Section 16: Neutral Axis and Parallel Axis Theorem 16-1 Geometry - PowerPoint PPT Presentation

Section 16: Neutral Axis and Parallel Axis Theorem 16-1

Geometry of deformation Geometry of deformation • We will consider the deformation of an ideal, isotropic prismatic beam – the cross section is symmetric about y-axis • All parts of the beam that were originally aligned with the longitudinal axis bend into circular arcs – plane sections of the beam remain plane and perpendicular to the beam s curved axis beam’s curved axis Note: we will take these directions for M 0 to be positive However they are positive. However, they are in the opposite direction to our convention (Beam 7), and we must remember to account for this at the end. 16-2 From: Hornsey

Neutral axis From: Hornsey 16-3

6.3 BENDING DEFORMATION OF A STRAIGHT MEMBER A STRAIGHT MEMBER • A neutral surface is where longitudinal fibers of the material will not undergo a change in length. g g g 16-4 From: Wang

6.3 BENDING DEFORMATION OF A STRAIGHT MEMBER A STRAIGHT MEMBER • Thus, we make the following assumptions: 1. Longitudinal axis x (within neutral surface) 1. Longitudinal axis x (within neutral surface) does not experience any change in length 2. All cross sections of the beam remain plane p and perpendicular to longitudinal axis during the deformation 3. Any deformation of the cross-section within its own plane will be neglected • In particular, the z axis, in plane of x-section and about which the x-section rotates, is called the neutral axis t l i 16-5 From: Wang

6.4 THE FLEXURE FORMULA 6 4 THE FLEXURE FORMULA • By mathematical expression, equilibrium equations of moment and forces, we get ∫ A y dA = 0 ∫ A y dA 0 Equation 6-10 Equation 6 10 σ σ max M = ∫ A y 2 dA Equation 6-11 c • The integral represents the moment of inertia of x- sectional area, computed about the neutral axis. We symbolize its value as I We symbolize its value as I. 16-6 From: Wang

6.4 THE FLEXURE FORMULA 6 4 THE FLEXURE FORMULA • Normal stress at intermediate distance y can be determined from My σ = − Equation 6-13 I • σ is -ve as it acts in the -ve direction (compression) • Equations 6-12 and 6-13 are often referred to as E ti 6 12 d 6 13 ft f d t the flexure formula . 16-7 From: Wang

*6 6 COMPOSITE BEAMS 6.6 COMPOSITE BEAMS • Beams constructed of two or more different materials are called composite beams • Engineers design beams in this manner to develop a more efficient means for carrying applied loads • Flexure formula cannot be applied directly to determine normal stress in a composite beam • Thus a method will be developed to “transform” a beam’s x-section into one made of a single material, th then we can apply the flexure formula l th fl f l 16-8 From: Wang

From: Hornsey 16-9

Moments of Inertia Moments of Inertia • Resistance to bending, Resistance to bending, twisting, compression or tension of an object is a function of its shape • Relationship of applied force to distribution of mass (shape) with respect to an axis respect to an axis. Figure from: Browner et al, Skeletal Trauma 2nd Ed, 16-10 From: Le Saunders, 1998.

Implant Shape Implant Shape • Moment of Inertia : Moment of Inertia : further away material is spread in an object, greater the stiffness • Stiffness and strength are proportional to radius 4 16-11 From: Justice

From: Hornsey 16-12

Moment of Inertia of an Area by Integration • Second moments or moments of inertia of S d f i i f an area with respect to the x and y axes, = = ∫ ∫ 2 ∫ ∫ 2 I y dA I x dA x x y y • Evaluation of the integrals is simplified by choosing d Α to be a thin strip parallel to one of the coordinate axes. one of the coordinate axes. • For a rectangular area, h = 2 2 = 2 2 = 3 3 ∫ ∫ I x y dA y bdy 1 bh 3 0 • The formula for rectangular areas may also be applied to strips parallel to the axes, = 3 = 2 = 2 dI 1 y dx dI x dA x y dx x y 3 16-13 From: Rabiei

Homework Problem 16.1 Determine the moment of Determine the moment of inertia of a triangle with respect to its base. 16-14 From: Rabiei

Homework Problem 16.2 a) Determine the centroidal polar moment of inertia of a circular area by direct integration area by direct integration. b) Using the result of part a , determine the moment of inertia of a circular area with respect to a of a circular area with respect to a diameter. 16-15 From: Rabiei

