Distributed Linear Loads
Replaced by resultant force at geometric center
If distribution is uniform (rectangular) geometric center is in
the middle; resultant force is (load)*(length)
If distribution is triangular geometric center is 1/3 of the way
from t

Method of Sections
Used when you only need to find the force in a few
members of a truss
If the truss is in equilibrium, any segment of the truss is
also in equilibrium so a FBD is drawn around part of it
The forces in each member act axially
Procedure

Frames and Machines
Frames support loads
Machines transmit and alter the effects of forces
Cannot use method of joints or method of sections
because not all members are 2-force members
Free Body Diagrams
Solving the problems will require drawing FBD f

SPE Eastern Regional Meeting Morgantown
October 13 15, 2015
Class will not meet Wednesday, October 14
Midterm Course Evaluations
Please complete online
Only 3 questions
Completely confidential
Equilibrium of a
Rigid Body
Chapter 4
Conditions for Equi

Force Vectors
Chapter 2
Scalars and Vectors Physical Quantities
Scalar Can be completely specified by its magnitude
length, mass, time, temperature
Add algebraically
Vector Requires both magnitude and direction
Statics uses primarily force, position,

Fall 2015
EGRG 311
Engineering Statics
Instructor: Craig Rabatin, PE
E-Mail: [email protected]
Phone: (740) 376-4759
Office: Brown 207
Office Hours: Tuesday/Thursday 9-11
Course Overview:
An introduction to the static analysis of mechanical syste

Engineering
Statics
EGRG 311
Statics is part of Mechanics
Mechanics Study of forces and their effects on bodies that are at rest or in
motion
Solid Mechanics
Statics (EGRG 311) - Bodies at rest or moving with a constant
velocity
Dynamics Accelerating bod

Practice: Problem 3-10
The hub of the wheel can be attached to
the axle either with negative offset or
with positive offset.
If the tire is subjected to both a normal
and radial load determine the resultant
moment about point O for both cases.
Solution
C