Frames and Machines Summary

Frames and Machines Summary - = C X = 350 lb + F Y = C Y...

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1 Chapter 6 Frames and Machines Review
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2
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3 FBD of Components
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4 Two Force Members z Page 226: When a member is subjected to no couple moments and forces are applied at only two points on the member, the member is a 2 FORCE MEMBER To maintain moment equilibrium, the forces must be collinear. The two forces must be of equal but opposite magnitude to prevent translation. Two Force Members
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5 Two Force Members Two Force Members
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6 EXAMPLE Given : The wall crane supports an external load of 700 lb. Find : The force in the cable at the winch motor W and the horizontal and vertical components of the pin reactions at A, B, C, and D. Plan: a) Draw FBDs of the frame’s members and pulleys. b) Apply the equations of equilibrium and solve for the unknowns. + F Y = 2 T – 700 = 0 T = 350 lb FBD of the Pulley E T T E 700 lb Necessary Equations of Equilibrium:
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7 + F X = C X – 350
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Unformatted text preview: = C X = 350 lb + F Y = C Y 350 = C Y = 350 lb + F X = B X + 350 350 sin 30 = 0 B X = 175 lb + F Y = B Y 350 cos 30 = B Y = 303 . 1 lb A FBD of pulley B B Y B X 30 350 lb 350 lb B A FBD of pulley C C 350 lb C Y C X 350 lb Please note that member BD is a two-force member. + M A = T BD sin 45 (4) 303 . 1 (4) 700 (8) = T BD = 2409 lb + F Y = A Y + 2409 sin 45 303 . 1 700 = A Y = 700 lb = 700 lb + F X = A X 2409 cos 45 + 175 350 = A X = 1880 lb A FBD of member ABC A X A Y A 45 T BD B 175 lb 303.11 lb 700 lb 350 lb 4 ft 4 ft 8 At D, the X and Y component are + D X = 2409 cos 45 = 1700 lb = 1700 lb + D Y = 2409 sin 45 = 1700 lb A FBD of member BD 45 2409 lb B 2409 lb D...
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This note was uploaded on 04/17/2008 for the course EGN 3311 taught by Professor Hudyma during the Spring '08 term at UNF.

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Frames and Machines Summary - = C X = 350 lb + F Y = C Y...

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