BYG140
KONSTRUKSJONSMEKANIKK 1
STATICS
Chapter 5 :Equilibrium of a Rigid Body
Revise:- 1.2 Fundamental Concepts
Idealizations (models)
Particle (Hook)
Particle
•Has a mass, but size that can be neglected.
Rigid body
•Can be considered as a large number of particles
in which all particles remain at a fixed distance
from one another, both before and after applying
a load (i.e. deformation is negligible compared to other
dimensions).
Rigid body
5.1 Conditions for Rigid-body Equilibrium
The body is in Equilibrium when resultant force and resultant
couple moment at any arbitrary point O is equal to zero.
Forces on a rigid body
Forces on a particle
𝐅R =
𝐌𝑅
𝑜
=
𝐅=0
𝐌𝑜 = 0
Necessary and sufficient for the
equilibrium of a rigid body.
Difference between “the forces on a particle” and “the forces on a rigid-body” is;
Particle: all forces pass through a common point
Rigid body: usually, all forces can not pass through a common point and may
cause rotation of the body (due to the moments created by the forces).
Equilibrium in Two Dimensions (2D)
5.2 Free Body Diagrams
Require a complete specification of all the known and unknown
external forces that act on the body.
Support Reactions
Roller
– Prevent
translation in
vertical direction.
Pin
– Prevent translation
in any direction
Fixed
– Prevent both
translation
and rotation.
Equilibrium in Two Dimensions (2D)
5.2 Free Body Diagrams (cont.)
Internal Forces
•
Require a complete specification of all the known
and unknown external forces that act on the body.
Weight and Center of Gravity
•
•
When a body is within a gravitational field, then each of its particles
has specified weight.
Resultant (Weight of the body) is acting through the center of gravity.
Idealized Models
•
For analyzing an actual physical system, need to create an idealized model.
Equilibrium in Two Dimensions (2D)
5.2 Free Body Diagrams (cont.)
Example (1) & (2) :(Example 5.1 and 5.3 in the text book)
(1) Draw free-body diagram of the uniform beam shown in Figure. The beam has a
mass of 100kg.
(2) Two smooth pipes, each having a mass of 300kg, are supported by the forked
tine of the tractor. Draw the free body diagram for each pipe and both pipe
together.
Idealized Model
Equilibrium in Two Dimensions (2D)
5.3 Equations of Equilibrium
Conditions for equilibrium in two dimensions
•
Most often use equation for coplanar equilibrium problems.
𝐹x = 0
•
𝐹y = 0
𝑀𝑜 = 0
Alternative sets of independent equilibrium.
𝐹x = 0
𝑀𝐴 = 0
𝑀𝐵 = 0
3 Independent equations
OR
𝑀𝐴 = 0
𝑀𝐵 = 0
𝑀𝐶 = 0
But A,B,C should not lie on
the same axis.
Equilibrium in Two Dimensions (2D)
5.3 Equations of Equilibrium (cont.)
Example (3) :(Example 5.5 in the text book)
(3 ) Determine the horizontal and vertical components of reaction on the beam
caused by the pin at B and the rocker at A as shown in Figure . Neglect the
weight of the beam.
Equilibrium in Two Dimensions (2D)
Procedure for Analysis
1) Free- Body Diagram
• Establish x,y coordinate axes.
• Draw an outlined shape of the body.
• Show all the forces and couple moments acting on the
body.
• Label all the loadings and specify their direction
relative to the x or y axis.
• Indicate the dimensions of the body.
2) Equation of Equilibrium
• Apply moment equation of equilibrium.
• When applying the force equilibrium equations, orient
the x and y axes along lines.
Equilibrium in Two Dimensions (2D)
5.4 Two and Three-force members
Two-Force Members
•
•
Forces applied at only two
points on the member.
Two forces acting on the
member must have same
magnitude, act in opposite
directions and have same line
of action.
Three-Force Members
•
•
Line of action must pass
through point O.
Special case of ,if force are
parallel point of intersection
will be infinity.
Some equilibrium problems can be
simplified by recognizing members that
are subjected to only two or three
forces.
