Vector Mechanics for Engineers: Statics

Ninth Edition

CHAPTER VECTOR MECHANICS FOR ENGINEERS: STATICS

Ferdinand P. Beer E. Russell Johnston, Jr. 9 Distributed Forces: Lecture Notes: Moments of Inertia J. Walt Oler Texas Tech University

Edition Ninth

Vector Mechanics for Engineers: Statics

Contents Introduction Sample Problem 9.6 Moments of Inertia of an Area Sample Problem 9.7 Moment of Inertia of an Area by Mohr’s Circle for Moments and Integration Products of Inertia Polar Moment of Inertia Sample Problem 9.8 Radius of Gyration of an Area Moment of Inertia of a Mass Sample Problem 9.1 Parallel Axis Theorem Sample Problem 9.2 Moment of Inertia of Thin Plates Parallel Axis Theorem Moment of Inertia of a 3D Body by Moments of Inertia of Composite Integration Areas Moment of Inertia of Common Sample Problem 9.4 Geometric Shapes Sample Problem 9.5 Sample Problem 9.12 Product of Inertia Moment of Inertia With Respect to an Principal Axes and Principal Arbitrary Axis Moments of Inertia Ellipsoid of Inertia. Principle Axes of Axes of Inertia of a Mass

Edition Ninth

Vector Mechanics for Engineers: Statics

Introduction • Previously considered distributed forces which were proportional to the area or volume over which they act. - The resultant was obtained by summing or integrating over the areas or volumes. - The moment of the resultant about any axis was determined by computing the first moments of the areas or volumes about that axis.

• Will now consider forces which are proportional to the area or volume over which they act but also vary linearly with distance from a given axis. - It will be shown that the magnitude of the resultant depends on the first moment of the force distribution with respect to the axis. - The point of application of the resultant depends on the second moment of the distribution with respect to the axis.

• Current chapter will present methods for computing the moments and products of inertia for areas and masses.

Edition Ninth

Vector Mechanics for Engineers: Statics

Moment of Inertia of an Area  • Consider distributed forces F whose magnitudes are proportional to the elemental areas A on which they act and also vary linearly with the distance of from a given axis. • Example: Consider a beam subjected to pure bending. Internal forces vary linearly with distance from the neutral axis which passes through the section centroid.  F  kyA

R  k y dA  0  y dA  Qx  first moment M  k y2dA  y2dA  second moment • Example: Consider the net hydrostatic force on a submerged circular gate. F  pA  yA R    y dA 2 M x    y dA

Edition Ninth

Vector Mechanics for Engineers: Statics

Moment of Inertia of an Area by Integration • Second moments or moments of inertia of an area with respect to the x and y axes, 2 2 I x   y dA I y   x dA

• Evaluation of the integrals is simplified by choosing dA to be a thin strip parallel to one of the coordinate axes.

• For a rectangular area, h I  y2dA  y2bdy  1 bh3 x   3 0

• The formula for rectangular areas may also be applied to strips parallel to the axes, dI  1 y3dx dI  x2dA  x2 y dx x 3 y

Edition Ninth

Vector Mechanics for Engineers: Statics

Polar Moment of Inertia

• The polar moment of inertia is an important parameter in problems involving torsion of cylindrical shafts and rotations of slabs. 2 J0   r dA

• The polar moment of inertia is related to the rectangular moments of inertia, 2 2 2 2 2 J0   r dA   x  y dA   x dA  y dA  I y  I x

Edition Ninth

Vector Mechanics for Engineers: Statics

Radius of Gyration of an Area • Consider area A with moment of inertia

Ix. Imagine that the area is concentrated in a thin strip parallel to

the x axis with equivalent Ix. I I  k 2 A k  x x x x A

kx = radius of gyration with respect to the x axis • Similarly, I I  k 2 A k  y y y y A J J  k 2 A k  O O O O A

