Page 212 - Handbook of Civil Engineering Calculations, Second Edition
P. 212

HANGERS, CONNECTORS, AND WIND-STRESS ANALYSIS  1.195

                                                                       2
                                                                            2
                                                                                2
                              an allowable stress of 22 kips/sq.in. (151.7 MPa). Thus, L   14 + 12   340 sq.ft. (31.6
                              sq.in.); L   18.4 ft (5.61 m); r min   (18.4   12)/300   0.74 in. (18.8 mm).
                                Try a 4   4   /4 in. (101.6   101.6   6.35 mm) angle; r   0.79 in. (20.1 mm); A
                                           1
                                             2
                              1.94 sq.in. (12.52 cm ); P max   1.94(22)   42.7 kips (189.9 kN).
                              4. Compute the wind drift if the assumed size of bracing is used
                                                                     3
                              By Eq. 6, P h   {196/[(340)(18.4)(12)]} l.94(29)(10)    147  kips (653.9  N). The
                              wind shear resisted by the columns of the bent is reduced by P h , and the wind drift is re-
                              duced proportionately.
                                From the previous calculation procedure, the following values are obtained: without
                              diagonal bracing,    0.382 in. (9.7 mm); with diagonal bracing,    0.382/(44 – P h )/44
                                0.382 – 1.28 . Solving gives    0.168 < 0.20 in. (5.1 mm), which is acceptable.
                              5. Check the axial force in the brace
                              Thus, P h   147(0.168)   24.7 kips (109.9 kN); P   P h L/a   24.7(18.4)/14   32.5 <
                              42.7 kips (189.9 kN), which is satisfactory. Therefore, the assumed size of the member
                              is satisfactory.

                              LIGHT-GAGE STEEL BEAM WITH
                              UNSTIFFENED FLANGE

                                                                        1
                              A beam of light-gage cold-formed steel consists of two 7   1 /2 in. (177.8   38.1 mm) by no.
                              12 gage channels connected back to back to form an I section. The beam is simply supported
                              on a 16-ft (4.88-m) span, has continuous lateral support, and carries a total dead load of 50
                              lb/lin ft (730 N/m). The live-load deflection is restricted to 1/360 of the span. If the yield-point
                              stress f y is 33,000 lb/sq.in. (227.5 MPa), compute the allowable unit live load for this member.


                              Calculation Procedure:

                              1. Record the relevant properties of the section
                              Apply the AISC Specification for the Design of Light Gage Cold-Formed Steel Structural
                              Members. This is given in the AISC publication Light Gage Cold-Formed Steel Design
                              Manual. Use the same notational system, except denote the flat width of an element by g
                              rather than w.
                                The publication mentioned above provides a basic design stress of 20,000 lb/sq.in.
                              (137.9 MPa) for this grade of steel. However, since the compression flange of the given
                              member is unstiffened in accordance with the definition in one section of the publication, it
                              may be necessary to reduce the allowable compressive stress. A table in the Manual gives
                              the dimensions, design properties, and allowable stress of each section, but the allowable
                              stress will be computed independently in this calculation procedure.
                                Let V   maximum vertical shear; M   maximum bending moment; w   unit load; f b
                              basic design stress; f c   allowable bending stress in compression; v   shearing stress;
                              maximum deflection.
                                                                                         4
                                Record the relevant properties of the section as shown in Fig. 23: I x   12.4 in (516.1
                                                   3
                                4
                                           3
                              cm ); S x   3.54 in (58.0 cm ); R   /16 in. (4.8 mm).
                                                        3
                              2. Compute f c
                              Thus, g   B/2 – t – R   1.1935 in. (30.3 mm); g/t   1.1935/0.105   11.4. From the
                              Manual, the allowable stress corresponding to this ratio is f c   l.667f b – 8640 – l( f b –
                              12,950)g/t]/15   1.667(20,000) – 8640 – (20,000 – 12,950)11.4/15   19,340 lb/sq.in.
                              (133.3 MPa).
   207   208   209   210   211   212   213   214   215   216   217