11.2.1. Column Shears

V = 15 = 2.5 k 5 (2)(3) V = 2(V ) = (2)(2.5) = 5 k 6 5 V = 2(V ) = (2)(2.5) = 5 k 7 5 V = V = 2.5 k 8 5 V = 15 + 30 = 7.5 k

This section provides the calculations for column shear forces. The first line states that for some load conditions, the shear force (V) in the column is calculated to be 15. It also specifies that with two bays (2.5 k each), the overall shear for the column can be summed or multiplied accordingly. This method illustrates how to derive shear forces from previous calculations. The shear values are further adjusted based on the total loads, arriving at a definitive column shear value.

Imagine a bridge holding up cars. If two cars go over one section, the load the bridge holds can increase significantly. Calculating shear forces in a similar way helps engineers ensure the bridge can support additional weight safely.

Mtop = V1H5 = (2.5)(14) = 17.5 k.ft Mbot = Mtop = 17.5 k.ft Mtop = V6H6 = (5)(14) = 35.0 k.ft Mbot = Mtop = 35.0 k.ft

This section analyzes the moments acting on the top column due to shear forces. The top moment (Mtop) is calculated by multiplying the shear force (V1) by the height (H5) to find the moment at the top of the column. Similarly, it shows that the bottom moment (Mbot) equals the top moment, maintaining equilibrium. This repetition demonstrates consistency within structural designs, ensuring forces remain balanced throughout the structure.

Think of a tall tree swaying in the wind. The forces acting on it cause it to momentarily bend. Engineers need to calculate the moments to ensure trees—and buildings—can withstand these forces without breaking.

Mtop = V1dwnH1 = (7.5)(16) = 60 k.ft Mbot = Mtop = 60 k.ft Mtop = V2dwnH2 = (15)(16) = 120 k.ft Mbot = Mtop = 120 k.ft

In this section, the changes in moments at the bottom of the column are accounted for. Similar to the calculations earlier, it applies the load (V1dwn) across a height (H1) to find the bottom moments (Mbot). This establishes a system of loading that confirms equilibrium through the structure by ensuring that upward and downward forces match.

Consider how you must balance a heavy box on a stick. If you push down on one end, the other end must push back with equal force or the stick will break. Similarly, these calculations ensure all forces in a structure adequately balance each other.

P = V = ( 1.75) k P = +V V = 1.75 ( 1.17) = 0.58 k

This part explains how to calculate the axial forces acting on the columns. It emphasizes that forces in different columns interact with the shear forces calculated earlier, leading to different tension or compression forces. Understanding these axial forces is crucial for determining if the columns can withstand the loads without failure.

Imagine a stack of plates; if you add a plate on top, the bottom plates have to carry the weight. If too many plates are stacked, the bottom plates might buckle. Similar calculations ensure columns can safely hold their 'plates' of load.

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