2 2 2 kO  kx  k y

Edition Ninth

Vector Mechanics for Engineers: Statics

Sample Problem 9.1

SOLUTION: • A differential strip parallel to the x axis is chosen for dA. 2 dIx  y dA dA  l dy

• For similar triangles, l h  y h  y h  y  l  b dA  b dy b h h h Determine the moment of inertia of a triangle with respect • Integrating dI from y = 0 to y = h, to its base. x h h 2 2 h  y b 2 3 Ix   y dA   y b dy   hy  y dy 0 h h 0 3 4 h b  y y  bh3  h   h 3 4 I x  0 12

Edition Ninth

Vector Mechanics for Engineers: Statics

Sample Problem 9.2

SOLUTION: • An annular differential area element is chosen,

2 dJO  u dA dA  2 u du r r 2 3 JO   dJO  u 2 u du  2 u du 0 0  J  r4 O 2

a) Determine the centroidal polar • From symmetry, I = I , moment of inertia of a circular x y  area by direct integration. J  I  I  2I r4  2I O x y x 2 x b) Using the result of part a, determine the moment of inertia  I  I  r4 of a circular area with respect to a diameter x 4 diameter.

Edition Ninth

Vector Mechanics for Engineers: Statics

Parallel Axis Theorem

• Consider moment of inertia I of an area A with respect to the axis AA’ I   y2dA

• The axis BB’ passes through the area centroid and is called a centroidal axis.

I   y2dA   y  d 2 dA   y2dA 2d  ydA d 2  dA

I  I  Ad 2 parallel axis theorem

Edition Ninth

Vector Mechanics for Engineers: Statics

Parallel Axis Theorem

• Moment of inertia IT of a circular area with respect to a tangent to the circle, I  I  Ad 2  1  r 4   r 2 r 2 T 4    5  r 4 4

• Moment of inertia of a triangle with respect to a centroidal axis, 2 I AA  I BB  Ad 2 I  I  Ad 2  1 bh3  1 bh 1 h BB AA 12 2 3   1 bh3 36

Edition Ninth

Vector Mechanics for Engineers: Statics

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

A1, A2, A3, ... , with respect to the same axis.

Edition Ninth

Vector Mechanics for Engineers: Statics

Moments of Inertia of Composite Areas

Edition Ninth

Vector Mechanics for Engineers: Statics

Sample Problem 9.4 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 The strength of a W14x38 rolled steel composite section centroidal axis. beam is increased by attaching a plate • Calculate the radius of gyration from the to its upper flange. moment of inertia of the composite Determine the moment of inertia and section. radius of gyration with respect to an axis which is parallel to the plate and passes through the centroid of the section.

Edition Ninth

Vector Mechanics for Engineers: Statics

Sample Problem 9.4 SOLUTION: • Determine location of the centroid of composite section with respect to a coordinate system with origin at the centroid of the beam section.

Section A, in 2 y, in. yA, in3 Plate 6.75 7.425 50.12 Beam Section 11.20 0 0  A  17.95  yA  50.12

 yA 50.12 in3 Y  A   yA Y    2.792 in.  A 17.95 in 2

Edition Ninth

Vector Mechanics for Engineers: Statics

Sample Problem 9.4 • Apply the parallel axis theorem to determine moments of inertia of beam section and plate with respect to composite section centroidal axis. 2 2 Ix,beam section  Ix  AY  385  11.202.792  472.3 in 4 3 I  I  Ad 2  1 9 3  6.75 7.425  2.792 2 x,plate x 12  4     145.2 in 4

Ix  Ix,beam section  Ix,plate  472.3 145.2

4 Ix  618 in

• Calculate the radius of gyration from the moment of inertia of the composite section. 4 Ix 617.5 in kx   kx  5.87 in. A 17.95 in 2

Edition Ninth

Vector Mechanics for Engineers: Statics

Sample Problem 9.5

SOLUTION: • 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 obtained by subtracting the moment of inertia of the half-circle from the moment of inertia of the rectangle. Determine the moment of inertia of the shaded area with respect to the x axis.

