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Frames and Machines Summary

# Frames and Machines Summary - = C X = 350 lb ↑ ∑ F Y =...

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1 Chapter 6 Frames and Machines Review

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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
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:
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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Frames and Machines Summary - = C X = 350 lb ↑ ∑ F Y =...

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