________________________________________________________________________ PROBLEM (12.74) Applying the unit-load method, determine the horizontal displacement of point C of the truss shown in Fig. P12.74. Each member has the same axial rigidity AE. Given: P = 40 kips E = 30 x 106 psi A = 1.2 in.2 . LAC = 12.806 ft LCB = 10 ft LAB = 16 ft SOLUTION C 41 8 ft P A 5 P 2 4 P FAC P 4 5 3 16 ft B P 2 A 4 5 FAB P/2 Joint A 41 P 8 ∑F = 0 : FAC = ∑F 3 = 0 : FAB = P 8 5 FBC = − P 8 y x Joint B For a horizontal unit load applied at C: 3 5 41 f AB = f BC = − f AC = 8 8 8 Therefore, 1 δC = ∑ f j Fj L j AE 12(40 ×103 ) 41 41 5 5 3 3 [( )( )(12.806) + (− )(− )(10) + ( )( )(16)] = 6 1.2(30 ×10 ) 8 8 8 8 8 8 = 0.191 in. → Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.