Statika dan Mekanika Bahan I

  1 D C B A

  1 = 2 t

  2 S

  3 E S

  L 6 = 6 m G F S

  1 = 3 m

  L 3 = 4 m L 2 = 4 m

  L L 7 = 3 m

L 5 = 4 m

L 4 = 5 m

  1 = 3,2 t/m P

  Tipe II

  2 = 1,5 t/m q

  3 m q

  1 =

  5 m h

  2 =

  h

• Hitunglah Reaksi Perletakan, Bidang Momen, Bidang Lintang, dan Bidang Normal pada konstruksi tersebut?

Menganalisa Konstruksi Gerber :

1 RVS

1 RVS

  B 1 C A q

  6 = 6 m

  1 = 3 m L

  2 = 4 m L

  3 = 4 m L

  4 = 5 m L

  5 = 4 m L

  7 = 3 m L

  2 L

  3 E S

  2 = 1,5 t/m S

  RHS

  1 M G RHA RHE RHG RVG RVS Batang S 1 – S

  1 q

  2 S

  3 m P 1 = 2 t S

  1 =

  3 D h

  5 m G F S

  2 =

  1 h

  2 RVS

  2 RVS

  3 RVE RVA RVD RHS

  3 RVS

  1 = 3,2 t/m S

2 Reaksi Perletakan

  Σ RVS

  1 M S2 = 0

  P = 2 t

  1 X

  1

  1

  2

  1 X

  2 RVS . 5 – P . 2,5 + RVS . 0 = 0

  . , RHS

  1 RVS = = 1 ton ( )

  1 S

  1 S

  2 RVS

  2 RVS

  1 L 4 = 5 m

  Σ RVS M = 0

  2 S1

• RVS . 5 + P . 2,5 + RVS . 0 = 0

  2

  1

  1 M

• + RVS = = 1 ton ( )

  . ,

  2

  2,5 tm 1 t 1 t

  Σ V = 0

• + Q -
• 1 t -1 t RVS

  1 + RVS 2 – P 1 = 0 1 + 1 – 2 = 0

  N

  0 = 0 Σ

  H = 0 RHS 1 = 0

  ≤ Bidang Momen

  Daerah S – P ( 0 ≤X 2,5 )

  2

  1

  2 ≤

  Daerah S 1 – P 1 ( 0 ≤X 1 2,5 ) QX 2 = - RVS 2 = - 1 MX = RVS . X = X

  X = 0 QS = - 1 ton

  1

  1

  1

  1

  2

  2 X = 0 MS = 0 tm X = 2,5 QP = - 1 ton

  1

  1

  2

  1 X 1 = 2,5 MP 1 = 2,5 tm Bidang Normal

  ≤ ≤

  Daerah S 2 – P 1 ( 0 ≤X 2 2,5 ) Daerah S 1 – P 1 ( 0 ≤X 1 2,5 ) MX = RVS . X = X

  NX = - RHS = 0

  2

  2

  2

  2

  1

  1 X 2 = 0 MS 2 = 0 tm

  X 1 = 0 NS 1 = 0 ton

  X 2 = 2,5 MP 1 = 2,5 tm

  X 1 = 2,5 NP 1 = 0 ton Bidang Lintang

  ≤ Daerah S – P ( 0 ≤X 2,5 )

  1

  1

  1 QX 1 = RVS 1 = 1 X = 0 QS = 1 ton

  1

  1 X 1 = 2,5 QP 1 = 1 ton

  X

2 Batang A – B – C – D – S

  1 (L ) + (h )²

  AB =

  X

  3 RVS 1 = 1 t q 1 = 3,2 t/m

  AB = (3) + (3)² RHS

  1 = 0 t

  AB = √18

  B C D S

  1 m

  ( q . 4 )

  1

3 AB = 4,24 m

  X

  1 =

  RVD

  1 h

  ( q . X )

