moment of inertia solved problems - montoguequiz.com
TRANSCRIPT
1 Β© 2020 Montogue Quiz
Quiz SM102
Moment of Inertia Lucas Montogue
Problems
PROBLEM βΆA
The moments of inertia of the area shown about the x-axis and the y-axis are:
A) πΌπΌπ₯π₯ = 213 (106) mm4 and πΌπΌπ¦π¦ = 457 (106) mm4
B) πΌπΌπ₯π₯ = 213 (106) mm4 and πΌπΌπ¦π¦ = 588 (106) mm4 C) πΌπΌπ₯π₯ = 331 (106) mm4 and πΌπΌπ¦π¦ = 457 (106) mm4 D) πΌπΌπ₯π₯ = 331 (106) mm4 and πΌπΌπ¦π¦ = 588 (106) mm4
PROBLEM βΆB
The polar moment of inertia of the area presented in the previous part about the origin of the coordinate frame is:
A) π½π½0 = 549 (106) mm4
B) π½π½0 = 670 (106) mm4 C) π½π½0 = 801 (106) mm4 D) π½π½0 = 919 (106) mm4 PROBLEM βΆC
The product of inertia of the area introduced in Part A with respect to the x- and y-axes is
A) πΌπΌπ₯π₯π¦π¦ = 155 (106) mm4
B) πΌπΌπ₯π₯π¦π¦ = 267 (106) mm4 C) πΌπΌπ₯π₯π¦π¦ = 372 (106) mm4
D) πΌπΌπ₯π₯π¦π¦ = 480 (106) mm4
2 Β© 2020 Montogue Quiz
PROBLEM β·
True or false?
1. ( ) The moment of inertia of the shaded area about the x-axis is Ix = (4 7β )ππβ3. 2. ( ) The moment of inertia of the shaded area about the y-axis is Iy = (2 15β )βππ3.
3. ( ) The moment of inertia of the shaded area about the x-axis is Ix = (4ππ4) (9ππ)β . 4. ( ) The moment of inertia of the shaded area about the y-axis is Iy = [ππ4(ππ2 β 2)] ππ3β .
PROBLEM βΈ
True or false?
1. ( ) The moment of inertia of the shaded area about the x-axis is Ix = ππππ04 8β . 2. ( ) The moment of inertia of the shaded area about the y-axis is Iy = ππππ04 4β .
3 Β© 2020 Montogue Quiz
3. ( ) The moment of inertia of the shaded area about the x-axis is greater than 1 m4.
4. ( ) The moment of inertia of the shaded area about the y-axis is greater than 1 m4.
PROBLEM βΉA
Determine the radius of gyration relative to the y-axis for the area shown.
A) π π π¦π¦ = 1.95 u.l.
B) π π π¦π¦ = 4.93 u.l. C) π π π¦π¦ = 8.44 u.l.
D) π π π¦π¦ = 12.1 u.l.
PROBLEM βΉB
Determine the radius of gyration relative to the y-axis for the shaded area.
A) π π π¦π¦ = 2.77 u.l.
B) π π π¦π¦ = 4.90 u.l. C) π π π¦π¦ = 8.20 u.l.
D) π π π¦π¦ = 11.8 u.l.
4 Β© 2020 Montogue Quiz
PROBLEM βΊ (Hibbeler, 2010, w/ permission)
Determine the moment of inertia of the area of the channel with respect to the y-axis.
A) πΌπΌπ¦π¦ = 273 in.4
B) πΌπΌπ¦π¦ = 388 in.4. C) πΌπΌπ¦π¦ = 476 in.4
D) πΌπΌπ¦π¦ = 545 in.4
PROBLEM β» (Bedford & Fowler, 2008, w/ permission)
Determine the moment of inertia of the section relative to the x-axis.