Edition Ninth

Vector Mechanics for Engineers: Statics

Sample Problem 9.5 SOLUTION: • Compute the moments of inertia of the bounding rectangle and half-circle with respect to the x axis. Rectangle: I  1 bh3  1 240 120 138.2106 mm4 x 3 3   

Half-circle: moment of inertia with respect to AA’, I  1r4  1 90 4  25.76106 mm4 AA 8 8   moment of inertia with respect to x’, 2 6 3 I   I   Aa  25.7610 12.7210  4r 490 x AA a    38.2 mm 6 4 3 3  7.2010 mm b  120 - a  81.8 mm moment of inertia with respect to x, 2 2 A  1 r  1  90 2 6 3 2 2 2 Ix  Ix  Ab  7.2010  12.7210 81.8 3 2  12.7210 mm  92.3106 mm4

Edition Ninth

Vector Mechanics for Engineers: Statics

Sample Problem 9.5 • The moment of inertia of the shaded area is obtained by subtracting the moment of inertia of the half-circle from the moment of inertia of the rectangle.

6 4 6 4 Ix  138.210 mm  92.310 mm

6 4 Ix  45.910 mm

Edition Ninth

Vector Mechanics for Engineers: Statics

Product of Inertia

• Product of Inertia:

I xy   xy dA

• When the x axis, the y axis, or both are an axis of symmetry, the product of inertia is zero.

• Parallel axis theorem for products of inertia:

I xy  I xy  xyA

Edition Ninth

Vector Mechanics for Engineers: Statics

Principal Axes and Principal Moments of Inertia • The change of axes yields I x  I y I x  I y I    cos 2  I sin 2 x 2 2 xy I x  I y I x  I y I    cos 2  I sin 2 y 2 2 xy I x  I y I    sin 2  I cos 2 x y 2 xy 2 2 • The equations for I and I are the Given I x   y dA I y   x dA x’ x’y’ parametric equations for a circle, I  xy dA xy  2 2 2 I x  Iave   I xy  R we wish to determine moments and product of inertia with I x  I y  I x  I y  2 Iave  R     I xy respect to new axes x’ and y’. 2  2  Note: x  x cos  y sin • The equations for Iy’ and Ix’y’ lead to the y  y cos  xsin same circle.

Edition Ninth

Vector Mechanics for Engineers: Statics

Principal Axes and Principal Moments of Inertia

• At the points A and B, Ix’y’ = 0 and Ix’ is a maximum and minimum, respectively.

Imax,min  Iave  R

2I xy tan 2m   I x  I y

• The equation for Qm defines two angles, 90o apart which correspond to the principal axes of the area about O.

• I and I are the principal moments I  I 2  I 2  R2 max min x ave xy of inertia of the area about O. I x  I y  I x  I y  2 Iave  R     I xy 2  2 

Edition Ninth

Vector Mechanics for Engineers: Statics

Sample Problem 9.6 SOLUTION: • Determine the product of inertia using direct integration with the parallel axis theorem on vertical differential area strips • Apply the parallel axis theorem to evaluate the product of inertia with respect to the centroidal axes.

Determine the product of inertia of the right triangle (a) with respect to the x and y axes and (b) with respect to centroidal axes parallel to the x and y axes.

