Section 16: Neutral Axis and Parallel Axis Theorem 161


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1 Section 16: Neutral Axis and Parallel Axis Theorem 161
2 Geometry of deformation We will consider the deformation of an ideal, isotropic prismatic beam the cross section is symmetric about yaxis 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 Note: we will take these directions for M 0 to be 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 From: Hornsey
3 Neutral axis 16 From: Hornsey
4 6. BENDING DEFORMATION OF A STRAIGHT MEMBER A neutral surface is where longitudinal fibers of the material will not undergo a change in length From: Wang
5 6. BENDING DEFORMATION OF Thus, A we STRAIGHT make the following MEMBER assumptions: 1. Longitudinal axis x (within neutral surface) does not experience any change in length. All cross sections of the beam remain plane and perpendicular to longitudinal axis during the deformation. Any deformation of the crosssection within its own plane will be neglected In particular, the z axis, in plane of xsection and about which the xsection rotates, is called the neutral axis 165 From: Wang
6 6.4 THE FLEXURE FORMULA By mathematical expression, equilibrium equations of moment and forces, we get Equation 610 A yda= 0 σ max Equation 611 M = c A y da The integral represents the moment of inertia of x sectional area, computed about the neutral axis. We symbolize its value as I From: Wang
7 6.4 THE FLEXURE FORMULA Normal stress at intermediate distance y can be determined from Equation 61 σ = My I σ is ve as it acts in the ve direction (compression) Equations 61 and 61 are often referred to as the flexure formula From: Wang
8 *6.66 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 xsection into one made of a single material, then we can apply the flexure formula 168 From: Wang
9 169 From: Hornsey
10 Moments of Inertia 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 From: Le Figure from: Browner et al, Skeletal Trauma nd Ed, Saunders, 1998.
11 Implant Shape Moment of Inertia: further away material is spread in an object, greater the stiffness Stiffness and strength are proportional to radius From: Justice
12 161 From: Hornsey
13 Moment of Inertia of an Area by Integration Second moments or moments of finertiai of an area with respect to the x and y axes, I x y da I y = = x da Evaluation of the integrals is simplified by choosing dα to be a thin strip parallel to one of the coordinate axes. For a rectangular area, h 0 I x = y da = y bdy = 1 bh The formula for rectangular areas may also be applied to strips parallel to the axes, di x y = 1 y dx di = x da = x y dx 161 From: Rabiei
14 Homework Problem 16.1 Determine the moment of inertia of a triangle with respect to its base From: Rabiei
15 Homework Problem 16. a) Determine the centroidal polar moment of inertia of a circular area by direct integration. b) Using the result of part a, determine the moment of inertia of a circular area with respect to a diameter From: Rabiei
16 Parallel Axis Theorem Consider moment of inertia I of an area A with respect to the axis AA I = y da The axis BB passes through the area centroid and is called a centroidal axis. I = = y y da = ( y + d ) da + d da y da + d da I = I + Ad parallel axis theorem From: Rabiei
17 Parallel Axis Theorem Moment of inertia I T of a circular area with respect to a tangent to the circle, I T = I + = 5 4 π r Ad ( ) = π r + π r r 4 Moment of inertia of a triangle with respect to a centroidal axis, I A A = I BB + I BB = I = AA 1 6 bh Ad Ad = 1 1 bh 1 bh ( 1 ) h From: Rabiei
18 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, A,..., with respect to the same axis From: Rabiei
19 Example: y 00 (Dimensions in mm) z Centroidal o Axis y = 89.6 mm 1 y = y' da n A A y = [ ( )( 15 ) + ( 10 0 )( 60 )] ( ) 1 y = + ( ) [ 50, ,000 ] 4, From: University of Auckland 94,000 = = mm 4,400 = m
20 y Example: 00 (Dimensions in mm) 10 What is I z? 0.4 What is maximum σ x? z o 0 1 bd I z, = I = I + n ( 0)( 89.6) z 00 Ay 5.4 bd Iz,1 = = = mm 4 bd ( 0 )( 0.4 ) Iz, = = = mm ( 00 10)( 5.4) = mm Ay = ( )( ) 160 University of Auckland 10
21 y Example: 00 (Dimensions in mm) 10 What is I z? 0.4 What is maximum σ x? z o I = I + n z Ay I = I + I + z z,1 z, I z, 0 I = z mm 6 mm 4 = m 6 m University of Auckland
22 Maximum Stress: y NA 40.4 M xz x 89.6 σ x = M xz I z y' σ σ x,max = M M xz I z y Max x,max = 10 xz ( 89.6 ) ( ) University of Auckland (N/m or Pa)
23 Homework Problem 16. The strength of a W14x8 rolled steel beam is increased by attaching a plate to its upper flange. Dt Determine the moment of finertia and radius of gyration with respect to an axis which is parallel to the plate and passes through the centroid of the section. 16 From: Rabiei SOLUTION: Determine location of the centroid of composite section with respect to a 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 composite section centroidal axis.
24 Homework Problem 16.4 SOLUTION: Compute the moments of inertia of the bounding rectangle and halfcircle with respect to the x axis. Determine the moment of inertia of the shaded area with respect to the x axis. The moment of finertia of fthe shaded hddarea is obtained by subtracting the moment of inertia of the halfcircle from the moment of inertia of the rectangle From: Rabiei
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