BAB 3 SISTIM RESULTAN GAYA - SISTEM RESULTAN GAYA

BAB 3 SISTIM RESULTAN GAYA Sasaran bab ini adalah : Mahasiswa Mampu untuk 1. Menjelaskan konsep moment satu gaya dan menghitungnya dalam 2 dan 3 dimensi.

  2. Menetapkan momen suatu gaya di suatu garis/sumbu.

  3. Mendefine the moment of a couple.

  4. Menghitung Resultan sistim gaya nonconcurrent/tidak bersamaan.

  5. Mengindikasikan bagaimana mengurangi beban yang terdistribusi ke dalam sisitim resultan gaya pada suatu lokasi tertentu.

  Definisi Moment Gaya

The moment of a force about a point or an axis provides a measure of the tendency

of the force to cause a body to rotate about the point or axis

  Momen di sumbu z akibat gaya pada sumbu x tegak lurus sumbu y F x - horizontal force d y - distance from point O to force

• - M o moment of force about point O (M - o ) z moment of force about axis z

  Momen di sumbu x akibat gaya pada sumbu z tegak lurus sumbu y Fz - horizontal force dy - distance from point O to force

  Mo - moment of force about point O (Mo)x - moment of force about axis z

  NO moment Magnitude of the moment

  M = Fd Direction of the moment Right Hand Rule

  Resultant Moment of a System of Coplanar Forces

• M = Fd

  R _{O ∑} Counterclockwise is positive by scalar sign convention

  CONTOH : For each case, find the moment of the force about the point O

  M = 100 N 2 m = 200 N m ⋅ ( )( )

  O

  

M =

  50 N 0.75m =

  75 N m ⋅ O ( )( ) o

  M = 40lb 4 2cos 30 ft = 229lb ft ⋅

• o

  O ( ) ( )

  M = 60lb 1sin 45 ft = 42.4lb ft ⋅

  O ( ) ( )

  CONTOH SOAL Determine the moment of the 800 N force about points A, B, C,

  and D

  JAWAB M = 800 N (2.5 m) = 2000 N m ⋅

  A

  M = 800 N (1.5 m) = 1200 N m ⋅

  B

  M = 800 N (0 m) =

  0 N m ⋅

  C

  M = 800 N (0.5 m) = 400 N m ⋅

  D

• ccw M = Fd

  ( ) R _{O ∑}

  = − + M

  50 N 2m

  60 N 0

  ( ) ( ) R

_O

• o o

  20 N 3sin 30 m −

  40 N 3cos 30 m

  ( ) ( ) CROSS PRODUCT Another method of vector multiplication

  r r r C = × A B

  Read as C equals A cross B

  C = ABsin θ

  Magnitude: Direction: Right Hand Rule

  1. Cross Product r r r r A × ≠ × B B A r r r r A B B A

  × = − ×

  ( ) Not Commutative.

  r r r r

2. Scalar Multiplication

  a A B aA B × = ×

  ( ) ( )

  r r = × A aB

  ( )

  r r = A × B a

  ( )

  r r r r r r r

  A × B D = A × + B A × D

• 3.

   Distributive Law: ( ) ( ) ( )

  × + × + × + × + × + × r r

  × = × = × = − × = − × = × =

  × = + + × + + = × + × + × +

  

A A A B B B

A B A B A B A B A B A B A B A B A B

  

x y z x y z

x x x y x z y x y y y z z x z y z z

  ˆ ˆ ˆ ˆ ˆ ˆ i i i j i k ˆ ˆ ˆ ˆ ˆ ˆ j i j j j k ˆ ˆ ˆ ˆ ˆ ˆ k i k j k k

  ˆ ˆ ˆ ˆ ˆ ˆ A B i j k i j k

  

( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

  × = × = − × =

  ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ k i j k j i k k θ = ⇒ θ =

  Unit Vektor Right Hand Rule Bentuk Cartesiannya : Operasi proses :

  1 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ i i i j k i k j ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ j i k j j j k i

  90 sin

  o

  

× = + + × + + =

  

