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Design Of Eccentrically Loaded Brick Masonry

Design Of Eccentrically Loaded Brick Masonry
Eccentricity -Offset distance from CG of member to CG of load Occurs in, i) Exterior wall -Bearing not sufficient ii) Flexible slab -Excessive deflection (timber) iii) When span is very large, code recommends to take some amount of eccentricity

Eccentricity -Offset distance from CG of member to CG of load

 

Occurs in,

 

i)             Exterior wall -Bearing not sufficient

 

ii)            Flexible slab -Excessive deflection (timber)

iii)          When span is very large, code recommends to take some amount of eccentricity

 


 

Guidelines given in Appendix A

 

1.Eccentricity in loading occurs in loading on walls due to various reasons causing reduction in stress. The cases where eccentricity needed to be assumed are given in Appendix A of IS1905-1987.

 

2.For an exterior wall, when span of roof is more than 30 times the thickness of wall, the eccentricity assumed is one sixth of the bearing width.


When bearing is not sufficient, eccentricity assumed is tw/12

3.When flexible floors are adopted, for full width bearing, eccentricity assumed is tw/6.

 

4.For interior walls, when there is unequal length of slabs on both sides and then the difference between the loading is greater than 15%, moment is generated for which, e=M/P.


Actual stress (?act) is the sum of direct compressive stress P/A and bending stress M/Z. The permissible stress given in Table 8 can be increased by 25% and modification factors applied on it.





1) Design an exterior wall of height 3.5m, which is unstiffened. The slab is light weight flexible slab of length 3.5m. Assume the unit weight of slab as 15kN/m3 with thickness 0.2m.

Half of the load from the slab comes to the wall and since the slab is flexible, eccentricity considered as per Appendix A of IS1905 -1987.

Assume 200mm thick wall with M1 mortar and brick of compressive strength of 10N/mm2.

Loading on wall:

 

Load from slab = 15 x 0.2 x 3.5/2 = 5.25 kN/m Self weight of wall = 0.2 x 3.5 x 20 = 14 kN/m

 

Total           = 19.25 kN/m

 

e = tw/6 = 33.33mm M = P x e

 

Moment due to eccentricity = 19.25 x 103 x 33.33 = 641.67 x 103 Nmm Z = bt2/6 = 1000 x 2002 / 6 = 6.67 x 106 mm3

M/Z = 0.096 N/mm2


?per :

 

Use M1 mortar and brick of compressive strength 10N/mm2,

?per = 0.96 N/mm2

 

Stress reduction factor (Kst)

 

Slenderness ratio (Least of le/t & he/t)

 

From Table 4,

 

he = 0.75 H = 0.75 x 3 = 2.25m     [Both ends fixed]

 

he/t = 2.625 / 0.2 = 13.125 < 27

 

Therefore, the stress reduction factor from T condition is,

 


Ksh for block of size 200 x 100 x 90 mm laid along 100mm side, from Table 10 for Height to Width ratio of 90 x 100,

 

Height / Width = 100/90 = 0.9

For 0.75xo 1

For 1xo 1.1

For 0.9xo 1.06

Ksh = 1.06

?per modified   = Kst.KA.Ksh. ?per

 

= 0.735 x 1 x 1.06 x 1.25 x 0.96 = 0.935 N/mm2 ?act [0.192N/mm2] < per1[.12517N/mm?2]

 

Note : For brick masonry columns laterally supported by beams,

He = H

 

Only when the column is not laterally supported (laterally unsupported), He = 2H

 

 

2.Design a masonry column tied effectively at top and bottom. Load from slab is 100kN. Assume size of column as 400 x 400mm

?act = 100x103 / 400x400  = 0.625 N/mm2

 

?per = 0.96 N/mm2

Heff = H = 2m

 

Lam=   5   <   6

Kst = 1 Ksh = 1.06 KA = 0.94

 

?per modified = 0.957 N/mm2

 

?act < per?

 

 

3.Design an interior wall of a single storeyed building supporting unequal concrete roof slabs. The plan is as shown in figure. Assume triangular pressure distribution and unit weight of slab is 10kN/m3. The height of the wall is 3.8m and the wall is fixed to the foundation block below.


Height = 3.8m


 


4) Design an exterior wall of height 4m, unstiffened and supports a flexible slab 150mm thick with unit weight 12 kN/m3. The length of the slab is 4m.


 

 


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