{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hw23solutions - AAE 221 Structures II Spring 2004 Chapter...

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

View Full Document Right Arrow Icon
AAE 221 – Structures II – Spring 2004 Chapter 2.3 Beam shearing 1. First the location of the centroid and the moments of inertia must be found ( ) ( ) ( ) t t t t t t A z A y i i cm 62 . 6 2 48 5 . 12 25 2 98 1 1 2 2 2 = + = = The moments of inertia are 4 2 2 3 2 2 3 2 4 3 2 2 3 2 6135 ) 48 ( ) 12 . 6 ( 12 ) ( 48 ) 25 ( ) 88 . 5 ( 12 ) 25 ( 2 39233 12 ) 48 ( ) 25 ( ) 5 . 24 ( 12 25 2 t t t t t t t t A y I I t t t t t t t A z I I i i zi z i i yi y = + + + = + = = + + = + = The axial displacement is 0 ) ( = ¢ ¢ x u EA with the boundary conditions F L u EA u = ¢ = ) ( 0 ) 0 ( The bending displacements are given by 0 ) ( 0 ) ( = ¢ ¢ ¢ ¢ = ¢ ¢ ¢ ¢ x w EI x v EI y z with the boundary conditions 0 ) ( 38 . 18 ( ) ( 0 ) 0 ( ) 0 ( = ¢ ¢ ¢ - = ¢ ¢ = ¢ = L v t F L v EI v o v F L w EI F t L w EI w w = ¢ ¢ ¢ = ¢ ¢ = ¢ = ) ( ) 25 ( ) ( 0 ) 0 ( 0 ) 0 ( By integrating and applying the boundary conditions, the displacements can be shown to be:
Image of page 1

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

View Full Document Right Arrow Icon
y y z z EI Fx EI x L t F x w EI x M x v EA Fx x u 3 2 2 6 1 ) 25 ( 2 1 ) ( 2 1 ) ( ) ( + - = - = = The rotation is found from 0 ) ( = ¢ ¢ x J f m o The boundary conditions are t M L J = ¢ = ) ( 0 ) 0 ( f m f with 4 3 3 3 98 ) 98 ( 3 1 3 1 ) 38 . 18 ( t t d b J t F M t = = = = x x Integrating and applying boundary conditions, Fx t x
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.