A) πΌπΌπ₯π₯ = 109.6 (109) mm4
B) πΌπΌπ₯π₯ = 163.6 (109) mm4 C) πΌπΌπ₯π₯ = 224.0 (109) mm4 D) πΌπΌπ₯π₯ = 298.5 (109) mm4
PROBLEM βΌ (Bedford & Fowler, 2008, w/ permission)
Determine the moment of inertia of the section relative to the x-axis.
A) πΌπΌπ₯π₯ = 6.0 (106) mm4
B) πΌπΌπ₯π₯ = 9.0 (106) mm4 C) πΌπΌπ₯π₯ = 12.0 (106) mm4 D) πΌπΌπ₯π₯ = 15.0 (106) mm4
5 Β© 2020 Montogue Quiz
PROBLEM β½ (Bedford & Fowler, 2008, w/ permission)
Determine the moment of inertia relative to the y-axis.
A) πΌπΌπ¦π¦ = 8945 in.4
B) πΌπΌπ¦π¦ = 16,565 in.4 C) πΌπΌπ¦π¦ = 24,860 in.4
D) πΌπΌπ¦π¦ = 32,235 in.4
PROBLEM βΎ (Hibbeler, 2010, w/ permission)
Consider the following rectangular beam section and the hypothetical axes u and v. True or false?
1. ( ) The moment of inertia Iu of the section about the u-axis is Iu = 52.5 in.4 2. ( ) The moment of inertia Iv of the section about the v-axis is Iv = 23.6 in.4 3. ( ) The product of inertia Iuv of the section relative to axes u and v is Iuv = 17.5 in.4
PROBLEM βΏ (Hibbeler, 2010, w/ permission)
Locate the centroid οΏ½Μ οΏ½π₯ and π¦π¦οΏ½ of the cross-sectional area and then determine the orientation of the principal axes, which have their origin at the centroid C of the area.
A) ππππ = 15o and ππππ = β15o
B) ππππ = 30o and ππππ = β30o C) ππππ = 45o and ππππ = β45o
D) ππππ = 60o and ππππ = β60o
6 Β© 2020 Montogue Quiz
Additional Information Figure 1 Moments of inertia for selected geometries
Solutions
P.1 β Solution
Part A: The moments of inertia in question are such that, for Ix,
πΌπΌπ₯π₯ = οΏ½ οΏ½ π¦π¦2πππ¦π¦πππ₯π₯β2π₯π₯
12
0
2
0= 2.13 m4 = 213 (106) mm4
and, for Iy,
πΌπΌπ¦π¦ = οΏ½ οΏ½ π₯π₯2 πππ¦π¦πππ₯π₯β2π₯π₯
12
0
2
0= 4.57 m4 = 457 (106) mm4
β The correct answer is A.
Part B: The polar moment of inertia, J0, is obtained with the equation
π½π½0 = οΏ½ππ2ππππ
π΄π΄
where r is the radial distance from the origin of the coordinate system to the elemental area dA. Instead of evaluating this integral, however, we could use the relation r2 = x2 + y2, so that
π½π½0 = οΏ½ππ2ππππ
π΄π΄= οΏ½ (π₯π₯2 + π¦π¦2)ππππ
π΄π΄= οΏ½π₯π₯2ππππ
π΄π΄+ οΏ½π¦π¦2ππππ
π΄π΄= πΌπΌπ¦π¦ + πΌπΌπ₯π₯
That is to say, the polar moment of inertia about the origin equals the sum of the moment of inertia with respect to the x-axis and the moment of inertia with respect to the y-axis. Using the results from Part A, we obtain
π½π½0 = 213 (106) + 457 (106) = 670 (106) mm4
β The correct answer is B.
7 Β© 2020 Montogue Quiz
Part C: The product of inertia of the area in question is given by
πΌπΌπ₯π₯π¦π¦ = οΏ½π₯π₯π¦π¦ππππ
π΄π΄= οΏ½ οΏ½ π₯π₯π¦π¦ πππ¦π¦πππ₯π₯
β2π₯π₯12
0
2
0= 267 (106)mm4
β The correct answer is B.
