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Distribution of lateral force among masonry walls

Short answer

Considering flexural and shear deformations, wall stiffnesses are computed; force shares are proportional to stiffness. The 1.0 m, 1.5 m, 2.0 m, and 3.0 m walls take 14 kN, 36 kN, 64 kN, and 126 kN, respectively.

A single-storey brick building has a rigid roof slab which transfers a total lateral seismic force of 240 kN to four parallel in-plane shear walls (piers), each 3.0 m high and 230 mm thick. The wall lengths are 1.0 m, 1.5 m, 2.0 m and 3.0 m. Treat each wall as fixed at the base and restrained against rotation at the top by the slab. Take E=2500E = 2500 N/mm² and G=0.4EG = 0.4E. Calculate the share of each wall (consider both flexural and shear deformations).

Principle

A rigid diaphragm gives the same horizontal displacement Δ\Delta to all walls, so each wall takes force in proportion to its stiffness k=V/Δk = V/\Delta.

For a wall fixed at both ends (top rotation prevented):

Δ=Vh312EI+1.2 VhGA⇒k=1h312EI+1.2hGA\Delta = \frac{Vh^3}{12EI} + \frac{1.2\,Vh}{GA} \quad\Rightarrow\quad k = \frac{1}{\dfrac{h^3}{12EI} + \dfrac{1.2h}{GA}}

where I=tL3/12I = tL^3/12 and A=tLA = tL (the factor 1.2 is the shape factor for rectangular shear).

Data: h=3000h = 3000 mm, t=230t = 230 mm, E=2500E = 2500 N/mm², G=1000G = 1000 N/mm².

Stiffness of each wall

Wall LL (m)II (mm⁴)AA (mm²)Flexural term h312EI\frac{h^3}{12EI}Shear term 1.2hGA\frac{1.2h}{GA}kk (N/mm)
1.01.917×10101.917\times10^{10}2.3×1052.3\times10^54.7×10−54.7\times10^{-5}1.565×10−51.565\times10^{-5}15 970
1.56.469×10106.469\times10^{10}3.45×1053.45\times10^51.39×10−51.39\times10^{-5}1.043×10−51.043\times10^{-5}41 070
2.01.533×10111.533\times10^{11}4.6×1054.6\times10^55.87×10−65.87\times10^{-6}7.83×10−67.83\times10^{-6}73 020
3.05.175×10115.175\times10^{11}6.9×1056.9\times10^51.74×10−61.74\times10^{-6}5.22×10−65.22\times10^{-6}143 750

Sum Σk=273 810\Sigma k = 273\,810 N/mm.

Shares of 240 kN

Vi=VkiΣkV_i = V \frac{k_i}{\Sigma k}
Wall LL (m)ki/Σkk_i/\Sigma kForce ViV_i (kN)
1.00.05814.0
1.50.15036.0
2.00.26764.0
3.00.525126.0
Total1.000240.0

Displacement of the slab: Δ=240 000/273 810=0.88\Delta = 240\,000/273\,810 = 0.88 mm (same for all walls).

Remarks

  • The longest wall (3.0 m) takes 52.5 percent of the force even though it is only 3 times as long as the shortest, because stiffness grows faster than length.
  • Slender walls are weaker in resisting lateral force. For these properties the flexural and shear terms are equal at L=h/3≈1.73L = h/\sqrt{3} \approx 1.73 m, so shorter piers (about L/h<0.58L/h < 0.58) are dominated by flexure. If the walls are treated as cantilevers (free at top), the shares become 8.8, 27.1, 57.2 and 147.0 kN, so the long wall attracts even more.
  • If only shear deformation were considered, the shares would be in proportion to the wall lengths.

Answer: Shares are 14 kN, 36 kN, 64 kN and 126 kN for the 1.0, 1.5, 2.0 and 3.0 m walls respectively.

✓ Answer checked by Claude Sonnet 5.5

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