Edition Ninth

Vector Mechanics for Engineers: Statics

Sample Problem 9.6 SOLUTION: • Determine the product of inertia using direct integration with the parallel axis theorem on vertical differential area strips  x   x  y  h1  dA  y dx  h1 dx  b   b  1 1  x  xel  x yel  y  h1  2 2  b 

Integrating dIx from x = 0 to x = b, 2 b  x  I  dI  x y dA  x 1 h2 1 dx xy  xy  el el  2   0  b  b 2 3 2 2 3 4 b 2  x x x  x x x   h    dx h      2 b 2  4 3b 2 0 2b   8b 0

I  1 b2h2 xy 24

Edition Ninth

Vector Mechanics for Engineers: Statics

Sample Problem 9.6 • Apply the parallel axis theorem to evaluate the product of inertia with respect to the centroidal axes. x  1 b y  1 h 3 3

With the results from part a,

Ixy  Ixy  xyA I  1 b2h2  1 b 1 h 1 bh xy 24 3 3 2 

I   1 b2h2 xy 72

Edition Ninth

Vector Mechanics for Engineers: Statics

Sample Problem 9.7 SOLUTION: • Compute the product of inertia with respect to the xy axes by dividing the section into three rectangles and applying the parallel axis theorem to each. • Determine the orientation of the principal axes (Eq. 9.25) and the principal moments of inertia (Eq. 9. 27). For the section shown, the moments of inertia with respect to the x and y axes 4 4 are Ix = 10.38 in and Iy = 6.97 in . Determine (a) the orientation of the principal axes of the section about O, and (b) the values of the principal moments of inertia about O.

Edition Ninth

Vector Mechanics for Engineers: Statics

Sample Problem 9.7 SOLUTION: • Compute the product of inertia with respect to the xy axes by dividing the section into three rectangles. Apply the parallel axis theorem to each rectangle,

Ixy  Ixy  xyA Note that the product of inertia with respect to centroidal axes parallel to the xy axes is zero for each rectangle.

Rectangle Area, in 2 x, in. y, in. xyA,in 4 I 1.5 1.25 1.75  3.28 II 1.5 0 0 0 III 1.5 1.25 1.75  3.28  xyA  6.56

4 Ixy   xyA  6.56 in

Edition Ninth

Vector Mechanics for Engineers: Statics

Sample Problem 9.7 • Determine the orientation of the principal axes (Eq. 9.25) and the principal moments of inertia (Eq. 9. 27).

2Ixy 2 6.56 tan 2m      3.85 Ix  I y 10.38  6.97

2m  75.4 and 255.4

m  37.7 and m 127.7

I  I I  I 2 I  10.38 in 4 x y  x y  2 x Imax,min      Ixy 4 2  2  I y  6.97 in 2 4 10.38  6.97 10.38  6.97 2 Ixy  6.56 in       6.56 2  2 

4 Ia  Imax 15.45 in 4 Ib  Imin 1.897 in

Edition Ninth

Vector Mechanics for Engineers: Statics

Mohr’s Circle for Moments and Products of Inertia

• The moments and product of inertia for an area are plotted as shown and used to construct Mohr’s circle,

I x  I y  I x  I y  2 Iave  R     I xy 2  2 

• Mohr’s circle may be used to graphically or analytically determine the moments and product of inertia for any other rectangular axes including the principal axes and principal moments and products of inertia.

Edition Ninth

Vector Mechanics for Engineers: Statics

Sample Problem 9.8 SOLUTION:

• Plot the points (Ix , Ixy) and (Iy ,-Ixy). Construct Mohr’s circle based on the circle diameter between the points. • Based on the circle, determine the orientation of the principal axes and the principal moments of inertia. The moments and product of inertia • Based on the circle, evaluate the

with respect to the x and y axes are Ix = moments and product of inertia with 4 4 7.24x106 mm , Iy = 2.61x106 mm , and respect to the x’y’ axes. 6 4 Ixy = -2.54x10 mm . Using Mohr’s circle, determine (a) the principal axes about O, (b) the values of the principal moments about O, and (c) the values of the moments and product of inertia about the x’ and y’ axes

Edition Ninth

Vector Mechanics for Engineers: Statics

Sample Problem 9.8 SOLUTION:

• Plot the points (Ix , Ixy) and (Iy ,-Ixy). Construct Mohr’s circle based on the circle diameter between the points. OC  I  1 I  I  4.925106 mm4 ave 2  x y  CD  1 I  I  2.315106 mm4 2  x y  R  CD2  DX 2  3.437106 mm4

