LECTURE+19+COE-3001-A

# The centroid locaon is quite obvious centroid

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 2 Q x = y 1 A1 + y 2 A2 = M. Mello/Georgia Tech Aerospace t2 ) t2 ) 20 Some ﬁnal notes on Centroids •  When a composite area divided into only 2 parts, the centroid lies along line joining centroids C1 and C2. •  Methodology for calcula\ng centroid can be applied to parts with “holes” •  Handle the absence of an area by subtrac\on •  Or treat the blue regions above as posi\ve areas and the gaps as nega\ve areas 2/22/13 M. Mello/Georgia Tech Aerospace 21 PROBLEM 4.3- 11 “DISTRIBUTED LIFT LOAD ACTING ON AIRPLANE WING” FREE- BODY DIAGRAMS/STATIC CALCULATIONS: 1600N/m M c V DETERMINE: (A)  SHEAR FORCE AT INBOARD END OF WING (B)  BENDING MOMENT AT INBOARD END OF WING SOLUTION: X 2.6m 2.6m 1m Fv = 0 V + 0.5(1600 V= 900N/m 900)(2.6) + (900)(5.2) + 0.5(900)(1) = 0 6.04kN (minus sign means guessed wrong sign for shear force) X Mc = 0 M + 0.5(700)(2.6)[2.6/3] + (900)(5.2)[5.2/2] + 0.5(900)(1)[5.2 + 1/3] = 0 M = 1.54 ⇥ 104 N m 2/22/13 M. Mello/Georgia Tech Aerospace 22 Homework problem 2/22/13 M. Mello/Georgia Tech Aerospace 23 Moment- Curvature Rela\onship Beam cross sec\on •  x- z plane coincident with neutral surface •  Beam axis running out of the page •  Normal stress distribu\on •  Beam axis running along x ²་  Balance of moments within the beam demands that the moment resultant induced by σx must equal the bending moment M. ²་  Assuming a posi\ve bending moment, we expect a compressive stress for y > 0 along the beam. ²་  The element of force σxdA ac\ng on the element dA is in the nega\ve direc\on when σx < 0. ²་  Hence, dM = 2/22/13 x ydA M. Mello/Georgia Tech Aerospace 24 Moment- Curvature Rela\onship (cont.) •  Integra\ng over all elements over the en\re cross sec\on: M = Z x ydA A Z •  Subs\tu\ng for stress x = E y then leads to: M = E y 2 dA A M = EI zz (5- 10) Z E y 2 dA •  Recall, = 1/⇢ and so equivalently we may write: M = ⇢A What is this? Z •  MOMENT OF INERTIA: Izz = y 2 dA (5- 11) (unique property of the cross sec\onal area) A •  Izz = moment of iner\a of the cross sec\onal area with respect to the z axis (neutral axis) •  We can carry this quan\ty as the variable I in the moment- curvatu...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online