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## COE 3001 Week 13

by: Eleanor Notetaker

40

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11

# COE 3001 Week 13 coe 3001 I

Eleanor Notetaker
Georgia Tech

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We figure out how to find I_yz and look at rotating the x and y axis on a beam. We then begin chapter 9 (deflection). We also work through several examples.
COURSE
Deformable Bodies
PROF.
Dr. Kennedy
TYPE
Class Notes
PAGES
11
WORDS
KARMA
25 ?

## Popular in Engineering and Tech

This 11 page Class Notes was uploaded by Eleanor Notetaker on Saturday November 14, 2015. The Class Notes belongs to coe 3001 I at Georgia Institute of Technology taught by Dr. Kennedy in Summer 2015. Since its upload, it has received 40 views. For similar materials see Deformable Bodies in Engineering and Tech at Georgia Institute of Technology.

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Date Created: 11/14/15
COE 3001: Mechanics of Deformable Bodies Week 13 Products of area: ❑ Iyz ∫zdA A Iyzmay be positive, negative, or zero. Iyz0 when cross­section is symmetric about y  or z axis Iyz0 Iy,Iz,Iyz  Rotation Parallel axis theorem: y=y +dy ' z=z +dz ❑ Iyz ∫zdA A ❑ I = ∫ y +dy )z +dz dA yz A ❑ ❑ ❑ ❑ ¿ y' z'dA+dy z' dA+dz y' dA+dydz dA ∫ ∫ ∫ ∫ A A A A ❑ dy∫ z dA=0 A ❑ dz∫ y' dA=0 A COE 3001: Mechanics of Deformable Bodies Week 13 IyzI +ycdzA Iyz0 Variation of y ,zI , yz I   with  θ: y =ycosθ+zsinθ 1 z =−ysinθ+zcosθ 1 ❑ I = ∫ dA z1 A 1 ❑ 2 Iy1 ∫ d1 A ❑ Iy1 1∫y z1d1 A ❑ ❑ 2 Iz= ∫ d1= ∫ (ycosθ+zsinθ d) 1 A A ❑ ❑ ❑ 2 2 2 2 ¿cos θ ∫ y dA+sin θ ∫ z dA+2s iθ oθ∫yzdA A A A ❑ ❑ ❑ I = y dAI = z dAI = yzdA z ∫ y ∫ yz ∫ A A A 2 2 Iz=cos θI +zin θI +2sy iθ oθIyz 1 ❑ ❑ 2 2 Iy1 ∫ d1= ∫ (−ysinθ+zcosθ dA) A A ❑ ❑ ❑ 2 2 2 2 ¿sin θ∫ y dA+cos θ ∫ z dA−2s iθ cθ∫yzdA A A A 2 2 Iy1sin θI +zos θ I −2y iθ oθIyz ❑ ❑ I = y z dA= (ycosθ+zsinθ)(−ysinθ+zcosθ)dA y1 1∫ 1 1 ∫ A A COE 3001: Mechanics of Deformable Bodies Week 13 ❑ ❑ ❑ ¿−siθ cθ y dA+s iθ θ z dA+(cos θ−sin θ) yzdA ∫A ∫A ∫A 2 2 Iy1 1−sniθ sθIz+iθ sθIy+(cos θ−sin θ)I yz cos θ= 1+cos2θ sin θ= 1−cos2θ iθ oθ= sin 2θ 2 2 2 (Iy+I z) (Iz−I y) Iz1 2 + 2 cos2θ+I siyzθ (Iy+I z) (Iy−I z) Iy1 + cos2θ−I siyzθ 2 2 (Iy−I z Iy1 1 sin 2θ+I yzs2θ 2 Principle axis:  Axes in which I y1r Iz1is maximum d I y=0=− I −( yin2z−2I cos2θ=0yz dθ −2 I tan2θ = yz p I yI z θp ≡ angle to the principle axis Iy1z1= 0 in principle axis Example: b=4∈¿ t=0.5∈¿ M =ok∗¿  Find the max stress 1. Find the centroid 2 2 2 1 1 t b t yc= [ ] y i =i 2[ (b−t )+ ] A i=1 2bt−t 2 2 COE 3001: Mechanics of Deformable Bodies Week 13 A 1t b−t ) A =bt 2 2 A=2bt−t t yc1 2 b yc2 2 y =1.18∈¿ c zc=−1.18∈¿ I = I +dy A + I +dy 2 z ( 1 1 1) ( z2 2) t (−t ) t 2 b t b 2 ¿( + ( ) t(b−t ) ( + ( )y c bt) 12 2 12 2 Iz=5.56¿ 4 4 Iy=5.56¿ Iyz Segment # dyi dzi 1 t (b−t) −y c 2 −z ct− 2 2 b t −y c −z c 2 2 I = ( +dy dz A + I) (y dz A ) yz y1 1 1 1 y2 2 2 2 t b−t ) b t 4 ¿0+ (2)y c(−z ct− 2 )t(b−t )+0+ (2)−y c(−t c 2)bt=3.267¿  Find stress at 0: COE 3001: Mechanics of Deformable Bodies Week 13 