VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS
VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS
11th Edition
ISBN: 9781259633133
Author: BEER
Publisher: MCG
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Chapter 9.5, Problem 9.118P

(a)

To determine

Find the mass moment of inertia (ICC') of the plate with respect to centroidal axis CC' that is perpendicular to the plate.

(a)

Expert Solution
Check Mark

Answer to Problem 9.118P

The mass moment of inertia (ICC') of the plate with respect to centroidal axis CC' that is perpendicular to the plate is 0.994ma2_.

Explanation of Solution

Given information:

The height of the section is 1.5a.

The width of the section is a.

The length of the section up to straight edge is 2a.

The length of the section in sloped edge is 2a.

Calculation:

Let divide the section into 1 and 2.

Show the section 1 and section 2 as in Figure 1.

VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS, Chapter 9.5, Problem 9.118P

Calculate the area (A1) of the section 1 as below:

A1=(2a)(a)=2a2

Calculate the area (A2) of the section 2 as below:

A2=12(2a)(a)=a2

Calculate the value (x¯A)1 of the section 1 as below:

(x¯A)1=2a2×A1

Substitute 2a2 for A1.

(x¯A)1=2a2×2a2=2a3

Calculate the value (x¯A)2 of the section 2 as below:

(x¯A)2=(2a+132a)×A2

Substitute a2 for A2.

(x¯A)2=(2a+132a)×a2=83a3

Calculate the value (z¯A)1 of the section 1 as below:

(z¯A)1=12a×A1

Substitute 2a2 for A1.

(z¯A)1=12a×2a2=a3

Calculate the value (z¯A)2 of the section 2 as below:

(z¯A)2=(13a)×A2

Substitute a2 for A2.

(z¯A)2=(13a)×a2=a33

Calculate the centroid (X¯) of the trapezoidal plate with respect to z axis using the relation:

X¯=x¯AAX¯=(x¯A)1+(x¯A)2A1+A2

Substitute 2a3 for (x¯A)1, 83a3 for (x¯A)2, 2a2 for A1 and a2 for A2.

X¯=2a3+83a32a2+a2=149a

Calculate the centroid (Z¯) of the trapezoidal plate with respect to x axis using the relation:

Z¯=AZ¯=(z¯A)1+(z¯A)2A1+A2

Substitute a3 for (z¯A)1, a33 for (z¯A)2, 2a2 for A1 and a2 for A2.

Z¯=a3+a332a2+a2=49a

Calculate the area (A) of the trapezoidal plate as below:

A=A1+A2=2a2+a2=3a2

Express the mass (m) as below:

m=ρtA

Here, ρ is density and t thickness of the plate.

Rewrite the above equation as,

ρt=mA

Substitute 3a2 for A.

ρt=m3a2

Express the mass moment of inertia (Imass) as below:

Imass=ρtIarea

Substitute m3a2 for ρt.

Imass=m3a2Iarea (1)

Calculate the moment of inertia of area (Ix,area)1 about x axis for section 1 as below:

(Ix,area)1=13(2a)(a3)=23a4

Calculate the moment of inertia of area (Ix,area)2 about x axis for section 2 as below:

(Ix,area)2=112(2a)(a3)=16a4

Calculate the moment of inertia (Ix,area) of area of trapezoidal plate with respect to x axis as below:

Ix,area=(Ix,area)1+(Ix,area)2

Substitute 23a4 for (Ix,area)1 and 16a4 for (Ix,area)2.

Ix,area=23a4+16a4=56a4

Calculate the mass moment of inertia (Ix,mass) of trapezoidal plate with respect to x axis as below:

Rewrite the equation (1) as,

Ix,mass=m3a2Ix,area

Substitute 56a4 for Ix,area.

