Adji Sutama

Tipe II

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

h2

= 5 m

h 1=

3 m

q2 = 1,5 t/mq1 = 3,2 t/mP1 = 2 t

L7 = 3 mL5 = 4 mL4 = 5 mL3 = 4 mL2 = 4 mL1 = 3 m L6 = 6 m

G

FS3ES2S1DC B

A

Adji Sutama

Menganalisa Konstruksi Gerber :

RHS1

MG

RHA

RHE

RHG

RVG

RVS3

RVS3RVE

RVA

RVD

RHS1

RVS2

RVS2

RVS1

RVS1

h2

= 5 m

G

FS3

D

h 1=

3 m

P1 = 2 t

S2S1

q1 = 3,2 t/m

S1C B

A

q2 = 1,5 t/m

S3ES2

L7 = 3 mL5 = 4 mL4 = 5 mL3 = 4 mL2 = 4 mL1 = 3 m L6 = 6 m

Adji Sutama

Batang S1 – S2 Reaksi Perletakan

RVS1 ΣMS2 = 0

RVS1 . 5 – P1 . 2,5 + RVS2 . 0 = 0

RVS1 =. ,

= 1 ton ( )

RVS2 ΣMS1 = 0

- RVS2 . 5 + P1 . 2,5 + RVS1 . 0 = 0

RVS2 =. ,

= 1 ton ( )

ΣV = 0

RVS1 + RVS2 – P1 = 0

1 + 1 – 2 = 0

0 = 0

ΣH = 0

RHS1 = 0

0 0

-1 t-1 t

1 t 1 t

0 0

2,5 tm

X2X1

-

+M

RHS1

RVS2RVS1

P1 = 2 t

S2S1

L4 = 5 m

Q

N

+

Adji Sutama

Bidang Momen

Daerah S1 – P1 ( 0 ≤ X1 ≤ 2,5 )

MX1 = RVS1 . X1 = X1

X1 = 0 MS1 = 0 tm

X1 = 2,5 MP1 = 2,5 tm

Daerah S2 – P1 ( 0 ≤ X2 ≤ 2,5 )

MX2 = RVS2 . X2 = X2

X2 = 0 MS2 = 0 tm

X2 = 2,5 MP1 = 2,5 tm

Bidang Lintang

Daerah S1 – P1 ( 0 ≤ X1 ≤ 2,5 )

QX1 = RVS1 = 1

X1 = 0 QS1 = 1 ton

X1 = 2,5 QP1 = 1 ton

Daerah S2 – P1 ( 0 ≤ X2 ≤ 2,5 )

QX2 = - RVS2 = - 1

X2 = 0 QS2 = - 1 ton

X2 = 2,5 QP1 = - 1 ton

Bidang Normal

Daerah S1 – P1 ( 0 ≤ X1 ≤ 2,5 )

NX1 = - RHS1 = 0

X1 = 0 NS1 = 0 ton

X1 = 2,5 NP1 = 0 ton

Adji Sutama

Batang A – B – C – D – S1 AB = (L ) + (h )²AB = (3) + (3)²AB = √18AB = 4,24 m

tan α =

tan α =

tan α = 1

α = tan-1 1 = 45º

Reaksi Perletakan

RVA ΣMD = 0

- RVA . 7 + RVS1 . 4 + ( q1 . 4 ) 2 + RVD . 0 = 0

RVA =. ( . )

= 4,23 ton ( )

RHS1 = 0 t

S1

( q1 . X2 )

X2

X3

X1

D

αRVA sin α RVA cos α

2,99 t

2,99 t

- 2,99 t

- 2,99 t

1 t

13,8 t

Q

- 4,23 t - 4,23 t

- 12,68 tm

0

- 29,6 tm

0M

0

+-

-

( q1 . 4 )

α = 45ºRHA

RVA

RVD

RVS1 = 1 t

h 1=

3 m

q1 = 3,2 t/m

C B

A

L3 = 4 mL2 = 4 mL1 = 3 m

N

Adji Sutama

RVD ΣMA = 0

- RVD . 7 + RVS1 . 11 + ( q1 . 4 ) 9 + RVA . 0 = 0

RVD =. ( . )

= 18,03 ton ( )

