Page 41 - Handbook of Civil Engineering Calculations, Second Edition
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1.24            STRUCTURAL STEEL ENGINEERING AND DESIGN


                              Then y m   2051/212.5   9.7 in. (246.4 mm). Since the area is symmetric with respect
                            to the y axis, this is also a centroidal axis. The intersection point G of the x and y axes is,
                            therefore, the centroid of the area.
                            4. Compute the distance between the centroidal axis and the
                            reference axis
                            Compute k, the distance between the horizontal centroidal axis of each element and the x
                            axis. Only absolute values are required. Thus k 1   9.7   4.0   5.7 in. (1448 mm); k 2
                            9.7   8.0   1.7 in. (43.2 mm); k 3   13.4   9.7   3.7 in. (94.0 mm).
                            5. Compute the moment of inertia of the entire area—x axis
                            Let I 0 denote the moment of inertia of an element with respect to its horizontal centroidal
                            axis and A its area. Compute the moment of inertia I x of the entire area with respect to the
                                                                  2
                            x axis by applying the transfer equation I x  I 0   AK . Thus

                                                                                   4
                                                                                 2
                                                    4
                                                        4
                            Element                                         I 0 , in (dm )                                                       Ak , in (dm )
                                                                                       4
                              1              1 /36(16)(6)   96 (0.40)       48(5.7)   1560  (6.49)
                                                    3
                                                                                 2
                                                                                 2
                              2              1 /12(16)(4)   85 (0.35)       64(1.7)   185  (0.77)
                                                    3
                              3               0.110(8)   451 (1.88)       100.5(3.7)   1376  (5.73)
                                                                                 2
                                                    4
                                                      _________
                                                                                   __________
                            Total                     632 (2.63)                   3121 (12.99)
                                                    4
                                                             4
                            Then, I x   632   3121   3753 in (15.62 dm ).
                            6. Determine the moment of inertia of the entire area—y axis
                            For this computation, subdivide element 1 into two triangles having the y axis as a base.
                            Thus
                                                                         4
                                                                      4
                                           Element                          I about y axis, in (dm )
                                             1          2( 1 / 12)(6)(8)   0512  (2.13)
                                                                3
                                                                3
                                             2           1 / 12(4)(16)   1365  (5.68)
                                             3          1 / 2(0.785)(8)   1607  (6.89)
                                                                4
                                                                   __________
                                                               I y   3484 (14.5)
                            7. Compute the polar moment of inertia of the area
                            Apply the equation for the polar moment of inertia J G with respect to G: J G   I x   I y
                                                     4
                                             4
                            3753   3484   7237 in (30.12 dm ).
                            8. Determine the moment of inertia of the entire area—w axis
                            Apply the equation in step 5 to determine the moment of inertia I w of the entire area with
                            respect to the horizontal axis w through A. Thus k   15.0   9.7   5.3 in. (134.6 mm); I w
                                                                     4
                                                    2
                                   2
                                                            4
                              I x   Ak   3753   212.5(5.3)   9722 in (40.46 dm ).
                            9. Compute the polar moment of inertia
                            Compute the polar moment of inertia of the entire area with respect to A. Then J A   I w
                                                          4
                                                 4
                            I y   9722 ± 3484   13,206 in (54.97 dm ).
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