z0=1.18∈¿ y0=−1.18∈¿ M zM 0 M =0 y Iy=I z5.56¿ 4 4 Iyz3.267¿ I yz o y y 0 σ xMz 2 =2570 psi ( IyIz−I yz ) −2I yz tan2θ =p I −I →−∞ y z −π 2θ p +nπ 2 −π π θ p +n 4 2 1 1 Iy' (Iz+Iy) ( yI cz)2θ −I spn2θyz p 2 2 I = 1 ( +I +) 1 ( −I co)2θ +I sin2θ z' 2 z y 2 z y p yz p (Iz+I y) Iyz ' sin2θ +p coyzθ p 2 π θ p 4 I =2.293¿ 4 y I =8.827¿ 4 z' Iyz=0 COE 3001: Mechanics of Deformable Bodies Week 13 Iyz=0 ' ' ' σ = −M yz 0+ M zy 0 x I' I' z y ' M 0 M =y √2 σ =2570 psi at0 x Chapter 9  We briefly covered deflection curves before  Now, we want to  compute deflection under applied loads v(x) ≡ deflection objective: compute v(x) given  q(x) and boundary  conditions/geometry Differential equation for the deflection curve: dV =−qx dx dM =V(x) dx 1 dθ κ= =ρ ds 1 dθ ds=ρ dθ=¿ = ρ ds=κ M κ= EI COE 3001: Mechanics of Deformable Bodies Week 13 ❑ 2 M=− yσ∫dA= x Eκd∫=E I κ z A ϵ x−yκ tanθ= dv dx  small deflections and small rotations: tanθ≈θ= dv dx dθ dθ κ= ≈ ds dx d v κ= 2 d x M d v d v κ= EI = 2=¿M=E I z 2 d x d x shear force 2 3 V( )= dM = d E Izd v =E I zd v=V (x) dx dx( d x2) d x3  For Ez  constant along length of the beam dV shear force)dx =−q(x) 2 2 d d v 2(EI z 2)−q x( ) d x d x For EIz constant d v E Iz 4=−q (x) d x  Solution for v(x) requires boundary conditions: COE 3001: Mechanics of Deformable Bodies Week 13 x=0 x=a v(0)=0 v(a)=a dv dv dx ¿x=00 dx ¿x=a0  Rollers/pin x=0 x=a v(0)=0 v(a)=0 M(0)=0 M(a)=0 2 2 d v ¿ =0 d v ¿ =0 d x2 x=0 d x2 x=0  Free end x=a M (a)=0 2 d v ¿ =0 d x2 x=0 3 V(a)=0=¿ d v=0 d x3  Joint between segments: COE 3001: Mechanics of Deformable Bodies Week 13 vL(x)=v Rx ) dv L¿ = dv R¿ dx x=a dx x=a  Forced Boundary Conditions: dv v(0)=0∨ dx=0  Natural Boundary Conditions M (0)=0∨V 0 =0 Method of Integration of Load: 2 2 d EI d v =−q x ( ) d x2( zd x2) EIz is constant 1. Determine the boundary conditions Displacement/forced boundary conditions: Natural Boundary Conditions: 2 d v 1 v 0 =0 (3)M L =0=¿ 2¿x=L=0 d x 2 3 d v d v 2 )d x2¿x=L=0 (4)V L =0=¿ d x3¿x=L=0 2. Integrate: d v ∫ EI z 4dx= ∫q dx0 d x d v E Iz 3=−q 0x−L )+c1 d x Boundary Condition 4 => c1=0 COE 3001: Mechanics of Deformable Bodies Week 13 2 E I d v = −q 0 x−L +c2 zd x2 2 ( ) 2 2 d v Boundary Condition 3 => c 20 =  2 d x c2=0 dv −q 0 3 E Izdx = 6 (x−L +) 3 2 d v Boundary Condition 2 =>  d x2¿x=0=0 −q 0 3 −q 0 3 0= (L +) =¿3 = 3 6 6 −q q L 3 E Izv( )= 0(x−L −) 0 x+c 4 24 6 v(0)=0 −q 0 4 (−L )c =4 24 q0L 4 c4= 24 q v( )= 0 (L −4L x+ x−( )) 24EI z   0−q 1 q1 q1 q(x =q 1 (L−0 x=q 1 L x= L(L−x) 1 Determine Boundary Conditions: d v 1 v 0 =0 3 )M (L)=0=¿ 2¿x=L=0 d x COE 3001: Mechanics of Deformable Bodies Week 13 3 dv d v 2 )dx ¿x=0=0 (4 V (L =0=¿ 3¿ x=L=0 d x d v q1 E Iz 4=−q x ( ) (x−L ) d x L d v q Integrate:E Iz = 1(x−L +) ¿01( )( ) d x 3 2L 2 E I d v = q 1(x−L +) ¿0 ( )( ) zd x2 6L 2 dv q1 4 EI = (x−L +) 3 dx 24 L q 1 4 −q 1 3 ( )= (L +) =¿3 = 3 L 24 L 24 q 1 5 q1 3 EIv ( ) (x−L −) L x+c 4 120L 24 q 5 q ( )¿ v ( )0= 1 (−L +) =¿4 = 4 1 L 4 120 L 120 q v( )= 1 ((x−L −)L x+L4 5) 120EIL

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