Question

Determine the maximum shear stress in a W12 X 40 steel I-beam using the simplified Shear Stress equation for steel shapes. It has been determined from Statics that the internal shear force is 30 kip. A18 APPENDIX C Properties of Rolled-Steel Shapes (U.S. Customary Units) Continued from page A17 W Shapes (Wide Flange Shapes Flange Web Thick Thick Width nessness by in. in. toin. Axis X-X Axis Y-Y Area Ain? Depth d. in. in Sy, in Fin. Designationt W12 X 96 50 833 597 391 307 982 21.1 146 11.7 103 8.79 7.65 6.48 4.71 19.7 12.3 192 11.9 19,5 12.3 122 12.3 120 12,2 0.900 120 0.670 8.04 0.640 8.01 0.515 6.56 0.500 6.590.440 6.49 0.380 0.495 3.90 0.265 00 0.430 0.370 0.095 0000 0 OGO 0.930 0900 0.260 0.920 Sin Fin. in 1315 .44 1970 974 531 195 64. 2 5 .18 563 515 5.13 44.1 45. 6 5 .95 94. 5 38. 6 5 .21 203 31.4 5.17 173 25.4 4.91 4.66 4.67 2.82 983 9:39 444 3.09 324 3.04 3.9 1.96 110 1,94 7473 .54 6 .94 1.52 3.34 1.51 2.310 .848 1.41 0.773 2014 156 103

Answer 1

giren S= 30 kip . tw=0.295 in dallagin, 2018- k = 8.01 in I=sotinu g=1119 6=0.295 Aj, - for a haded area obove AA A = (0.515) (8.01 = 4.125 in flison f1.603) of 5.u3rxo.rar - 5.7281 in ² y, for shaded Area Ji = 0.51548.00) 05.435+ 0.09460.545)45:436) (oryo 240 295 (0.515x8.01) + costs x 5.435) 5 = 23.4824 & zerot u.35t 9.120 + 200 1.603) = 4.8599 in man ot N.A= SAG ab 30) (5.7283) (4.8599) 307x0.295 Tmax = 9.2217 kpline