Ix,mass=m3a2×56a4=518ma2

Calculate the moment of inertia of area (Iz,area)1 about z axis for section 1 as below:

(Iz,area)1=13(a)(2a)3=83a4

Calculate the centroid (x¯) of the section 2 with respect to z axis as below:

x¯=2a+13×2a

Express the moment of inertia (Iz') of section to about z axis as below:

Iz'=136(a)(2a)3

Calculate moment of inertia of area (Iz,area)2 about z axis for section 2 using the relation:

(Iz,area)2=Iz'+A2x¯2

Substitute 136(a)(2a)3 for Iz', 12(2a)(a) for A2 and 2a+13×2a for x¯.

(Iz,area)2=136(a)(2a)3+(12(2a)(a))×(2a+13×2a)2=836a4+(a2×(83a)2)=836a4+(649a4)

Calculate the moment of inertia (Ix,area) of area of trapezoidal plate with respect to z axis as below:

Iz,area=(Iz,area)1+(Iz,area)2

Substitute 83a4 for (Iz,area)1 and 836a4+(649a4) for (Iz,area)2.

Iz,area=83a4+836a4+(649a4)=10a4

Calculate the mass moment of inertia (Iz,mass) of trapezoidal plate with respect to z axis as below:

Rewrite the equation (1) as,

Iz,mass=m3a2Iz,area

Substitute 10a4 for Ix,area.

Iz,mass=m3a2×10a4=103ma2

Calculate the mass moment of inertia (Iy,mass) of trapezoidal plate with respect to y axis as below:

Iy,mass=Ix,mass+Iz,mass

Substitute 103ma2 for Iz,mass and 518ma2 for Ix,mass.

Iy,mass=518ma2+103ma2=6518ma2

Write the expression for mass moment of inertia (Iy,mass) with respect to y axis as below:

Iy,mass=ICC',mass+m(X¯2+Z¯2)

Rewrite the above equation as,

ICC',mass=Iy,massm(X¯2+Z¯2) (2)

Calculate the mass moment of inertia (ICC') of the plate with respect to centroidal axis CC' that is perpendicular to the plate as below:

Substitute 6518ma2 for Iy,mass, 149a for X¯ and 49a for Z¯ in equation (2).

ICC',mass=6518ma2m((149a)2+(49a)2)=6518ma221281ma2=0.994ma2

Therefore, the mass moment of inertia (ICC') of the plate with respect to centroidal axis CC' that is perpendicular to the plate is 0.994ma2_.

(b)

To determine

The mass moment of inertia (Iy,mass) of the plate with respect to axis AA' that is parallel to the x axis and is located at a distance 1.5a from the plate.

(b)

Expert Solution
Check Mark

Answer to Problem 9.118P

The mass moment of inertia (Iy,mass) of the plate with respect to axis AA' that is parallel to the x axis and is located at a distance 1.5a from the plate is 2.33ma2_.

Explanation of Solution

Given information:

The height of the section is 1.5a.

The width of the section is a.

The length of the section up to straight edge is 2a.

The length of the section in sloped edge is 2a.

Calculation:

Write the expression for mass moment of inertia (Ix,mass) with respect to x axis as below:

(Ix,mass)=IBB',mass+m(Z¯2)IBB',mass=Ix,massm(Z¯2) (3)

Calculate the mass moment of inertia (IAA',mass) with respect to AA' axis as below:

IAA',mass=IBB',mass+m(1.5a)2

Substitute Ix,massm(Z¯2) for IBB',mass.

IAA',mass=(Ix,massm(Z¯2))+m(1.5a)2IAA',mass=Ix,mass+m((1.5a)2(Z¯2))

Substitute 518ma2 for Ix,mass and 49a for Z¯.

IAA',mass=518ma2+m((1.5a)2(49a)2)=518ma2+m(2.05a2)=2.33ma2

Therefore, the mass moment of inertia (Iy,mass) of the plate with respect to axis AA' that is parallel to the x axis and is located at a distance 1.5a from the plate is 2.33ma2_.

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Chapter 9 Solutions

VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS

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