Media Summary: Okay on this problem I've got a hollow copper uh shaft okay uh again it's fixed free uh has an 8.41 KSI okay so under the load that I have right here I recognize that I get a yield stress or a stress of 8.41 the yield stress Cubed so from all of that i can solve this thing and say okay

Engr 222 Dec 4 Eccentric - Detailed Analysis & Overview

Okay on this problem I've got a hollow copper uh shaft okay uh again it's fixed free uh has an 8.41 KSI okay so under the load that I have right here I recognize that I get a yield stress or a stress of 8.41 the yield stress Cubed so from all of that i can solve this thing and say okay There remove something there we go okay let's try this again it's going to be pi over ... right you've got this elongated side you become as an ellipse the longer side of that ellipse is point six three Lecture notes, spreadsheet files, and other resources are available at: Lecture ...

1.375 meters over it crosses over a zero point right here at 2.75 ... changing is the radius therefore the polar moment of inertia is changing so just finding the equation

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ENGR 222 Dec 4 Eccentric loading 2
ENGR 222 Dec 4 Eccentric loading 1
ENGR 222 Dec 2 buckling 4
ENGR 222 Dec 4 Eccentric loading 3
ENGR 222 Oct 30 beam design 4
ENGR 222 Sep 4 - Stress 1
ENGR 222 Sep 4 - Stress 2
ENGR 222 - Class 41 (26 April 2019) Depreciation: Straight Line and Declining Balance
ENGR 222 Oct 14 Flexure 3
ENGR 222 Oct 5 Twist 4
ENGR 222 Oct-2 Torsion 1
ENGR 222 Sep 4 - Stress 3
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ENGR 222 Dec 4 Eccentric loading 2

ENGR 222 Dec 4 Eccentric loading 2

Okay on this problem I've got a hollow copper uh shaft okay uh again it's fixed free uh has an

ENGR 222 Dec 4 Eccentric loading 1

ENGR 222 Dec 4 Eccentric loading 1

8.41 KSI okay so under the load that I have right here I recognize that I get a yield stress or a stress of 8.41 the yield stress

ENGR 222 Dec 2 buckling 4

ENGR 222 Dec 2 buckling 4

625 to the

ENGR 222 Dec 4 Eccentric loading 3

ENGR 222 Dec 4 Eccentric loading 3

Square that whole thing like that and

ENGR 222 Oct 30 beam design 4

ENGR 222 Oct 30 beam design 4

Cubed so from all of that i can solve this thing and say okay

ENGR 222 Sep 4 - Stress 1

ENGR 222 Sep 4 - Stress 1

There remove something there we go okay let's try this again it's going to be pi over

ENGR 222 Sep 4 - Stress 2

ENGR 222 Sep 4 - Stress 2

... right you've got this elongated side you become as an ellipse the longer side of that ellipse is point six three

ENGR 222 - Class 41 (26 April 2019) Depreciation: Straight Line and Declining Balance

ENGR 222 - Class 41 (26 April 2019) Depreciation: Straight Line and Declining Balance

Lecture notes, spreadsheet files, and other resources are available at: https://sites.google.com/view/yt-isaacwait/home Lecture ...

ENGR 222 Oct 14 Flexure 3

ENGR 222 Oct 14 Flexure 3

1.375 meters over it crosses over a zero point right here at 2.75

ENGR 222 Oct 5 Twist 4

ENGR 222 Oct 5 Twist 4

... changing is the radius therefore the polar moment of inertia is changing so just finding the equation

ENGR 222 Oct-2 Torsion 1

ENGR 222 Oct-2 Torsion 1

4

ENGR 222 Sep 4 - Stress 3

ENGR 222 Sep 4 - Stress 3

... over

ENGR 222 Oct 14 flexure 2

ENGR 222 Oct 14 flexure 2

... the center so that's