11.2. Horizontal Loads

1. Column Shears

V = 15 = 2.5 k V = 2(V ) = (2)(2.5) = 5 k V = 2(V ) = (2)(2.5) = 5 k V = V = 2.5 k V = 15 + 30 = 7.5 k V = 2(V ) = (2)(7.5) = 15 k

In this part, we calculate the shear force at various points in a building's columns. We start by using basic equilibrium principles. The total shear at each column is influenced by the loads applied above it. This calculation provides the necessary values for designing the column strength to resist lateral forces such as wind or seismic activity.

Think of the columns as the legs of a table. If you push the table from one side, the legs on that side will bear more weight (shear). Understanding how much load each leg (column) experiences helps ensure that the table (building) does not tip over.

2. Top Column Moments

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 M6 top = V7upH6/7 = (5)(14) = 35.0 k.ft Mbot = Mtop = 35.0 k.ft M7 top = V8upH7/8 = (2.5)(14) = 17.5 k.ft Mbot = Mtop = 17.5 k.ft

Here, we calculate the bending moments (Mtop and Mbot) at the top and bottom of the columns. The formulas given show how the moment is affected by the shear force acting on the column and the height (H) of the column. These moments are crucial in ensuring that the columns can resist bending and maintain structural integrity.

Imagine squeezing a tube of toothpaste from the bottom. The pressure you exert creates a moment at the top of the tube. Similarly, the applied forces in a building create moments at the columns, affecting how they bend under load. Understanding these moments helps engineers design columns that can withstand bending without failing.

3. Bottom Column Moments

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 Mtop = V3dwnH3 = (15)(16) = 120 k.ft Mbot = Mtop = 120 k.ft Mtop = V4dwnH4 = (7.5)(16) = 60 k.ft Mbot = Mtop = 60 k.ft

In this part, we calculate similar bending moments at the bottom part of the columns. This time, we focus on the downward forces acting on these columns. By solving these equations, we ensure that the bottom of the columns can also withstand the bending caused by the loads they support, thus providing stability to the entire structure.

Continuing with the toothpaste analogy, if the top of the tube is successfully squeezed to hold its shape, the bottom must also support that shape under pressure. This analogy reflects how columns must be designed to handle forces from above without bending excessively or collapsing.

4. Top Girder Moments

lft top M = M = 17.5 k.ft rgt lft M = M = 17.5 k.ft lft rgt top M = M + M = 17.5 + 35 = 17.5 k.ft rgt lft M = M = 17.5 k.ft

Top girder moments are essential for understanding the forces acting on horizontal beams (girders) at the top of the structure. These calculations help determine how the girders will flex under the loads. It is vital for ensuring that girders can support loads without bending excessively.

Think of a long, flexible stick that someone is trying to bend by pressing down on it from both ends. The moments calculated here help us understand how much bending the stick (girder) can sustain before it starts to fail.

6. Top Girder Shear

lft V = 2Mlft/(12L12-20)=1.75 k rgt lft V = +V = 1.75 k lft V = 2Mlft/(13L13-30)=1.17 k rgt lft V = +V = 1.17 k lft V = 2Mlft/(14L14-24)=1.46 k rgt lft V = +V = 1.46 k

The top girder shear calculations provide insights into the vertical forces that the girders need to resist. By calculating the shear forces across different points, engineers can ensure that girders are properly supported to prevent shear failure.

Imagine cutting a sandwich with a knife. The force exerted by the knife creates shear forces on the sandwich. Similarly, in a building, the weights from above create shear forces on girders. By understanding these, engineers can design safer structures.

7. Bottom Girder Shear

lft V = 2Mlft/(12L12-20)=7.75 k rgt lft V = +V = 7.75 k lft V = 2Mlft/(10L10-30)=5.17 k rgt lft V = +V = 5.17 k lft V = 2Mlft/(11L11-24)=6.46 k rgt lft V = +V = 6.46 k

The bottom girder shear calculations are similar to the top girder calculations but focus on the lower part of the girders. These values help ensure that the foundation and lower parts of the structure can support not just the weight but also forces acting on the building from different directions.

If we think of the building as a giant sandcastle under pressure from waves, the bottom girders need to be strong enough to withstand those waves, just as you must ensure the base of your sandcastle is solid to hold it together against the tide.

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