P.2 β Solution
1. False. The area of the differential element parallel to the x-axis is
ππππ = (β β π¦π¦)πππ₯π₯ = οΏ½β ββππ2 π₯π₯
2οΏ½ πππ₯π₯
Applying the equation for Ix and performing the integration, we have
πΌπΌπ₯π₯ = οΏ½π¦π¦2ππππ
π΄π΄= οΏ½ π¦π¦2 Γ
ππββ
π¦π¦12πππ¦π¦
β
0
β΄ πΌπΌπ₯π₯ =ππββ
Γ2π¦π¦7 2β
7 οΏ½π¦π¦=0
π¦π¦=β
=ππ
β12
Γ2β7 2β
7
β΄ πΌπΌπ₯π₯ =27ππβ
3
2. True. The area of the differential element parallel to the y-axis is
ππππ = (β β π¦π¦)πππ₯π₯ = οΏ½β ββππ2 π₯π₯
2οΏ½ πππ₯π₯
so that, substituting in the expression for Iy and performing the integration, we get
πΌπΌπ¦π¦ = οΏ½ π₯π₯2 οΏ½β ββππ2 π₯π₯
2οΏ½ πππ₯π₯
π΄π΄= οΏ½ οΏ½βπ₯π₯2 β
βππ2 π₯π₯
4οΏ½ πππ₯π₯ππ
0
β΄ πΌπΌπ¦π¦ = οΏ½βπ₯π₯3
3 ββ
5ππ2 π₯π₯5οΏ½οΏ½
π¦π¦=0
π¦π¦=ππ
=βππ3
3 ββππ3
5
β΄ πΌπΌπ¦π¦ =2
15 βππ3
3. True. The differential area element parallel to the y-axis is shown in blue.
8 Β© 2020 Montogue Quiz
The contribution of this element to the moment of inertia Ix is given by
πππΌπΌπ₯π₯ =π¦π¦3
12πππ₯π₯ + (π¦π¦πππ₯π₯) οΏ½π¦π¦2οΏ½
2=π¦π¦3
12πππ₯π₯ +π¦π¦3
4 πππ₯π₯ =13 π¦π¦
3πππ₯π₯
or, substituting y,
πππΌπΌπ₯π₯ =13 οΏ½ππ sin
ππππ π₯π₯οΏ½
3πππ₯π₯ =
ππ3
3 sin3ππππ π₯π₯ πππ₯π₯
Integrating on both sides, we have
οΏ½πππΌπΌπ₯π₯ = οΏ½ππ3
3 sin3ππππ π₯π₯
ππ
0πππ₯π₯ =
ππ3
3 οΏ½ sinππππ π₯π₯ οΏ½1 β cos2
ππππ π₯π₯οΏ½ πππ₯π₯
ππ
0
Let cos(ππ ππβ )π₯π₯ = t. Differentiating on both sides, we have
βοΏ½ππππ sin
ππππ π₯π₯οΏ½
πππ₯π₯ππππ = 1 β sin οΏ½
ππππ π₯π₯οΏ½ πππ₯π₯ = β
ππππππππ
The bounds of the integral change from x = 0 to t = 1 and from x = a to t = - 1. Backsubstituting in the previous integral gives
πΌπΌπ₯π₯ =ππ3
3 οΏ½ (1 β ππ2) οΏ½βππππ πππποΏ½
β1
1= β
ππ4
3πποΏ½(1 β ππ2)ππππ
