{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

lecture_7a

# lecture_7a - Torsion Properties for Line Segments and...

This preview shows pages 1–3. Sign up to view the full content.

Y x2 y2 x1 area3 x0 x1 x1 Torsion Properties for Line Segments and Computational Scheme for Piecewise Straight Section Calculations Closed Thin walled Sections the new material consists of the "corrections for ω and Q ω A = enclosed area definition of 4 A 2 J = J 1 ω c = h c ds J s 1 ds = 2 A s 1 ds as b 2 A t b t 1 ds d c = h c 2 A t ds = d ω c 0 1 ds 0 t t 0 0 b 1 circ_integral = ds = X i ( 2 Y i ( 2 + ) ) = l i if we define l i = 0 t i t i t i segment X i ( 2 Y i ( 2 + ) ) i length: we now need calculation of the enclosed area A in this expression area of triangle determined by two points and the origin: area = area1 + area2 area3 y area x0,y0 x1,y1 y x0,y0 x1,y1 y area1 area2 x0,y0 x1,y1 y area3 x0,y0 x1,y1 0,0 x 0,0 x 0,0 x 0,0 1 1 1 area1 := 2 y1 area2 := (y0 + y1) (x0 ) area3 := 2 y0 ( ) 2 area := area1 + area2 1 1 1 area simplify y0 x1 + 2 y1 x0 area_2_pts_origin := (y1 x0 y0 ) 2 2 area between three points: x0,y0 x1,y1 x2,y2 area 1 1 area := (y1 x0 y0 x1) (y1 x2 x1) 2 2 1 1 area := (y1 x0 y0 x1) + (y2 x1 y1 ) 2 2 1 etc ......... area_enclosed := 2 ( X i i 1 X i 1 Y i ) + + i 1 notes_15a_tor_prop_clsd_calc.mcd x

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
for a straight line segment ρ c = constant ∆ω c = ρ c L 2area_enclosed l i and is linear along line circ_integral t i
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern