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Int: 3 rd Session: ENGINEERS MECHANICS ENGINEERS MECHANICS STATICS STATICS CHAPTER3CHAPTER3 Lecture Notes: Professor A. Salam Al-Ammri Suhad Ibraheem Mohammed Al-Khwarizmi College of Engineering University of Baghdad 2/3/2012 Professor Dr. A.Salam Al-Ammri & Suhad Ibraheem Mohammad

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Int: 3 rd Session: Contents 2/3/2012 Professor Dr. A.Salam Al-Ammri & Suhad Ibraheem Mohammad

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Int: 3 rd Session: Introduction Current chapter describes replacement of three dimensional forces exerted on a rigid body by an equivalent system consisting of one force acting at a given point and one couple. 2/3/2012 Professor Dr. A.Salam Al-Ammri & Suhad Ibraheem Mohammad

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Int: 3 rd Session: Varignon’s Theorem The moment about a give point O of the resultant of several concurrent forces is equal to the sum of the moments of the various moments about the same point O. Varigon’s Theorem makes it possible to replace the direct determination of the moment of a force F by the moments of two or more component forces of F. 2/3/2012 Professor Dr. A.Salam Al-Ammri & Suhad Ibraheem Mohammad

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Int: 3 rd Session: System of Forces: Reduction to a Force and Couple A system of forces may be replaced by a collection of force-couple systems acting a given point O The force and couple vectors may be combined into a resultant force vector and a resultant couple vector, The force-couple system at O may be moved to O’ with the addition of the moment of R about O’, 2/3/2012Professor Dr. A.Salam Al-Ammri & Suhad Ibraheem Mohammad

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Int: 3 rd Session: Sample Problem Three cables are attached to the bracket as shown. Replace the forces with an equivalent force-couple system at A. 2/3/2012Professor Dr. A.Salam Al-Ammri & Suhad Ibraheem Mohammad

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Int: 3 rd Session: Sample Problem Compute the equivalent force, Compute the equivalent couple, 2/3/2012Professor Dr. A.Salam Al-Ammri & Suhad Ibraheem Mohammad

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Int: 3 rd Session: Sample Problem The building slab has four columns. F1 and F2 = 0. Find: The equivalent resultant force and couple moment at the origin O. Also find the location (x, y) of the single equivalent resultant force. F RO = -50 k – 20 k = -70 k kN M RO = (10 i) x (-20 k) +(4 i + 3 j) x (-50 k) = 200 j j – 150 i = = -150 i j kN· m The location of the single equivalent resultant force is given as: x = M Ryo /F Rzo = 400/(70) = 5.71 m y = M Rxo /F Rzo = (150)/(70) = 2.14 m 2/3/2012 Professor Dr. A.Salam Al-Ammri & Suhad Ibraheem Mohammad

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Int: 3 rd Session: Sample Problem The special purpose milling cutter is subjected to the force of 1200 N and a couple of 240 N.m as shown. Determine the moment of this system about O. 2/3/2012Professor Dr. A.Salam Al-Ammri & Suhad Ibraheem Mohammad

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