Unidad 3 Estatica Del Solido Rigido (1)

8/16/2019 Unidad 3 Estatica Del Solido Rigido (1)

http://slidepdf.com/reader/full/unidad-3-estatica-del-solido-rigido-1 1/25

EQUILIBRIO DE CUERPOS RIGIDOS

Mecánica para Ingenieros



4 - 2

ContentsIntroduction

Free-Body Diagram

Reactions at Supports and Connectionsfor a Two-Dimensional Structure

Equilibrium of a Rigid Body in TwoDimensions

Statically Indeterminate Reactions

Sample roblem !"#

Sample roblem !"$

Sample roblem !"!

Equilibrium of a Two-Force Body

Equilibrium of a T%ree-Force Body

Sample roblem !"&

Equilibrium of a Rigid Body in T%reeDimensions

Reactions at Supports and Connections for aT%ree-Dimensional Structure

Sample roblem !"'



4 - 3

Introduction

• T%e necessary and sufficient condition for t%e static equilibrium of a body are t%at t%e resultant force and couple from all e(ternal forces

form a system equi)alent to *ero+

• Resol)ing eac% force and moment into its rectangular components

leads to & scalar equations w%ic% also e(press t%e conditions forstatic equilibrium+

• For a rigid body in static equilibrium+ t%e e(ternal forces and

moments are balanced and will impart no translational or rotationalmotion to t%e body"

( )∑ ∑ =∑ ×== ,, F r M F O

∑ =∑ =∑ =

∑ =∑ =∑ =

,,,

,,,

z y x

z y x

M M M

F F F



4 - 4

Free-Body Dia ra!First ste" in t#e static e$ui%i&riu! ana%ysis o' a

ri id &ody is identi'ication o' a%% 'orces actin ont#e &ody (it# a free-body dia ra!)

• Select t%e e(tent of t%e free-body and detac% itfrom t%e ground and all ot%er bodies"

• Include t%e dimensions necessary to computet%e moments of t%e forces"

• Indicate point of application and assumeddirection of un nown applied forces" T%ese

usually consist of reactions t%roug% w%ic% t%eground and ot%er bodies oppose t%e possiblemotion of t%e rigid body"

•Indicate point of application+ magnitude+ anddirection of e(ternal forces+ including t%e rigid

body weig%t"



4 - *

Reactions at Su""orts and Connections 'or a+(o-Di!ensiona% Structure

• Reactions equi)alent to aforce wit% nown line ofaction"



4 - ,

Reactions at Su""orts and Connections 'or a +(o-Di!ensiona% Structure

• Reactions equi)alent to aforce of un nown directionand magnitude"

• Reactions equi)alent to aforce of un nowndirection and magnitudeand a couple"of un nownmagnitude



4 -

E$ui%i&riu! o' a Ri id Body in +(o Di!ensions• For all forces and moments acting on a two-

dimensional structure+

• Equations of equilibrium become

w%ere A is any point in t%e plane of t%estructure"

• T%e $ equations can be sol)ed for no more

t%an $ un nowns"

• T%e $ equations can not be augmented wit%additional equations+ but t%ey can be replaced

O z y x z M M M M F ==== ,,

∑ ∑ ∑ === ,,, A y x

M F F

∑ ∑ ∑ === ,,, B A x M M F



4 - .

Statica%%y Indeter!inate Reactions

• .ore un nowns t%anequations

• Fewer un nowns t%anequations+ partially

constrained

• Equal number un nownsand equations butimproperly constrained



4 - /

Sa!"%e Pro&%e! 4)0

1 'i ed crane #as a !ass o' 0 and is used to %i't a 24

crate) It is #e%d in "%ace &y a "inat A and a roc er at B) +#ecenter o' ra5ity o' t#e crane is%ocated at G)

Deter!ine t#e co!"onents o'

t#e reactions at A and B)

S/01TI/23

• Create a free-body diagram for t%e crane"

• Determine B by sol)ing t%e equationfor t%e sum of t%e moments of all forcesabout A" 2ote t%ere will be no

contribution from t%e un nownreactions at A"

• Determine t%e reactions at A bysol)ing t%e equations for t%e sum ofall %ori*ontal force components andall )ertical force components"

• C%ec t%e )alues obtained for t%ereactions by )erifying t%at t%e sum oft%e moments about B of all forces is*ero"



4 - 0

Sa!"%e Pro&%e! 4)0

• Create t%e free-body diagram"