  1

  2 α = 45º RHA A α

  RVA sin α

  tan α =

  RVA cos α RVA L = 3 m L 2 = 4 m L 3 = 4 m

  1

  tan α =

• 29,6 tm
• 12,68 tm

  tan α = 1

• - -1

  α = tan 1 = 45º

  M

  Reaksi Perletakan

  13,8 t

  Σ RVA M = 0 +

  1 t D

• - - 4,23 t
• 4,
• RVA . 7 + RVS . 4 + ( q . 4 ) 2 + RVD . 0 = 0

  1

  1

• 2,99 t

  Q . ( . )

  RVA = = 4,23 ton ( )

  2,99 t

• 2,99 t

  N 2,99 t

• – D ( 0 ≤X
• RVD . 7 + RVS 1 . 11 + ( q

  2

  2 = 4 MD = - 29,6 tm

  MX 2 = - RVS 1 . X 2 – ( q 1 . X 2 ) ½ X 2 = - X 2 – 1,6 X

  2 X

  2 = 0 MS

  1 = 0 tm

  X

  = 18,03 ton ( ) Σ

  V = 0

  1

  . ( . )

  1 . 4 ) 9 + RVA . 0 = 0 RVD =

  M A = 0

  RVD Σ

  Letak Mmax pada Qx = 0 = 0

  2 ≤ 4 )

• RVA + RVD – ( q

• 4,23 + 18,03 – ( 3,2 . 4 ) – 1 = 0 0 = 0 Σ
• RVS 1 – 3,2 X

  1 = - 4,23 cos 45º . X

  3 ≤ 7 )

  X 3 = 7 MC = 0 tm

  3 = 4 MB = - 12,68 tm

  X

  3 = 0 MD = - 29,6 tm

  3 X

  3 MX 3 = - 1 ( 4 + X 3 ) – ( 3,2 . 4 ) ( 2 + X 3 ) + 18,03 . X

  3 ) + RVD . X

  1 . 4 ) ( 2 + X

  3 ) – ( q

  1 ( 4 + X

  3 = - RVS

  MX

  = - 0,3125 m ( Tidak Memenuhi ) Daerah D – B – C ( 0 ≤X

  1 = 0 MA = 0 tm

  X 2 =

  2 = 0

  1 . 4 ) – RVS

  1 = 0

• – ,

  H = 0 RHA – RHS 1 = 0 RHA = RHS

  1 = 0 ton

  Bidang Momen Daerah A – B ( 0 ≤X

  1 ≤

  4,24 ) MX

  1 = - RVA cos α . X

1 X

  X 1 = 4,24 MB = - 12,68 tm Daerah S

  1 ≤

  4,24 ) NX

  1 = RVA sin α = 4,23 sin 45º

  X 1 = 0 NA = 2,99 ton

  X

  1 = 4,24 NB = 2,99 ton

• – D ( 0 ≤X

  Daerah S

• – D ( 0 ≤X

  1

  2 ≤ 4 )

  NX 2 = - RHS 1 = 0

  X 2 = 0 NS 1 = 0 ton

  X

  2 = 4 ND = 0 ton

  Daerah D – B – C ( 0 ≤X

  3 ≤ 7 )

  NX 3 = - RHS 1 = 0

  X

  3 = 0 ND = 0 ton

  X 3 = 4 NB = 0 ton

  Bidang Normal Daerah A – B ( 0 ≤X

  X

  3 = 7 QC = - 4,23 ton

  1

  Bidang Lintang Daerah A – B ( 0 ≤X

  1 ≤

  4,24 ) QX

  1 = - RVA cos α = - 4,23 cos 45º

  X 1 = 0 QA = - 2,99 ton

  X

  1 = 4,24 QB = - 2,99 ton

  Daerah S

  2 ≤ 4 )

  X 3 = 4 QB = - 4,23 ton

  QX 2 = RVS 1 + ( q 1 . X 2 ) = 1 + ( 3,2 X 2 )

  X 2 = 0 QS 1 = 1 ton

  X

  2 = 4 QD = 13,8 ton

  Daerah D – B – C ( 0 ≤X

  3 ≤ 7 )

