 Ch 4. Moments/Equivalent Systems Multimedia Engineering Statics Moment2-D Scalar Moment3-D Scalar Moment3-D Vector Couples and Equiv. System Distributed Loads, Intro
 Chapter 1. Basics 2. Vectors 3. Forces 4. Moments 5. Rigid Bodies 6. Structures 7. Centroids/Inertia 8. Internal Loads 9. Friction 10. Work & Energy Appendix Basic Math Units Sections Search eBooks Dynamics Statics Mechanics Fluids Thermodynamics Math Author(s): Kurt Gramoll ©Kurt Gramoll STATICS - CASE STUDY SOLUTION Joint A Joint A The line of action of the force is 30 cm from joint A. Therefore, the magnitude of the bending moment is      MA = F dA            = (80 N) (30 cm) = 2,400 N-cm Joint B Joint B The distance from joint B to the line of action of F is found from the Pythagorean theorem:      dB = (302 + 452)0.5 = 54.08 cm Substitute and solve for the magnitude of the moment:      MB = (80 N) (54.08 cm) = 4,327 N-cm Joint C Joint C Calculate the moment arm and the moment at joint C in the same manner as for joint B:      dC = (152 + 452)0.5 = 47.43 cm      MC = (80 N) (47.43 cm) = 3,795 N-cm Joint D Joint D and E Calculate the moment of F at joints D and E in the same manner as for joints B and C:      dD = (152 + 152)0.5 = 21.21 cm      MD = (80 N) (21.21 cm) = 1,697 N-cm      dE = (302 + 152)0.5 = 33.54 cm      ME = (80 N) (33.54 cm) = 2,683 N-cm Joint E Maximum Moment The joint with the largest moment is joint B, which has a moment of 4,327 N cm, or in more common units 43.27 N-m.

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