Question 3.CS.1: The Vlooybergtoren tower in Tielt-Winge, Belgium, was constr...
The Vlooybergtoren tower in Tielt-Winge, Belgium, was constructed to provide a distinctive and unique platform for visitors to view the Kabouterbos “fairytale forest” (CS Photo 3.1). The staircase and observation deck is supported by a structural steel frame (CS Photo 3.2) that is clad in weathering steel (which oxidizes to produce the reddish-orange hue shown, forming a protective layer that inhibits further corrosion). Overall, this cantilever structure rises 11.3 m above the ground and weighs approximately 130 kN.* The base of the tower is supported against overturning by the anchor points shown in CS Photo 3.2.
Considering only the self-weight of the tower, let’s estimate the resulting equivalent force-couple applied at the support that prevents uplift (i.e., the anchor toward the rear of the tower). We will then use this equivalent force-couple to determine the total uplift force acting on the support.
STRATEGY: Use a two-dimensional model to represent the tower, and simple supports (a pin and a roller) to model the actual base conditions. Replace the self-weight load with an equivalent force-couple at the uplift anchor support. Then, by replacing the moment of the couple with vertical forces applied at the two support locations, the overall uplift force can be determined.
MODELING: CS Fig. 3.1 provides the geometry assumed for the tower, with supports A and B located at the tower’s base and below the ground surface as shown. (Support A reflects the anchorage subject to uplift.) The 135-kN dead load is applied at the structure’s center of gravity, a concept that will be examined in detail in Chap. 5. (For demonstration purposes, we will assume a center of gravity G approximated as shown in CS Fig. 3.1. It has been positioned closer to the left end than the right because the supporting structure becomes increasingly heavier toward the base of the tower.)
ANALYSIS:
a. Force-Couple System at A.
To replace the given 135-kN force, move the force F = −(135 kN)j to A, and at the same time, add a couple of moment \pmb{M_A} that is equal to the moment about A of the force in its original position (CS Fig. 3.2a):
\pmb{M}_A=\overrightarrow{AG} × F =[(8.6 m)i +(4.7 m)j]×(−135 kN)j
= −(1161 kN⋅m) k
b. Vertical Forces Equivalent to Moment of Couple \pmb{M_A}.
To replace the moment of couple \pmb{M_A} with two equal and opposite vertical forces at support locations A and B, separated by perpendicular distance d = 1.5 m, divide M_A by this perpendicular distance. The resulting magnitude of each force is
F=\frac{M_A}{d}=\frac{1161 kN⋅m}{1.5 m}=774 kNThese forces are directed as shown in CS Fig. 3.2b.
c. Uplift Force at Anchor A.
Combining the forces acting at A in CS Fig. 3.2b, the result shown in CS Fig. 3.2c is obtained. Being a complete system that is equivalent to the original load, the pair of forces in CS Fig. 3.2c represents the total load exerted by the tower’s self-weight on these supports. Thus, the total uplift exerted on the anchorage at A is the equivalent force at this point, or
639 kN ↑
(Because the structural frame consists of two equal sides, this total uplift force would be divided equally over both sides.)
REFLECT and THINK: The equivalent forces exerted on the supports, as shown in CS Fig. 3.2c, are equal and opposite to the support reactions acting on the structure at these points. Such reactions can be determined more directly by the principles of rigid-body equilibrium that we will examine in Chap. 4.
*Source: “What’s Cool in Steel?” Modern Steel Construction, Chicago, IL: The American Institute of Steel Construction, August 2016, pp. 42–43.
