Fundamental period of RC buildings with infill walls in Nepal

Abstract

The presence of infill walls in a building increases the stiffness and mass of the building leading to significant changes in the fundamental period. If this increase in stiffness is not considered, it may cause severe damages to the buildings during the earthquakes. This paper presents an analysis of the fundamental period of buildings considering the effect of infill walls. For this, a computer program was created to generate buildings with different configurations and calculate the fundamental period. It was found that besides the geometric parameters: number of stories and the height of the building, the bay span also had a significant effect on the fundamental period. Thus, a formula that takes into account those parameters was formulated for infilled reinforced concrete (RC) frames. The building codes must take parameters like bay span and the presence of infill walls into account for estimating the fundamental period.

Appendices

Appendix

Sample calculation performed by the computer program

To elaborate on the method that the computer program uses, a sample calculation is presented here:

In the following steps, the fundamental period of the building with the below-mentioned geometry will be calculated.

Geometry

Number of stories ($N = 3$), story height ($h = 3$), bay span ($B_{s} = 5$) and two bays along each axis ($B_{x} = 2,B_{y} = 2$).

Member sizes and strength

Column size 0.3 m by 0.3 m, Beam Depth 0.500 m, Beam Width 0.300 m, Wall Thickness 0.230 m, Slab thickness 0.150 m, Column Elasticity $5000\sqrt {25} {\text{MPa}} = 2.500*10^{10} {\text{Pa}}$, Wall Elasticity $2.65*10^{9}$ Pa, Concrete Unit Weight 25 kN, Masonary Unit Weight 20 kN or Concrete Density = 2548.420 kg/m³ and Masonry Density = 2038.736 kg/m³.

Mass matrix

Mass calculation for Floor 1 and 2.

Total extra load of (3.5 kN/m² = 356.779 kg/m² (live load + floor finish)) on the floor is added on slab and subtracted from column and wall area.

• 1. Each column = 2548.420*0.300*0.300*2.850 − 356.779*0.300*0.300 = 621.560 kg

• 2. Columns total = 621.560*9 = 5594.04 kg

• 3. Each beam span = 2548.420*4.700*0.300*0.350 = 1257.645 kg

• 4. Beams total = 1257.645*12 = 15,091.74 kg

• 5. Each wall span = 2038.736*4.700*0.230*2.500 − 356.779*4.700*0.230 = 5124.006 kg

• 6. Walls total = 5124.006*12 = 61,488.072 kg

• 7. Slab = 2548.420*10.300*10.300*0.150 + 10.300*10.300*356.779 = 78,404.944 kg

Total mass = 160,578.796 kg.

And from the distribution of mass from the floor plan, the centroids and the mass moment of inertia about the centroid is calculated to be:

XC	5.000	X-coordinate of centroid
YC	5.000	Y-coordinate of centroid
MASS	160,578.798	Total mass
IXX	1,734,595.484	Moment of inertia along X-direction from the centroid
IYY	1,734,595.484	Moment of Inertia along Y-direction from the centroid
IPC	3,469,190.968	Polar Moment of Inertia

Mass calculation for 3rd floor

• Each column = 2548.420*0.300*0.300*1.350 = 309.633

• Each beam span = 2548.420*4.700*0.300*0.350 = 1257.645

• Each wall = 2038.736*4.700*0.230*1.000 = 2203.874

• Slab = 2548.420*10.300*10.300*0.150 + 10.300*10.300*0.000 = 40,554.283

MASS	84,879.207
IPC	1,838,611.448

Finally, we get the mass matrix:

160,578.798	0.0	0.0	0.0	0.0	0.0	0.0	0.0	0.0
0.0	160,578.7981	0.0	0.0	0.0	0.0	0.0	0.0	0.0
0.0	0.0	3,469,190.968	0.0	0.0	0.0	0.0	0.0	0.0
0.0	0.0	0.0	160,578.7981	0.0	0.0	0.0	0.0	0.0
0.0	0.0	0.0	0.0	160,578.7981	0.0	0.0	0.0	0.0
0.0	0.0	0.0	0.0	0.0	3,469,190.968	0.0	0.0	0.0
0.0	0.0	0.0	0.0	0.0	0.0	84,879.2070	0.0	0.0
0.0	0.0	0.0	0.0	0.0	0.0	0.0	84,879.2070	0.0
0.0	0.0	0.0	0.0	0.0	0.0	0.0	0.0	1,838,611.448

