/m (> 0.881kNm/m) thre fore OKMaximum characteristic wind for this thickness

m = M

0.0806 Wk x 1.2 x 4.52 = 0903

(Wk = 0.903

0.0806 x 1.2 x 4.52

= 0.46 kN/m2 > 0.45kN/m2Slenderness LimitsPanel simply supported on more than one side, therefore from clause 32.3

h x L ( 1350 tef2since wall is single leaf tef = t

15therefore

In addition no dimension shall exceed 50 tef. since h and L both = 4.5 then

16Design for ShearConsider the wind load to be distributed to the supports as shown below

Then total load to support

=f Wk loaded area

Thus the total shear along base

=1.2 x 0.45 x (4.5 x 2.25) = 2.734 kN

2

Assuming that this load is uniformly distributed along base, representing the design shear force per metre run:

The design shear stress (vh) is therefore

=0.608 x 103 = 0.0028 N/mm2

215 x 1000

The characteristic shear strength (fv) from clause 25

= fvko + 0.6 gA

Where fvko is characteristic initial sheer strength in N/mm2 generally taken as 0.35N/mm2

gA design vertical load per unit area at base of wall

= 4.5 x 1200 x 9.81 x 10-6 x 0.9 = 0.048kN/m2So including self weight along the base characteristic strength fv

= 0.35 + 0.048 = 0.0398N/mm2The design shear stress (vh) must be limited so that

In this case vh = 0.0025 N/mm2 and

Therefore shear resistance is adequate along base,Total shear to each vertical support

= 1.2 x 0.45 x 2.25(2.25 + 4.5) = 4.10 kN

2

Again, considering load to be uniformly distributed along support, then design shear

force per metre run

=4.10 = 0.91 kN

4.5

Using 2 mm thick anchors into dovetail slots in column.

Characteristic strength of each tie = 4.5 kN

Placing ties at 900 mm centres and taking the partial safety factor for material as 3.5. The design load resistance per metre run of wall

=4.5x1000 = 1.43 kN

3.5

900

This is greater than the design shear force and therefore adequate.

No enhancement due to self weight exists.

Solution - Problem 21. = 0.3/0.9 = 0.33. Assuming worst flexural strength s for clay units.

2. Aspect ratio h/L = 5.625/4.5 = 1.25

3. Support conditions Simple

Using BS5628:Part 1, Table 8A, for h/L = 1.25, = 0.095 + 2/5(0.097 0.095) = 0.0958

And using the transformation technique, {enhance moment coefficient by window ht/ panel ht}

Window ht hw = 1.0m, panel ht hp = 5.625

Therefore, hw/hp = 1.0/5.625 = 0.18

Modification to moment coefficients due to line loads = 1 + 0.18 = 1.18

Hence modified = 0.095 x 1.18 = 0.113

Now Wk = 0.7kN/m2,

f = 1.2,

L = 4.5m

Then applied design moment per unit height

m = 0.113 x 0.7 x 1.2 x 4.52 = 1.92kNm

Design moment of resistance M = (fkpZ)/m = (0.9 x 1000 x t2)/6

Therefore required thickness [(1.92 x 3.0 x 6 x 106)/(0.9 x 1000)]0.5 196mmSince some strength will be obtained from window surround, use 215mm wall.

This example may be continued by considering what thickness would be required for the basic wall panel without the window and what thickness would be required had the panes been taken as the full height of 6.625m.

For the basic wall panel, 5.625 x 4.5m, bending moment coefficient, as determined previously is 0.0958 which gives a required thickness of :

196 x 0.0958/0.113 = 166.2mm {i.e. Neglecting the impact of the line load from the window}

Had the panel been 6.625 x 4.5m then the aspect ratio is 6.625/4.5 = 1.47

And from BS5628 Table 8A, for h/L = 1.5 (very close), = 0.1004 by interpolation.

Hence required thickness = 196 x 0.1004/0.113 = 174mm

This latter value is useful in comparing to the original case with window opening since both panels are subject to a similar total wind load but in the case of the panel with openings, the total length of yield line isles and hence a slightly greater panel thickness. A 215 is justified.Alternative design.

The loading from the window could be considered to be simply carried by a strip of masonry running parallel to the opening. Had this method been adopted, them an additional loading as shown hatched in the figure below could be assumed to be carried by the top 1.125m of the wall. (i.e. 0.25L as given in the notes.

Equivalent uniformly distributed load transferred to strip from window = [0.7 x 0.5 x (3.5 + 4.5)]/(1.125 x 4.5 x 2) = 0.28kN/m2Total uniformly distributed loading on strip, therefore

= 0.7 x 0.28 0.98kN/m2Designing strip as simply supported span

M = 0.125 x 0.98 x 1.2 x 4.52 {wL2/8} = 2.98kNm

So trequired [(2.98 x 3.0 x 6 x 106)/(0.9 x 1000)]0.5 = 244mm

If moment were assessed as WkfL2/10 then trequired reduces to 218mm

Considering this and that there will be some stiffness in the window frame, then a wall thickness as originally suggested of 215mm is adequate.