β1
1
β΄ πΌπΌπ₯π₯ = βππ4
3ππ οΏ½ππ βππ3
3 οΏ½οΏ½π‘π‘=1
π‘π‘=β1
= βππ4
3ππ οΏ½β23 β
23οΏ½ =
4ππ4
9ππ
4. False. Proceeding similarly with the moment of inertia about the y-axis, the following integral is proposed,
πΌπΌπ¦π¦ = οΏ½π₯π₯2ππππ = πποΏ½ π₯π₯2 sinππππ π₯π₯ πππ₯π₯
ππ
0
β΄ πΌπΌπ¦π¦ = ππ οΏ½2 οΏ½πππποΏ½π₯π₯ sin οΏ½πππποΏ½π₯π₯ + οΏ½2β οΏ½πππποΏ½
2π₯π₯2οΏ½ cos οΏ½πππποΏ½π₯π₯
οΏ½πππποΏ½3 οΏ½οΏ½
π₯π₯=0
π₯π₯=ππ
Accordingly,
πΌπΌπ¦π¦ = πποΏ½οΏ½2οΏ½πππποΏ½ ππ sin οΏ½πππποΏ½ππ + οΏ½2β οΏ½πππποΏ½
2π₯π₯2οΏ½ cos οΏ½πππποΏ½ ππ
οΏ½πππποΏ½3 οΏ½
β οΏ½2 οΏ½πππποΏ½0 sin οΏ½πππποΏ½0 + οΏ½2 β οΏ½πππποΏ½
2π₯π₯2οΏ½ cos οΏ½πππποΏ½0
οΏ½πππποΏ½3 οΏ½οΏ½
β΄ πΌπΌπ¦π¦ =ππ4[(ππ2 β 2) β 2]
ππ3 =ππ4(ππ2 β 4)
ππ3
P.3 β Solution
1. True. A circular cross-section such as the one considered here begs the use of polar coordinates. The area of the differential area element shown below is ππππ =(ππππππ)ππππ.
In polar coordinates, the y-coordinate is transformed as π¦π¦ = ππ sinππ. We can then set up the integral as follows,
πΌπΌπ₯π₯ = οΏ½π¦π¦2ππππ
π΄π΄= οΏ½ οΏ½ ππ2 sin2 ππ ππππππππππ
ππ0
0
ππ 2β
βππ 2β
β΄ πΌπΌπ₯π₯ = οΏ½ οΏ½ ππ3 sin2 ππ ππππππππππ0
0
ππ 2β
βππ 2β= οΏ½ οΏ½ ππ3 sin2 ππ ππππππππ
ππ0
0
ππ 2β
βππ 2β
9 Β© 2020 Montogue Quiz
β΄ πΌπΌπ₯π₯ =ππ04
4 οΏ½ sin2 ππ ππππππππππ 2β
βππ 2β
At this point, note that sin2 2ππ β‘ (1 2β )(1β cos 2ππ). Accordingly,
πΌπΌπ₯π₯ =ππ04
4 οΏ½ sin2 ππ ππππππππππ 2β
βππ 2β=ππ04
8 οΏ½ππ βsin 2ππ
2 οΏ½οΏ½βππ 2β
ππ 2β
=ππππ04
8
2. False. In polar coordinates, the x-coordinate is transformed as π₯π₯ = ππ cosππ. Substituting this quantity into the equation for the moment of inertia and integrating, it follows that
