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EQUILIBRIO DE CUERPOS RIGIDOS
Mecánica para Ingenieros
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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 !"'
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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
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Free-Body Dia ra!First ste" in t#e static e$ui%i&riu! ana%ysis o' a
ri id &ody is identi'ication o' a%% 'orces actin ont#e &ody (it# a free-body dia ra!)
• Select t%e e(tent of t%e free-body and detac% itfrom t%e ground and all ot%er bodies"
• Include t%e dimensions necessary to computet%e moments of t%e forces"
• Indicate point of application and assumeddirection of un nown applied forces" T%ese
usually consist of reactions t%roug% w%ic% t%eground and ot%er bodies oppose t%e possiblemotion of t%e rigid body"
•Indicate point of application+ magnitude+ anddirection of e(ternal forces+ including t%e rigid
body weig%t"
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Reactions at Su""orts and Connections 'or a+(o-Di!ensiona% Structure
• Reactions equi)alent to aforce wit% nown line ofaction"
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Reactions at Su""orts and Connections 'or a +(o-Di!ensiona% Structure
• Reactions equi)alent to aforce of un nown directionand magnitude"
• Reactions equi)alent to aforce of un nowndirection and magnitudeand a couple"of un nownmagnitude
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E$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
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Statica%%y Indeter!inate Reactions
• .ore un nowns t%anequations
• Fewer un nowns t%anequations+ partially
constrained
• Equal number un nownsand equations butimproperly constrained
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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"
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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
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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"
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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
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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.
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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
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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"
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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"
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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"
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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
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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
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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
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Reactions at Su""orts and Connections 'or a+#ree-Di!ensiona% Structure
4 22
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Reactions at Su""orts and Connections 'or a+#ree-Di!ensiona% Structure
4 23
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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
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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*
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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$#4lb$"#,#
−+=
==