A A A B B B

A B A B A B A B AB AB

A B -AB A B AB A B A B

  ˆ ˆ ˆ ˆ ˆ ˆ k j k i j i ˆ ˆ ˆ ( )i ( )j ( )k x y z x y z

x y x z y x y z z x z y

y z z y x z z x x y y x

  ( ) ( ) ˆ ˆ ˆ ˆ ˆ ˆ A B i j k i j k

• + − − + + − =

  − − + − r r

Formula ekivalen : Moment of a Force -Vector Formulation PRINSIP TRANSMISIBLE

  r vector can be taken to any point on line of action of F ˆ ˆ ˆ i j k

  B B B

  M r F r F r F r F

  O A B

  = θ = θ =

  O M rF sin F(r sin ) Fd

  O M r F = × r r r

  = −

  B B B

  x y z x y y x x y z For Element : A A A A B A B

  ˆ ˆ ˆ i j k ˆk ( )

  = − − ˆk

  x y z x z z x x y z For Element : A A A A B A B

  A B x y z x y z

  ˆ ˆ ˆ i j k ˆj ( )

  = − ˆj

  B B B

  x y z y z z y x y z For Element A A A A B A B

  ˆi ( )

  ˆi : ˆ ˆ ˆ i j k

  × = r r

  B :

  A r

  B B m

  Determin A A ant fo

  = × = × = × = × r r r r r r r r r Bentuk Cartesiannya dan Formulasi vektor cartesian :

  r

  ˆ ˆ ˆ ˆ

  M = ( r F -r F )i

  i j k

  O y z z y

  r r r M = × = r F r r r ˆ

  O x y z − ( r F − r F ) j x z z x

  F F F

  ˆ ( )k r F − r F x y y x

• x y z

  MOMENT RESULTAN MOMEN DARI SUATU SISTIM GAYA

  r r r M r F = ×

  R _{O ∑ ( )} r r r r r r

  = + × + r × F r F r × F

  1

  1

  2

  2

  3

  3 ( ) ( ) ( )

  Contoh soal : Cari moment di titik A

  r r 1. r and r

  Find vectors A B

2. Force vector is 60 N times a unit vector in direction of 3.

  1

  = − − + = × = + + × − − + r r r r r r

  = + + = + +

  ˆ ˆ ˆ ˆ ˆ ˆ M r F (1i 3j 2k)m ( 40i 20 j 40k) N

  ˆ ˆ ˆ F ( 40i 20 j 40k) N

  ˆ ˆ ˆ r (1i 3j 2k)m ˆ ˆ ˆ r (3i 4 j 0k)m

  B C A B

  = = − − + r r r

  − − + = − − + = = − + − + = − − +

  3 ˆ F (60 N) u ˆ ˆ ˆ F ( 40i 20 j 40k) N = − − +

  3

  3

  2 ˆ ˆ ˆ ˆu i j k

  Moment Vektor posisi : Vektor Gaya : Vektor momen

  CB

  ˆ ˆ ˆ r 2i 1j 2k ˆ ˆ ˆ ˆu 2i 1j 2k r ( 2) ( 1) (2)

  2 CB CB CB ˆ ˆ ˆ r 2i 1j 2k

  2

  2

  CB CB CB

  = − = − + − + − = − − + r r r r r r r

  = = + + = = + +

  ˆ ˆ ˆ r (1 3)i (3 4) j (2 0)k ˆ ˆ ˆ r 2i 1j 2k

  ˆ ˆ ˆ r r (1i 3j 2k)m ˆ ˆ ˆ r r (3i 4 j 0k)m r r r

  B BA C CA CB B C CB CB

  M r F M r F or = × = × r r r r r r

  

A A A B

  ˆu

  2 r r r ˆ ˆ ˆ ˆ ˆ ˆ

• M r F (1i 3j 2k)m ( 40i 20 j 40k) N = × = + + × − −

  A B ˆ ˆ ˆ i j k

  M =

  1

  3

2 A

• 40 -20

  40 ˆ ˆ ˆ = [3(40) − − + 2( 20)]i − [(1(40) − − 2( 40)]j [1( 20) − − − 3( 40)]k = 160i 120 j 100k − N m ⋅

• ˆ ˆ ˆ

  ( )

