Torsion from an Asymmetric Shear Wall Layout
A three-story, 100 ft × 60 ft building is braced in the north–south direction by two reinforced concrete shear walls on the east and west end walls, at x = 0 and x = 100 ft. Story height is 12 ft and the floors are 6 in concrete on metal deck. Ground-story shear is 300 kip. Because the east wall has a large service opening, its stiffness is 800 kip/in versus 2,000 kip/in for the west wall. The mass is essentially symmetric about x = 50 ft. A younger engineer has distributed the 300 kip to the walls in proportion to stiffness and called it done: 86 kip to the east wall.
1. Compute the center of rigidity, the design eccentricity, and the actual shear in each wall for the 300 kip story shear.
Read the worked answer
Center of rigidity (measured from the west wall, x = 0):
- x_CR = (2000 × 0 + 800 × 100)/(2000 + 800) = 80,000/2,800 = 28.6 ft
- Center of mass at 50 ft → inherent eccentricity e = 50 − 28.6 = 21.4 ft
- Add accidental torsion, 0.05 × 100 = 5 ft → design e = 26.4 ft
Torsional stiffness about the CR:
- J = 2000 × (28.6)² + 800 × (71.4)² = 1.63×10⁶ + 4.08×10⁶ = 5.71×10⁶ kip-ft²
Rotation and wall displacements:
- θ = Ve/J = 300 × 26.4/5.71×10⁶ = 1.39×10⁻³ rad
- δ_trans = 300/2800 = 0.107 in
- δ_west = 0.107 − 1.39×10⁻³ × 28.6 = 0.068 in
- δ_east = 0.107 + 1.39×10⁻³ × 71.4 = 0.206 in
Wall shears:
- West: 2000 × 0.068 = 135 kip
- East: 800 × 0.206 = 165 kip (check: 135 + 165 = 300)
The flexible east wall takes 55% of the story shear while having only 29% of the stiffness. The stiffness-proportional answer of 86 kip is 48% low - the wall would be under-designed by nearly a factor of two.
2. Is the stiffness-proportional distribution acceptable? What irregularity and analysis requirements apply?
Read the worked answer
No. It ignores torsion, and the torsion here works against the wall that is already the weak link.
Irregularity. Displacement ratio = δmax/δavg = 0.206/((0.068 + 0.206)/2) = 1.51. That exceeds 1.2 (torsional irregularity, Type 1a) and exceeds 1.4, which makes it an extreme torsional irregularity (Type 1b).
Analysis requirements that follow:
- A three-dimensional analysis with the inherent plus accidental torsion is required; a two-dimensional, stiffness-proportional hand distribution is not permitted.
- The torsional amplification factor Ax = (δmax/1.2 δavg)² = (1.51/1.2)² ≈ 1.58 applies to the accidental torsional moment (ASCE 7 provisions), so the accidental 5 ft becomes about 7.9 ft in the design model. Re-running with e = 21.4 + 7.9 = 29.3 ft pushes the east wall to roughly 173 kip.
- Because the plan irregularity is present, check the vertical irregularity provisions as well (the ground story is also the one with the reduced wall) and verify the diaphragm is genuinely rigid - a 6 in slab on deck at this size is, but say so in the calc rather than assuming it.
The teaching point: the stiff wall does not take the most load. Torsion can make the flexible element the critical one, and a stiffness-proportional spreadsheet will never show it.
3. What is the most effective way to reduce the east wall's demand, and what would you check after changing it?
Read the worked answer
Stiffen the east wall. Adding length or thickness at x = 100 does two things at once:
- It raises k_east, so the wall takes more of the direct shear but deforms less.
- It pulls the center of rigidity from 28.6 ft toward the mass at 50 ft, shrinking the inherent eccentricity that is driving the torsion. It also raises J, since the added stiffness is at the far lever arm.
Run the numbers before believing it: if k_east goes from 800 to 1,600 kip/in, x_CR = 160,000/3,600 = 44.4 ft, e = 5.6 + 5 = 10.6 ft, J = 2000 × 44.4² + 1600 × 55.6² = 3.94×10⁶ + 4.95×10⁶ = 8.89×10⁶. θ = 300 × 10.6/8.89×10⁶ = 3.58×10⁻⁴; δ_east = 300/3600 + 3.58×10⁻⁴ × 55.6 = 0.083 + 0.020 = 0.103 in → F_east = 165 kip. The demand does not fall - the wall earns back in stiffness what it loses in torsion, and it ends up with the same force at half the drift.
What to check after any change: δmax/δavg again (irregularity can persist), the drift limit against the 12 ft story height, the wall's own flexural and shear capacity with the new force, and the foundation overturning that follows the increased force at the base.