EWPAA Structural Plywood and LVL Design Manual - Engineered ...

force at some point P (see FIGURE 12.15 (a)) on the edge, due to arch action, will be thesame as the force applied by the cable at that point. Also, because they act at equal anglesto the edge, but in opposite senses, there is no force component perpendicular to theedge member. Therefore, this double system of forces can be resolved into a series ofshear forces along the edge requiring a perimeter beam to carry them as shown in FIGURE12.15 (c).Since arching action is associated with compression forces, which in turn relates tobuckling, a limit must be placed on the ratio of the rise of the diagonal / span of thediagonal.Single shell support can be effected by providing suitable restraint at two support points,e.g. A & C in FIGURE 12.15 (a) being the most common. Accumulation of the membranalshears into the intersecting perimeter members at A & C results in larger thrusts having to beresisted at these two locations. This can be done by suitably designed buttresses or a tieacross AC which, although it is the most economical, detracts from appearance and reducesheadroom. Alternatively, the two high points (D & B) can be supported resulting in theperimeter members being in tension and the resultant force being inwards rather thanoutwards.12.13 Hypar Design - MethodologyThere are several methods available for determining the forces in a hypar shell the onefollowed herein is that presented in the Western Wood Products Technical Guide; HyperbolicParaboloid Shells.For symmetrical loading of the hypar shown in FIGURE 12.16 the vertical reactions (R) arehalf the sum of the vertical load (W). The horizontal thrust (H) can be determined byconsidering the triangle of base (/2), height (h) and hypotenuse (k). Since the total load(W) can be assumed to act vertically at (O) along the line of (h), and if the resultant of (H)and (R) is assumed to have its line of action (k), then summation of the moments of theforces to the left about (O) results in:Hence:M L ∑ 0 = 0= R. / 2 − HhR H=h / 2HR=2h(12.9)(12.8)Taking moments of the resultant force (F) and the vertical reaction (R) about (D) results in:R F=h k (12.10)Giving :Fk= Rh170