  QX 3 = RVS 1 + ( q 1 . 4 ) – RVD QX

  3 = 1 + ( 3,2 . 4 ) – 18,03 = - 4,23

  X 3 = 0 QD = - 4,23 ton

  X 3 = 7 NC = 0 ton Batang S 2 – E – S

  3 Reaksi Perletakan

  X

  2

  1 X

  Σ RVE M S3 = 0

  RVS = 1 t

  2 RVE . 6 – RVS . 10 – ( q . 10 ) 5 + RVS . 0 = 0

  2

  2

  3

  q = 1,5 t/m

  2 . ( . )

  RVE = = 14,17 ton ( )

  S

  2 S

  3 E ( q . 10 )

  2 RHE ( q 2 . X 1 ) RVE

  ( q 2 . X 2 ) RVS

  3 L 5 = 4 m L 6 = 6 m

  Σ RVS

  3 M E = 0

  1 m 5 m

• RVS . 6 + ( q . 10 ) 1 - RVS . 4 + RVE . 0 = 0

  3

  2

  2

  2,44 m

• 16 tm

  1,22 m . ( . )

  RVS = = 1,83 ton ( )

  3

• - M
• + Σ

  1,12 tm V = 0 7,17 t

• RVS 2 + RVE + RVS

  3 – ( q 2 . 10 ) = 0

• + - 1 + 14,17 + 1,83 – ( 1,5 . 10 ) = 0
• - - 1 t
• - - 1,83 t

  0 = 0

  Q

• 7 t

  Σ H = 0

  N

  RHE = 0 Bidang Momen Letak Mmax pada Qx = 0

  ≤ Daerah S 2 – E ( 0 ≤X 1 4 )

  = 0 MX = - RVS . X – ( q . X ) ½ X

  1

  2

  1

  2

  1

  1

  2 RVS 3 – 1,5 X 2 = 0 MX = - X – 0,75 X

  1

  1

  1

  ,

  X 1 = 0 MS 2 = 0 tm X = = 1,22 m

  2

  ,

  X = 4 ME = - 16 tm

  1 X = 1,22 M = 1,12 tm 2 max Letak Mmax pada Qx = 0

  Letak Momen Mx = 0

  2 = 0

  1,83 X – 0,75 X = 0

  2

  2

  2

• 0,75 X 2 + 1,83 X

  2 = 0

• RVS 2 – 1,5 X

  1 = 0 X = = - 0,667 m ( Tidak Memenuhi )

  1

• – ,

  ± –

  X 1,2 = ≤

  Daerah S 3 – E ( 0 ≤X 2 6 )

  , ± ( , ) – ( , ) ( )

  MX = RVS . X – ( q . X ) ½ X

  2

  3

  2

  2

  2

  2

2 X 1,2 =

  MX = 1,83 X – 0,75 X

  2

  2

  2

  ( , )

  X 2 = 0 MS 3 = 0 tm

  , ± ,

  X = 1,2

  X = 6 ME = -16 tm

• – ,

  2

  , ,

  X = = 0 m

  1

• – , , ,

  X = = 2,44 m

  2

• – ,

  Bidang Lintang ≤

  Daerah S – E ( 0 ≤X 4 )

  2

1 QX = - RVS – ( q . X ) = - 1 – 1,5 X

  1

  2

  2

  1

  1 X 1 = 0 QS 2 = - 1 ton X = 4 QE = - 7 ton

  1 ≤

  Daerah S – E ( 0 ≤X 6 )

  3

  2 QX 2 = - RVS 3 + ( q 2 . X 2 ) = - 1,83 + 1,5 X

  2 X 2 = 0 QS 3 = - 1,83 ton X = 6 QE = 7,17 ton

  2 X 2 = 1,22 Q max = 0 ton

1 Y

  1 M G M G G M G RHG RVG RVS 3 = 1,83 t h

  1

  X

  3 L 7 = 3 m

  5 m F S

  2 =

  3 L

7 = 3 m

G RHG RVG h

  5 m F S

  2 =

  3 L 7 = 3 m G RHG RVG h

  5 m F S

  2 =

  1 Y

  1 X

• 5,49 tm
• 5,
• 1,

• -

  -

  N Q M Y

  49 tm

  49 tm

  83 t

  83 t

  1

  X

  3 = 1,83 t

  3 = 1,83 t RVS

  RVS

  Batang S 3 – F – G

• - 1,83 t - 1,83 t

• 5,
• 1,

  MY 1 = RHG . Y 1 – MG = - 5,49 Y

  V = 0

  1 ≤ 5 )

