Issue 51
G. Cocchetti et alii, Frattura ed Integrità Strutturale, 51 (2020) 356-375; DOI: 10.3221/IGF-ESIS.51.26 360 self-weight distribution along the arch ( CCR = 1, M = 1), respectively leading to a “linear” , a “quadratic” and a “cubic problem” in the algebraic solution for unknown triplet A ( ), ( ), h ( ). This is originally and extensively derived, described and discussed in [6]. All what will be considered in the following will stick to classical hypothesis M = 0, i.e. by assuming a self-weight distribution along the geometrical centreline of the arch, in classical Heyman sense, though with correct evaluation CCR = 1 of the tangency condition in Eqn. (1) c . An extensive discussion on the differences between the three arising solutions has been reported in [6], including about the spreading discrepancies appearing for over-complete (horseshoe) circular masonry arches. Known Heyman “linear” solution may finally be obtained from Eqns. (1), for CCR = 0, M = 0, and can be classically represented as: ( )( ) ( ) 2 2 2 cos sin cos sin cot cot 2 2 cos sin cos sin cos sin 1 cos 2 1 cos cot H H A h h + + = = + − − − = = + = = (3) Going to properly correct CCR solution ( CCR = 1, M = 0), for the complete semi-circular arch, i.e. = A = /2, system (1) renders the following characteristic CCR solution triplet for the purely-rotational collapse mode to be recorded: 0.951141 54.4963 , 0.107426, 0.621772 ( / 2) r r r rad h = = = = = (4) More generally, at variable half-opening angle of the circular masonry arch (thus at variable A = cot ( /2)), the solution of system (1) for CCR = 1, M = 0 can be analytically represented in closed-form as follows [6]: 2 2 2 2 2 2 2 (2 ) 2 2 2 2 f f gS S A g S g S g S f gS S f g f S f gS S h S − + = − − − + = + − − + = with 2 ( ) ( sin )' ( ) ( ) sin ( ) cos f f S C g g S Cf SC S S C C = = = + = = + = + = = = = (5) Indeed, triplet A ( ), ( ), h ( ) becomes a double-valued function of inner hinge position , coming from the solution of the following “quadratic problem” [6]: 2 2 2 2 (2 ) 2 0 ( ) 4( ) 4( ) 0 2 2( ) 0 S g S A fg A g f g g S g f S h f S h g f − − + = + − − + − = − − + − = (6) The above three quadratic equations in A , , h can be obtained from system (1) by eliminating in turn couples ( , h ), ( A , h ) and ( A , ). Notice that solutions (5) b and (5) c for and h can be obtained from a 2×2 subsystem formed by Eqns. (1) a and (1) c , namely those independent on A ( ) in source system (1). Functions A ( ), ( ), h ( ) become single-valued at = s = 1.12909 rad = 64.6918°, namely at the value of setting to zero the term under square roots
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