Dynamics 14th edition by r c hibbeler chapter 10

The moment of inertia of the differential element parallel to the x axis shown shaded in Fig... a Differential Element: The area of the differential element parallel to y axis is The mom

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y1n The area of the differential element parallel

to the x axis shown shaded in Fig a is dA = (a - x)dy = aa - a

b

y x n

a b n

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Differential Element The area of the differential element parallel to the y axis

shown shaded in Fig a is dA = ydx = a b n x n dx

Moment of Inertia Perform the integration,

I y =

LA x2dA = L

a

0x2 aa b n x n dxb =

b

y x n

a b n

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Differential Element Here x = 250y1 The area of the differential element parallel

to the x axis shown shaded in Fig a is dA = 2x dy = 2250y1dy

I x =

LA y2dA = L

200 mm

0 y2c 2250y1dyd = 2250L

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I y = 53.3(106) mm4

Solution

Differential Element Here x = 250 y1 The moment of inertia of the differential

element parallel to x axis shown in Fig a about y axis is

dIy = 121 (dy)(2x)3 = 23 3dy = 23(250y1

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Ix = 0.133 m4

Solution

shown shaded in Fig a is dA = ydx The moment of inertia of this element about

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shown shaded in Fig a is dA = ydx = x1dx

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Differential Element Here x = 2(1 - y2) The area of the differential element

parallel to the x axis shown shaded in Fig a is dA = xdy = 2(1 - y2)dy.

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Differential Element Here x = 2(1 - y2) The moment of inertia of the differential

element parallel to the x axis shown shaded in Fig a about the y axis is

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(a) Differential Element: The area of the differential element parallel to y axis is

The moment of inertia of this element about x axis is

Moment of Inertia: Performing the integration, we have

Ans.

(b) Differential Element: Here, The area of the differential

element parallel to x axis is

Moment of Inertia: Applying Eq 10–1 and performing the integration, we have

Determine the moment of inertia of the area about the x

axis Solve the problem in two ways, using rectangular

differential elements: (a) having a thickness dx and

(b) having a thickness of dy.

y

x

y = 2.5 – 0.1x2

5 ft2.5 ft

Ans:

Ix = 23.8 ft4

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Differential Element Here, x = 2y1 The area of the differential element parallel to

the x axis shown shaded in Fig a is dA = xdy = 2y1 dy

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Differential Element The area of the differential element parallel to the y axis,

shown shaded in Fig a, is dA = (8 - y)d x= a8 - 18 3bdx

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Ix = p8 m4

Determine the moment of inertia about the x axis.

Solution

Differential Element Here, y = 12 24 - x2 The moment of inertia of the differential

element parallel to the y axis shown shaded in Fig a about x axis is

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Differential Element Here, y = 12 24- x2 The area of the differential element

parallel to the y axis shown shaded in Fig a is dA = ydx = 12 24 - x2dx

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Differential Element The area of the differential element parallel with the x axis

shown shaded in Fig a is dA = x dy = y2 dy

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Determine the moment of inertia for the shaded area about

the y axis.

Solution

Differential Element The moment of inertia of the differential element parallel to

the x axis shown shaded in Fig a about the y axis is

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the x axis.

Solution

Differential Element The moment of inertia of the differential element parallel to

the y axis shown shaded in Fig a about the x axis is

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the y axis.

Solution

shown shaded in Fig a is dA = ydx = h

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the x axis.

Solution

Differential Element Here y = (1 - x)1 The moment of inertia of the

differential element parallel to the y axis shown shaded in Fig a about the x axis is

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the y axis.

Solution

Differential Element Here x = 1 - y2 The moment of inertia of the differential

element parallel to the x axis shown shaded in Fig a about the y axis is

dIy = dI y′ + dAx∼2

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the x axis.

Solution

Differential Element Here x2 = y and x1 = 12 2 The area of the differential element

parallel to the x axis shown shaded in Fig a is dA = (x2 - x1)dy = ay - 12 2bdy.

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the y axis.

Solution

Differential Element Here, y2 = 22x1 and y1 = x The area of the differential

element parallel to the y axis shown shaded in Fig a is dA = (y2 - y1) dx

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b2y2 Thus, the area of the

differential element parallel to the x axis shown shaded in Fig a is dA = (x2 - x1) dy

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a2 x2 Thus, the area of the

differential element parallel to the y axis shown shaded in Fig a is dA = (y2- y1)dx

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Determine the moment of inertia of the composite area

about the x axis.

