3. A W350x90 A36 steel section with a Fy = 248 MPa is used as a simply supported beam G(1) m long. The beam carries three equal concentrated loads at every quarter points. It also carries a uniform distributed load of L(2) kN/m (including its own weight). The allowable bending stress is 0.66F and allowable shear stress is 0.40Fy. Properties of W350x90: Flange width, br = 250 mm Flange thickness, tr = 16.40 mm Overall depth, d = 350 mm Web thickness, tw = 9.50 mm flange Moment of Inertia, I, = 266x 10° mm* Moment of Inertia, Iy = 44.54x10° mm“ Area, A = 11550 mm? web a. Determine the maximum value of each concentrated load based on flexure. b. Determine the maximum value of each concentrated load based on W Section shear.
3. A W350x90 A36 steel section with a Fy = 248 MPa is used as a simply supported beam G(1) m long. The beam carries three equal concentrated loads at every quarter points. It also carries a uniform distributed load of L(2) kN/m (including its own weight). The allowable bending stress is 0.66F and allowable shear stress is 0.40Fy. Properties of W350x90: Flange width, br = 250 mm Flange thickness, tr = 16.40 mm Overall depth, d = 350 mm Web thickness, tw = 9.50 mm flange Moment of Inertia, I, = 266x 10° mm* Moment of Inertia, Iy = 44.54x10° mm“ Area, A = 11550 mm? web a. Determine the maximum value of each concentrated load based on flexure. b. Determine the maximum value of each concentrated load based on W Section shear.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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G=5
M=9
L=6

Transcribed Image Text:3. A W350x90 A36 steel section with a Fy = 248 MPa is used as a simply supported beam G(1) m long. The beam
carries three equal concentrated loads at every quarter points. It also carries a uniform distributed load of L(2)
kN/m (including its own weight). The allowable bending stress is 0.66F and allowable shear stress is 0.40FY.
Properties of W350x90:
Flange width, b = 250 mm
Flange thickness, tr= 16.40 mm
Overall depth, d = 350 mm
Web thickness, tw = 9.50 mm
Moment of Inertia, I = 266x10° mm“
Moment of Inertia, Iy = 44.54x10 mm“
Area, A = 11550 mm?
flange
web
a. Determine the maximum value of each concentrated load based on
flexure.
b. Determine the maximum value of each concentrated load based on
W Section
shear.
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