Issue 51

E. Mousavian et alii, Frattura ed Integrità Strutturale, 51 (2020) 336-355; DOI: 10.3221/IGF-ESIS.51.25 351 For such a convex model specialized for the hemispherical dome, the constraints on all the six internal forces preventing the mentioned structural failures are as follows: to avoid the block separation, f n must be in compression ( f n ≤ 0); two tangential forces are considered along the locks f t 1 , and normal to the locks f t 2 ; f t 1 must satisfy the Coulomb’s friction law ( f t1 ≤ μ f n ), otherwise the block slides along the locks; f t 2 and torsion moment t r must be less than the lock shear, bending and torsional resistance, as explained later; lastly, since the flow of forces, within the dome thickness, passes through the contact points of the convex model, bn 1 and bn 2 are zero and ignored (Fig. 16c). (a) (b) (c) Figure 16 : a) Contact point at the interface centroid according to the convex contact model in [7]; b) contact point as the intersection point of the interface and the network of forces according to the proposed convex contact model for the hemispherical dome; c) internal force on the contact point of the proposed convex contact model. The lock bending, shear, and torsional resistance can be found as follows: first of all, following [2] bending failure is neglected at the fracture plane through g ≤ h , where g and h are the thickness and height of the lock as already introduced above (Fig. 1). This means that the lock is thick enough so that, when subjected to a lateral force, it does not bend generating crack where the lock is connected to the main body of the lock. Besides, a conservative formulation to avoid both shear and torsional failures at locks is herein proposed. The tangential forces normal to a lock can be distributed uniformly or non-uniformly. If the force is uniformly distributed on lock k (Fig. 17a), its limiting shear value T ky equals ( t y b k ), where t y = ( g  ) is the yielding force per unit length,  is the material shear strength, g is the lock thickness and b k is length of the lock, calculated through Eqns. (25) to (31). When the force is non-uniformly distributed on lock k (Fig. 17b), the lock may slide and/or twist after reaching the limiting value. For this case, the shear resistance of the lock is lower than the limiting value, since for a non-uniformly distributed force, whose resultant equals t y b k , there would be at least one point at which the force per unit length t b is more than t y . For such a non-uniformly distributed force, the overall shear resistance of the lock can be reduced to T ky = ( cf k t y b k ), where cf k is a coefficient less than one (Fig. 17b). Applying this coefficient, a combination of shear and torsion of the lock is considered, which becomes pure torsion for cf k = 0 and pure shear for cf k = 1 [7]. Given f t 2 , the tangential force at each lock can be considered as a component of f t 2 ( f t 2 k ) subjected to an intersection point of f t 2 and the centreline of the k th lock (Fig. 17c and d). Then, the following scalar summation represents the relation between f t 2 and its components: 1 2 2 2 0 m k t t k f f     (32) When the intersection point of f t 2 and the centreline of the k th lock is not the centreline mid-point, the equivalent distributed force of f t 2 k is spread along the centreline non-uniformly. Therefore, f t 2 k ≤ cf k t y b k . In sum, the interlocking interface in Fig. 18 can be designed so that the fracture occurs on a flat face, i.e. fracture occurs only at one of the two interlocked blocks where its locks are attached to the main body of that block (Fig. 18), considering an overall shear resistance as: 1 1 2 2 2 2 0 0 m m k t t y k k k k f f t b cf         (33)