SOLUTION

Composite Parts: The composite area can be subdivided into three segments as

shown in Fig a The perpendicular distance measured from the centroid of each

segment to the x axis is also indicated.

Moment of Inertia: The moment of inertia of each segment about the x axis can be

determined using the parallel-axis theorem Thus,

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Determine the moment of inertia of the composite area

about the y axis.

SOLUTION

Composite Parts: The composite area can be subdivided into three segments as

shown in Fig a The perpendicular distance measured from the centroid of each

segment to the x axis is also indicated.

Moment of Inertia: The moment of inertia of each segment about the y axis can be

determined using the parallel-axis theorem Thus,

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The polar moment of inertia for the area is

J C= 642 (106) mm4, about the z′ axis passing through the

centroid C The moment of inertia about the y′ axis is

264 (106) mm4, and the moment of inertia about the x axis is

938 (106) mm4 Determine the area A.

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Determine the location of the centroid of the channel’s

cross-sectional area and then calculate the moment of

inertia of the area about this axis

Moment of Inertia: The moment of inertia about the axis for each segment can be

determined using the parallel-axis theorem Ix¿ = Ix¿ + Ady

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Determine y, which locates the centroidal axis x′ for the

cross-sectional area of the T-beam, and then find the

moments of inertia I x′ and I y′

Solution

Centroid Referring to Fig a, the areas of the segments and their respective centroids

are tabulated below

Moment of Inertia The moment of inertia about the x ′ axis for each segment can

be determined using the parallel axis theorem, I x′ = I x′ + Ad2 Referring to Fig b,

Iy′ = 5.725(106) mm4

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Determine the moment of inertia for the beam’s

cross-sectional area about the x axis.

Solution

Moment of Inertia The moment of inertia about x axis for each segment can be

determined using the parallel axis theorem, I x = I x′ + Ad2 Referring to Fig a,

Segment A i(in 2 ) (d y)i (in.) (I x′)i (in 4 ) (Ad y)i (in 4 ) (I x)i (in 4 )

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Moment of Inertia The moments of inertia about the y axis for each segment can

be determined using the parallel axis theorem, I x = I x′ + Ad2 Referring to Fig a,

Segment A i(in 2 ) (dx) i (in.) (I y' )i (in 4 ) (Adx2 )i (in 4 ) (I y)i (in 4 )

cross-sectional area about the y axis.

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Moment of Inertia The moment of inertia about the x axis for each segment can

be determined using the parallel axis theorem, I x = I x′ + Ad2y Referring to Fig a

Ix = Σ(I x)i = 1.715(109) mm4 = 1.72(109) mm4 Ans.

*10–32.

Determine the moment of inertia I x of the shaded area

about the x axis.

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Moment of Inertia The moment of inertia about the y axis for each segment can be

determined using the parallel-axis theorem, I y = I y′ + Ad2 Referring to Fig a

I y = Σ(I y)i= 2.033(109) mm4 = 2.03(109) mm4 Ans.

Determine the moment of inertia I x of the shaded area

about the x axis.

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Determine the moment of inertia of the beam’s

SOLUTION

Moment of Inertia: The dimensions and location of centroid of each segment are

shown in Fig a Since the y axis passes through the centroid of both segments, the

moment of inertia about y axis for each segment is simply

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Determine which locates the centroidal axis for the

cross-sectional area of the T-beam, and then find the

moment of inertia about the x¿ axis

x¿y,

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Moment of Inertia Since the x axis passes through the centroids of the two segments,

Fig a,

I x = 121 (300)(4003) - 121 (280)(3603) = 511.36(106) mm4

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Determine the moment of inertia about the y axis.

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be determined using the parallel-axis theorem, I x = I x′ + Ad y Referring to Fig a,

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Moment of Inertia The moment of inertia about the y axis for each segment can

be determined using the parallel-axis theorem, I y = I y′ + Ad x Referring to Fig a,

Segment A i(in 2 ) (d x)i (in.) (I y′)i (in 4 ) (Ad x)i (in 4 ) (I y)i (in 4 )

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Centroid Referring to Fig a, the areas of the segments and their respective centroids

are tabulated below

Segment A(in 2 ) y∼(in.) y∼A(in 3 )

Thus y = Σy∼A ΣA = 19.00 in3

Moment of Inertia The moment of inertia about the x ′ axis for each segment can

be determined using the parallel axis theorem, I x′ = I x′ + Ad y Referring to Fig b,

Segment A i(in 2 ) (d y)i (in.) (I x′)i (in 4 ) (Ad y)i (in 4 ) (I x′)i (in 4 )

Determine the distance y to the centroid of the beam’s

cross-sectional area; then find the moment of inertia about

the centroidal x′ axis.

