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
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Please answer with complete details. Thank you.
L = 18
N = 219
M = 311.9
Transcribed Image Text:A W 10 x 45 is used as a beam with two lines
Determine the effective net area that
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
Use
section through the holes.
Ae = U An U= 0.90
Use LRFD Method.
® Determine the design strength due to
fracture in the net section where bolt or
respect to each other. Use LRFD Method.
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
br = 203.71 mm
tr = 15.75 mm
Fy = N MPa
Fu = M MPa
to prevented excessive
tw = 8.89 mm
elongation of the member. ø = 0.90
%3D
(4) Determine the allowable strength
203.71 mm
due to fracture. Use ASD Method.
(5) Determine the allowable strength
due to yielding. Use ASD Method.
15.75 mm
Diam of hole = L +3
Diam of hole = mm.
d= 256.64 mm
15.75 mm
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