Using the assigned wide-flange section in the picture having Fy=260 MPa, determine the value of “Pn” (plastic distributed load) using the concept of collapse mechanism. FlangeWebMassDepth(mm)AreaWidthThicknessthickness(kg/m)(mm²)(mm)(mm)(mm)3144010078538433.519.7Axis X-XAxis Y-YS = I/c(10' mm')S lfe(10' mm')r= VIJA(to mm*)(mm)(10 mm)(mm)4 29010 900328315164088.6 0.6PnPn10 ft10 ft10 ft(a)

Solution: Given: fy = 260 MPa Span = 30 ft

Flange Web Mass Depth (mm) Area Width Thickness thickness (kg/m) (mm²) (mm) (mm) (mm) 314 40100 785 384 33.5 19.7 Axis X-X Axis Y-Y S = I/c (10' mm') S lfe (10' mm') r= VIJA (to mm*) (mm) (10 mm) (mm) 4 290 10 900 328 315 1640 88.6 0.6Pn Pn 10 ft 10 ft 10 ft (a)

Flange Web Mass Depth (mm) Area Width Thickness thickness (kg/m) (mm²) (mm) (mm) (mm) 314 40100 785 384 33.5 19.7 Axis X-X Axis Y-Y S = I/c (10' mm') S lfe (10' mm') r= VIJA (to mm*) (mm) (10 mm) (mm) 4 290 10 900 328 315 1640 88.6 0.6Pn Pn 10 ft 10 ft 10 ft (a)

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

Using the assigned wide-flange section in the picture having Fy=260 MPa, determine the value of “Pn” (plastic distributed load) using the concept of collapse mechanism.

Flange
Web
Mass
Depth
(mm)
Area
Width
Thickness
thickness
(kg/m)
(mm²)
(mm)
(mm)
(mm)
314
40100
785
384
33.5
19.7
Axis X-X
Axis Y-Y
S = I/c
(10' mm')
S lfe
(10' mm')
r= VIJA
(to mm*)
(mm)
(10 mm)
(mm)
4 290
10 900
328
315
1640
88.6

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