3 in

Ans:

y = 1.36 in Ix′ = 18.9 in4

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Moment of Inertia The moment of inertia about the y axis for each segment can

Segment A i(in 2 ) (d x)i (in.) (I y′)i (in 4 ) (Ad x)i (in 4 ) (I y)i (in 4 )

3 in

Ans:

Iy = 341 in4

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cross-sectional area about the x axis.

Ans:

Ix = 154(106) mm4

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Ans:

Iy = 91.3(106) mm4

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Determine the distance to the centroid C of the beam’s

cross-sectional area and then compute the moment of

inertia Ix¿about the axis.x¿

Ans:

y- = 80.7 mm I- x′ = 67.6(106) mm4

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Determine the distance to the centroid C of the beam’s

cross-sectional area and then compute the moment of

inertia Iy¿about the axis.y¿

Ans:

x = 61.6 mm

Iy= = 41.2(106) mm4

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be determined using the parallel-axis theorem, I x = I x′ + Ad y Referring to Fig a,

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Segment A i(in 2 ) (d x)i (in.) (I x′)i (in 4 ) (Ad x)i (in 4 ) (I y)i (in 4 )

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Determine the moment of inertia of the parallelogram

about the x′ axis, which passes through the centroid C of

' y

Ans:

I- x′ = 121 a3b sin3 u

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Determine the moment of inertia of the parallelogram

about the y′ axis, which passes through the centroid C of

the area

SOLUTION

x = a cos u + b - a cos u2 = 12 (a cos u + b)

Iy′ = 2J361 (a sin u)(a cos u)3 + 12 (a sin u)(a cos u) ab2 + a2 cos u - 23 cos ub2R

+ 121 (a sin u)(b - a cos u)3

' y

Ans:

I y′ = ab sin u12 (b2 + a2 cos2 u)

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Locate the centroid of the cross section and determine the

moment of inertia of the section about the axis.x¿

y

SOLUTION

Centroid: The area of each segment and its respective centroid are tabulated below.

0.2 m0.05 m0.4 m

0.2 m 0.3 m 0.2 m 0.2 m

x' –y

Thus,

Ans.

Moment of Inertia: The moment of inertia about the axis for each segment can be

determined using the parallel-axis theorem Ix¿ = Ix¿ + Ady

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cross-sectional area about the axis passing through the centroid

Cof the cross section

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Determine the moment of inertia of the area about the

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u

y

x l

t

Determine the product of inertia of the thin strip of area

with respect to the and axes The strip is oriented at an

angle from the axis Assume that

Ans:

I xy = 13 tl3sin 2u

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Determine the product of inertia of the shaded area with

respect to the x and y axes.

SOLUTION

Differential Element: The area of the differential element parallel to the y axis

this element are x c = x and Thus, the product of inertia of this

element with respect to the x and y axes is

Product of Inertia: Performing the integration, we have

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Determine the product of inertia for the shaded portion of

the parabola with respect to the x and y axes.

SOLUTION

Differential Element: Here, The area of the differential element

for this element are , Then the product of inertia for this element is

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respect to the x and y axes, and then use the parallel-axis

theorem to find the product of inertia of the area with

respect to the centroidal and axes

SOLUTION

Differential Element: The area of the differential element parallel to the y axis

element with respect to the x and y axes is

Ans.

Using the information provided on the inside back cover of this book, the location of

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Determine the product of inertia for the parabolic area with

y

x a

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SOLUTION

Differential Element: The area of the differential element parallel to the y axis is

The coordinates of the centroid for this element are Then the product of inertia for this element is

Ans.

= 280a4

= 12 x4 +4 a22x2+ 2ax3 - 85a3x5 - 87a1x7 `

0 a

' = x,

dA= ydx = Aa1 - x1B2

dx

x y

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t ro

f airen

i o tcuorpeh

t nehTe

ra

2 24 - x2.x

' = x,

dA = ydx = 24 - x2dx

y = 24 - x2

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SOLUTION

Differential Element: Here, x = 2y1 The area of the differential element parallel

to the x axis is dA = xdy = 2y12 dy The coordinates of the centroid for this element

are x = x2 = y1, y = y Then the product of inertia for this element is

dI xy = dI x′y′ + dAx y

= 0 + a2y1dyb(y1)(y) = 2y2 dy

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