Determine the moment of inertia Ic with respect to the horizontal axis C–C through the centroid C of the beam cross section shown in Fig. 12-16. (The position of the centroid C was determined previously in Example 12-2 of Section 12.3.)
Note: From beam theory (Chapter 5), we know that axis C–C is the neu-tral axis for bending of this beam, and therefore the moment of inertia $I_{c}$ must be determined in order to calculate the stresses and deflections of this beam.

Question

Determine the moment of inertia Ic with respect to the horizontal axis C–C through the centroid C of the beam cross section shown in Fig. 12-16. (The position of the centroid C was determined previously in Example 12-2 of Section 12.3.)
Note: From beam theory (Chapter 5), we know that axis C–C is the neu-tral axis for bending of this beam, and therefore the moment of inertia [latex]I_{c}[/latex] must be determined in order to calculate the stresses and deflections of this beam.

Accepted Answer

We will determine the moment of inertia [latex]I_{c}[/latex] with respect to axis C–C by applying the parallel-axis theorem to each individual part of the composite area. The area divides naturally into three parts: (1) the cover plate, (2) the wide-flange section, and (3) the channel section. The following areas and centroidal distances were obtained previously in Example 12-2:

[latex]\begin{gathered}A_{1}=37.5 \mathrm{~cm}^{2} \quad A_{2}=178 \mathrm{~cm}^{2} \quad A_{3}=75.8 \mathrm{~cm}^{2} \\bar{y}_{1}=227.5 \mathrm{~mm} \quad \bar{y}_{2}=0 \quad \bar{y}_{3}=246 \mathrm{~mm}\quad \bar{c}=34.73 \mathrm{~mm}\end{gathered}[/latex]

The moments of inertia of the three parts with respect to horizontal axes through their own centroids [latex]C_{1}[/latex] , [latex]C_{2}[/latex] , and [latex]C_{1}[/latex] are as follows:

[latex]\begin{gathered}I_{1}=\frac{b h^{3}}{12}=\frac{1}{12}(25 \mathrm{~cm})(1.5 \mathrm{~cm})^{3}=7.031 \mathrm{~cm}^{4} \I_{2}=63720 \mathrm{~cm}^{4} \quad I_{3}=597 \mathrm{~cm}^{4}\end{gathered}[/latex]

The moments of inertia [latex]I_{2}[/latex] and [latex]I_{3}[/latex] are obtained from Tables E-1 and E-3, respectively, of Appendix E.