• C%ec t%e )alues obtained"

• Determine B by sol)ing t%e equation for t%e

sum of t%e moments of all forces about A"

• Determine t%e reactions at A by sol)ing t%eequations for t%e sum of all %ori*ontal forcesand all )ertical forces"

( ) ( )

( ) ,m&24"5$

m52'#"6m4"#3,

=−

−∑ += B M A

2#"#,7+= B

,3, =+=∑ B A F x x

2#"#,7−= x A,24"5$2'#"63, =−−=∑ y y A F

2$"$$+= y A



4 - 00

Sa!"%e Pro&%e! 4)3

1 %oadin car is at rest on aninc%ined trac ) +#e ross (ei #t o't#e car and its %oad is ** %&6 andit is a""%ied at at G) +#e cart is#e%d in "osition &y t#e ca&%e)

Deter!ine t#e tension in t#e ca&%e

and t#e reaction at eac# "air o'(#ee%s)

S/01TI/23

• Create a free-body diagram for t%e carwit% t%e coordinate system alignedwit% t%e trac "

• Determine t%e reactions at t%e w%eels

by sol)ing equations for t%e sum ofmoments about points abo)e eac% a(le"

• Determine t%e cable tension bysol)ing t%e equation for t%e sum offorce components parallel to t%e trac "

• C%ec t%e )alues obtained by )erifyingt%at t%e sum of force components

perpendicular to t%e trac are *ero"



4 - 02

Sa!"%e Pro&%e! 4)3

• Create a free-body diagram

• Determine t%e reactions at t%e w%eels"

•Determine t%e cable tension"( )

( )

lb5$5,

54sinlb44,,

lb!6',

54coslb44,,

−=

−=

+=

+=

y

x

W

W

( ) ( )( ) ,,in"4

in"&lb6',!in"54lb5$5,3,

5 =+

−−=∑

R

M A

lb#74'5 = R

( ) ( )( ) ,,in"4

in"&lb6',!in"54lb5$5,3,# =−

−+=∑ R

M B

lb4&5# = R

,Tlb!6',3, =−+=∑ x F

lb!6',+=T



4 - 03

Sa!"%e Pro&%e! 4)4

+#e 'ra!e su""orts "art o' t#e roo'o' a s!a%% &ui%din ) +#e tension in

t#e ca&%e is 0* 7)Deter!ine t#e reaction at t#e 'i edend E )

S/01TI/23

• Create a free-body diagram for t%eframe and cable"

• Sol)e $ equilibrium equations for t%ereaction force components and

couple at E.



4 - 04

Sa!"%e Pro&%e! 4)4

• Create a free-body diagram for

t%e frame and cable"

• Sol)e $ equilibrium equations for t%e

reaction force components and couple"( ) ,2#4,

4"74"!

3, =+=∑ x x E F

2,"6,−= x E

( ) ( ) ,2#4,4"7&25,!3, =−−=∑ y y E F

25,,+= y E

∑ = 3, E M ( ) ( )

( ) ( )

( ) ,m4"!2#4,4"7

&

m'"#25,m&"$25,

m!"425,m7"525,

=+−

++

++

E M

m-2,"#', ⋅= E M



4 - 0*

E$ui%i&riu! o' a +(o-Force Body• Consider a plate sub8ected to two forces F 1 and F 2

• For static equilibrium+ t%e sum of moments about A must be *ero" T%e moment of F 2 must be *ero" Itfollows t%at t%e line of action of F 2 must pass

t%roug% A"

• Similarly+ t%e line of action of F 1 must pass t%roug% B for t%e sum of moments about B to be *ero"

• Requiring t%at t%e sum of forces in any direction be*ero leads to t%e conclusion t%at F 1 and F 2 must%a)e equal magnitude but opposite sense"



4 - 0,

E$ui%i&riu! o' a +#ree-Force Body• Consider a rigid body sub8ected to forces acting at

only $ points"

• 9ssuming t%at t%eir lines of action intersect+ t%emoment of F 1 and F 2 about t%e point of intersectionrepresented by D is *ero"

• Since t%e rigid body is in equilibrium+ t%e sum of t%emoments of F 1+ F 2+ and F 3 about any a(is must be*ero" It follows t%at t%e moment of F 3 about D must

be *ero as well and t%at t%e line of action of F 3 must

pass t%roug% D"• T%e lines of action of t%e t%ree forces must be

concurrent or parallel"