6 4 Ix  7.2410 mm • Based on the circle, determine the orientation of the 6 4 principal axes and the principal moments of inertia. I y  2.6110 mm DX I  2.54106 mm4 tan 2   1.097 2  47.6   23.8 xy m CD m m

6 4 Imax  OA  Iave  R Imax  8.3610 mm

6 4 Imin  OB  Iave  R Imin 1.4910 mm

Edition Ninth

Vector Mechanics for Engineers: Statics

Sample Problem 9.8 • Based on the circle, evaluate the moments and product of inertia with respect to the x’y’ axes. The points X’ and Y’ corresponding to the x’ and y’ axes are obtained by rotating CX and CY counterclockwise through an angle Q  2(60o) = 120o. The angle that CX’ forms with the x’ axes is f = 120o - 47.6o = 72.4o.

o Ix'  OF  OC  CX cos  Iave  Rcos 72.4 6 4 Ix  5.9610 mm o I y'  OG  OC  CYcos  Iave  Rcos72.4 6 4 I y  3.8910 mm o Ixy'  FX   CYsin  Rsin 72.4

Edition Ninth

Vector Mechanics for Engineers: Statics

Moment of Inertia of a Mass • Angular acceleration about the axis AA’ of the small mass m due to the application of a couple is proportional to r2m. r2m = moment of inertia of the mass m with respect to the axis AA’

• For a body of mass m the resistance to rotation about the axis AA’ is 2 2 2 I  r1 m  r2 m  r3 m    r 2dm  mass moment of inertia

• The radius of gyration for a concentrated mass with equivalent mass moment of inertia is I I  k 2m k  m

Edition Ninth

Vector Mechanics for Engineers: Statics

Moment of Inertia of a Mass

• Moment of inertia with respect to the y coordinate axis is 2 2 2 I y   r dm   z  x dm

• Similarly, for the moment of inertia with respect to the x and z axes, 2 2 I x   y  z dm 2 2 I z   x  y dm

• In SI units, I   r 2dm  kg m2  In U.S. customary units,  lb s2  I  slug  ft 2   ft 2   lb ft  s2        ft 

Edition Ninth

Vector Mechanics for Engineers: Statics

Parallel Axis Theorem • For the rectangular axes with origin at O and parallel centroidal axes, 2 2 2 2 I x   y  z dm   y  y  z  z dm   y2  z2 dm  2y ydm 2z zdm  y 2  z 2  dm

2 2 I x  I x  my  z  2 2 I y  I y  mz  x  2 2 I z  I z  mx  y 

• Generalizing for any axis AA’ and a parallel centroidal axis, I  I  md 2

Edition Ninth

Vector Mechanics for Engineers: Statics

Moments of Inertia of Thin Plates • For a thin plate of uniform thickness t and homogeneous material of density r, the mass moment of inertia with respect to axis AA’ contained in the plate is 2 2 I AA   r dm  rt r dA  rt I AA,area

• Similarly, for perpendicular axis BB’ which is also contained in the plate,

I BB  rt I BB,area

• For the axis CC’ which is perpendicular to the plate,

ICC  rt JC,area  rt I AA,area  I BB,area 

 I AA  I BB

Edition Ninth

Vector Mechanics for Engineers: Statics

Moments of Inertia of Thin Plates • For the principal centroidal axes on a rectangular plate,

I  rt I  rt 1 a3b  1 ma2 AA AA,area 12  12

I  rt I  rt 1 ab3  1 mb2 BB BB,area 12  12

I  I  I  1 m a2  b2 CC AA,mass BB,mass 12  

• For centroidal axes on a circular plate, I  I  rt I  rt 1  r 4  1 mr 2 AA BB AA,area 4  4

I  I  I  1 mr 2 CC AA BB 2

Edition Ninth

Vector Mechanics for Engineers: Statics

Moments of Inertia of a 3D Body by Integration • Moment of inertia of a homogeneous body is obtained from double or triple integrations of the form I  r r 2dV

• For bodies with two planes of symmetry, the moment of inertia may be obtained from a single integration by choosing thin slabs perpendicular to the planes of symmetry for dm.

