Please answer with complete details. Thank you.L = 18N = 219M = 311.9 A W 10 x 45 is used as a beam with two linesDetermine the effective net area thatcan be assumed to resist tension of theof L mm diameter boplts in each flange usingA 36 steel. There are at least three bolts ineach line and the bolts are not staggered withUsesection through the holes.Ae = U An U= 0.90Use LRFD Method.® Determine the design strength due tofracture in the net section where bolt orrespect to each other. Use LRFD Method.Prop. of W 10 x 45A = 8581 mm2d = 256.54 mmrivet holes are present. o = 0.75Use LRFD Method.® Determine the design strength due toyielding in the gross section which isintededbr = 203.71 mmtr = 15.75 mmFy = N MPaFu = M MPato prevented excessivetw = 8.89 mmelongation of the member. ø = 0.90%3D(4) Determine the allowable strength203.71 mmdue to fracture. Use ASD Method.(5) Determine the allowable strengthdue to yielding. Use ASD Method.15.75 mmDiam of hole = L +3Diam of hole = mm.d= 256.64 mm15.75 mm

Given: W10X45 section, A36 steel 18mm dia bolts, Fy=219 MPa, Fu=311.9 MPa A=8581 mm2 Dia of hole…

45 is used as a beam with lihes can be assumed to resist tension of the of L mm diameter boplts in each flange using A 36 steel. There are at least three bolts in each line and the bolts are not staggered with section through the holes. Ae = U An U = 0.90 respect to each other. Use LRFD Method. Use LRFD Method. Determine the design strength due to fracture in the net section where bolt or Prop. of W 10 x 45 A = 8581 mm2 d 256.54 mm rivet holes are present. o = 0.75 Use LRFD Method. Determine the design strength due to yielding in the gross section which is inteded %3D Di = 203.71 mm = 15.75 mm w = 8.89 mm Fy = N MPa Fu = M MPa to prevented excessive elongation of the member. ø = 0.90 (4) Determine the allowable due to fracture. Use ASD Me b, = 203.71 mm (5) Determine the allowable due to yielding. Use ASD Me 15.75 mm am of hole = L +3 am of hole = mm. d= 256.64 mm 575

45 is used as a beam with lihes can be assumed to resist tension of the of L mm diameter boplts in each flange using A 36 steel. There are at least three bolts in each line and the bolts are not staggered with section through the holes. Ae = U An U = 0.90 respect to each other. Use LRFD Method. Use LRFD Method. Determine the design strength due to fracture in the net section where bolt or Prop. of W 10 x 45 A = 8581 mm2 d 256.54 mm rivet holes are present. o = 0.75 Use LRFD Method. Determine the design strength due to yielding in the gross section which is inteded %3D Di = 203.71 mm = 15.75 mm w = 8.89 mm Fy = N MPa Fu = M MPa to prevented excessive elongation of the member. ø = 0.90 (4) Determine the allowable due to fracture. Use ASD Me b, = 203.71 mm (5) Determine the allowable due to yielding. Use ASD Me 15.75 mm am of hole = L +3 am of hole = mm. d= 256.64 mm 575

Steel Design (Activate Learning with these NEW titles from Engineering!)

6th Edition

ISBN:9781337094740

Author:Segui, William T.

Publisher:Segui, William T.

Chapter3: Tension Members

Section: Chapter Questions

Problem 3.6.3P

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