TABLE E-1
Properties of European Wide-Flange Beams

Designation	Mass per meter	Area of section	Depth of section	Width of section	Thickness		Strong axis 1-1			Weak axis 2-2
	G	A	h	b	[latex]t_{w}[/latex]	[latex]t_{t}[/latex]	[latex]I_{1}[/latex]	[latex]S_{1}[/latex]	[latex]r_{1}[/latex]	[latex]I_{2}[/latex]	[latex]S_{2}[/latex]	[latex]r_{2}[/latex]
	Kg/m	[latex]{cm}^{2}[/latex]	mm	mm	mm	mm	[latex]{cm}^{4}[/latex]	[latex]{cm}^{3}[/latex]	cm	[latex]{cm}^{4}[/latex]	[latex]{cm}^{3}[/latex]	cm
HE 1000 B HE 900 B HE 700 B HE 650 b HE 600 B	314 291 241 225 212	400 371.3 306.4 286.3 270	1000 900 700 650 600	300 300 300 300 300	19 18.5 17 16 15.5	36 35 32 31 30	644700 494100 256900 210600 171000	12890 10980 7340 6480 5701	40.15 36.48 28.96 27.12 25.17	16280 15820 14440 13980 13530	1085 1054 962.7 932.3 902	6.38 6.53 6.87 6.99 7.08
HE 550 B HE 600 A HE 450 B HE 550 A HE 360 B HE 450 A	199 178 171 166 142 140	254.1 226.5 218 211.8 180.6 178	550 590 450 540 360 440	300 300 300 300 300 300	15 13 14 12.5 12.5 11.5	29 25 26 24 22.5 21	136700 141200 79890 111900 43190 63720	4971 4787 3551 4146 2400 2896	23.2 24.97 19.14 22.99 15.46 18.92	13080 11270 11720 10820 10140 9465	871.8 751.4 781.4 721.3 676.1 631	7.17 7.05 7.33 7.15 7.49 7.29
HE 340 B HE 320 B HE 360 A HE 340 A	134 127 112 105	170.9 161.3 142.8 133.5	340 320 350 330	300 300 300 300	12 11.5 10 9.5	21.5 20.5 17.5 16.5	36660 30820 33090 27690	2156 1926 1891 1678	14.65 13.82 15.22 14.4	9690 9239 7887 7436	646 615.9 525.8 495.7	7.53 7.57 7.43 7.46
HE 320 A HE 260 B HE 240 B HE 280 A HE 220 B HE 260 A HE 240 A	97.6 93 83.2 76.4 71.5 68.2 60.3	124.4 118.4 106 97.26 91.04 86.82 76.84	310 260 240 270 220 250 230	300 260 240 280 220 260 240	9 10 10 8 9.5 7.5 7.5	15.5 17.5 17 13 16 12.5 12	22930 14920 11260 13670 8091 10450 7763	1479 1148 938.3 1013 735.5 836.4 675.1	13.58 11.22 10.31 11.86 9.43 10.97 10.05	6985 5135 3923 4763 2843 3668 2769	465.7 395 326.9 340.2 258.5 282.1 230.7	7.49 6.58 6.08 7 5.59 6.5 6
HE 180 B HE 160 B HE 140 B HE 120 B HE 140 A	51.2 42.6 33.7 26.7 24.7	65.25 54.25 42.96 34.01 31.42	180 160 140 120 133	180 160 140 120 140	8.5 8 7 6.5 5.5	14 13 12 11 8.5	3831 2492 1509 864.4 1033	425.7 311.5 215.6 144.1 155.4	7.66 6.78 5.93 5.04 5.73	1363 889.2 549.7 317.5 389.3	151.4 111.2 78.52 52.92 55.62	4.57 4.05 3.58 3.06 3.52
HE 100 B HE 100 A	20.4 16.7	26.4 21.24	100 96	100 100	6 5	10 8	449.5 349.2	89.91 72.76	4.16 4.06	167.3 133.8	33.45 26.76	2.53 2.51
Note: Axes 1-1 and 2-2 are principal centroidal axes.

TABLE E-3
Properties of European Standard Channels

Designation	Mass per meter	Area of section	Depth of section	Width of section	Thickness		Strong axis 1-1			Weak axis 2-2
	G	A	h	b	[latex]t_{w}[/latex]	[latex]t_{t}[/latex]	[latex]I_{1}[/latex]	[latex]S_{1}[/latex]	[latex]r_{1}[/latex]	[latex]I_{2}[/latex]	[latex]S_{2}[/latex]	[latex]r_{2}[/latex]	c
	Kg/m	[latex]{cm}^{2}[/latex]	mm	mm	mm	mm	[latex]{cm}^{4}[/latex]	[latex]{cm}^{3}[/latex]	cm	[latex]{cm}^{4}[/latex]	[latex]{cm}^{3}[/latex]	cm	cm
UPN 400	71.8	91.5	400	110	14	18	20350	1020	14.9	846	102	3.04	2.65
UPN 380 UPN 350 UPN 320 UPN 300	63.1 60.6 59.5 46.2	80.4 77.3 75.8 85.5	380 350 320 300	102 100 100 100	13.5 14 14 10	16 16 17.5 16	15760 12840 10870 8030	829 734 679 535	14 12.9 12.1 11.7	615 570 597 495	78.7 75 80.6 67.8	2.77 2.72 2.81 2.9	2.38 2.4 2.6 2.7
UPN 280 UPN 260 UPN 240 UPN 220 UPN 200	41.8 37.9 33.2 29.4 25.3	53.3 48.3 42.3 37.4 32.2	280 260 240 220 200	96 90 85 80 75	10 10 9.5 9 8.5	15 14 13 12.5 11.5	6280 4820 3600 2690 1910	448 371 300 245 191	10.9 9.99 9.22 8.48 7.7	399 317 248 197 148	57.2 47.7 39.6 33.6 27	2.74 2.56 2.42 2.3 2.14	2.53 2.36 2.23 2.14 2.01
UPN 180 UPN 160 UPN 140 UPN 120 UPN 100	22 18.8 16 13.4 10.6	28 24 20.4 17 13.5	180 160 140 120 100	70 65 60 55 50	8 7.5 7 7 6	11 10.5 10 9 8.5	1350 925 605 364 206	150 116 86.4 60.7 412	6.95 6.21 5.45 4.62 3.91	114 85.3 62.7 43.2 29.3	22.4 18.3 14.8 11.1 8.49	2.02 1.89 1.75 1.59 1.47	1.92 1.84 1.75 1.6 1.55
UPN 80	8.64	11	80	45	6	8	106	26.5	3.1	19.4	6.36	1.33	1.45
Notes: 1. Axes 1-1 and 2-2 are principal centroidal axes. 2. The distance c is measured from the centroid to the back of the web. 3. For axis 2-2, the tabulated value of S is the smaller of the two section moduli for this axis.