Stiffness matrix

• Area moment of inertia of a column $I_{c} = \frac{1}{12}0.3^{4}$ = 6.75*10^–4

• K of column = $12\frac{{E_{c} I_{c} }}{{h^{3} }}$ = 75*10⁵

Infill wall (length l = 5–0.3 = 4.7 m, height = 3–0.5 = 2.5 m).

• $L_{ds}$ = 5.323

• $\theta$ = 0.489

• $\alpha$ = 3.289

• $w_{ds}$ = 0.579

• $t_{ds} = \max \left( {\frac{h}{12},\frac{l}{12},\;b = 0.23} \right)$ = 0.392

• Strut’s stiffness $K = \frac{{t_{ds} w_{ds} E_{w} }}{{l_{ds} }}\cos^{2} \theta$ = 8.794*10⁷

For a single frame in X-direction (single floor).

• Stiffness from walls = 8.794*10⁷*2 = 17.58*10⁷

• Stiffness from columns = 75*10⁵ * 3 = 2.25 * 10⁷

• Total stiffness = ${k}_{1}={k}_{2}={k}_{3}=19.84*{10}^{7}$

Local stiffness matrix ($K_{lx1}$).

396,755,259.2	− 198,377,629.6	0.0
− 198,377,629.6	396,755,259.2	− 198,377,629.6
0.0	− 198,377,629.6	198,377,629.6

Local coordinate to the global coordinate transformation matrix ${A}_{x1}$ (for the first frame along X-axis).

1.0	0.0	5.0	0.0	0.0	0.0	0.0	0.0	0.0
0.0	0.0	0.0	1.0	0.0	5.0	0.0	0.0	0.0
0.0	0.0	0.0	0.0	0.0	0.0	1.0	0.0	5.0

Finally, Stiffness matrix is obtained $K = \sum \left[ {A_{xi}^{T} K_{lxi} A_{xi} + A_{yi}^{T} K_{lyi} A_{yi} } \right]$.

1,190,265,777.6	0.0	− 0.0000009537	− 595,132,888.8	0.0	0.00000047684	0.0	0.0	0.0
0.0	1,190,265,777.6	0.0	0.0	− 595,132,888.8	0.0	0.0	0.0	0.0
− 0.0000009537	0.0	39,675,525,920	0.00000047684	0.0	− 19,837,762,960	0.0	0.0	0.0
− 595,132,888.8	0.0	0.00000047684	1,190,265,777.6	0.0	− 0.0000009537	− 595,132,888.8	0.0	0.00000047684
0.0	− 595,132,888.8	0.0	0.0	1,190,265,777.6	0.0	0.0	− 595,132,888.8	0.0
0.00000047684	0.0	− 19,837,762,960	− 0.0000009537	0.0	39,675,525,920	0.00000047684	0.0	− 19,837,762,960
0.0	0.0	0.0	− 595,132,888.8	0.0	0.00000047684	595,132,888.81	0.0	− 0.0000004768
0.0	0.0	0.0	0.0	− 595,132,888.8	0.0	0.0	595,132,888.81	0.0
0.0	0.0	0.0	0.00000047684	0.0	− 19,837,762,960	− 0.0000004768	0.0	19,837,762,960

Once we have the stiffness and mass matrix we can solve the eigenvalue problem $\left( {K - M\Omega^{2} } \right)\Phi = 0$.

• Eigenvalues $\left\{ {\omega_{i}^{2} } \right\}$ = 13,587.9, 7273.8, 974.4, 20,947.9, 11,212.6, 1502.0, 974.4, 13,587.9, 7273.8

• Time-periods (sorted) $T_{i} = 2\pi /\omega_{i}$ = 0.2012, 0.2012, 0.1621, 0.0736, 0.0736, 0.0593, 0.0539, 0.0539, 0.0434,

Hence, the fundamental period is 0.2012s.