πΌπΌπ¦π¦ = οΏ½π₯π₯2ππππ
π΄π΄= οΏ½ οΏ½ ππ2 cos2 ππ ππππππππππ
ππ0
0
ππ 2β
βππ 2β
β΄ πΌπΌπ¦π¦ = οΏ½ οΏ½ ππ3 cos2 ππ ππππππππππ0
0
ππ 2β
βππ 2β= οΏ½ οΏ½ ππ3 cos2 ππ ππππππππ
ππ0
0
ππ 2β
βππ 2β
β΄ πΌπΌπ¦π¦ =ππ04
4 οΏ½ cos2 ππ ππππππππππ 2β
βππ 2β
However, cos2 ππ β‘ (1 2β )(cos 2ππ + 1). Thus,
πΌπΌπ¦π¦ =ππ04
8 οΏ½ (cos 2ππ + 1)ππππππ 2β
βππ 2β=ππ04
8 οΏ½ππ βsin 2ππ
2 οΏ½οΏ½βππ 2β
ππ 2β
=ππππ04
8
3. True. Consider the following illustration of the shaded area.
The elemental contribution of the strip to the overall moment of inertia is
πππΌπΌπ₯π₯ = πππΌπΌοΏ½Μ οΏ½π₯ + πππππ¦π¦οΏ½
which, upon substitution of the pertaining variables, becomes
πππΌπΌπ₯π₯ =πππ₯π₯ Γ π¦π¦3
12 + π¦π¦ Γ πππ₯π₯ Γπ¦π¦2
2 =π¦π¦3πππ₯π₯
3
The overall moment of inertia follows by integrating the relation above, that is,
πΌπΌπ₯π₯ =13οΏ½ π¦π¦3πππ₯π₯
1
0=
13οΏ½ οΏ½πππ₯π₯2οΏ½
3πππ₯π₯
1
0=
13οΏ½ ππ3π₯π₯2πππ₯π₯
1
0
There is no simple antiderivative available to evaluate the integral above. Thus, a numerical procedure is in order. Let us apply Simpsonβs rule,
οΏ½ ππ(π₯π₯)πππ₯π₯π₯π₯ππ
π₯π₯0ββ3
[π¦π¦0 + 4(π¦π¦1 + π¦π¦3 + β―+ π¦π¦ππβ1) + 2(π¦π¦2 + π¦π¦4 + β―+ π¦π¦ππβ2) + π¦π¦ππ]
The bounds of the integral are x = 0 and x = 1. Let the number of intervals be equal to 6, so that the width h of each area element is written as
β =upper limit β lower limit
6 =1 β 0
6 β 0.167
Then, the following table is prepared.
10 Β© 2020 Montogue Quiz
We are now ready to evaluate the integral via Simpsonβs rule, giving
β« ππ3π₯π₯2πππ₯π₯π₯π₯πππ₯π₯0
β (1 6β )3
[1.000 + 4(1.087 + 2.123 + 8.098) + 2(1.397 + 3.814) +20.086] = 4.263
Moment of inertia Ix is then
πΌπΌπ₯π₯ = 13 β« ππ3π₯π₯2πππ₯π₯1
0 = 13
Γ 4.263 = 1.421 m4
which is greater than 1 m4, thus implying that statement 3 is true.
4. False. As before, consider an elemental area of thickness dx, as shown.
The area of the elemental strip is dA = ydx. Then, the moment of inertia of the area about the y-axis is
πΌπΌπ¦π¦ = οΏ½π₯π₯2ππππ
π΄π΄= οΏ½ π₯π₯2π¦π¦πππ₯π₯
1
0
Substituting y = πππ₯π₯2, we have
πΌπΌπ¦π¦ = οΏ½ π₯π₯2πππ₯π₯2πππ₯π₯1
0
This integral, much like the previous one, does not lend itself to simple analytical methods. We shall evaluate it numerically using Simpsonβs rule. Let the number of intervals be equal to 6, so that the width h of each area element becomes h = 1/6 β 0.167. The following table is prepared.
We now have enough information to evaluate the integral; that is,
β« π₯π₯2πππ₯π₯2πππ₯π₯π₯π₯πππ₯π₯0
β (1 6β )3
[0 + 4(0.029 + 0.323 + 1.400) + 2(0.125 + 0.697) +2.718] = 0.632 m4
which is less than 1 m4, thus implying that statement 4 is false.