Solution - Problem 3Using the transformation method indicated in the notes, then fixity to edges is reduced by a quarter = 0.25, i.e. = 0.75 {Use the notation : = 1.0 for full fixity, = 0.0 for simple support}

Hence the bending moment coefficient is determined from 8E and 8I. In order to transform, we reduce the panel fixity to a value intermediate between full fixity and simple support. i.e We assume partial fixity.

Now = 0.25/0.75 = 0.33 (Table 3)

h/L = 4/4 = 1.0

For the pinned edge condition, Table E ( = 0), then = 0.066 (by linear interpolation)

For the pinned edge condition, Table I ( = 1), then = 0.033 (by linear interpolation)

So including effect fro transformation, = 0.75, = 0.041

Hence m = 0.041 x 0.60 x 1.2 x 42 = 0.47kNm/m

Hence trequired = [(0.47 x 3.0 x 6 x 106)/(0.75 x 1000)]0.5 = 106mm

Solution - Problem 4Outer leaf.

No information given on water absorption of bricks, so take WA > 12 which gives the worst case.

For brick fkb = 0.3,

fkp = 0.9,

wall density = 1700kg/m3Consider wall as simply supported at base, built in up the sides and include self weight.

Self weight stress at mid height = 0.5 x 3.375 x 1700 x 9.81 x 10-6 x 0.9 = 0.025N/mm2Orthogonal ratio including self weight = [0.3 + (0.025 x 3.0)]/0.9 = 0.417

Therefore bending moment coefficient (Table 8c) for an aspect ratio = 3.375/4.5 = 0.75,

= 0.044 (by interpolation)

Consider thickness as 103mm,

Design moment capacity of wall M = [fkp x Z]/m

= [fkpbt2]/(6m) = {0.9 x 1000 x 1032]/[3.0 x 6 x 106]

= 0.531kNm per m

Applied design moment m = WkfL2 and setting this as equal to the design moment capacity of the wall :

Wk = M/[fL2] = 0.531/[0.044 x 1.2 x 4.52] = 0.499kN/m2Total characteristic wind pressure = 0.6kN/m2 therefore, excess force to be carried by inner leaf :

is 0.6 0.499 = 0.101kN/m2Design of inner leaf.

Try 3.6N/mm2 blocks, 100m thick. Aspect ratio of panel = 1.25

Consider self weight as 1000kg/m3,Then self weight stress at mid height = 0.025 x 1000/1700 = 0.015N/mm2Modified orthogonal strength ratio = [0.25 + 0.015 x 3.0]/0.45 = 0.66

Now considering simple support to all s=edges then from BS5628 par 1 Table 8A,

= 0.066 + 4/10(0.069 0.066) = 0.0672 (by interpolation)

Thickness is 100mm so

M = [0.45 x 1000 x 1002]/(3.0 x 6 x 106) = 0.25kNmThere fore characteristic wind force that can be carried

Wk = 0.25/[0.0672 x 1.2 x 4.52] = 0.153kN/m2Total capacity of cavity wall is acceptable.

It should be noted that the wall panel could be shown to carry a greater wind force since some degree of fixity could be allowed at the base of both walls. In addition, if mortar jointing were used between the inner blockwall and the columns, then some degree of fixity might effectively be taken at the support due to arching action in the wall. In the example, the total wind load was determined based on the assumption that both leaves attain their ultimate strength simultaneously as inferred by the first paragraph to clause 32.4.5 of BS5628 : Part 1. If, however, the wind load were shared between the two leaves, in proportion to their design moments of resistance, as previously given then the stress in the leaves would be as follows:

Mbrick = 0.532,Mblock = 0.250 kNm per m

Thus Wk would be shared in the ratio :

0.531/0.781 x 0.6 = 0.41 kN/m2 to brick

And 0.250/0.781 x 0.6 = 0.19kN/m2 to block

This then gives :

fkp in bricks = [0.41 x 1.2 x 0.044 x 4.52 x 3.0 x 6 x 106]/(1000 x 1032) = 0.744kN/m2fkb in blocks = [0.19 x 1.2 x 0.0672 x 4.52 x 3.0 x 6 x 106]/(1000 x 1002) = 0.559kN/m2This suggests an overstress in the block. However, as we are working in ultimate terms, and since the wind pressure will be distributed as suction and positive pressure on both leaves, then it would appear reasonable to accept the first method for design purposes. In this instance, the additional allowance for self weight fixity could be shown to prove the wall adequate.

Wall ties.

Wall ties used to be specified by strength. Now Annex C of Bs 5628 : Part 1 provides guidance. No data on the use of the building is provided so a conservative choice must be made. Select type 1 wall ties. Provide 2.5 ties per m2 and in addition, 3 4 ties per m at unbonded edges. The ties must be bedded at least 50mm into the mortar. Table C3 gives the tensile and compressive load capacity of each type 1 tie as = 2.5kN. Hence design transfer potential per tie = 2.5/3.0 = 0.8kN. {m wall ties = 3.0}Hence at a density of 2.5 ties per m2, the transfer capacity of the ties :

= 0.8 x 2.5 = 2.0kN/m2 > 0.6kN/m2

h

L

Free edge

Simple supports

4.5m

4.5m

45o

4.5m

5.625m

1.25m

1.0m

Loading transmitted to wall

_1266642604.unknown