This is the basis of the portal method assumption. So, it may be reasonable to assume that, since the interior columns are approximately twice as big, and therefore twice as stiff, as the exterior columns, those interior columns will take twice as much shear as the exterior columns. A column that is twice as stiff will take twice as much load for the same lateral displacement. If we then have three columns in parallel as shown and they all share the total lateral load at the top as shown, then they will resist the total load in shear in proportion to their relative stiffness. If the interior columns are twice as strong, they may also be approximately twice as stiff (as shown in the diagram at the top right of Figure 7.3). So, if we combine all of these individual portal frames together, our interior column (the sum of the two individual portal frame columns) will need to be twice as strong as the exterior columns. When we join these all together into a stacked system, we can see, as in the figure, that the interior columns have two portal frame columns each since they need to take axial force from the left and from the right (whereas the exterior columns only take gravity loads from the left or right). The columns on either end of each individual portal frame are likely similar size because they would each equally share the gravity load from above. Let's consider our multi-storey, multi-bay frame as a series of stacked single storey moment frames as shown at the top of Figure 7.3. Figure 7.3: Portal Method for the Approximate Analysis of Indeterminate Frames

Portal frame calculate