Ch. 5 Equilibrium of Bodies! §5.1 Equations of equilibrium.! Looking ahead:! ! §5.1 Equations of equilibrium.! §5.2 Equilibrium of bodies in 2–D.! §5.3 Additional topics:! !• More mechanical elements: Cables, pulleys, springs, gears! !• Statically determinate and statically indeterminate problems ! !• 2-force and 3-force members.! §5.4 Equilibrium of bodies in 3–D.! ! §5.2 Equilibrium of bodies in 2-D.! For a 2-D problem, of the six equilibrium equations for a body in 3-D, three of these are automatically satisfied, leaving:! We will drop the z subscript! for problems in 2-D.! Summary:! A body in 3-D has six equilibrium equations.! A body in 2-D (i.e., a 2-D problem) has three equilibrium equations.! ! Procedure for drawing FBDs! 1) Decide on the body whose equilibrium you want to analyze.! ! 2) Imagine that this body is "cut" completely free (separated) from the structure and/or its environment.! ! • In 2-D, think of a closed line that completely encircles the body.! • In 3-D, think of a closed surface that completely surrounds the body.! ! 3) Sketch the body.! ! 4) Sketch the loads:! !• Loads from the environment (e.g., weight).! !• Loads where a cut passes through a structural member.! !• Loads where a cut passes through a support.! ! 5) Select a coordinate system, show dimensions.! A body in 3-D has six equilibrium equations.! or ...! Free Body Diagram! A Free Body Diagram (FBD) is a sketch of a body that shows all of the loads that are applied to the body. A FBD is an essential aid for the application of Newton s laws! The loads that are typically applied to a body have a number of sources, including:! ! • Loads from the environment (e.g., weight, wind force, etc.), also called external loads.! ! • Loads from structural members that are attached to (or contain) the body, also called internal loads.! ! • Loads from supports (these are called support reactions, or simply reactions).! Support reactions! Example 1: A steel I beam in the roof of a building is supported by a pin at point A, a vertical link CD, and a cable applies a 2000 lb vertical force at point B. The beam's cross section is a W14x26 shape which has a weight of 26 lb/ft. Determine the support reactions. ! Original! structure:! Write and solve the equilibrium equations:! ∑ Fx = 0: Ax = 0 ⇒ Ax = 0.! ∑ Fy = 0: Ay + Cy – 286 lb – 2000 lb = 0.! ∑ MA = 0: – (286 lb)(5 ft) – (2000 lb)(7 ft) + Cy (10 ft) = 0.! ! !⇒ Cy =15,430 ft-lb/10 ft = 1543 lb,! ! !⇒ Ay = 743 lb.! ! Note: we could have used the equations ∑ Fx , ∑ MA and ∑ MC to ! FBD:! ! ! Weight of beam = ! (26 lb/ft)(11 ft) = ! 286 lb.! obtain our solution Alternative sets of equilibrium Eqs.! ! ∑ Fx = 0! ∑ Fy = 0! ∑ MA = 0! ! ∑ Fx = 0! ∑ MA = 0! ∑ MB = 0! ! ∑ MA = 0! ∑ MB = 0! ∑ MC = 0! No restrictions on Points A and B Points A, B and C the location of cannot have the cannot lie on the point A.! same x coordinate.! same line.! These rules will provide three equilibrium equations that are linearly independent.! (verify this for yourself ... the answers are the same)!! Example 2: Draw the FBD for each structure and determine the support reactions. ! !