Multistorey plane frame systems

The following project <B_531> of a multistorey plane frame shows the interstorey fluctuation of the apparent stiffness factor k_j and the moment distribution factor a_j in both middle and end columns.

The apparent stiffness factors k_j vary from 4 to 9, while the moment distribution factors a_j from 0.40 to 0.60.

The apparent stiffnesses are independent of the seismic force magnitude. They depend only on the distribution of forces along the height of the frame. To this end, in the following examples, both forces, shear forces, bending moments, and deformations are unitless.

The characteristic value of the seismic action is the total horizontal force, henceforth called "base shear" because it is equal to the sum of all column shear forces at the base of the building. In the specific example, where the triangular distribution [*] NoteThe triangular distribution of seismic accelerations usually approximates the one derived from the response spectrum analysis. The distribution of seismic forces is triangular when all floor masses are identical. is used, the base shear is equal to 50. If equal horizontal forces V_base=50 are applied by means of orthogonal distribution, i.e. horizontal force H=12.5 at each floor, the results differ by a small percentage of the order of 5%.

In case of a basement, seismic forces developed at its roof level have a zero value. However, fixed end condition applies only to the base of the columns along the basement perimeter walls.

If the frame also comprises walls, as presented in the following paragraph, the stiffnesses and the moment distribution of the walls differ between them. This difference becomes more distinctive as the number of stories increase.

The total joint displacements and column stress resultants (shear forces and bending moments) are obtained from the frame analysis. Quantities K, k and a derive from the previous results. The apparent stiffness K_i of storey i derives from the expression K_i=V_i/δ_i, while the apparent stiffness of column j of storey i from the expression K_i,j=V_i,j/δ_i.

Example 5.3.1 (3^rd storey) :

Column stiffnesses: K_3,1=9.6/0.823=12, K_3,2=15.8/0.823=19, K_3,3=9.6/0.823=12.

Storey stiffness: Κ ₃ = (20+15)/0.823=43. The same value is obtained if calculated as the summation of the column stiffnesses of the storey, i.e. K₃= K _3,1+K_3,2+ K_3,3=43.

If shear effect is taken into account (Shear effect=ON), the displacements are δ ₁ =0.85, δ ₂ =1.05, δ ₃ =0.84, δ ₄ =0.50 mm, i.e. the difference is insignificant.

If rigid bodies effect is taken into account (Rigid body=ON), the displacements are δ ₁ =0.80, δ ₂ =0.97, δ ₃ =0.77, δ ₄ =0.45 mm, i.e. both decrease of displacements and increase of stiffnesses are small but measurable.