• The moment of inertia with respect to a particular axis for a composite body may be obtained by adding the moments of inertia with respect to the same axis of the components.

Edition Ninth

Vector Mechanics for Engineers: Statics

Moments of Inertia of Common Geometric Shapes

Edition Ninth

Vector Mechanics for Engineers: Statics

Sample Problem 9.12

SOLUTION: • With the forging divided into a prism and two cylinders, compute the mass and moments of inertia of each component with respect to the xyz axes using the parallel axis theorem. • Add the moments of inertia from the components to determine the total moments of inertia for the forging.

Determine the moments of inertia of the steel forging with respect to the xyz coordinate axes, knowing that the specific weight of steel is 490 lb/ft3.

Edition Ninth

Vector Mechanics for Engineers: Statics

Sample Problem 9.12 SOLUTION: cylinders a 1in., L  3in., x  2.5in., y  2in.: • Compute the moments of inertia I  1 ma2  my2 of each component with respect x 2 to the xyz axes. 2 2  1 0.0829 1  0.0829 2 2  12  12  2.59103lb  ft s2

I  1 m 3a2  L2  mx2 y 12   2 2 2  1 0.08293 1    3    0.08292.5 12  12 12  12  4.17103lb  ft s2

each cylinder : 1 2 2 2 2 I y  m3a  L  mx  y  V 490lb/ft3  12  3in3 12 m   2 2 2 2 g 1728in3 ft3 32.2ft s2  1 0.08293 1    3    0.08292.5   2      12  12 12   12 12  2 m  0.0829 lb s ft  6.48103lb  ft s2

Edition Ninth

Vector Mechanics for Engineers: Statics

Sample Problem 9.12 prism (a = 2 in., b = 6 in., c = 2 in.): 2 2 I  I  1 mb2  c2  1 0.211 6    2   x z 12 12  12 12   4.88103 lb  ft s2

2 2 I  1 mc2  a2  1 0.211 2    2   y 12 12  12 12   0.977103 lb  ft s2 • Add the moments of inertia from the components to determine the total moments of inertia. 3 3 Ix  4.8810  22.5910  3 2 prism : Ix 10.0610 lb  ft s V 490lb/ft3 2 2 6in3 I  0.977103  2 4.17103 m     y   3 3 2 g 1728in ft 32.2ft s  3 2 I y  9.3210 lb  ft s 2 m  0.211lb s ft 3 3 Iz  4.8810  26.4810  3 2 Iz 17.8410 lb  ft s

Edition Ninth

Vector Mechanics for Engineers: Statics

Moment of Inertia With Respect to an Arbitrary Axis

• IOL = moment of inertia with respect to axis OL 2   2 IOL   p dm     r dm   • Expressing  and r in terms of the vector components and expanding yields 2 2 2 IOL  I xx  I yy  I zz

 2I xyxy  2I yzyz  2I zxzx

• The definition of the mass products of inertia of a mass is an extension of the definition of product of inertia of an area I xy   xy dm  I xy  mxy

I yz   yz dm  I yz  myz

I zx   zx dm  I zx  mzx

Edition Ninth

Vector Mechanics for Engineers: Statics

Ellipsoid of Inertia. Principal Axes of Inertia of a Mass • Assume the moment of inertia of a body has been computed for a large number of axes OL and that point Q is plotted on each axis at a distance OQ  1 IOL • The locus of points Q forms a surface known as the ellipsoid of inertia which defines the moment of inertia of the body for any axis through O. • x’,y’,z’ axes may be chosen which are the principal axes of inertia for which the products of inertia are zero and the moments of inertia are the principal moments of inertia.