Parallel Axis Theorem • Consider moment of inertia I of an area A with respect to the axis AA’ = ∫ 2 I y dA • The axis BB’ passes through the area centroid and is called a centroidal axis. ( ) ′ = 2 2 = + 2 2 ∫ ∫ I y dA y d dA ′ ′ = 2 + + 2 ∫ ∫ ∫ y dA d y dA d dA 2 = + 2 parallel axis theorem I I Ad 16-16 From: Rabiei

Parallel Axis Theorem • Moment of inertia I T of a circular area with respect to a tangent to the circle, ( ( ) 2 ) = + 2 2 = π 4 4 + π 2 2 2 I T I Ad 1 1 r r r 4 = π 4 5 r 4 • Moment of inertia of a triangle with respect to a centroidal axis, id l i = + 2 I I Ad ′ ′ A A B B ( ( ) ) 2 2 = − 2 2 = 3 3 − I I Ad 1 1 bh 1 1 bh 1 1 h ′ ′ B B A A 12 2 3 = 3 1 bh 36 16-17 From: Rabiei

Moments of Inertia of Composite Areas • The moment of inertia of a composite area A about a given axis is obtained by adding the moments of inertia of the component areas A 1 , A 2 , A 3 , ... , with respect to the same axis. 16-18 From: Rabiei

Example: (Dimensions in mm) y 200 10 z z Centroidal C t id l o Axis 125 120 = y 89 . 6 mm 60 1 ∫ = ⋅ y y ' dA n A A 20 ) ] ) ] ( ( )( )( [ [ 1 ( ( )( )( ) ) + + 120 × 200 × 200 × y = = 120 20 20 60 60 10 10 125 125 y ( ) × + × 200 10 120 20 1 ) [ ) [ ] ] 394 , 000 = + + = = y y 250 250 , , 000 000 144 144 , , 000 000 89 89 . . 55 55 mm mm ( ( 4 , 400 4 , 400 = × − 3 89 . 6 10 m 16-19 From: University of Auckland

y Example: (Dimensions in mm) 200 10 • What is I ? What is I z ? • What is maximum σ x ? 30.4 o z 20 2 = + I I I I A A y 1 n z 89.6 200 10 20 3 20 2 30.4 35.4 ( )( ) 3 3 bd 20 89 . 6 6 mm = = = × 4 I 4 . 79 10 z , 1 3 3 1 89.6 ( ( )( )( ) ) bd 3 3 3 3 bd 20 20 30 30 . 4 4 6 mm = = = × 4 I 0 . 19 10 z , 2 3 3 20 ( ( )( )( ) ) 3 3 bd 200 10 ( ( )( )( ) ) 2 6 mm = + + = + × = × 2 2 6 4 4 I I A A y y 200 200 10 10 35 35 . 4 4 3 3 . 28 28 10 10 z , 3 12 12 16-20 University of Auckland

y Example: (Dimensions in mm) 200 10 • What is I ? What is I z ? • What is maximum σ x ? 30.4 o z 20 2 = + I I I I A A y 1 n z 89.6 200 10 20 3 20 2 30.4 35.4 1 89.6 = + + I I I I I I I I z z , 1 z , 2 z , 3 20 6 mm 6 m − ⇒ ⇒ = = × × = = × × 4 4 I I 8 8 . 26 26 10 10 mm 8 8 . 26 26 10 10 m z 16-21 University of Auckland

Maximum Stress: y 40.4 M xz x NA 89.6 M xz σ = − ⋅ xz y y ' ' x I z M σ = − ⋅ xz xz y x , Max Max I z ) ( ) ( ) ) M (N/m 2 or Pa) − ⇒ σ = − ⋅ − × 3 ( ) ( ( xz 89 . 6 10 × − x x , Max Max 6 6 8 . 26 10 16-22 University of Auckland

Homework Problem 16.3 SOLUTION: • Determine location of the centroid of composite section with respect to a it ti ith t t coordinate system with origin at the centroid of the beam section. • Apply the parallel axis theorem to determine moments of inertia of beam section and plate with respect to The strength of a W14x38 rolled steel The strength of a W14x38 rolled steel composite section centroidal axis. beam is increased by attaching a plate to its upper flange. Determine the moment of inertia and D t i th t f i ti d radius of gyration with respect to an axis which is parallel to the plate and passes through the centroid of the passes through the centroid of the section. 16-23 From: Rabiei

Homework Problem 16 4 Homework Problem 16.4 SOLUTION: • Compute the moments of inertia of the Compute the moments of inertia of the bounding rectangle and half-circle with respect to the x axis. • The moment of inertia of the shaded area is Th t f i ti f th h d d i obtained by subtracting the moment of inertia of the half-circle from the moment of inertia of the rectangle of inertia of the rectangle. Determine the moment of inertia of the shaded area with respect to the x axis. 16-24 From: Rabiei