n x n y n
0 0 1.0001 0.167 1.0872 0.334 1.3973 0.501 2.1234 0.668 3.8145 0.835 8.0986 1 20.086
n x n y n
0 0 0.0001 0.167 0.0292 0.334 0.1253 0.501 0.3234 0.668 0.6975 0.835 1.4006 1 2.718
11 Β© 2020 Montogue Quiz
P.4 β Solution
Part A: The problem involves a straightforward application of the integral formula for moment of inertia, followed by use of the definition of radius of gyration with respect to the y-axis,
π π π¦π¦ = οΏ½πΌπΌπ¦π¦ππ
where Iy is the moment of inertia about the y-axis and A is the area of the surface. Before applying this formula, however, we require the points at which the curve intercepts the x-axis, namely,
π¦π¦ = β14π₯π₯2 + 4π₯π₯ β 7 = 0 β π₯π₯ = 2 and 14
That is, the parabola in question intercepts the x-axis at x = 2 and x = 14. We can now determine the moment of inertia Iy,
πΌπΌπ¦π¦ = οΏ½ οΏ½ π₯π₯2πππ¦π¦πππ₯π₯β14π₯π₯
2+4π₯π₯β7
0
14
2= 5126 [u. l. ]4
The area A of the surface is, in turn,
ππ = οΏ½ οΏ½β14 π₯π₯
2 + 4π₯π₯ β 7οΏ½πππ₯π₯14
2= 72 [u. a. ]
Finally, the radius of gyration Ry is such that
π π π¦π¦ = οΏ½512672 = 8.44 [u. l. ]
β The correct answer is C.
Part B: First, we need to locate the points at which the curve intersects the horizontal line,
5 = β14π₯π₯
2 + 4π₯π₯ β 7 β β14π₯π₯
2 + 4π₯π₯ β 12 = 0
π₯π₯ = 4 and 12
That is, the curve intersects the line at x = 4 and x = 12. We can now produce the integral for the moment of inertia,
πΌπΌπ¦π¦ = οΏ½ οΏ½ π₯π₯2πππ¦π¦πππ₯π₯β14π₯π₯
2+4π₯π₯β7
5
12
4= 1433.6 [u. l]4
We also require area A1 = A β A2, as illustrated below.
The total area A is such that
ππ = οΏ½ οΏ½β14 π₯π₯
2 + 4π₯π₯ β 7οΏ½πππ₯π₯12
4= 61.33 [u. a. ]
In addition, A2 = (12 β 4)Γ5 = 40 u.a., so that A1 = A β A2 = 61.33 β 40 = 21.33 u.a. Finally, the radius of gyration Ry is such that
π π π¦π¦ = οΏ½πΌπΌπ¦π¦ππ1
= οΏ½143421.33
= 8.20 [u. l. ]
β The correct answer is C.
12 Β© 2020 Montogue Quiz
P.5 β Solution
The section can be represented as the sum of a solid rectangular area and a negative rectangular area, as shown.
Since the y-axis passes through the centroid of both rectangular segments, we have
πΌπΌπ₯π₯ = πΌπΌ1 β πΌπΌ2
Here, I1 is such that
πΌπΌ1 =12
(6.5)(14)3 = 1486.3 in.4
and I2 equals
πΌπΌ2 =1
12(6)(13)3 = 1098.5 in.4
Finally, the required moment of inertia is
πΌπΌπ₯π₯ = πΌπΌ1 β πΌπΌ2 = 388 in.4
β The correct answer is B.
P.6 β Solution
The area is divided into 3 rectangles, as shown below.
The total moment of inertia is the sum of the contributions of each part of the section,
πΌπΌπ₯π₯ = πΌπΌ1 + πΌπΌ2 + πΌπΌ3
I1 is determined as
πΌπΌ1 =(600)(200)3
12 = 400 (106) mm4
I2 can be obtained with the parallel-axis theorem,
πΌπΌ2 =ππβ3
12 + πππ¦π¦2ππ =(200)(600)3
12 + 5002(200)(600) = 33,600 (106) mm4
The same applies to I3, which is calculated as
πΌπΌ3 =ππβ3
12 + πππ¦π¦2ππ =(800)(200)2
12 + 9002(200)(800) = 129,600 (106) mm4
Finally, moment of inertia Ix equals
πΌπΌπ₯π₯ = πΌπΌ1 + πΌπΌ2 + πΌπΌ3 = (400 + 33,600 + 129,600)(106) mm4
= 163,600 (106) mm4 = 163.6 (109) mm4
β The correct answer is B.
