After determining the shear per floor, the next step is to distribute the shear into the frames. This procedure would require the configuration of columns in each frame.
Horizontal distribution is required for the two axes - among the x-frames and the y-frames. With both axes, the same value of shear per floor will be distributed.
The shear would be distributed accordingly: direct shear + torsional shear.
- Direct shear, Vd, is solved like the vertical distribution of the base shear. The procedure would only account for the stiffness of each column in a frame whereby the percentage distribution of this stiffness is used to distribute the base shear computed previously.
- Torsional shear, Vt, on the other hand, needs a longer computation. This part of the shear would put the center of gravity and center of rigidity into account. The distance between the Cg and Cr is computed and is used in determining the torsion of the building when subjected to base shear. This distance is known the eccentricity.
Center of gravity - the point where the lateral loads is applied same as how discussion in physics are discussed.
Center of rigidity - the point where the structure will not be able to rotate, if force is applied.
Eccentricity - the distance between the center of gravity and the center of rigidity. This is taken as the moment arm to which moment is observed when a force is applied to the structure.
COMPUTATION OF DIRECT SHEAR, Vd
Going back to the previous discussion on how the base shear is vertically distributed among the floors of a structure, the procedure in distributing direct shear is similar. However, instead of floors, the components used are now the frames. Each frame may differ from the others, thus apart from the idea that the frames are not located equally from the Cg, the stiffness of each frame should be considered. Stiffness has been derived as
[ for both ends fixed]
where:
I = moment of inertia
L = length of column
The relative stiffness of each column in a frame is then simplified to k=I taken into account that all lengths of columns in one floor are the same. Thus the stiffness formula would be:
and 
| X-Frames | ksx | %ksx | Vd |
|---|
| 4 | -- | 0.-- | -- |
| 3 | -- | 0.-- | -- |
| 2 | -- | 0.-- | -- |
| 1 | -- | 0.-- | -- |
| Σ | | 1 | Vf |
| Y-Frames | ksy | %ksy | Vd |
|---|
| A | -- | 0.-- | -- |
| B | -- | 0.-- | -- |
| C | -- | 0.-- | -- |
| D | -- | 0.-- | -- |
| Σ | | 1 | Vf |
where: ksx and ksy = the summation of moment of inertia of columns in one frame
COMPUTATION FOR TORSIONAL SHEAR, Vt
- For torsional shear, Vt, the center of rigidity is already considered in order to determine the eccentricity (functioning as the moment arm). This eccentricity is used to solve for the moment per floor by multiplying it with the base shear.
NOTE: As per code the minimum eccentricity of any structure is taken as 5% of the length in the direction analyzed. Thus the standard eccentricity, ex, is computed as:
- The moment applied on the floor is then computed as the base shear on the floor, Vf, multiplied by the moment arm, ex.
 "M_{f}=V_{f}\left ( e_{x} \right )")
- The distribution of this moment will be as follows:
| X-frames | ksx | c | c/c1 | ksx' | M' | %M | Applied M | Vtx |
|---|
| 4 | -- | -- | -- | -- | -- | 0.-- | -- | -- |
| 3 | -- | -- | -- | -- | -- | 0.-- | -- | -- |
| 2 | -- | -- | -- | -- | -- | 0.-- | -- | -- |
| 1 | -- | -- | -- | -- | -- | 0.-- | -- | -- |
| | | | | Σ | 1 | M | |
| Y-frames | ksy | c | c/c1 | ksy' | M' | %M | Applied M | Vty |
|---|
| A | -- | -- | -- | -- | -- | 0.-- | -- | -- |
| B | -- | -- | -- | -- | -- | 0.-- | -- | -- |
| C | -- | -- | -- | -- | -- | 0.-- | -- | -- |
| D | -- | -- | -- | -- | -- | 0.-- | -- | -- |
| | | | | Σ | 1 | M | |
From the column of ksx (ksy), take the location of the center of rigidity with the formula:
- c is determined by computing the distance between the frame to the
.
- In most cases, the reference line is taken as the left-most frame or the bottom frame.
is taken as the distance of from the reference lines.
 "k_{sy}'=k_{sy}\left ( \frac{c}{c_{1}} \right )")
 "M'=k_{sy}'\left ( c \right )")
 "AppliedM=PercentageM\left ( M_{f} \right )")
- Torsional shear,
Summary of shear for each frame is as follows:
| X-Frames | Vd | Vt | Vtot |
|---|
| 4 | -- | -- | -- |
| 3 | -- | -- | -- |
| 2 | -- | -- | -- |
| 1 | -- | -- | -- |
| Y-Frames | Vd | Vt | Vtot |
|---|
| A | -- | -- | -- |
| B | -- | -- | -- |
| C | -- | -- | -- |
| D | -- | -- | -- |
which in units 
. This pressure (distributed load) is still to be converted into a force in order to be applied in the frame, thus 
