University:
Michigan State University
Course:
ERG 160 | Success in Science, Technology, Engineering, and Mathematics
Academic year:

2019
Views:

132

Pages:

2
Author:

Quintin Ware

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 127. Free-Body Diagram: Express forces, weight in terms of rectangular components: uuur CA = − (1.2 m ) i + (1.2 m ) j − ( 0.6 m ) k uuur CB = (1.2 m ) i + (1.2 m ) j − ( 0.6 m ) k By symmetry FCA = FCB , and at the load corresponding to buckling FCA = FCB = 1.8 kN FCA uuur CA = FCA = (1.8 kN ) CA −1.2 i + 1.2 j − 0.6 k ( −1.2 )2 + (1.2 )2 + ( − 0.6 )2 FCA = − (1.2 kN ) i + (1.2 kN ) j − ( 0.6 kN ) k FCB uuur CB = FCB = (1.8 kN ) CB 1.2 i + 1.2 j − 0.6 k (1.2 )2 + (1.2 )2 + ( − 0.6 )2 FCB = (1.2 kN ) i + (1.2 kN ) j − ( 0.6 kN ) k continued COSMOS: Complete Online Solutions Manual Organization System ΣM D = 0: ( 2.4 m ) i × E + ( 2.4 m ) i + (1.2 m ) j × FCB + (1.2 m ) j × FCA + (1.2 m ) i + ( 0.6 m ) j × Pk = 0 i or ( 2.4 m ) Ex j k i j k 0 0 + 2.4 1.2 0 kN ⋅ m E y Ez 1.2 1.2 − 0.6 i j k i + 0 1.2 0 kN ⋅ m + (1.2 m ) −1.2 1.2 − 0.6 0 j k ( 0.6 m ) 0 = 0 0 P Setting the coefficient of the unit vector i equal to zero: (a) i : P(0.6 m) − ( 0.6 )(1.2 ) kN ⋅ m − ( 0.6 )(1.2 ) kN ⋅ m = 0 P = 2.4000 kN or P = 2.40 kN (b) By symmetry, Dz = Ez ΣFz = 0: Dz + Dz + 2.4 kN − 0.6 kN − 0.6 kN = 0 Dz = Ez = − 0.60000 kN Therefore: E z = − ( 0.600 kN ) k

COSMOS Chapter 4 Solution 127

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