ΣV = 0

- RVA + RVD – ( q1 . 4 ) – RVS1 = 0

- 4,23 + 18,03 – ( 3,2 . 4 ) – 1 = 0

0 = 0

ΣH = 0

RHA – RHS1 = 0

RHA = RHS1 = 0 ton

Bidang Momen

Daerah A – B ( 0 ≤ X1 ≤ 4,24 )

MX1 = - RVA cos α . X1 = - 4,23 cos 45º . X1

X1 = 0 MA = 0 tm

X1 = 4,24 MB = - 12,68 tm

Daerah S1 – D ( 0 ≤ X2 ≤ 4 )

MX2 = - RVS1 . X2 – ( q1 . X2 ) ½ X2 = - X2 – 1,6 X22

X2 = 0 MS1 = 0 tm

X2 = 4 MD = - 29,6 tm

Letak Mmax pada Qx = 0

= 0

- RVS1 – 3,2 X2 = 0

X2 = – , = - 0,3125 m ( Tidak Memenuhi )

Daerah D – B – C ( 0 ≤ X3 ≤ 7 )

MX3 = - RVS1 ( 4 + X3 ) – ( q1 . 4 ) ( 2 + X3 ) + RVD . X3

MX3 = - 1 ( 4 + X3 ) – ( 3,2 . 4 ) ( 2 + X3 ) + 18,03 . X3

X3 = 0 MD = - 29,6 tm

X3 = 4 MB = - 12,68 tm

X3 = 7 MC = 0 tm

Adji Sutama

Bidang Lintang

Daerah A – B ( 0 ≤ X1 ≤ 4,24 )

QX1 = - RVA cos α = - 4,23 cos 45º

X1 = 0 QA = - 2,99 ton

X1 = 4,24 QB = - 2,99 ton

Daerah S1 – D ( 0 ≤ X2 ≤ 4 )

QX2 = RVS1 + ( q1 . X2 ) = 1 + ( 3,2 X2 )

X2 = 0 QS1 = 1 ton

X2 = 4 QD = 13,8 ton

Daerah D – B – C ( 0 ≤ X3 ≤ 7 )

QX3 = RVS1 + ( q1 . 4 ) – RVD

QX3 = 1 + ( 3,2 . 4 ) – 18,03 = - 4,23

X3 = 0 QD = - 4,23 ton

X3 = 4 QB = - 4,23 ton

X3 = 7 QC = - 4,23 ton

Bidang Normal

Daerah A – B ( 0 ≤ X1 ≤ 4,24 )

NX1 = RVA sin α = 4,23 sin 45º

X1 = 0 NA = 2,99 ton

X1 = 4,24 NB = 2,99 ton

Daerah S1 – D ( 0 ≤ X2 ≤ 4 )

NX2 = - RHS1 = 0

X2 = 0 NS1 = 0 ton

X2 = 4 ND = 0 ton

Daerah D – B – C ( 0 ≤ X3 ≤ 7 )

NX3 = - RHS1 = 0

X3 = 0 ND = 0 ton

X3 = 4 NB = 0 ton

X3 = 7 NC = 0 ton

Adji Sutama

Batang S2 – E – S3 Reaksi Perletakan

RVE ΣMS3 = 0

RVE . 6 – RVS2 . 10 – ( q2 . 10 ) 5 + RVS3 . 0 = 0

RVE =. ( . )

= 14,17 ton ( )

RVS3 ΣME = 0

- RVS3 . 6 + ( q2 . 10 ) 1 - RVS2 . 4 + RVE . 0 = 0

RVS3 =. ( . )

= 1,83 ton ( )

ΣV = 0

- RVS2 + RVE + RVS3 – ( q2 . 10 ) = 0

- 1 + 14,17 + 1,83 – ( 1,5 . 10 ) = 0

0 = 0

ΣH = 0

RHE = 0

Q

00

--

+

- 1,83 t- 1 t

- 7 t

7,17 t

0

00

1,12 tm

- 16 tm

0

1,22 m

2,44 m

+

-

( q2 . X2 )( q2 . X1 )

5 m1 m

( q2 . 10 )RHE

RVS3RVE

RVS2 = 1 t

q2 = 1,5 t/m

S3ES2

L5 = 4 m L6 = 6 m

X2X1

M

N

Adji Sutama

Bidang Momen

Daerah S2 – E ( 0 ≤ X1 ≤ 4 )

