G=5M=9L=6 3. A W350x90 A36 steel section with a Fy = 248 MPa is used as a simply supported beam G(1) m long. The beamcarries 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 mmFlange thickness, tr= 16.40 mmOverall depth, d = 350 mmWeb thickness, tw = 9.50 mmMoment of Inertia, I = 266x10° mm“Moment of Inertia, Iy = 44.54x10 mm“Area, A = 11550 mm?flangeweba. Determine the maximum value of each concentrated load based onflexure.b. Determine the maximum value of each concentrated load based onW Sectionshear.

Flexural members are those that experience essentially bowing burdens, like bars. A shaft is an…

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.

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Question

G=5

M=9

L=6

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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