4 - 0

Sa!"%e Pro&%e! 4),

1 !an raises a 0 8oist6 o'%en t# 4 !6 &y "u%%in on aro"e)

Find t#e tension in t#e ro"eand t#e reaction at A)

S/01TI/23

• Create a free-body diagram of t%e 8oist" 2ote t%at t%e 8oist is a $ force body actedupon by t%e rope+ its weig%t+ and t%ereaction at A"

• T%e t%ree forces must be concurrent forstatic equilibrium" T%erefore+ t%e reaction

R must pass t%roug% t%e intersection of t%elines of action of t%e weig%t and ropeforces" Determine t%e direction of t%e

reaction force R"• 1tili*e a force triangle to determine t%e

magnitude of t%e reaction force R"



4 - 0.

Sa!"%e Pro&%e! 4),• Create a free-body diagram of t%e 8oist"

• Determine t%e direction of t%e reactionforce R"

( )

( )

( )

&$&"#!#!"#

$#$"5tan

m5"$#$m4#4",'5'"5

m4#4",5,tanm!#!"#:5,!4cot;m!#!"#

m'5'"5!4cosm!!4cos

5#

===

=−=−=

==+=

===

===

AE

CE

BD BF CE

CD BD AF AE CD

AB AF

α

&"4'=α



4 - 0/

Sa!"%e Pro&%e! 4),• Determine t%e magnitude of t%e reaction

force R"

$'"&sin

2#"6'

##,sin!"$#sin== RT

2'"#!7

26"'#

=

=

R

T



4 - 2

E$ui%i&riu! o' a Ri id Body in +#reeDi!ensions

• Si( scalar equations are required to e(press t%e

conditions for t%e equilibrium of a rigid body in t%egeneral t%ree dimensional case"

• T%ese equations can be sol)ed for no more t%an &un nowns w%ic% generally represent reactions at supportsor connections"

• T%e scalar equations are con)eniently obtained by applying t%e

)ector forms of t%e conditions for equilibrium+

∑ =∑ =∑ =

∑ =∑ =∑ =

,,,

,,,

z y x

z y x

M M M

F F F

( )∑ ∑ =∑ ×== ,, F r M F O

20



4 - 20

Reactions at Su""orts and Connections 'or a+#ree-Di!ensiona% Structure

4 22



4 - 22

Reactions at Su""orts and Connections 'or a+#ree-Di!ensiona% Structure

4 23



4 - 23

Sa!"%e Pro&%e! 4).

1 si n o' uni'or! density (ei #s2 %& and is su""orted &y a &a%%-

and-soc et 8oint at A and &y t(oca&%es)

Deter!ine t#e tension in eac#ca&%e and t#e reaction at A)

S/01TI/23• Create a free-body diagram for t%e sign"

• 9pply t%e conditions for staticequilibrium to de)elop equations fort%e un nown reactions"

4 24



4 - 24

Sa!"%e Pro&%e! 4).

• Create a free-body diagram for t%esign"

Since t%ere are only 4 un nowns+t%e sign is partially constrain" It isfree to rotate about t%e x a(is" It is+%owe)er+ in equilibrium for t%e

gi)en loading"

( )

( )k jiT

k jiT

r r r r

T T

k jiT

k jiT

r r

r r T T

EC

EC

E C

E C EC EC

BD

BD

B D

B D

BD BD

75

7$

7&

$5

$#

$5

75$&

#5'!'

++−=

++−=

−

−=

−+−=

−+−=

−

−=

4 2*



4 - 2*

Sa!"%e Pro&%e! 4).

• 9pply t%e conditions forstatic equilibrium tode)elop equations for t%eun nown reactions"

So%5e t#e * e$uations 'or t#e * un no(ns6

( )

( ) ( )

,lb#,',47#"5&&7"53

,7#!"#$$$"43

,lb57,ft!

,3

,lb57,3

,3

,lb57,

75

$5

7$

$#

7&

$5

=−+

=−

=−×+×+×=∑

=+−

=−++

=−−

=∑ −++=

EC BD

EC BD

EC E BD B A

EC BD z

EC BD y

EC BD x

EC BD

T T k

T T j

jiT r T r M

T T Ak

T T A j

T T Ai

jT T A F

( ) ( ) ( )k ji A

T T EC BD

lb55"4lb#,#"5lb$$'

lb$#4lb$"#,#

−+=

==

Unidad 3 Estatica Del Solido Rigido (1)

Documents

Transcript of Unidad 3 Estatica Del Solido Rigido (1)