contoh soal mencari luas bidang momen

Soal no: 1

∑ 

∑ 

= 0

 . q . L   =  . 4 . 6   = 12 ton = 

6 RA

RB . 6

2

6 RA RA

Kontrol :

2

=

∑

RB

= 0

 . q . L   = 

2

=

= 12 ton

= 72 – 72 – 72  72 = 0

Momen MA

= 0

M max

= RA . X -

 . q . X 

2

= 12 . 3 -

= RA – RA – qX

= 18 tm

= 12 – 12 – 4  4 X X= MB

 

=3m

= RA . 6 – 4  – 4 . = 0

 . q . L 

2

 . 4 . 3 

2

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Gaya Lintang DA1

= 0

DA2

= RA

= -12 + 12

= 12 ton

= 0

DB1

= DA2 – q  – q . 6 = 12 – 12 – 24  24 = - 12 ton

Gambar diagram M .D .N

DB2

= DB1 + RB

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PERHITUNGAN LUAS dan TITIK BERAT BIDANG MOMEN

Cara INTEGRAL

Cara Pendekatan L1

= 2/3 x 3 x 18 = 36 Tm2

Titik berat L1 terhadap A = 5/8 x 3

= 1,875 m

∫    =           

L1

=

L1

= 36 Tm 2

TITIK BERAT

       ( )          =            TBLI =

 = 1,875 m

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Cara pendekatan L2

= 2/3 x 3 x 18 = 36 Tm 2

Titik berat L2 terhadap A = (3/8 x 3 )+ 3 =4,125m

Cara INTEGRAL L2

L2

∫    =           

=

= 36 Tm 2

TITIK BERAT L2 dari A

              TB L2

    ( )                  =    = 216 – 216 – 67,5  67,5 =

 = 4,125 m 

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Soal no : 2

∑ 

= 0 = 2 . 3 – RB  – RB . 6 +

RB

=

∑

=

 = 13 ton 

 . 4 . 6 

2

RB

= 13 ton

∑ 

= -

RA

 . 4 . 6  – 2  – 2 . 3 + RA . 6  2

= 13 ton

0

= 26 – 26 – 26  26 = 0 ………….. ( ok ) Moment MA

= 0

MC

= RA . 3 -

 . 4 . 3 

2

= 21 tm MB

= RA . 6 -

 . 4 . 6  – 2  – 2 . 3  2

2

= 0 tm Gaya Lintang DA1

= 0

DA2

= RA

DC1

= DA2 – q  – q . 3

DB1

= DC2 – 4 . 3 = - 13 ton

DB2

= DB1 + RB

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Gambar diagram M. D .N

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PERHITUNGAN LUAS BIDANG MOMEN

PERHITUNGAN LUAS BIDANG MOMEN Cara integral Mx

= RA .X  – P . ( X-3 )  – 4/2 .q 2

L1

=

∫    =          2

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PERHITUNGAN LUAS BIDANG MOMEN

L2

∫     =    (  )        

=

= 81 – 81 – 40,5  40,5 L2

= 40,5 Tm 2

TITIK BERAT L2 TERHADAP A Metode Pendekatan L2

= 2/3 x 3 x 21 = 42 Tm2

TITIK BERAT L 2 thd. A TB L1 = (3/8 x 3)+ 3 = 4,125 m

 ()) ∫ .TB L 1 = ∫   ( .TB L 1 = ∫ (  (   (  )   )  .TB L 1 =                .TB L 1 = 252 – 252 – 171/2  171/2 .TB L 1 = 333/2 TB L 1

= 37/9 m

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Soal no 3 :

∑ 

= 0 =2 . 2 + 2 . 4 +

RB

=

 

 q . L  – RB.   – RB. 6 2

= 14 + 14 - 28 =

0 …………… ( ok )

Moment MA

= 0

MC

= RA . 2 = 20 tm

=

 

= 14 ton

= 0

0

= 0 =-2 . 2 - 2 . 4 -

RB

= 14 ton

∑

∑ 

4.2 

2

 q . L  + RB . 6  2

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Momen max

Mx

= RA . X – P  – P1 . ( X – X – 2  2 ) -

.q.X 

2

14 . 3 – 2  – 2 ( 3 – 3  – 2  2 ) -

= RA – RA – P  P –  – q  q X

= 22 tm

= 3m Gaya Lintang DA1

= 0

DA2

= RA

DC1

= DA2 – 4  – 4 . 2

DD2

= - 6 ton DB1

= 6 ton DC2

DD1

= DC1 – 2  – 2

= DD1 – 2  – 2

= DD2 – 4  – 4 . 2 = - 14 ton

DB2

= DB1 + DB2

= - 4 ton

= - 14 – 14 – 14  14

= DC2 – 4  – 4 . 2

= 0 ton

= - 4 ton

.4.3 

2

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Gambar diagram M ,D ,N

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Perhitungan Perhitung an LUAS BIDANG MOMEN

Perhitungan LUAS 1

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Perhitungan LUAS 2

L2

= 1 X 20 = 20 Tm2

Titik berat L2 terhadap A L2

=(½x1)+2 = 2,5 m

Perhitungan LUAS 3

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Perhitungan LUAS 4

L4

= 2/3 x (Mmax – (Mmax  – MD  MD ) = 2/3 x ( 22 -20 ) = 4/3 Tm2

Titik berat L4 terhadap A L4

= 3 + ( 3/8 x 1 ) = 3,375 m

Perhitungan LUAS 5

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Perhitungan LUAS 6 Titik berat L6 terhadap A

Metode pendekatan L6

= 2/3 x 2 x 20 =80/3 Tm2

Titik berat L6 terhadap A

∫     ∫  (  )        68/3 TB L6 = ∫   (    (   )  (    ) )   = ∫                       )    =                           = 288 -

 

 /     =  =

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