3620736

BMayer@ChabotCollege.edu • ENGR-36_Lec-10_FBDs_2D_3D_Systems.pptx
1
Bruce Mayer, PE
Engineering-36: Engineering Mechanics - Statics
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Engineering 36
Chp 5: FBDs
2D/3D Systems

2
Bruce Mayer, PE
Free Body Diagrams (FBDs)
 A free-body diagram is a sketch of an object
of interest with all the surrounding objects
stripped away to reveal all of the forces
acting on the body
 The purpose of a free-body force diagram is to
assist with determination of the Net Force
and/or Moment acting on a body
Space
Diagram
Free
Body
Diagram

3
Bruce Mayer, PE
Constructing a free-body diagram
 Select an object or group of objects to focus on as the "body“:
i.e., the system.
 Sketch the body by itself, "free" of its surroundings
 Draw only those forces/moments that are acting directly on the body.
• Include both the magnitude and the direction of these forces.
 Do not include any forces that the body exerts on it surroundings,
they do NOT act ON the body.
• However, there is always an equal reaction force acting on the body.
 For a compound body (e.g. Trusses, Machines) you do NOT need to
include any INTERNAL forces acting between the body's SUBPARTS
• these internal forces come in action-reaction pairs which cancel out each
other because of Newton's Third Law.
 Choose a coordinate system and sketch it on the free-body diagram.
 Often choose one of the axes to be parallel one or more forces
• it can sometimes simplify the equations to be solved.

4
Bruce Mayer, PE
Structural Supports
 NonMoving Structures are typically
Connected to Some Sort of
Supporting Base
 The connection between the Structure
and Base are usually Called
“Structural Supports”
 The Force and/or moments exerted on
the Structure Base are usually called
“Structural Reactions” (RCNs for Short)

5
Bruce Mayer, PE
Structural Supports
 A Support that Prevents Linear Motion
(sliding, translating) of the structure then
exerts a Force on the structure
 A Support that Prevents Rotating
Motion (twisting, turning) of the structure
then exerts a Couple Moment on the
structure

6
Bruce Mayer, PE
Recall SLIDING & FREE Vectors
 Forces are SLIDING Vectors;
They can applied at
ANY-POINT on the
Vector Line of Action (LoA)
 COUPLE-Moments are
FREE Vectors; They
can be applied at
ANY Point,
On or Off the Body

7
Bruce Mayer, PE
2D Support ReActions
 Cable can only Generate TENSION
 WeightLess Link is 2-Force Element

8
Bruce Mayer, PE
 Note that in BOTH these Cases the Support
ReAction is NORMAL (Perpendicular) to the
Supporting Surface
 RCN can only PUSH, and NOT PULL

9
Bruce Mayer, PE
 Note that in BOTH these Cases the Support
ReAction is NORMAL (Perpendicular) to the
Supporting Surface
 RCN can PUSH or PULL

10
Bruce Mayer, PE
 Only the Supports (9) & (10) Can Generate a
Couple-Moment ReAction

11
Bruce Mayer, PE
Center of Gravity
 If the Weight of the Rigid
Body is Not Negligible, then
the Entire Weight of the
Body can be concentrated
at a Single Point Called the
Center of Gravity (CG)
• Many times the CG location
is Given
– Can Calculate using Centroid
Methods which will be covered
later

12
Bruce Mayer, PE
2D Free-Body Diagram
 First Steps for Static
Rigid-Body Equilibrium
Analysis
• Identification of All
Forces & Moments
Acting on the Body
• Formulation of the
Free-Body Diagram
 Free Body Diagram
Construction Process
• See next slide

13
Bruce Mayer, PE
2D Free-Body Diagram cont
1. Select the extent of the
free-body and detach it
from the ground and other
bodies
2. Indicate for external loads:
• Point of application
• Magnitude & Direction Of
External Forces
– Including The Body Weight.
3. Indicate point of application
and ASSUMED direction of
UNKNOWN applied forces

14
Bruce Mayer, PE
2D Free-Body Diagram cont.2
• The Unknown Forces
Typically Include
REACTIONS through which
the GROUND and OTHER
BODIES oppose the possible
motion of the rigid body
4. Include All dimensions
Needed to Calculate the
Moments of the Forces

15
Bruce Mayer, PE
Example: Truss Structure
 Consider Rocker &
Pin Supported Truss
 Analyze Loading
• Four External Force
Loads as shown
• Truss Weight, W
• RCN at Pt-A by
Rocker
– Expect NORMAL to
support Pad
• RCN at Pt-B by Pin
– Expect
 in plane of Truss
 Arbitrarily Directed
 Draw the FBD for
this Structure

16
Bruce Mayer, PE
Example: Truss – Draw FBD
W
RA RBy
RBx
RB
 This Dwg is,
in fact, a Full
Free Body
Diagram
φ

17
Bruce Mayer, PE
 Same as 2D ReActions of this Type

18
Bruce Mayer, PE
 Ball-n-Socket is the 3D analog to the
2D Smooth Pin or Hinge

19
Bruce Mayer, PE
 This configuration is Commonly
Known as a
“Pillow Block Bearing”.
 Type of support is (obviously)
designed to allow the shaft to
SPIN FREELY on its AXIS

20
Bruce Mayer, PE
 The Sq-Shaft Bearing System does NOT Allow the
shaft to spin completely freely, Thus the My

21
Bruce Mayer, PE
 These supports are (obviously) designed to allow the
Free Spin on the Pin Axis

22
Bruce Mayer, PE
 This Type of support is
commonly Known as a
CANTILEVER.
• Generates the Maximum
Amount of Unknowns for
3D systems