13 Β© 2020 Montogue Quiz
P.7 β Solution
The area is subdivided into a rectangle (section 1), a semicircle (2), and a void circle (3), as shown.
Using tabulated results, the moment of inertia of part 1 about the x-axis is
πΌπΌ1 =(120)(80)3
12 = 5.12 (106) mm4
Moment of inertia of part 2 about the x-axis is
πΌπΌ2 =πππ π 4
8 =ππ(40)4
8 = 1.01 (106) mm4
Finally, we have the moment of inertia of part 3, which, being a void area, must be subtracted from the total moment of inertia,
πΌπΌ3 = βπππ π 4
4 = βππ(40)4
4 = β0.126 (106) mm4
The total moment of inertia, Ix, is given by
πΌπΌπ₯π₯ = πΌπΌ1 + πΌπΌ2 β πΌπΌ3 = (5.12 + 1.01β 0.126) Γ 106 = 6.0 (106) mm4
β The correct answer is A.
P.8 β Solution
We divide the composite area into a triangle (section 1), a rectangle (2), a half-circle (3), and a circular cutout (4); see below.
The contribution to the moment of inertia due to triangle 1 is
πΌπΌ1 =14 Γ 12 Γ 83 = 1536 in.4
14 Β© 2020 Montogue Quiz
The contribution owing to rectangle 2, in turn, follows from the parallel-axis theorem,
πΌπΌ2 =1
12 Γ 12 Γ 83 + 122 Γ 8 Γ 12 = 14,336 in.4
The contribution relative to half-circle 3 is
πΌπΌ3 = οΏ½ππ8 β
89πποΏ½ (6)4 + οΏ½16 +
4 Γ 63ππ οΏ½
2
Γ12ππ(6)2 = 19,593 in.4
Finally, the contribution from circular cutout 4 is
πΌπΌ4 = β14 Γ ππ(2)4 + (16)2 Γ ππ Γ 22 = β3230 in.4
which, being a hole, is associated with a negative sign. The final moment of inertia is
πΌπΌπ¦π¦ = πΌπΌ1 + πΌπΌ2 + πΌπΌ3 + πΌπΌ4 = 1536 + 14,336 + 19,593β 3230 = 32,235 in.4
β The correct answer is D.
P.9 β Solution
1. False. Since the rectangular beam cross-sectional area is symmetrical about the x- and y-axes, the product of inertia equals zero, Ixy = 0. As for the moments of inertia relative to the x- and y-axes, we have
πΌπΌπ₯π₯ =(3)(6)3
12 = 54 in.4 ; πΌπΌπ¦π¦ =(6)(3)3
12 = 13.5 in.4
We can then apply the following equations for the moments of inertia with respect to the u and v axes.
πΌπΌπ’π’ =πΌπΌπ₯π₯ + πΌπΌπ¦π¦
2 +πΌπΌπ₯π₯ β πΌπΌπ¦π¦
2 cos 2ππ β πΌπΌπ₯π₯π¦π¦ sin 2ππ
πΌπΌπ£π£ =πΌπΌπ₯π₯ + πΌπΌπ¦π¦
2 βπΌπΌπ₯π₯ β πΌπΌπ¦π¦
2 cos 2ππ + πΌπΌπ₯π₯π¦π¦ sin 2ππ
Substituting Ix = 54 in.4, Iy = 13.5 in4, Ixy = 0, and ππ = 30o, it follows that
πΌπΌπ’π’ =54 + 13.5
2 +54 β 13.5
2 cos 2(30)β 0 sin 2(30) = 43.9 in.4
2. True. Indeed,
πΌπΌπ£π£ =54 + 13.5
2 β54 β 13.5
2 cos 2(30) + 0 sin 2(30) = 23.6 in.4
3. True. Indeed,
πΌπΌπ’π’π£π£ =πΌπΌπ₯π₯βπΌπΌπ¦π¦2
sin 2ππ + πΌπΌπ₯π₯π¦π¦ cos 2ππ = 54β13.52
sin 2(30) + 0 cos 2(30) = 17.5 in4
P.10 β Solution
The area is subdivided into two rectangles, as illustrated below.