  Reaksi Perletakan Σ

• RVS
• RVG = 0

  3

• 1,83 + RVG = 0 RVG = 1,83 ton ( ) Σ

  Bidang Lintang Daerah S 3 – F ( 0 ≤X

  1 = 5 QF = 0 ton

  Y

  1 = 0 QG = 0 ton

  QY 1 = - RHG = 0 Y

  1 ≤ 5 )

  X 1 = 3 QF = - 1,83 ton Daerah G – F ( 0 ≤Y

  X 1 = 0 QS 3 = - 1,83 ton

  3 = - 1,83

  1 = - RVS

  QX

  1 ≤ 3 )

  Y

  1 = 5 MF = - 5,49 tm

  H = 0 RHG = 0 Σ

  1 = 0 MG = - 5,49 tm

• RVS 3 . 3 + RVG . 0 + RHG . 0 + MG = 0

  X 1 = 3 MF = - 5,49 tm Daerah G – F ( 0 ≤Y

  3 = 0 tm

  1 = 0 MS

  MX 1 = - RVS 3 . X 1 = - 1,83 X

  1 ≤ 3 )

  Bidang Momen Daerah S 3 – F ( 0 ≤X

  3 . 3 = 1,83 . 3 = 5,49 tm ( )

  MG = RVS

  M = 0

1 X

  Bidang Normal ≤

  Daerah G – F ( 0 ≤Y 1 5 ) NY = - RVG = - 1,83

1 Y = 0 NG = - 1,83 ton

  1 Y 1 = 5 NF = - 1,83 ton Gambar Keseluruhan Bidang Momen (M)

  P 1 = 2 t q 1 = 3,2 t/m q 2 = 1,5 t/m C B

  E S

  3 F m

  3 =

  1 h h

  2 =

  A 5 m

  G L 2 = 4 m L 3 = 4 m L

4 = 5 m L

5 = 4 m L 6 = 6 m L 7 = 3 m

  L 1 = 3 m

• 29,6 tm

  2,44 m

• 16 tm

  1,22 m

• 12,68 tm
• 5,49 tm
• 5,

  49 tm 1,12 tm 2,5 tm

• 5,

  49 tm C B 3 E A

  F S

  7 = 3 m L

  6 = 6 m G

  1 = 3 m L

  2 = 4 m L

  3 = 4 m L

  4 = 5 m L

  5

= 4 m L

  L

  Gambar Keseluruhan Bidang Lintang (Q)

  1 = 3,2 t/m P 1 = 2 t

  3 m q 2 = 1,5 t/m q

  1 =

  5 m h

  2 =

  1,22 m h

• 1 t -1 t 1 t 1 t
>1,83 t 7,17 t
• 1,8>7 t
• 2,>2,99 t 13,8 t
• 4,23 t - 4,23 t
Gambar Keseluruhan Bidang Normal (N)

  P 1 = 2 t q 1 = 3,2 t/m q 2 = 1,5 t/m C B

  E S

  3 F m

  3 =

  1 h h

  2 =

  A 5 m

  G L 2 = 4 m L 3 = 4 m L 4 = 5 m L 5 = 4 m L 6 = 6 m L 7 = 3 m

  L 1 = 3 m 2,99 t

• 1,

  83 t 2,99 t

• 1,

  83 t

  Penampang Balok Pot. I-I Pot. I-I

  P 1 = 2 t h

  2 =

  5 m h

  1 =

  3 m q

  2 = 1,5 t/m q

  1 = 3,2 t/m

  L L 7 = 3 m

L 5 = 4 m

L 4 = 5 m

  L 3 = 4 m L 2 = 4 m

  1 = 3 m

  L 6 = 6 m G F S

  3 E S

  2 S

  1 D C B A Pot. II-II Pot. II-II

  

10 cm 10 cm

30 cm 10 cm 50 cm
• Hitunglah besar Tegangan Lentur, Tegangan Geser, dan Tegangan Normal pada potongan I-I dan II-II dengan penampang balok berbentuk U serta gambarkan?

