32. Case Study III: MAGAZINI GENERALI

This storage house, built by Maillart in Chiasso in 1924, provides a good example of the marriage between aesthetic and engineering.

The most striking feature of the Magazini Generali is not the structure itself, but rather the shape of its internal supporting frames, Fig. 32.1.

The frame can be idealized as a simply supported beam hung from two cantilever column supports. Whereas the beam itself is a simple structural idealization, the overhang is designed in such a way as to minimize the net moment to be transmitted to the supports (foundations), Fig. 32.2.

This section introduces the Magazini Generali storage house, explaining its significance in combining beauty with structural engineering. It highlights the unique internal frame shape, indicating that while the frame functions like a simple beam supported at both ends, it has been carefully designed to reduce forces acting on its foundational supports. This reduction of moment helps in stabilizing the structure and enhances its aesthetic value.

Think of a well-designed hanging sign at a store: it not only needs to look good but must also be securely held up by the brackets. Just like how a sign needs to distribute its weight evenly to avoid falling, the internal frames of Magazini Generali are engineered to ensure stability while looking elegant.

The load applied on the frame is from the weights of the roof slab and the frame itself. Given the space between adjacent frames is 14.7 ft, the roof load is 98 psf, and the total frame weight is 13.6 kips, the total uniform load becomes:

q = (98 psf)(14.7 ft) = 1.4 k/ft (32.1-a)

q = (13.6 k)/(63.6 ft) = 0.2 k/ft (32.1-b)

q = 1.4 + 0.2 = 1.6 k/ft (32.1-c) total.

This part focuses on how to calculate the total loads acting on the structure. Here, we consider both the roof slab's weight and the weight of the frame itself. By using the given dimensions and weights, we calculate the load from the roof and the load from the frame, which together give us a total uniform load on the beams. This total load is crucial for understanding how much weight the beams must support.

Imagine carrying a backpack. First, you weigh the bag itself, and then you consider how many books you're putting in it. You need to know the total weight to ensure you can comfortably carry it. Just as you add the weight of the bag to the weight of the books, in engineering, we sum the weights of various structural components to find out the total load that needs to be supported.

Reactions for the beam are determined first taking advantage of symmetry:

W = (1.6 k/ft)(63.6 ft) = 102 k (32.2-a)

R = W/2 = 102 k/2 = 51 k (32.2-b)

We note that these reactions are provided by the internal shear forces.

In this section, the reactions at the supports of the beam are calculated. We leverage the symmetry of the structure, which simplifies our calculations since the load is evenly distributed. Knowing the total load allows us to easily find the reaction forces at each support of the beam. These reactions are essential as they show how the forces from the load are transferred to the supports, ensuring structural integrity.

Consider a seesaw. When two children of equal weight sit at opposite ends, the seesaw balances perfectly in the middle. Each side pushes against the ground with an equal force. Similarly, in our beam, the distribution of weight leads to equal reactions at the supports, which are crucial for maintaining balance and stability.

The internal forces are primarily the shear and moments. Those can be easily determined for a simply supported uniformly loaded beam. The shear varies linearly from 51 kips to -51 kips with zero at the center, and the moment diagram is parabolic with the maximum moment at the center, equal to:

Mmax = (q*L^2)/(8) = (1.6 k/ft)(63.6 ft^2)/8 = 808 k.ft (32.3)

This section describes the internal forces acting within the beam, specifically the shear forces and moments. The shear force varies linearly, indicating that it is strongest at the supports and zero at the midpoint. It explains how the moment, which is the rotational effect caused by the forces, reaches its peak at the center of the beam. Understanding these forces is crucial in ensuring that the beam can withstand the loads applied to it effectively without failing.

Think about bending a stick. When you press down in the middle, the ends push up; that's how shear works. The maximum bending occurs where you apply the pressure the most—in the center—just as you’ve seen in a seesaw or a diving board when the weight is at the furthest point from the support.

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