Equilibrium in Three Dimensions (3D)
5.5 Free–Body Diagrams
Support Reactions
Equilibrium in Three Dimensions (3D)
5.6 Equations of Equilibrium
Vector Equations of Equilibrium
•
Vector sum of all the external forces acting on the body should be
equal to zero. 𝐅 = 𝟎
•
Sum of the couple moments and the moments of all the forces
about any point O should be equal to zero. 𝐌𝑜 = 0
Scalar Equations of Equilibrium
𝐅=
𝐌𝒐 =
F𝑥 𝐢 +
𝑀𝑥 𝐢 +
F𝑦 𝐣 +
𝑀𝑦 𝐣 +
F𝑧 𝐤 = 𝟎
𝑀𝑧 𝐤 = 𝟎
Fx = 0
𝑀𝑥 = 0
Fy = 0
𝑀𝑦 = 0
Scalar equilibrium equations may be used to solve
for at most six unknowns on the free body diagram.
Fz = 0
𝑀𝑧 = 0
Equilibrium in Three Dimensions (3D)
5.7 Constraints and Statical Determinacy
Redundant Constraints
•
•
When a body has redundant supports (more support than
necessary to hold it in equilibrium) it becomes statically
indeterminate.
Statically indeterminate means that there will be more
unknown loadings on the body than no of equations of
equilibrium.
Consider in Structural Mechanics II.
Equilibrium in Three Dimensions (3D)
5.7 Constraints and Statical Determinacy
Improper Constraints (unstable bodies/structures)
•
A body is considered improperly constrained if all the reactive
forces intersect at a common point or pass through a common
axis.
(1)
(2)
(3)
(4)
Equilibrium in Three Dimensions (3D)
Procedure for Analysis
1) Free- Body Diagram
• Establish x,y,z coordinate axes.
• Draw an outlined shape of the body.
• Show all the forces and couple moments acting on the
body.
• Establish the origin x,y,z axes at a convenient point.
• Label all the loadings and specify their directions relative
to the x, y or z axis.
• Indicate the dimensions of the body.
2) Equation of Equilibrium
• If the x,y,z force and moment components seem easy to
determine, apply the six scalar equations of equilibrium
otherwise use vector equations. Apply moment equation
of equilibrium.
Equilibrium in Three Dimensions (3D)
Example (4) & (5):(Example 5.15)
(4) The homogeneous plate shown in Figure ① has a mass of 100kg and is
subjected to a force and couple moment along its edges. If it is supported in the
horizontal plane by a rollar at A, a ball-and-socket joint at B, and a cord at C,
determine the components of reaction at these supports.
(5) Determine the Horizontal and vertical components of reaction at pin A and
force in the cable BC shown in Figure ② Neglect the thickness of the members.
①
②
Example (1) & (2) :(Example 5.1 and 5.3 in the text book)
(1) Draw free-body diagram of the uniform beam shown in Figure. The beam has a
mass of 100kg.
(2) Two smooth pipes, each having a mass of 300kg, are supported by the forked
tine of the tractor. Draw the free body diagram for each pipe and both pipe
together.
Idealized Model
Example (1) & (2) Solutions: (Example 5.1 and 5.3 in the text book)
(1)
(2)
FBD for pipe A
FBD for pipe B
FBD for both pipe
combined
Example (3) :(Example 5.5 in the text book)
(3 ) Determine the horizontal and vertical components of reaction on the beam
caused by the pin at B and the rocker at A as shown in Figure . Neglect the
weight of the beam.
Example (3) Solution:(Example 5.5 in the text book)
3 Unknowns
3 Equations
Example (4) & (5):(Example 5.15)
(4) The homogeneous plate shown in Figure ① has a mass of 100kg and is
subjected to a force and couple moment along its edges. If it is supported in the
horizontal plane by a rollar at A, a ball-and-socket joint at B, and a cord at C,
determine the components of reaction at these supports.
(5) Determine the Horizontal and vertical components of reaction at pin A and
force in the cable BC shown in Figure ② Neglect the thickness of the members.
①
②
Example (4) Solution :Example 5.15 in the text book)
Example (5) Solution:
Summary- Chapter 5
5.1. Conditions for Rigid body Equilibrium
Equilibrium in 2 Dimensions
5.2. Free-Body Diagrams (support reactions)
5.3. Equations of Equilibrium
5.4. Two- and Three-force Members
Equilibrium in 3 Dimensions
5.5. Free-Body Diagrams(support reactions)
5.6. Equations of Equilibrium
5.7. Constraints and Statical Determinacy