4.2 Second Moments of Area
4.2.2 Parallel-Axis Theorem
110 4 Bending of Beams
The moment of inertiaIz is obtained fromIy by exchanginga andb:
Iz=π
4 b a3. (b)
The coordinate axesy andz are axes of symmetry. Therefore, Iyz= 0. The polar moment of inertia is determined from (4.6c):
Ip=Iy+Iz= π
4 a b(a2+b2).
The area of an ellipse is given byA=π a b. Thus, the Equations (4.7) yield the radii of gyration:
rgy= b
2, rgz =a
2, rgp =1 2
a2+b2.
Then the second moments of area with respect to the ¯y, ¯z-system are given by
Iy¯=
¯
z2dA=
(z+ ¯zc)2dA=
z2dA+ 2 ¯zc
zdA+ ¯zc2 dA, Iz¯=
¯
y2dA=
(y+ ¯yc)2dA=
y2dA+ 2 ¯yc
ydA+ ¯yc2 dA, I¯y¯z =−
¯
yzdA¯ =−
(y+ ¯yc)(z+ ¯zc)dA=−
y zdA−y¯c zdA
−z¯c
ydA−y¯cz¯c dA . The first moments of area
zdA and
ydA about the axes y andz through the centroidCare zero (see Volume 1, Chapter 4).
WithA=
dAandIy =
z2dAetc we thus obtain Iy¯=Iy+ ¯z2cA ,
Iz¯=Iz+ ¯y2cA , I¯y¯z =Iyz−y¯cz¯cA .
(4.13)
These are the relations between the moments of inertia with re- spect to the axes through the centroidCand the moments of iner- tia with respect to axes which are parallel to them. The Equations (4.13) are known as theparallel-axis theorem. Note that when ap- plying the parallel-axis theorem one of the two axes must be a centroidal axis. The terms ¯zc2Aand ¯y2cAare always positive. The- refore, the moments of inertia with respect to centroidal axes are always smaller than those with respect to axes parallel to them.
The term ¯ycz¯cA which appears in the expression forI¯y¯z may be positive, negative or zero, depending on the location of the respec- tive axes.
As an example we determine the moments of inertia with re- spect to the axes ¯y,z¯for the rectangle shown in Fig. 4.4a. Since the moments of inertia with respect to centroidal axes are given by (4.8), we obtain from (4.13)
Iy¯= b h3 12 +
h 2
2
b h=b h3 3 ,
112 4 Bending of Beams
Table4.1.MomentsofInertia AreaIyIzIyzIpI¯y Rectangle y
b Ch ¯yz
bh3 12hb3 120bh 12(h2 +b2 )bh3 3 Square yCa ¯y
a z
a4 12a4 120a4 6a4 3 Triangle a zCh b
y ¯y
bh3 36bh 36(b2 −ba+a2 )−bh2 72(b−2a)bh 36(h2 +b2 −ba+a2 )bh3 12
Circle zC y ¯y
R πR4 4πR4 40πR4 25π 4R4 ThinCircularRing t<<Rm y ¯y
CRm t z
πR3 mtπR3 mt02πR3 mt3πR3 mt Semi-Circle z
RC y ¯y
R4 72π(9π2 −64)πR4 80R4 36π(9π2 −32)πR4 8 Ellipse C yb z
a ¯y
π 4ab3π 4ba3 0πab 4(a2 +b2 )5π 4ab3
114 4 Bending of Beams
Iz¯= h b3 12 +
b 2
2
b h= h b3 3 , Iy¯¯z= 0− b
2 h
2b h=−b2h2 4 .
E4.2 Example 4.2 Determine the moments of inertia for the I-profile shown in Fig. 4.8a. Simplify the results ford, t≪b, h.
a b
d
t
y y
h/2 h/2
C3 (h+t)/2
z C2
C1
z b/2 b/2
t (h+t)/2
2
3 1
Fig. 4.8 Solution We consider the area to be composed of three rectangles (Fig. 4.8b). According to (4.13), the moments of inertia of each part consist of the moments of inertia with respect to the centro- idal axes (see (4.8)) and the corresponding additional terms:
Iy= d h3 12 + 2
b t3 12 +
h 2 + t
2 2
b t
= d h3 12 +b t3
6 +h2b t 2 +t2h b+b t3
2 = d h3
12 +2b t3
3 +h2b t
2 +t2h b , Iz= h d3
12 + 2t b3 12 .
