885_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

885_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Unformatted text preview: 11Ch11.qxd 9/27/08 3:39 PM Page 879 879 SECTION 11.5 Columns with Eccentric Axial Loads Problem 11.5-13 A frame ABCD is constructed of steel wide-flange members (W 8 * 21; E 30 * 106 psi) and subjected to triangularly distributed loads of maximum intensity q0 acting along the vertical members (see figure). The distance between supports is L 20 ft and the height of the frame is h 4 ft. The members are rigidly connected at B and C. A D h E B C q0 q0 E L (a) Calculate the intensity of load q0 required to produce a maximum bending moment of 80 k-in. in the horizontal member of BC. (b) If the load q0 is reduced to one-half of the value calculated in part (a), what is the maximum bending moment in member BC? What is the ratio of this moment to the moment of 80 k-in. in part (a)? Section E-E Solution 11.5-13 Frame with triangular loads (a) LOAD q0 TO PRODUCE M max 80 k-in. Substitute numerical values into Eq. (1). Units: pounds and inches PL2 M max 80,000 lb-in. A 4EI 0.1170093 2P (radians) P resultant force eccentricity q0h h e 2 3 5,000 P A EI P A 4EI P q0 Pe sec PL I2 E 30 * 106 psi L h 4 ft 48 in. h 16 in. 3 e 9.77 in.4 (from Table E-1a) 20 ft 240 in. 2P h 186 lb/in. 2230 lb/ft ; (1) NUMERICAL DATA W 8 * 21 I (2) 0 4461.9 lb 2 ‹ M max P sec (0.0070093 1P) SOLVE EQ. (2) NUMERICALLY MAXIMUM BENDING MOMENT IN BEAM BC kL From Eq. (11-56): M max Pe sec 2 k P(16 in.) [sec (0.0070093 1P)] 5,000 [cos (0.0070093 1P)] 80,000 P e (b) LOAD q0 IS REDUCED TO ONE-HALF ITS VALUE ‹ P is reduced to one-half its value. 1 (4461.9 lb) 2231.0 lb P 2 Substitute numerical values into Eq. (1) and solve for Mmax. 37.75 k-in. ; M max 37.7 0.47 Ratio: 80 k-in. 80 M max ; This result shows that the bending moment varies nonlinearly with the load. ...
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