Can you show to solve the Mn= 424.063 kn.mHow it gets the answer the 424.063 , the green highlights belowPlease show solution Civil engineering- steel design The beam shown is lateraily braced only at the ends. The180kN load is a service live load. Use Fy= 345 MPa anddetermine whether a W310 x 74 is adequate.180 kN2.5mSection Properties:d = 310mmtw = 9.40mmAg= = 9420 mm²lx = 163x106mmªly = 23.4x105mmªHo=295mm2.5mbf = 205mmtf = 16.3 mmJ=712x10³mm*Zx = 1.18x10 mm³Cw=505x10 mm5rts = 57.2mm C=1-0Sx = 1xсc= d = 3102mQ=7 = 2052(14.3)точу345x118x106163x106IssAp=038Section comportسالIssLb= 5000mm46642p=1.76 16₂1 €Z2x100745 0-773453Sxho6-28038 BLp= 2112.03 mmfy 34527: 185rbs € 30 in hans fes del ] ²2r= 1.95 E$xho0.7ysxnoE828157-02X105=1.05.16 x106عاد23.4×100942049.84712x10²³ (1) = 2.205 X10³ → ⒸAx 295ال2r=1.85 (57.2][3][√2] =√1+√ √ HU 74 [1/ + =]6.762x = 7209 07311205< 500-32 720403cb=1.32Mm- cb [Mp - (mp - 0.773 5x ) [ 10-22]]Mn = 42 4. 063 kn. m

Section properties of W310 x 74 is givenFy = 345 MPa Load = 180 kN

Answered: Can you show to solve the Mn= 424.063…

Can you show to solve the Mn= 424.063 kn.m How it gets the answer the 424.063 , the green highlights below Please show solution Civil engineering- steel design

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

Can you show to solve the Mn= 424.063 kn.m

How it gets the answer the 424.063 , the green highlights below

Please show solution

Civil engineering- steel design

The beam shown is lateraily braced only at the ends. The
180kN load is a service live load. Use Fy= 345 MPa and
determine whether a W310 x 74 is adequate.
180 kN
2.5m
Section Properties:
d = 310mm
tw = 9.40mm
Ag= = 9420 mm²
lx = 163x106mmª
ly = 23.4x105mmª
Ho=295mm
2.5m
bf = 205mm
tf = 16.3 mm
J=712x10³mm*
Zx = 1.18x10 mm³
Cw=505x10 mm5
rts = 57.2mm

C=1-0
Sx = 1x
с
c= d = 310
2
m
Q=
7 = 205
2(14.3)
точу
345x118x106
163x106
Iss
Ap=038
Section comport
سال
Iss
Lb= 5000mm
4664
2p=1.76 16₂1 €
Z
2x10
0745 0-77345
3
Sxho
6-28
038 B
Lp= 2112.03 mm
fy 345
27: 185rbs € 30 in hans fes del ] ²
2r= 1.95 E
$xho
0.7y
sxno
E
828157-0
2X105
=
1.05.16 x106
عاد
23.4×100
9420
49.84
712x10²³ (1) = 2.205 X10³ → Ⓒ
Ax 295
ال
2r=1.85 (57.2][3][√2] =√1+√ √ HU 74 [1/ + =]
6.76
2x = 7209 073
11205< 500-32 720403
cb=1.32
Mm- cb [Mp - (mp - 0.773 5x ) [ 10-22]]
Mn = 42 4. 063 kn. m

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