Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected]

[email protected] • ENGR-36_Lec-10_FBDs_2D_3D_Systems.pptx1

Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Engineering 36

Chp 5: FBDs2D/3D

Systems



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

Spa

ceD

iagr

amFree B

odyD

iagram



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.



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)



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



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



2D Support ReActions

Cable can only Generate TENSION WeightLess Link is 2-Force Element




Note that in BOTH these Cases the Support ReAction is NORMAL (Perpendicular) to the Supporting Surface

RCN can only PUSH, and NOT PULL




Note that in BOTH these Cases the Support ReAction is NORMAL (Perpendicular) to the Supporting Surface

RCN can PUSH or PULL




Only the Supports (9) & (10) Can Generate a Couple-Moment ReAction



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

http://www.m0a.com/wordpress/wp-content/uploads/2009/09/cg11.jpg



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



2D Free-Body Diagram cont1. 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



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



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



Example: Truss – Draw FBD

WRA RBy

RBx

RB

This Dwg is, in fact, a Full Free Body Diagram

φ




Same as 2D ReActions of this Type




Ball-n-Socket is the 3D analog to the 2D Smooth Pin or Hinge




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




The Sq-Shaft Bearing System does NOT Allow the shaft to spin completely freely, Thus the My




These supports are (obviously) designed to allow the Free Spin on the Pin Axis




This Type of support is commonly Known as a CANTILEVER.• Generates the Maximum

Amount of Unknowns for 3D systems



ROUGH SURFACE ReActions Friction on a Rough Surface will

Generate RCNs Parallel to the Supporting Surface• 2D

• 3D



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



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



Example: cont.

MAz

FAz

FAx

FAyMAy

FBy*FBx*

FBz*

• If the Weight of the Bar is negligible, then All Forces are accounted for and this is, in fact, the FBD



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



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



Example: Pole Draw in the BASE

ReActions

FAzFAy

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



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



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



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)



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



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

FOrFOr



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

WPFOrFOr

Pai

r

Pfluid



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

WPFOrFOr

Pai

r

Pfluid



Example – Cont.

9.81kN

Additional Symplifications• The SideWall Forces on

the O-Ring must cancel if the Cylinder is Balanced

FOrFOr

Pai

r

Pfluid

• The AIR pressure is negligible compared to the FLUID pressure



Example – Cont.

9.81kN

Thus in the NonMoving Simplified System the Fluid Pressure balances the Counter Weight.

Mathematically

Pfluid

kNAreaP pistonfluid 819.



WhiteBoard Work

None Today;Did by

PowerPointW

7kN

57°47°

A

B

C

D



Bruce Mayer, PERegistered Electrical & Mechanical Engineer

[email protected]

Engineering 36

Appendix

Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected]

Documents

pe engineering

freebody force diagram

body spacediagramfree

compound body

freebody diagramselect

force element

free vectorsforces

structure base