  2

  2 M = (160) + − ( 120) (100) = 224 N m ⋅ A

• 2

  Contoh soal : Determine the resultant moment at O and the coordinate direction angles for the moment. r r

  ˆ r = r = (5j)ft

  A OA

  r r ˆ ˆ ˆ r = r = (4i + − 5j 2k)ft

  B OB

  Vektor posisi (lihat gambar 04.17.b)

  r ˆ ˆ ˆ

  = − +

• Vektor gayanya : F ( 60i 40 j 20k) lb

  1 r ˆ

  F = (50 j) lb

  2 r

• ˆ ˆ ˆ F = (80i 40 j 30k) lb −

  3 r r r r r r r r r

  Vektor momennya :

  1 A

  2 B

  3 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ i j k i j k i j k

  M r F r F r F r F = × = + × + × × R A _{O ∑ ( ) ( ) ( ) ( )}

  5

  4 5 −

  2

  5 + + =

  ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ i j k i j k i j k r

  M

  5

  5

  4

  5

  2 = − + +

  R _O

• 60

  40 20 0 50

  80

  40

  30 −

  ˆ ˆ ˆ = − + [5(20) 40(0)]i − [0]j [0(40) − 60(5)]k

  ˆ ˆ ˆ [0]i [0]j [0]k − + +

  ˆ ˆ ˆ [5( 30) 40( 2)]i [4 30 80 2 ]j [4(40) 80(5)]k

• − − − − − + − −

  ( ) ( )

  ˆ ˆ ˆ = − ⋅

• 30i 40 j 60k lb ft r

  ( )

  ˆ ˆ ˆ _O ( )

• M = 30i − 40 j 60k lb ft ⋅

  R

  2

2 M = + −

  30

  40 60 lb ft ⋅

• 2

  ( ) ( ) R _{O ( )}

  M = 78.10 lb ft ⋅

  R _O

  r ˆ ˆ ˆ

  30i − + 40 j 60k lb ft ⋅ M

  ˆu = =

  M 78.10 lb f ⋅

  R _O

  R ( _O )

  ˆ ˆ ˆ

• = 0.3841i − 0.5121j 0.7682k

  Sudut Arahnya :

• ˆ ˆ ˆ ˆu = 0.3841i − 0.5121j 0.7682k

  o cos α = 0.3841 α =

  

67.4

o cos 0.5121 121 β = − β = o cos γ = 0.7682 γ =

39.8 Principle of Moments

  The moment of a force about a point is equal to the sum of the moments of the force’s components about the point.

  F F ¹ F ² A Contoh Soal : Determine the moment of the force about A Moment of a Force About a Specified Axis 1.

  Sometimes need the component of a moment about a particular axis 2. Scalar analysis: ma=Fda where da is the ⊥ or shortest distance from the force line of action to the axis of interest

  This called the triple scalar product

  ( )( ) ( ) o A

  A CB d 100cos 45 70.71mm 0.07071m M Fd 200N 0.07071m

  14.1N m ˆ M 14.1k N m = = = =

  = = = ⋅ = ⋅

r

3. Vector analysis: ma= ua • (r x F) 4.

  In general case, to find r moment of

  F, at point , O about axis a ' a ' , we project

r r

M of about onto F O a ' a ' .

_O

r r r

  ˆ ˆ ˆ

  M = r × F _{O A}

  i j k

  ˆ ˆ

  M = ( u i u j u k) r ⋅ r r _{a a a a x y z x y z} r r r

• Then
• ˆ

  ˆ ˆ F F F M = M ⋅ u = u ⋅ r × F x y z _{a O A A A}

  ( ) u u u u u u _{a a a a a a x y z x y z} M = r r r M = r r r _{a x y z a x y z} F F F F F F _{x y z x y z} u u u ⇒ _{a a a x y z} x,y,z components of unit vector directed along the axis. r r r ⇒ _{x y z} x,y,z components of position vector directed from point O on the axis to any point along the line of action of the force.

  F F F ⇒ _{x y z} x,y,z components of the force vector

  1. Ma is a scalar with positive or negative sign.

  2. Positive sign indicates the sense is the same as unit vector ua.

  3. Negative sign indicates the sense is opposite the sense of unit vector ua.