In the case of d, t≪b, h the terms which containd, tquadra- tically or to the third power may be neglected as compared with the terms that are linear indandt:
Iy≈ d h3
12 +h2b t
2 , Iz ≈t b3 6 .
One can see that in the case of a thin-walled profile the moments of inertia 2bt3/12 of the flanges about their respective centroidal axis may be neglected when calculatingIy. CalculatingIz, we may neglect the moment of inertiah d3/12 of the web.
4.2.3 Rotation of the Coordinate System, Principal Moments of Inertia
Let us now consider two coordinate systemsy,zandη,ζ(Fig. 4.9).
Their relative orientation is given by the angleϕ. In order to derive relations between the respective moments of inertia, we use the geometrical relations
η=ycosϕ+zsinϕ, ζ=−ysinϕ+zcosϕ
(see Fig. 4.9). Then the moments of inertia with respect to the axesη,ζ are given by
Iη =
ζ2dA= sin2ϕ
y2dA+ cos2ϕ z2dA
−2 sinϕcosϕ y zdA, Iζ =
η2dA= cos2ϕ
y2dA+ sin2ϕ z2dA + 2 sinϕcosϕ
y zdA, Iηζ =−
η ζdA= sinϕcosϕ
y2dA−cos2ϕ y zdA + sin2ϕ
y zdA−sinϕcosϕ z2dA . With the moments of inertia about the axes y,z according to (4.6) and the trigonometrical relations sin2ϕ = 12(1−cos 2ϕ),
y z
ϕ 0
ζ η
y
z
0 y
η
y z ζ z
ϕ ϕ zsinϕ
ycosϕ
zcosϕ
ysinϕ ϕ
dA η η
ζ ζ
ϕ
Fig. 4.9
116 4 Bending of Beams
cos2ϕ = 12(1 + cos 2ϕ) and 2 sinϕcosϕ = sin 2ϕ we obtain the transformation equations
Iη = 12(Iy+Iz) +12(Iy−Iz) cos 2ϕ+Iyzsin 2ϕ , Iζ = 12(Iy+Iz)−12(Iy−Iz) cos 2ϕ−Iyzsin 2ϕ , Iηζ = −12(Iy−Iz) sin 2ϕ+Iyzcos 2ϕ .
(4.14)
If the moments of inertia with respect to they, z-coordinate sys- tem are known, the moments of inertia with respect to the inclined systemη,ζcan be determined from (4.14).
If we add the first two equations in (4.14) and use (4.6c), we obtain
Iη+Iζ =Iy+Iz=Ip. (4.15)
Hence, the sum of the moments of inertia Iy and Iz (= polar moment of inertia) is independent of the angleϕ. Therefore,Iη+Iζ
is referred to as aninvariantof the coordinate transformation (=
rotation of the coordinate system). One may verify that another invariant is given by [12(Iη−Iζ)]2+Iηζ2 .
According to (4.14), the magnitude of a moment of inertia de- pends on the angleϕ. The momentsIη or Iζ have a maximum or a minimum value if the conditions dIη/dϕ= 0 or dIζ/dϕ= 0 are satisfied. Both conditions lead to the same result:
−1
2(Iy−Iz) sin 2ϕ+Iyzcos 2ϕ= 0.
