154_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

# 154_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Unformatted text preview: Sec_2.4.qxd 9/25/08 148 11:38 AM Page 148 CHAPTER 2 Axially Loaded Members Problem 2.4-16 A rigid bar ABCD is pinned at point B and sup- a = 250 mm ported by springs at A and D (see figure). The springs at A and D have stiffnesses k1 10 kN/m and k2 25 kN/m, respectively, and the dimensions a, b, and c are 250 mm, 500 mm, and 200 mm, respectively. A load P acts at point C. If the angle of rotation of the bar due to the action of the load P is limited to 3°, what is the maximum permissible load Pmax? A b = 500 mm B C D P c = 200 mm k 2 = 25 kN/m k1 = 10 kN/m Solution 2.4-16 Rigid bar supported by springs EQUATION OF EQUILIBRIUM MB 0 FA(a) P(c) FD(b) a 250 mm EQUATION OF COMPATIBILITY dA dD a b FORCE-DISPLACEMENT RELATIONS FA FD dD dA k1 k2 b 500 mm 200 mm k1 10 kN/m k2 25 kN/m Substitute (3) and (4) into Eq. (2): FA FD ak1 bk2 (Eq. 1) SOLUTION OF EQUATIONS c 0 NUMERICAL DATA umax 3° p rad 60 FREE-BODY DIAGRAM AND DISPLACEMENT DIAGRAM (Eq. 2) (Eqs. 3, 4) (Eq. 5) SOLVE SIMULTANEOUSLY EQS. (1) AND (5): ack1P bck2P FA FD 2 2 2 a k1 + b k2 a k1 + b2k2 ANGLE OF ROTATION FD dD bcP cP dD u k2 b a2k1 + b2k2 a2k1 + b2k2 MAXIMUM LOAD u2 (a k1 + b2k2) P c Pmax umax 2 (a k1 + b2k2) c ; SUBSTITUTE NUMERICAL VALUES: Pmax p/60 rad [(250 mm)2(10 kN/m) 200 mm + (500 mm)2(25 kN/m)] 1800 N ; ...
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