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Unformatted text preview: 4-2Determine the following for the connection shown assuming the bolts are ¾” diameter and the angle and plate is A36. Ignore block shear. 1) Tensile capacity of the angle 2) Required gusset plate thickness to develop the tensile capacity of the angle determined in part 1 above From AISC Table 2-3:From AISC Table 1-7:42Ag= 2.86 in.2Fy= 36ksi .in13.1xg32F = 58ksi to 80 ksi (use F = 58ksi)uut= 0.375 in. Net area of the angle: tg4stdAA2hgng54g14g54g16g32)375.(8143)86.2(g14g184g185g183g168g169g167g14g16= = 2.53 in.2Effective area of the angle: )33(13.11g14g16g32g34x1Ug16g32= 0.811 Alternatively, U = 0.60 from Table 4-1. The larger value is permitted to be used, so U = 0.792. 2Ae= A U =(2.53)(0.811) = 2.05 in.nDesign strength of angle: = (0.9)(36)(2.86) = 92.6kips(gross area)g73P = g73FnyAgg73Pn= g73FuAe= (0.75)(58)(2.05) = 89.3k ips (net area)g197controlsb) Plate thickness: )58)(75.(3.89FPAuueg32g73g32)36)(9.(6.92FPAyugg32g73g32= 2.86 in....
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This note was uploaded on 12/07/2011 for the course CES 4605 taught by Professor Prevatt during the Fall '11 term at University of Florida.
- Fall '11