[PPT] - R F 0 y y rigid body M 0 M equilibrium Rigid Body PowerPoint Presentation

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five

rigid body equilibrium

Rigid Body Equilibrium 1 Lecture 5 Architectural Structures ARCH 331

lecture ARCHITECTURAL STRUCTURES: FORM, BEHAVIOR, AND DESIGN

ARCH 331

DR. ANNE NICHOLS

FALL 2018

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 

x x

F R  

y y

F R  M M

Equilibrium

rigid body

– doesn’t deform – coplanar force systems

static:

Rigid Body Equilibrium 2 Lecture 5 Foundations Structures ARCH 331

A C B

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Free Body Diagram

FBD (sketch)
tool to see all forces on a body or a

point including

– external forces – weights – force reactions – external moments – moment reactions – internal forces

Rigid Body Equilibrium 3 Lecture 5 Foundations Structures ARCH 331 S2010abn

Free Body Diagram

determine body
FREE it from:

– ground – supports & connections

draw all external forces

acting ON the body

– reactions – applied forces – gravity

Rigid Body Equilibrium 4 Lecture 5 Foundations Structures ARCH 331

mg + weight 100 lb 100 lb

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Free Body Diagram

sketch FBD with relevant geometry
resolve each force into components

– known & unknown angles – name them – known & unknown forces – name them – known & unknown moments – name them

are any forces related to other forces?
for the unknowns
write only as many equilibrium equations as

needed

solve up to 3 equations

Rigid Body Equilibrium 5 Lecture 5 Foundations Structures ARCH 331 S2010abn

Free Body Diagram

solve equations

– most times 1 unknown easily solved – plug into other equation(s)

common to have unknowns of

– force magnitudes – force angles – moment magnitudes

Rigid Body Equilibrium 6 Lecture 5 Foundations Structures ARCH 331 S2010abn

Reactions on Rigid Bodies

result of applying force
unknown size
connection or support type

– known direction – related to motion prevented

Rigid Body Equilibrium 7 Lecture 5 Foundations Structures ARCH 331

no vertical motion no translation no translation no rotation

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Supports and Connections

Rigid Body Equilibrium 8 Lecture 5 Foundations Structures ARCH 331

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Supports and Connections

Rigid Body Equilibrium 9 Lecture 5 Foundations Structures ARCH 331 S2010abn

FBD Example

500 lb known
pin – Ax, Ay
smooth surface –

B at 4:3

3 equations
sum moments at

– A? – B?

Rigid Body Equilibrium 10 Lecture 5 Foundations Structures ARCH 331

(Bx)

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Fx 



Fy 



M

1 



F 



M1 



M

2 



M1 



M2 



M

3 



Moment Equations

sum moments at intersection where the

most forces intersect

multiple moment equations may not be

useful

combos:

Rigid Body Equilibrium 11 Lecture 5 Foundations Structures ARCH 331 S2010abn

Recognizing Reactions

Rigid Body Equilibrium 12 Lecture 5 Foundations Structures ARCH 331

F F

unknowns

3

weight mg unknowns

3

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Recognizing Reactions

Rigid Body Equilibrium 13 Lecture 5 Foundations Structures ARCH 331

unknowns

3

unknowns for 2 bodies

6

unknowns

2

mg weight F1 F2 weight F1 F2 mg not independent

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Constraints

completely constrained

– doesn’t move – may not be statically determinate

improperly or partially constrained

– has  unknowns – can’t solve

Rigid Body Equilibrium 14 Lecture 5 Foundations Structures ARCH 331 S2010abn

Constraints

overconstrained

– won’t move – can’t be solved with statics – statically indeterminate to nth degree

Rigid Body Equilibrium 15 Lecture 5 Foundations Structures ARCH 331

A C B 200 lb-ft 60 lb 55 A 5’ 9’ C B 200 lb-ft 55 60 lb

Ax A

y

Cy MRA

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Partial Constraints

Rigid Body Equilibrium 16 Lecture 5 Foundations Structures ARCH 331

100 N 1 m 0.75 m 30 A B 100 N 1 m 0.75 m 30 A B

A B W 500 mm 200 mm B W

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Cable Reactions

equilibrium:

– more reactions (4) than equations – but, we have slope relationships – x component the same everywhere

Rigid Body Equilibrium 17 Lecture 5 Foundations Structures ARCH 331

A C 45 kN 4 m 2 m B 6 m 45 kN

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Two Force Rigid Bodies

equilibrium:

– forces in line, equal and opposite

Rigid Body Equilibrium 18 Lecture 5 Foundations Structures ARCH 331

A B C

A F2 B F1 d A F2 B F1 d A F2 B F1 a

(no) (no)

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Three Force Rigid Bodies

equilibrium:

– concurrent or parallel forces

Rigid Body Equilibrium 19 Lecture 5 Foundations Structures ARCH 331

A B C

F1 F2 A B C F3 F2 A F1 B C F3 d1 d2 F2 A F1 B C F3 a

(no)

beams!

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Concentrated Loads

Rigid Body Equilibrium 20 Lecture 5 Foundations Structures ARCH 331

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Distributed Loads

Rigid Body Equilibrium 21 Lecture 5 Foundations Structures ARCH 331 S2010abn

Beam Supports

statically determinate
statically indeterminate

Rigid Body Equilibrium 22 Lecture 5 Foundations Structures ARCH 331

L L L simply supported (most common)

verhang

cantilever

L continuous (most common case when L1=L2) L L L Propped Restrained

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loading loading L

A dA wdx W   

 

Equivalent Force Systems

replace forces by resultant
place resultant where M = 0
using calculus and area centroids

Rigid Body Equilibrium 23 Lecture 5 Foundations Structures ARCH 331

dx w(x) x L

dx y x

el

x

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Load Areas

area is width x “height” of load
w is load per unit length
W is total load

Rigid Body Equilibrium 24 Lecture 5 Foundations Structures ARCH 331

x x/2 W x/2 x 2x/3 W/2 x/3 x x/2 W x/6 x/3 W/2 W x w   w w 2 2 W x w   w 2w

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Method of Sections

relies on internal forces being in

equilibrium on a section

cut to expose 3 or less members
coplanar forces  M = 0 too

Rigid Body Equilibrium 25 Lecture 5 Foundations Structures ARCH 331

A B C P F E D P . A By AC AB

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Method of Sections

joints on or off the section are good to

sum moments

quick for few members
not always obvious where to cut or sum

Rigid Body Equilibrium 26 Lecture 5 Foundations Structures ARCH 331

A B C P F E D P . A By AC AB B .