Cee section shape

Results SummarySummary﻿
Design Check
Parameter
Utilisation
Status
Shear
Vn = 8164.64 force_pound
6.12
🟢
Flexural Moment
Mn = 41679.42 foot * force_pound
0.3
🟢
1. Properties1.1 Material
﻿﻿fy
:33.00 ksi​
﻿
﻿﻿G
:80.00 GPa​
﻿
﻿﻿E
:200.00 GPa​
﻿
﻿﻿u﻿
:0.30​
﻿
1.2 Geometry
﻿﻿Section
:C4.0x2.25GA16​
﻿
﻿﻿L
:5.00 ft​
﻿
﻿
4.00 in​
2.25 in​
0.63 in​
0.06 in​
﻿﻿A
:0.5524590429849433​
﻿
Advanced Properties
﻿
1.46 in^4​
0.39 in^4​
0.80 in​
1.45 in​
2.00 in​
0.73 in^3​
0.49 in^3​
1.62 in​
0.84 in​
-1.08 in​
2.00 in​
1.88 in​
-0.00 in​
0.00 in^4​
1.41 in^6​
0.73 in^3​
0.73 in^3​
﻿
﻿
2. Design Actions﻿﻿Vu
:3,000.0 lbf​
﻿
﻿﻿Mu
:1,500.0 lbf ft​
﻿
﻿﻿MA
:300.0 lbf ft​
﻿
﻿﻿MB
:250.0 lbf ft​
﻿
Note, ﻿MA​
 and ﻿MB​
 are the flexural moments and the end of the unbraced length of the purlin/girt. 
3. Design Checks3.1 Flexural Strength
The nominal flexural strength [resistance], ﻿Mne​
, for yielding and global (lateral-torsional) buckling considering capacity up to first yield shall be calculated in accordance with:
﻿
Mne=SfFn≤My    (Eq. F2.1−1)M_{ne}=S_fF_n\le M_y \space \space \space \space (Eq. \space F2.1-1)Mne​=Sf​Fn​≤My​    (Eq. F2.1−1)
Where:
﻿﻿Mne​
: Nominal flexural strength [resistance] for yielding and global buckling.
﻿﻿Sf​
: Elastic section modulus of full unreduced section relative to extreme compression fiber
﻿﻿Fn​
: Flexural Stress
﻿﻿My​=Sfy​Fy​
﻿
﻿﻿Sfy​
: Elastic section modulus of full unreduced cross-section relative to extreme fiber in first yielding.
﻿﻿Fy​
: Yield Stress
﻿﻿Fcre​
: Critical elastic lateral-torsional buckling stress
﻿﻿	σe,x
:209.91 ksi​
﻿
﻿﻿σe,x​=π2E/(Rx​kt​L​)2
﻿
﻿
﻿﻿M1
:250.0 ft lbf​
﻿
﻿﻿M2
:300.0 ft lbf​
﻿
﻿﻿CTF
:0.267​
﻿
﻿﻿CTF={0.6+0.4M2​M1​​0.6−0.4M2​M1​​​if MA​⋅MB​<0,otherwise.​
﻿
﻿
﻿﻿Fcre
:95.60 ksi​
﻿
﻿﻿Fcre1​=CTFSx​Aσe,x​​(scx​+scx2​+rox2​σe,x​σt,x​​​)
﻿
Section. F2.1.1 to 5
﻿﻿Fn
:33.00 ksi​
﻿
﻿﻿Fn​=⎩⎨⎧​Fy​Fcre​910​Fy​(1−36Fcre​10Fy​​)​if Fcre​≥2.78Fy​if Fcre​≤0.56Fy​otherwise​
﻿
Eq. F2.1-3 to 5
﻿﻿Sfy
:0.73 in^3​
﻿
﻿
﻿﻿My
:1,804.57 lbf ft​
﻿
﻿﻿My​=Sfy​Fy​
﻿
Eq. F2.1-2
﻿﻿Mne
:1,804.57 lbf ft​
﻿
﻿﻿Mne​=Sf​Fn​≤My​
﻿
Eq. F2.1-1
﻿﻿Flexural utilization
:0.83​
﻿
﻿﻿Flexural check
:🟢​
﻿
﻿﻿Mu​≤Mne​
﻿
﻿
﻿
3.2 Shear Strength
The nominal shear strength [resistance], ﻿Vn​
, of flexural members without holes in the web shall be calculated in accordance with:
﻿
Vn={Vyfor λv≤0.8150.815VcrVyfor 0.815<λv≤1.227Vcrfor λv>1.227V_n =\begin{cases}     V_y & \text{for } \lambda_v \leq 0.815  \\    0.815 \sqrt{V_{cr} V_y} & \text{for } 0.815 < \lambda_v \leq 1.227 \\    V_{cr} & \text{for } \lambda_v > 1.227\end{cases}Vn​=⎩⎨⎧​Vy​0.815Vcr​Vy​​Vcr​​for λv​≤0.815for 0.815<λv​≤1.227for λv​>1.227​
Where:
﻿﻿λv​=Vcr​Vy​​​
﻿
﻿﻿Vy​=0.6Aw​Fy​
: Yield shear for of cross-section
﻿﻿Vy
:4,290.96 lbf​
﻿
﻿﻿Vy​=0.6Aw​Fy​
﻿
Eq. G2.1-5
﻿﻿	λv
:0.72​
﻿
﻿﻿λv​=Vcr​Vy​​​
﻿
Eq. G2.1-4
﻿﻿Vn
:4,290.96 lbf​
﻿
﻿﻿Vn​=⎩⎨⎧​Vy​0.815Vcr​Vy​​Vcr​​for λv​≤0.815for 0.815<λv​≤1.227for λv​>1.227​
﻿
Eq. G2.1-1 to 2
﻿﻿Shear utilization
:0.70​
﻿
﻿﻿Shear check
:🟢​
﻿
﻿﻿Vu​≤Vn​
﻿
﻿
﻿
﻿
Design Check	Parameter	Utilisation	Status
Shear	Vn = 8164.64 force_pound	6.12	🟢
Flexural Moment	Mn = 41679.42 foot * force_pound	0.3	🟢
	1.46 in^4
	0.39 in^4
	0.80 in
	1.45 in
	2.00 in
	0.73 in^3
	0.49 in^3
	1.62 in
	0.84 in
	-1.08 in
	2.00 in
	1.88 in
	-0.00 in
	0.00 in^4
	1.41 in^6
	0.73 in^3
	0.73 in^3