  3 . y

  2 = 5 cm y

  3 y = . . . y = ( ) . ( ) . ( ) .

  ( ) ( ) ( )

  y = y = 11,67 cm

  x y x 3 = 45 cm x

  2 = 25 cm x

  1 = 5 cm y

  3 = 20 cm y

  2

  1

= 20 cm

F

  2 F

  3 F

  1

  2

  3

  1 10 cm

10 cm 10 cm

• F
• F
• F
• F
• F
• F
• F
• F

  2 . y

  Menentukan Garis Netral (GN) Statis momen terhadap serat bawah

  2

  ( F

  1

  2

  3 ) x = F

  1 . x

  1

  2 . x

  3 . x

  1

  3 x = . . . x = ( ) . ( ) . ( ) .

  ( ) ( ) ( )

  x = x = 25 cm ( F

  1

  2

  3 ) y = F

  1 . y

  30 cm

50 cm

1 GN GN

  (25, 11,67) x = 25 cm F

  Momen Inersia

  I total

  = I

  20 cm

50 cm

  10 cm y atas = 18,33 cm y bawah = 11,67 cm

10 cm 10 cm

  3

  2

  1

  3 F

  3 = 8,33 cm a

1 = 8,33 cm

F

  2 a 2 = 6,67 cm a

• I
• I
• F
• F
• F

  2 . h

  3

  1

  2

  3 I total = ( 1/12 . b

  1 . h

  1

  3

  1 . a

  1

  2 ) + ( 1/12 . b

  4

  = 6666,67 + 13877,78 + 4166,67 + 22244,45 + 6666,67 + 13877,78 I total = 67500,02 cm

  I total

  2 )

  2 ) + ( 1/12 . 10 . 20

  2

  3

  2 ) + ( 1/12 . 50 . 10

  3

  I total = ( 1/12 . 10 . 20

  2 )

  3

  3

  3

  3 . h

  2 ) + ( 1/12 . b

  2

  2 . a

  3

  3 . a

• 200 . 8,33
• 500 . 6,67
• 200 . 8,33

  

a a a

F ^{1 a} F ¹ X ¹ X ¹ I

  I I

  I 18,33 cm

  GN GN

  III

  III

  III

  III 1,67 cm

  X ^{3 3} X F ³

c c

c c

F ³

  10 cm

II II

  ₂ X F ²

b b

10 cm 30 cm 10 cm

  ≤ Potongan I – I Daerah a – a s.d. GN – GN ( 0 ≤X 18,33 )

1 SX = 10 . X ( 18,33 – ½ X ) + 10 . X ( 18,33 – ½ X )

  1

  1

  1

  1

  1

  3

  τ

  X 1 = 0 S a-a = 0 cm a-a = =

  ,

  3 X 1 = 18,33 S GN-GN = 10 . 18,33 ( 18,33 – ½ 18,33 ) + 10 . 18,33 ( 18,33 – ½ 18,33 ) = 3359,889 cm

  , τ

  GN-GN = =

  ,

  ≤ Potongan II – II Daerah b – b s.d. c – c ( 0 ≤X 10 )

  2 SX 2 = 50 . X 2 ( 11,67 – ½ X 2 )

  3

  τ

  X 2 = 0 S b-b = 0 cm b-b = =

  ,

  3 X = 10 S = 50 . 10 ( 11,67 – ½ 10 ) = 3335 cm 2 c-c

  τ

  = = c-c

  ,

  ≤ Potongan III – III Daerah c – c s.d. GN – GN ( 0 ≤X 3 1,67 ) SX = 50 . 10. 6,67 + 10 . X ( 1,67 – ½ X ) + 10 . X ( 1,67 – ½ X )

