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. !
!