23
Bruce Mayer, PE
ROUGH SURFACE ReActions
 Friction on a Rough Surface will
Generate RCNs Parallel to the
Supporting Surface
• 2D
• 3D

24
Bruce Mayer, PE
Example: Hinge & Rough-Surf
 Given Bar supported
by Hinge at Pt-A
and rests on the
Rough x*y*z*
Surface at Pt-B
 Analyze Rcn at Pt-A.
By 5.2-(9) the Single
Axially Constrained
Hinge will
• Provide Lateral (y &
z) and Axial (x)
Support
• Resist twisting about
the y and z axes

25
Bruce Mayer, PE
Example: Hinge & Rough-Surf
 PUSH (not PULL)
Normal to the
Surface
• In this case the y*
direction is normal to
the supporting plane
 Resist Sliding in any
direction WITHIN
the supporting plane
 Analyze Rcn at Pt-B.
Support Leg on a
rough surface will

26
Bruce Mayer, PE
Example: cont.
MAz
FAz
FAx
FAy
MAy
FBy*
FBx*
FBz*
• If the Weight of the Bar is
negligible, then All Forces
are accounted for and this
is, in fact, the FBD

27
Bruce Mayer, PE
Symmetry City
 If We’re Lucky enough to have a Plane of
Symmetry for BOTH Loading and Structural
GEOMETRY then we can treat real world 3D
problems as 2D
• OtherWise we need to Operate in full 3D
Can Treat as 2D Must Treat as 3D
No Symmetry

28
Bruce Mayer, PE
Example: Utility Pole
 Consider Leaning
Utility Pole
 Determine the Loads
acting on the BASE
of the Pole
 Analyze Rcn at Base
• This is a FIXED
support which is often
call a CantiLever
• Cantilever supports
resist both forces and
moments in ALL 3
Spatial Directions

29
Bruce Mayer, PE
Example: Pole
 Draw in the BASE
ReActions
FAz
FAy
FAx
MAz
MAy
 This Diagram is
NOT a FBD as it
does not account for
these forces acting
on the pole
• Pole Weight
• Cable Tension
MAx

30
Bruce Mayer, PE
Distributed Forces/Loads
 In Some Cases Forces are concentrated
at Points; this is simplest case
 Often times a Load cannot be identified
with a single point; Instead the Load is
Spread Out over a supporting surface
• Such Forces are Called “Distributed”
 Distributed Loads
are indicated with
a Load Profile

31
Bruce Mayer, PE
Distributed Force Profiles
 A uniformily Dist Load
Has the same action
at every point on it’s
region of application.
• It’s profile is “Flat”
 NonUniform Loads are also common
• They may be kinked, curved, or arbitrary

32
Bruce Mayer, PE
Distributed-Force Equivalent
 In Chp4 we discussed how to Replace a
Distributed-Load with an Equivalent
Point-Load placed at a Specific Location
 Units for Distributed Forces
• 2D → Force per Length (lb/ft, lb/in N/m)
• 3D → Force Per Area (Pa, PSI, PSF)

33
Bruce Mayer, PE
Example: Hydraulic Cylinder
 The Hydraulic
Cylinder Pumps
Fluid in & out of the
Cylinder Reservoir
as Shown at Right
 Draw The loads on
the Piston Assy
 Game Plan:
• Isolate Piston Assy as Free Body
• CareFully Account for all Pt-Force and
PRESSURES acting on the Piston Assy

34
Bruce Mayer, PE
Example – Cont.
9.81kN
 Load-1 = 100 kg (220lb,
9.81kN) CounterWt
 Load-2 = Weight of the
Piston Rod
WR
 Load-3 = Weight of the
Piston
WP
 Load-4 = Lateral Restraining
Forces Exerted by the
Cylinder Wall on the O-Ring
FOr
FOr

35
Bruce Mayer, PE
Example – Cont.
9.81kN
 Load-5 = The Air
Pressure on Top of the
Piston
 Load-6 = The Hydraulic
Fluid Pressure on the
Bottom of the Cylinder
WR
WP
FOr
FOr
Pair
Pfluid

36
Bruce Mayer, PE
Example – Cont.
9.81kN
 We can SIMPLIFY the
analysis by making
assessments about the
relative significance of
the loads
• The Weight of the Rod
and Piston are likely
negligible compared to
the Counter Weight
WR
WP
FOr
FOr
Pair
Pfluid

37
Bruce Mayer, PE
Example – Cont.
9.81kN
 Additional
Symplifications
• The SideWall Forces on
the O-Ring must cancel
if the Cylinder is
Balanced
FOr
FOr
Pair
Pfluid
• The AIR pressure is
negligible compared to
the FLUID pressure

38
Bruce Mayer, PE
Example – Cont.
9.81kN
 Thus in the NonMoving
Simplified System the
Fluid Pressure balances
the Counter Weight.
 Mathematically
Pfluid
kN
Area
P piston
fluid 81
9.



39
Bruce Mayer, PE
WhiteBoard Work
None Today;
Did by
PowerPoint
W
7
kN
57°
47°
A
B
C
D

40
Bruce Mayer, PE
Bruce Mayer, PE
Registered Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Engineering 36
Appendix

3620736

Recommended

Recommended

More Related Content

Similar to 3620736

Similar to 3620736 (20)

More from abrhsh abadi

More from abrhsh abadi (8)

Recently uploaded

Recently uploaded (20)

3620736