The coordinates of the centroid are determined as
π₯π₯ =βπ₯π₯οΏ½ππβππ =
0.25(0.5)(6)οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½Area 1
+ 3.25(5.5)(0.5)οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½Area 2
0.5(6) + 5.5(0.5) = 1.68 in
π¦π¦ =βπ¦π¦οΏ½ππβππ =
0.25(0.5)(5.5) + 3(6)(0.5)0.5(5.5) + 6(0.5) = 1.68 in.
15 Β© 2020 Montogue Quiz
Hence, the location of the centroid C is πΆπΆ(οΏ½Μ οΏ½π₯,π¦π¦οΏ½) = (1.68, 1.68). We are now ready to determine the moments of inertia relative to the axes with origin at the centroid. For Ix, we have
πΌπΌπ₯π₯ = πΌπΌ1 + πΌπΌ2
where I1 pertains to cross-sectional area 1 and I2 pertains to cross-sectional area 2; that is,
πΌπΌ1 = οΏ½(0.5)(6)3
12 + 0.5(6)(3β 1.68)2οΏ½ = 14.23 in.4
πΌπΌ2 = οΏ½(5.5)(0.5)3
12 + 5.5(0.5)(1.68β 0.25)2οΏ½ = 5.68 in.4
so that πΌπΌπ₯π₯ = 14.23 + 5.68 = 19.91 in.4 Similarly, for Iy, we have
πΌπΌπ¦π¦ = πΌπΌ1 + πΌπΌ2
where I1 and I2 are given by
πΌπΌ1 = οΏ½6(0.5)3
12 + 0.5(6)(1.68β 0.25)2οΏ½ = 6.20 in.4
πΌπΌ2 = οΏ½0.5(5.5)3
12 + 0.5(5.5)(1.57)2οΏ½ = 13.71 in.4
Therefore, πΌπΌπ¦π¦ = 6.20 + 13.71 = 19.91 in.4 We also require the product of inertia Ixy,
πΌπΌπ₯π₯π¦π¦ = 6(0.5)(β1.435)(1.32) + 5.5(0.5)(1.57)(β1.435) = β11.92 in.4
Finally, we establish the orientation of the principal axes,
tan 2ππππ = β2πΌπΌπ₯π₯π¦π¦
οΏ½πΌπΌπ₯π₯ β πΌπΌπ¦π¦οΏ½= β
2(β11.92)(19.91β 19.91)οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½
=0
β β
Clearly, then, 2ππππ = arctan(β) β 90o andβ 90o, which implies that
2ππππ = 90o β ππππ = 45o andβ 45o
β The correct answer is C.
Answer Summary
Problem 1 1A A 1B B 1C B
Problem 2 T/F Problem 3 T/F
Problem 4 4A C 4B C
Problem 5 B Problem 6 B Problem 7 A Problem 8 D Problem 9 T/F
Problem 10 C
References HIBBELER. R. (2010). Engineering Mechanics: Statics. 12th edition. Upper Saddle
River: Pearson. HIBBELER, R. (2013). Engineering Mechanics: Statics. 13th edition. Upper Saddle
River: Pearson. BEDFORD, A. and FOWLER, W. (2008). Engineering Mechanics: Statics. 5th
edition. Upper Saddle River: Pearson.
Got any questions related to this quiz? We can help! Send a message to [email protected] and weβll
answer your question as soon as possible.