  3

  3

  3

  3

  3

  3

  τ

  X = 0 S = 3335 cm = = 3 c-c c-c

  ,

  3 X = 1,67 S = 3335 + 10 . 1,67 ( 1,67 – ½ 1,67 ) + 10 . 1,67 ( 1,67 – ½ 1,67 ) = 3362,889 cm

  3 GN-GN

  , τ

  = = GN-GN

  , Momen potongan I-I Tinjau Kiri : M I-I = RVS 1 . 2,5 M

  I-I = 1 . 2,5 = 2,5 tm

  2 ( Tarik + )

  = 67,89 kg/cm

  2 ( Tekan - )

  σ

  bawah = =

  ,

  = 43,22 kg/cm

  L

  atas = =

  4 = 5 m Pot. I-I

  Pot. I-I RHS

  1 RVS

  2 RVS

  1 P 1 = 2 t S

  2 S

  ,

  σ

  M I-I

  σ

  = 250000 kgcm Lintang potongan I-I Tinjau Kiri : Q I-I = RVS

  1 Q I-I

  = 1 = 1 ton Q I-I = 1000 kg

  Normal potongan I-I Tinjau Kiri : N I-I = - RHS

  1 N I-I

  = 0 kg Tegangan Lentur

  = W atas

  3

  = =

  , ,

  = 3682,49 cm

  3 W bawah = =

  , ,

  = 5784,06 cm

  1 Catatan : Jika tampang balok mengalami lenturan positif, maka tegangan tekan terjadi di serat atas dan tegangan tarik di serat bawah. Tegangan Geser ≤

  Potongan I – I Daerah a – a s.d. GN – GN ( 0 ≤X 1 18,33 ) SX = 10 . X ( 18,33 – ½ X ) + 10 . X ( 18,33 – ½ X )

  1

  1

  1

  1

  1

  3

  2

  τ

  X = 0 S = 0 cm = = = 0 kg/cm 1 a-a a-a

  ,

  3 X = 18,33 S = 10 . 18,33 ( 18,33 – ½ 18,33 ) + 10 . 18,33 ( 18,33 – ½ 18,33 ) = 3359,889 cm

  1 GN-GN

  ,

  2

  τ

  = = = 2,49 kg/cm GN-GN

  ,

  ≤ Potongan II – II Daerah b – b s.d. c – c ( 0 ≤X 10 )

2 SX = 50 . X ( 11,67 – ½ X )

  2

  2

  2

  3

  2

  τ

  X = 0 S = 0 cm = = = 0 kg/cm 2 b-b b-b

  ,

  3 X 2 = 10 S c-c = 50 . 10 ( 11,67 – ½ 10 ) = 3335 cm

  2

  τ

  c-c = = = 0,99 kg/cm

  ,

  ≤ Potongan III – III Daerah c – c s.d. GN – GN ( 0 ≤X 1,67 )

  3 SX 3 = 50 . 10. 6,67 + 10 . X 3 ( 1,67 – ½ X 3 ) + 10 . X 3 ( 1,67 – ½ X 3 )

  3

  2

  τ

  X 3 = 0 S c-c = 3335 cm c-c = = = 2,47 kg/cm

  ,

  3 X = 1,67 S = 3335 + 10 . 1,67 ( 1,67 – ½ 1,67 ) + 10 . 1,67 ( 1,67 – ½ 1,67 ) = 3362,889 cm

  3 GN-GN

  ,

  2

  τ

  = = = 2,49 kg/cm GN-GN

  ,

  Tegangan Normal

  σ

  =

  σ

  =

  σ

  =

  2

  σ

  = 0 kg/cm Momen potongan II-II Tinjau Kiri : M

  II-II = - RVS

  2 ( Tekan - )

  atas = =

  ,

  = - 434,49 kg/cm

  2 ( Tarik + )

  σ

  bawah = =

  ,

  = - 276,62 kg/cm

  L

  3

  5 = 4 m L

  6 = 6 m Pot. II-II

  Pot. II-II RHE RVS

  3 RVE RVS

  2 q

  2 = 1,5 t/m S

  3 E S

  2 ( q 2 . 4 )