MX1 = - RVS2 . X1 – ( q2 . X1 ) ½ X1

MX1 = - X1 – 0,75 X12

X1 = 0 MS2 = 0 tm

X1 = 4 ME = - 16 tm

Letak Mmax pada Qx = 0

= 0

- RVS2 – 1,5 X1 = 0

X1 = – , = - 0,667 m ( Tidak Memenuhi )

Daerah S3 – E ( 0 ≤ X2 ≤ 6 )

MX2 = RVS3 . X2 – ( q2 . X2 ) ½ X2

MX2 = 1,83 X2 – 0,75 X22

X2 = 0 MS3 = 0 tm

X2 = 6 ME = -16 tm

Letak Mmax pada Qx = 0

= 0

RVS3 – 1,5 X2 = 0

X2 =, , = 1,22 m

X2 = 1,22 Mmax = 1,12 tm

Letak Momen Mx = 0

1,83 X2 – 0,75 X22 = 0

- 0,75 X22 + 1,83 X2 = 0

X1,2 =± –

X1,2 =, ± ( , ) – ( , ) ( )( , )

X1,2 =, ± ,– ,

Adji Sutama

X1 =, ,– , = 0 m

X2 =, ,– , = 2,44 m

Bidang Lintang

Daerah S2 – E ( 0 ≤ X1 ≤ 4 )

QX1 = - RVS2 – ( q2 . X1 ) = - 1 – 1,5 X1

X1 = 0 QS2 = - 1 ton

X1 = 4 QE = - 7 ton

Daerah S3 – E ( 0 ≤ X2 ≤ 6 )

QX2 = - RVS3 + ( q2 . X2 ) = - 1,83 + 1,5 X2

X2 = 0 QS3 = - 1,83 ton

X2 = 6 QE = 7,17 ton

X2 = 1,22 Qmax = 0 ton

Adji Sutama

Batang S3 – F – G

RVS3 = 1,83 tRVS3 = 1,83 t

X1

0000

-1,83 t

-1,83 t- 1,83 t- 1,83 t

-5,49 tm

-5,49 tm

- 5,49 tm

-

-

NQM

Y1

X1

Y1Y1

MGMG

G

MG

RHG

RVG

RVS3 = 1,83 t

h2

= 5 m

FS3

L7 = 3 m

GRHG

RVGh

2=

5 m

FS3

L7 = 3 m

GRHG

RVG

h2

= 5 m

FS3

L7 = 3 m

X1

Adji Sutama

Reaksi Perletakan

ΣV = 0

- RVS3 + RVG = 0

- 1,83 + RVG = 0

RVG = 1,83 ton ( )

ΣH = 0

RHG = 0

ΣM = 0

- RVS3 . 3 + RVG . 0 + RHG . 0 + MG = 0

MG = RVS3 . 3 = 1,83 . 3 = 5,49 tm ( )

Bidang Momen

Daerah S3 – F ( 0 ≤ X1 ≤ 3 )

MX1 = - RVS3 . X1 = - 1,83 X1

X1 = 0 MS3 = 0 tm

X1 = 3 MF = - 5,49 tm

Daerah G – F ( 0 ≤ Y1 ≤ 5 )

MY1 = RHG . Y1 – MG = - 5,49

Y1 = 0 MG = - 5,49 tm

Y1 = 5 MF = - 5,49 tm

Bidang Lintang

Daerah S3 – F ( 0 ≤ X1 ≤ 3 )

QX1 = - RVS3 = - 1,83

X1 = 0 QS3 = - 1,83 ton

X1 = 3 QF = - 1,83 ton

Daerah G – F ( 0 ≤ Y1 ≤ 5 )

QY1 = - RHG = 0

Y1 = 0 QG = 0 ton

Y1 = 5 QF = 0 ton

Adji Sutama

Bidang Normal

Daerah G – F ( 0 ≤ Y1 ≤ 5 )

NY1 = - RVG = - 1,83

Y1 = 0 NG = - 1,83 ton

Y1 = 5 NF = - 1,83 ton

Adji Sutama

Gambar Keseluruhan Bidang Momen (M)

-5,49 tm

-5,49 tm

- 5,49 tm

1,22 m

2,44 m

0

1,12 tm

- 16 tm

00

- 12,68 tm

- 29,6 tm

0 0

2,5 tm

0

h2

= 5 m

h 1=

3 m

q2 = 1,5 t/mq1 = 3,2 t/mP1 = 2 t

L7 = 3 mL5 = 4 mL4 = 5 mL3 = 4 mL2 = 4 mL1 = 3 m L6 = 6 m

G

FS3ES2S1DC B

A

Adji Sutama

Gambar Keseluruhan Bidang Lintang (Q)

0

0

- 1,83 t

7,17 t

- 1,83 t

- 7 t- 2,99 t

- 2,99 t

13,8 t

- 4,23 t - 4,23 t

-1 t -1 t

1 t1 t

0

0

0

1,22 m

h2

= 5 m

h 1=

3 m

q2 = 1,5 t/mq1 = 3,2 t/mP1 = 2 t

L7 = 3 mL5 = 4 mL4 = 5 mL3 = 4 mL2 = 4 mL1 = 3 m L6 = 6 m

G

FS3ES2S1DC B

A

Adji Sutama

Gambar Keseluruhan Bidang Normal (N)

0

0 0

-1,83 t

-1,83 t

2,99 t

2,99 t

h2

= 5 m

h 1=

3 m

q2 = 1,5 t/mq1 = 3,2 t/mP1 = 2 t

L7 = 3 mL5 = 4 mL4 = 5 mL3 = 4 mL2 = 4 mL1 = 3 m L6 = 6 m

G

FS3ES2S1DC B

A

Adji Sutama

Penampang Balok

Pot. I-I

Pot. I-I

P1 = 2 t

h2

= 5 m

h 1=

3 m

q2 = 1,5 t/mq1 = 3,2 t/m

L7 = 3 mL5 = 4 mL4 = 5 mL3 = 4 mL2 = 4 mL1 = 3 m L6 = 6 m

G

FS3ES2S1DC B

A

Pot. II-II

Pot. II-II

Adji Sutama

Hitunglah besar Tegangan Lentur, Tegangan Geser, dan Tegangan Normal pada potongan I-I dan II-II dengan

penampang balok berbentuk U serta gambarkan?

10 cm

10 cm10 cm

30 cm

50 cm

Adji Sutama

Menentukan Garis Netral (GN)

Statis momen terhadap serat bawah

( F1 + F2 + F3 ) x = F1 . x1 + F2 . x2 + F3 . x3

x =. . .

x =( ) . ( ) . ( ) .( ) ( ) ( )

x =

x = 25 cm

( F1 + F2 + F3 ) y = F1 . y1 + F2 . y2 + F3 . y3

y =. . .

y =( ) . ( ) . ( ) .( ) ( ) ( )

y =

y = 11,67 cm

x

y

x3 = 45 cmx2 = 25 cm

x1 = 5 cm

y2 = 5 cm

y3 = 20 cmy1 = 20 cm

F2

F3F1

2

31

10 cm

10 cm10 cm

30 cm

50 cm

Adji Sutama

Momen Inersia

I total = I1 + I2 + I3

I total = ( 1/12 . b1 . h13 + F1 . a1

2 ) + ( 1/12 . b2 . h23 + F2 . a2

2 ) + ( 1/12 . b3 . h33 + F3 . a3

2 )

I total = ( 1/12 . 10 . 203 + 200 . 8,332 ) + ( 1/12 . 50 . 103 + 500 . 6,672 ) + ( 1/12 . 10 . 203 + 200 . 8,332 )

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

I total = 67500,02 cm4

(25, 11,67)

x = 25 cm

F2

a2 = 6,67 cm

a3 = 8,33 cma1 = 8,33 cm

F3F1

2

31

GNGN

10 cm

y atas = 18,33 cm

y bawah = 11,67 cm

10 cm10 cm

20 cm

50 cm

Adji Sutama

Potongan I – I Daerah a – a s.d. GN – GN ( 0 ≤ X1 ≤ 18,33 )

SX1 = 10 . X1 ( 18,33 – ½ X1 ) + 10 . X1 ( 18,33 – ½ X1 )

X1 = 0 S a-a = 0 cm3 τ a-a = = ,X1 = 18,33 S GN-GN = 10 . 18,33 ( 18,33 – ½ 18,33 ) + 10 . 18,33 ( 18,33 – ½ 18,33 ) = 3359,889 cm3

τ GN-GN = =, ,

F3ccF3cc

F2bb

aaF1aa

III

F1

IIII

IIIIII III

I II I

X3 X3

X2

X1 X1

1,67 cm

10 cm10 cm 30 cm

GNGN

18,33 cm

10 cm

Adji Sutama

Potongan II – II Daerah b – b s.d. c – c ( 0 ≤ X2 ≤ 10 )

SX2 = 50 . X2 ( 11,67 – ½ X2 )

X2 = 0 S b-b = 0 cm3 τ b-b = = ,X2 = 10 S c-c = 50 . 10 ( 11,67 – ½ 10 ) = 3335 cm3

τ c-c = = ,Potongan III – III Daerah c – c s.d. GN – GN ( 0 ≤ X3 ≤ 1,67 )

SX3 = 50 . 10. 6,67 + 10 . X3 ( 1,67 – ½ X3 ) + 10 . X3 ( 1,67 – ½ X3 )

X3 = 0 S c-c = 3335 cm3 τ c-c = = ,X3 = 1,67 S GN-GN = 3335 + 10 . 1,67 ( 1,67 – ½ 1,67 ) + 10 . 1,67 ( 1,67 – ½ 1,67 ) = 3362,889 cm3

τ GN-GN = =, ,

Adji Sutama

Momen potongan I-I

Tinjau Kiri :

M I-I = RVS1 . 2,5

M I-I = 1 . 2,5 = 2,5 tm

M I-I = 250000 kgcm

Lintang potongan I-I

Tinjau Kiri :

Q I-I = RVS1

Q I-I = 1 = 1 ton

Q I-I = 1000 kg

Normal potongan I-I

Tinjau Kiri :

N I-I = - RHS1

N I-I = 0 kg

Tegangan Lentur

σ =

W atas = =,, = 3682,49 cm3

W bawah = =,, = 5784,06 cm3

σ atas = = , = 67,89 kg/cm2 ( Tekan - )

σ bawah = = , = 43,22 kg/cm2 ( Tarik + )

L4 = 5 m

Pot. I-I

Pot. I-I

RHS1

RVS2RVS1

P1 = 2 t

S2S1

Catatan : Jika tampang balok mengalami lenturan positif, maka tegangan

tekan terjadi di serat atas dan tegangan tarik di serat bawah.

Adji Sutama

Tegangan Geser

Potongan I – I Daerah a – a s.d. GN – GN ( 0 ≤ X1 ≤ 18,33 )

SX1 = 10 . X1 ( 18,33 – ½ X1 ) + 10 . X1 ( 18,33 – ½ X1 )

X1 = 0 S a-a = 0 cm3 τ a-a = = , = 0 kg/cm2

X1 = 18,33 S GN-GN = 10 . 18,33 ( 18,33 – ½ 18,33 ) + 10 . 18,33 ( 18,33 – ½ 18,33 ) = 3359,889 cm3

τ GN-GN = =,, = 2,49 kg/cm2

Potongan II – II Daerah b – b s.d. c – c ( 0 ≤ X2 ≤ 10 )

SX2 = 50 . X2 ( 11,67 – ½ X2 )

X2 = 0 S b-b = 0 cm3 τ b-b = = , = 0 kg/cm2

X2 = 10 S c-c = 50 . 10 ( 11,67 – ½ 10 ) = 3335 cm3

τ c-c = = , = 0,99 kg/cm2

Adji Sutama

Potongan III – III Daerah c – c s.d. GN – GN ( 0 ≤ X3 ≤ 1,67 )

SX3 = 50 . 10. 6,67 + 10 . X3 ( 1,67 – ½ X3 ) + 10 . X3 ( 1,67 – ½ X3 )

X3 = 0 S c-c = 3335 cm3 τ c-c = = , = 2,47 kg/cm2

X3 = 1,67 S GN-GN = 3335 + 10 . 1,67 ( 1,67 – ½ 1,67 ) + 10 . 1,67 ( 1,67 – ½ 1,67 ) = 3362,889 cm3

τ GN-GN = =,, = 2,49 kg/cm2

Tegangan Normal

σ =

σ =

σ =

σ = 0 kg/cm2

Adji Sutama

Momen potongan II-II

Tinjau Kiri :

M II-II = - RVS2 . 4 – ( q2 . 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 = - RVS1 – ( q2 . 4 )

Q II-II = - 1 – ( 1,5 . 4 )= - 7 ton

Q II-II = - 7000 kg

Normal potongan II-II

Tinjau Kiri :

N II-II = - RHE

N II-II = 0 kg

Tegangan Lentur

σ =

W atas = =,, = 3682,49 cm3

W bawah = =,, = 5784,06 cm3

σ atas = = , = - 434,49 kg/cm2 ( Tarik + )

σ bawah = = , = - 276,62 kg/cm2

( Tekan - )

L5 = 4 m L6 = 6 mPot. II-II

Pot. II-II

RHE

RVS3RVE

RVS2 q2 = 1,5 t/m

S3ES2

( q2 . 4 )

Catatan : Jika tampang balok mengalami lenturan negatif, maka tegangan tarik

terjadi di serat atas dan tegangan tekan di serat bawah.

Adji Sutama

Tegangan Geser

Potongan I – I Daerah a – a s.d. GN – GN ( 0 ≤ X1 ≤ 18,33 )

SX1 = 10 . X1 ( 18,33 – ½ X1 ) + 10 . X1 ( 18,33 – ½ X1 )

X1 = 0 S a-a = 0 cm3 τ a-a = = , = 0 kg/cm2

X1 = 18,33 S GN-GN = 10 . 18,33 ( 18,33 – ½ 18,33 ) + 10 . 18,33 ( 18,33 – ½ 18,33 ) = 3359,889 cm3

τ GN-GN = =,, = - 17,42 kg/cm2

Potongan II – II Daerah b – b s.d. c – c ( 0 ≤ X2 ≤ 10 )

SX2 = 50 . X2 ( 11,67 – ½ X2 )

X2 = 0 S b-b = 0 cm3 τ b-b = = , = 0 kg/cm2

X2 = 10 S c-c = 50 . 10 ( 11,67 – ½ 10 ) = 3335 cm3

τ c-c = = , = - 6,92 kg/cm2

Adji Sutama

Potongan III – III Daerah c – c s.d. GN – GN ( 0 ≤ X3 ≤ 1,67 )

SX3 = 50 . 10. 6,67 + 10 . X3 ( 1,67 – ½ X3 ) + 10 . X3 ( 1,67 – ½ X3 )

X3 = 0 S c-c = 3335 cm3 τ c-c = = , = - 17,29 kg/cm2

X3 = 1,67 S GN-GN = 3335 + 10 . 1,67 ( 1,67 – ½ 1,67 ) + 10 . 1,67 ( 1,67 – ½ 1,67 ) = 3362,889 cm3

τ GN-GN = =,, = - 17,42 kg/cm2

Tegangan Normal

σ =

σ =

σ =

σ = 0 kg/cm2

Adji Sutama

Gambar Tegangan Lentur, Tegangan Geser, dan Tegangan Normal

Tegangan Normal (σ)

0

0

0

τ GN-GN= 2,49 kg/cm2

τ c-c= 2,47 kg/cm2

τ c-c= 0,99 kg/cm2

τ b-b= 0 kg/cm2

τ a-a= 0 kg/cm2

σ = 43,22 kg/cm2

σ = 67,89 kg/cm2

cccc

bb

aaaa

GN GN

1,67 cm

18,33 cm

10 cm

10 cm10 cm 30 cm

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

0

τ GN-GN= - 17,42 kg/cm2

τ c-c= - 17,29 kg/cm2

τ c-c= - 6,92 kg/cm2

τ b-b= 0 kg/cm2

τ a-a= 0 kg/cm2

σ = - 276,62 kg/cm2

σ = - 434,49 kg/cm2

cccc

bb

aaaa

GN GN

1,67 cm

18,33 cm

10 cm

10 cm10 cm 30 cm Tegangan Lentur (σ) Tegangan Geser (τ)