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Extra info for About Face 3: The Essentials of Interaction Design
Number of shear connectors over jirst stub of length 2050 mm NI = 2050/(95/2) = 44 Longitudinal shear force transfer R4 - 44 x 72 = 3168 kN > R,,, 45 P118: Design of Stub Girders Discuss me ... -- Rev. Job No. 7 kNm. Created on 30 March 2011 This material is copyright - all rights reserved. 4 k N m This exceeds the applied moment. Number of shear connectors over second stub N2 - 76 - 44 = 32 Nominal spacing of shear connectors over central stub, singly at 95 mm spacing N3 - = 10 1000/95 Total number of shear connectors in span - 46 76 X 2 + 10 = 162 P118: Design of Stub Girders Discuss me ...
Table 2: Model factors for different modes of failure in tests MODEL FACTORS’ Created on 30 March 2011 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement Test No. 75 I Mode of failure Shear failure Longitudinal ofstubsbending failure of shear connectors I I Failure of slab at outer stubs I Notes: 1. Model factor = Failure load / failure load determined from analysis of mode consideredand using measured properties.
Even allowing for this effect, equation (12) appears to under-predict the observed behaviour. Created on 30 March 2011 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement Only in Test 3 , with simple transverse reinforcement, would BS 5950 and/or the Chien and Ritchie methods appear to he directly applicable. Both methods under-predict the longitudinal shear transfer of the slab. However, calculations (not presented in this report) which included the shear resistance of the slab and reinforcement ahead (that is, on the centre-line side) of the stub, and the transverse tying actionof the secondary beam ahead of the stub, could remove most of the discrepancy shown in Table 3 between predicted and observed behaviour.