Now we can use the parallel-axis theorem to calculate the moments of inertia about axis C–C for each of the three parts of the composite area

[latex]\begin{aligned}&\left(I_{c} ight)_{1}=I_{1}+A_{1}\left(\bar{y}_{1}+\bar{c} ight)^{2}=7.031\mathrm{~cm}^{4}+\left(37.5 \mathrm{~cm}^{2} ight)(26.22 \mathrm{~cm})^{2}=25790 \mathrm{~cm}^{4} \&\left(I_{c} ight)_{2}=I_{2}+A_{2} \bar{c}^{2}=63720 \mathrm{~cm}^{4}+\left(178 \mathrm{~cm}^{2} ight)(34.73\mathrm{~cm})^{2}=65870 \mathrm{~cm}^{4} \&\left(I_{c} ight)_{3}=I_{3}+A_{3}\left(\bar{y}_{3}-\bar{c} ight)^{2}=597 \mathrm{~cm}^{4}+\left(75.8 \mathrm{~cm}^{2} ight)(21.13 \mathrm{~cm})^{2}=34430\mathrm{~cm}^{4}\end{aligned}[/latex]

The sum of these individual moments of inertia gives the moment of inertia of the entire cross-sectional area about its centroidal axis C–C:

[latex]I_{c}=\left(I_{c} ight)_{1}+\left(I_{c} ight)_{2}+\left(I_{c} ight)_{3}=1.261 imes 10^{5} \mathrm{~cm}^{4}[/latex]

This example shows how to calculate moments of inertia of composite areas by using the parallel-axis theorem.

Question 12.5: Determine the moment of inertia Ic with respect to the horiz...

Related Answered Questions

Determine the product of inertia Ixy of the Z-section shown in Fig. 12-23. The section has width b, height h, and constant thickness t. ...

Verified Answer:

The cross section of a steel beam is constructed of a HE 450A wide-flange section with 25 cm × 1.5 cm a cover plate welded to the top flange and a UPN 320 channel section welded to the bottom flange (Fig. 12-8). Locate the centroid C of the cross-sectional area. ...

Verified Answer:

The parabolic semisegment OAB shown in Fig. 12-15 has base b and height h. Using the parallel-axis theorem, determine the moments of inertia and with respect to the centroidal axes xc and yc I . ...

Verified Answer:

A parabolic semisegment OAB is bounded by the x axis, the y axis, and a par-abolic curve having its vertex at A (Fig. 12-5). The equation of the curve is y=f(x)=h (1-x²/b²) (a) in which b is the base and h is the height of the semisegment. Locate the centroid C of the semisegment. ...

Verified Answer:

Determine the moments of inertia Ix and Iy for the parabolic semiseg -ment OAB shown in Fig. 12-12. The equation of the parabolic boundary is y=f(x)=h(1-x^2/b^2)(a)(This same area was considered previously in Example 12-1.) ...

Verified Answer:

Determine the orientations of the principal centroidal axes and the magnitudes of the principal centroidal moments of inertia for the cross-sectional area of the Z-section shown in Fig. 12-28. Use the following numerical data: height h = 200 mm, width b = 90 mm, and constant thickness t = 15 mm. ...

Verified Answer: