3Pn Pn 1m Зт 2m
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question
100%
Determine the value of the plastic load Pn using the concept of collapse mechanism.
For the properties of the steel section, use the given designated wide-flange beam section below. (Assume
?? = 415 ???).
![Pn
3Pn
2m
3m
1m](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F775cedf0-0582-41d7-ada2-259cf43429db%2F78dae68c-5616-4c90-b534-d9caca079715%2Fv60he9o_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Pn
3Pn
2m
3m
1m
![Flange
Web
Mass
Area
Width
Thickness
thickness
Depth
(mm)
(kg/m)
(mm²)
(mm)
(mm)
(mm)
314
40100
785
384
33.5
19.7
Axis X-X
Axis Y-Y
S= Ilc
(10³ mm³)
r = IJA
(mm)
S = I|c
(10' mm³)
= VIJA
r =
(10 mm*)
(10 mm4)
(mm)
4 290
10 900
328
315
1 640
88.6](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F775cedf0-0582-41d7-ada2-259cf43429db%2F78dae68c-5616-4c90-b534-d9caca079715%2Fq713zni_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Flange
Web
Mass
Area
Width
Thickness
thickness
Depth
(mm)
(kg/m)
(mm²)
(mm)
(mm)
(mm)
314
40100
785
384
33.5
19.7
Axis X-X
Axis Y-Y
S= Ilc
(10³ mm³)
r = IJA
(mm)
S = I|c
(10' mm³)
= VIJA
r =
(10 mm*)
(10 mm4)
(mm)
4 290
10 900
328
315
1 640
88.6
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