Therefore, the angleϕ=ϕ∗ which makes the moments of inertia a maximum or a minimum is given by
tan 2ϕ∗= 2Iyz
Iy−Iz
. (4.16)
Because of tan 2ϕ∗= tan 2 (ϕ∗+π/2) there exist two perpendicu- lar axes with the directions given by the anglesϕ∗ andϕ∗+π/2
for which the moments of inertia become an extremum. These axes are calledprincipal axes. We obtain the correspondingprin- cipal moments of inertia if we insert (4.16) into (4.14). Using the trigonometrical relations
cos 2ϕ∗= 1
1 + tan22ϕ∗ = Iy−Iz
(Iy−Iz)2+ 4Iyz2 ,
sin 2ϕ∗= tan 2ϕ∗
1 + tan22ϕ∗ = 2Iyz
(Iy−Iz)2+ 4Iyz2 we find
I1,2=Iy+Iz
2 ±
Iy−Iz
2 2
+Iyz2 . (4.17) The positive (negative) sign corresponds to the maximum (mini- mum) moment of inertia.
We will now determine the angle for which the product of iner- tiaIηζ vanishes. If we introduce the conditionIηζ = 0 into (4.14) we obtain the same angle ϕ∗ as given by (4.16). Thus, the pro- duct of inertia with respect to the principal axes is zero. As already mentioned in Section 4.2.1, the product of inertia is zero with re- spect to an axis of symmetry. Therefore, an axis of symmetry and the axis perpendicular to it are principal axes.
As an example we consider the rectangular area in Fig. 4.10.
Since Iyz = 0 (see (4.8c)), the axes y and z are principal axes and the moments of inertiaIy=b h3/12 andIz =h b3/12 are the
Fig. 4.10 y
b
z C h η
ζ ϕ
118 4 Bending of Beams
principal moments of inertia. The moments of inertia with respect to theη,ζ-coordinate system are determined from (4.14):
Iη =b h
24[(h2+b2) + (h2−b2) cos 2ϕ], Iζ =b h
24[(h2+b2)−(h2−b2) cos 2ϕ], Iηζ =−b h
24(h2−b2) sin 2ϕ .
In the special case of a square (h=b) we obtainIη=Iζ =h4/12 andIηζ = 0. These results are independent of the angleϕ. Hence, in the case of a square each inclined coordinate system represents principal axes.
Note that the moments of inertia are components of a tensor.
Therefore, the transformation relations (4.14) and the resulting equations (4.15) - (4.17) are analogous to the corresponding equa- tions for the stress tensor (see Section 2.2). In analogy to Mohr’s circle for the stresses, a circle for the moments of inertia can be constructed. In doing so, the normal stresses σx, σy have to be replaced with Iy, Iz and the shear stress τxy has to be replaced with the product of inertiaIyz.
E4.3 Example 4.3 Determine the principal axes and the principal mo- ments of inertia for the thin-walled area with constant thickness t(t≪a) shown in Fig. 4.11a.
a b
principal axis 1 (I1)
principal axis 2 (I2) z
a
t
y
z y
a a
a a
a
ϕ∗2 ϕ∗1
Fig. 4.11
Solution First we determine the moments of inertia about the axesy andz. If we divide the area into several rectangular parts and neglect terms of higher order in the thicknesst, we obtain
Iy= 1
12 t(2a)3+ 2
"1
3t a3+a2(a t)
#
=10 3 t a3, Iz= 2
"1
3t a3+a2(a t)
#
=8 3t a3, Iyz = 2$
−%a 2a(a t)&
−% aa
2(a t)&'
=−2t a3. The orientation of the principal axes follows from (4.16)
tan 2ϕ∗= 2Iyz
Iy−Iz =−2·2·3 10−8 =−6 which leads to
ϕ∗1=−40.3◦, ϕ∗2=ϕ∗1+ 90◦= 49.7◦. (a) The principal moments of inertia are calculated from (4.17):
I1,2= t a3 3
⎡
⎣10 + 8
2 ±
10−8 2
2
+ (−6)2
⎤
⎦= (
3±
√37 3
) t a3
→ I1= 5.03t a3, I2= 0.97t a3. (b) The principal axes are shown in Fig. 4.11b. In order to decide which principal moment of inertia (b) corresponds to which direc- tion (a), we may insert (a) into (4.14). However, in this example one can see by inspection that the maximum moment of inertia I1belongs to the angleϕ∗1since the distances of the area elements from the corresponding axis are larger than those for the angle ϕ∗2.