  σ

  = 5784,06 cm

  2 . 4 – ( q

  Normal potongan II-II Tinjau Kiri : N

  2 . 4 ) 2

  M

  II-II = - 1 . 4 – ( 1,5 . 4 ) 2 = - 16 tm M

  II-II = - 1600000 kgcm

  Lintang potongan II-II Tinjau Kiri : Q

  II-II = - RVS 1 – ( q 2 . 4 ) Q

  II-II = - 1 – ( 1,5 . 4 )= - 7 ton Q

  II-II = - 7000 kg

  II-II = - RHE N

  , ,

  II-II = 0 kg

  Tegangan Lentur

  σ

  = W atas

  = =

  , ,

  = 3682,49 cm

  3 W bawah = =

  Catatan : Jika tampang balok mengalami lenturan negatif, maka tegangan tarik terjadi di serat atas dan tegangan tekan di serat bawah. Tegangan Geser ≤

  Potongan I – I Daerah a – a s.d. GN – GN ( 0 ≤X 1 18,33 ) SX = 10 . X ( 18,33 – ½ X ) + 10 . X ( 18,33 – ½ X )

  1

  1

  1

  1

  1

  3

  2

  τ

  X = 0 S = 0 cm = = = 0 kg/cm 1 a-a a-a

  ,

  3 X = 18,33 S = 10 . 18,33 ( 18,33 – ½ 18,33 ) + 10 . 18,33 ( 18,33 – ½ 18,33 ) = 3359,889 cm

  1 GN-GN

  

,

  2

  τ

  = = = - 17,42 kg/cm GN-GN

  

,

  ≤ Potongan II – II Daerah b – b s.d. c – c ( 0 ≤X 10 )

2 SX = 50 . X ( 11,67 – ½ X )

  2

  2

  2

  3

  2

  τ

  X = 0 S = 0 cm = = = 0 kg/cm 2 b-b b-b

  ,

  3 X 2 = 10 S c-c = 50 . 10 ( 11,67 – ½ 10 ) = 3335 cm

  2

  τ

  c-c = = = - 6,92 kg/cm

  ,

  ≤ Potongan III – III Daerah c – c s.d. GN – GN ( 0 ≤X 1,67 )

  3 SX 3 = 50 . 10. 6,67 + 10 . X 3 ( 1,67 – ½ X 3 ) + 10 . X 3 ( 1,67 – ½ X 3 )

  3

  2

  τ

  X 3 = 0 S c-c = 3335 cm c-c = = = - 17,29 kg/cm

  ,

  3 X = 1,67 S = 3335 + 10 . 1,67 ( 1,67 – ½ 1,67 ) + 10 . 1,67 ( 1,67 – ½ 1,67 ) = 3362,889 cm

  3 GN-GN

  

,

  2

  τ

  = = = - 17,42 kg/cm GN-GN

  

,

  Tegangan Normal

  σ

  =

  σ

  =

  σ

  =

  2

  σ

  = 0 kg/cm Gambar Tegangan Lentur, Tegangan Geser, dan Tegangan Normal Tegangan Normal (σ)

  τ GN-GN

  = 2,49 kg/cm ² τ c-c = 2,47 kg/cm ² τ c-c

  = 0,99 kg/cm ² τ b-b = 0 kg/cm ² τ a-a

  = 0 kg/cm ² σ = 43,22 kg/cm

²

σ

  = 67,89 kg/cm ² c c c c b b a a a a

  GN GN 1,67 cm 18,33 cm

  10 cm 10 cm 10 cm 30 cm

  Tegangan Lentur (σ) Tegangan Geser (τ) Tegangan Normal (σ)

  τ GN-GN = - 17,42 kg/cm ²

  τ c-c = - 17,29 kg/cm ² τ c-c = - 6,92 kg/cm ² τ b-b = 0 kg/cm ² τ a-a = 0 kg/cm ²

  σ = - 276,62 kg/cm ² σ = - 434,49 kg/cm ² c c c c b b a a a a

  GN GN 1,67 cm 18,33 cm

  10 cm 10 cm 10 cm 30 cm

  Tegangan Lentur (σ) Tegangan Geser (τ)