P = 68 kN

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answerhappygod
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P = 68 kN

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P = 68 kN
P 68 Kn 1
P 68 Kn 1 (77.62 KiB) Viewed 7 times
P 68 Kn 2
P 68 Kn 2 (29.02 KiB) Viewed 7 times
Directions: Show your solutions explicitly, i.e., do not just write the final answer. Always simplify your final answer. Write legibly in engineering lettering. Problem I. A steel beam cross section is assembled as shown by bolting the channel, C200 x 27.9, through its web on the wide flange, W200 x 35.9. The beam is 4 m in span and is simply supported loaded with a uniformly distributed load of intensity 20 kN/m throughout its span and a concentrated load, P (refer to the given table), at the midspan. If the diameter of the bolts is 22 mm, a) Compute for the moment of Inertia of the cross section b) Determine the maximum bending stress in the beam. c) Calculate the spacing of bolts (pitch) if the allowable shearing stress for a bolt is 100 MPa. 1 O
C200 x 27.9 properties Area Depth, d Flange (mm²) (mm) 3550 203 W200 x 35.9 properties Area Depth, d Flange (mm²) (mm) 4570 201 Width, br (mm) 64.3 Width, br (mm) 165 Thickness, tr(mm) 9.91 Thickness, tr(mm) 10.2 Web Thickness, tr(mm) 12.4 Web Thickness, tr(mm) 6.22 lx ly (x10 mm) (x108 mm²) 18.3 0.820 lx ly (x108 mm²) (x10³ mm²) 34.4 7.62 S (mm) 14.4
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