laporan tugas ujian mid beton 1

8/18/2019 Laporan Tugas Ujian Mid Beton 1

1/44

Struktur

Beton 1

1

LAPORAN TUGAS UJIAN MID

STRUKTUR BETON 1

Oleh :

Nama : Desi Ariyo Nurhidayat

NIM : 30201203259

FAKULTAS TEKNIK JURUSAN TEKNIK SIPIL

UNIVERSITAS ISLAM SULTAN AGUNG

SEMARANG

2014


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Struktur

Beton 1

2

SOAL

1.

Balok Penahan Momen

Hitung luas tulangan tarik dan jumlah tulangan tarik balok persegi beton bertulang dengan data sebagai berikut :

fc'=200 kg/cm2

(20 MPa); fy=4000 kg/cm2 (400 MPa); d=1,9 cm (19 mm)

a. b=30cm; h=50cm; Mu=20 tm

b. b=35cm; h=50cm; Mu=20 tm

c. b=30cm; h=50cm; Mu=10 tm

d. b=40cm; h=70cm; Mu=40 tm

2. Balok Penahan Lintang

Hitung tegangan geser & kuat geser beton dengan data sebagai berikut :

a. Gaya Lintang (Vu) = 20 ton

Mutu Beton (fc’) = 250 kg/cm

2

Mutu Baja sengkang (fyv) = 2200 kg/cm2

Tulangan sengkang (ds) =

∅12 cmLebar (b) =35 cm

Tinggi (h) =65 cm

= 0,9*65

= 59 cm

b. Gaya Lintang (Vu) = 35 ton

Mutu Beton (fc’

) = 200 kg/cm2


Tulangan sengkang (ds) = ∅14 cm

Lebar (b) =30 cm


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Beton 1

3

Tinggi (h) =55 cm

= 0,9*55

= 50 cm

c. Gaya Lintang (Vu) = 20 ton

Mutu Beton (fc’) = 345 kg/cm

2



Lebar (b) =20 cm

Tinggi (h) =40 cm

= 0,9*40

= 36 cm

d. Gaya Lintang (Vu) = 22 ton

Mutu Beton (fc’) = 400 kg/cm2



Lebar (b) =20 cm

Tinggi (h) =35 cm

= 0,9*35

= 32 cm


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Beton 1

4

3. Balok Penahan Kolom

Hitung dan gambar prosedur perencanaan tulangan pokok kolom beton bertulang 4 sisi dengan

ketentuan sebagai berikut :

a.

b = 50 cmh = 50 cm

fc’ = 200 kg/cm2

fy = 4000 kg/cm2

Mu = 40 tm

Pu = 100 t

Diameter Tulangan = 19 mm

b. b = 55 cm

h = 55 cm

fc’ = 250 kg/cm2

fy = 4000 kg/cm2

Mu = 28 tm

Pu = 200 t


c. b = 60 cm

h = 60 cm

fc’

= 150 kg/cm2

fy = 4000 kg/cm2

Mu = 10 tmPu = 250 t


d. b = 30 cm

h = 30 cm

fc’

= 150 kg/cm2

fy = 4000 kg/cm2

Mu = 8 tm

Pu = 40 t



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Struktur

Beton 1

5

JAWABAN

1. Balok Penahan Momen

1.a Diketahui : fc’ = 200 kg/cm2

( 20 Mpa )

fy = 4000 kg/cm2

( 400 Mpa )

b = 30 cm

h = 50 cm

d = 0.9 x 50 = 45 cm

Mu = 20 tm = 2000000 kg/cm

Ditanya : Luas Tulangan (As)

Jumlah Tulangan (n)

Penyelesaian :

1. Koefisien penampang

Rn =2^***85.0*

' d b fc

Mu

=2^45*30*200*85.0*8.0

2000000

=8262000

2000000

= 0,24207

2. Indeks tulangan ( n )

n = 1- )*21( Rn

= 1 - )24207,0*21(

= 1 – 0,71823

= 0,28177


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Beton 1

6

3. Rasio tulangan ( )

= n fy

fc'*85.0

= 0,218774000

200*85.0

= 0,01198

ρ min =14

=14

4000

= 0,00350

ρ max =0,75 .0,85 .0,85 .fc ′

fy .6000 / (6000+fy)

=0,75 .0,85 .0,85 .200

4000 .6000 / (6000+4000)

= 0,01626

4. Luas Tulangan (As)

As = *b*d

= 0,001198 * 30 *45

= 16,1665 cm

2

5. Jumlah tulangan (nb)

Nb = As/Ab (dibulatkan keatas)

Dimana Ab = ¼*3.14*1,92

Misal dipakai diameter tulangan 19 mm

Ab = ¼*3.14*1,92

= 2,83385 cm

2

nb = As/Ab

= 16,1665 cm2

/ 2,83385 cm2

= 5,704 dibulatkan menjadi 6 batang


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Struktur

Beton 1

7

Perhitungan menggunakan dsgwin

DATA:

Beam Width, Bw = 30.00 cm

Beam Height, Ht = 50.00 cm

Factored Moment, Mu = 20000.00 kg.m

Factored Shear, Vu = 0.00 kg

Factored Normal, Pu = 0.00 kg

Factored Torsion, Tu = 0.00 kg.m

Concrete Compressive Strength fc1 = 200.00 kg/cm2

Concrete Crack Strength fcr = 150.00 kg/cm2

Concrete Initial Strength fci = 120.00 kg/cm2

Main Rebar Yield Strength fy = 4000.00 kg/cm2

Stirrups Rebar Yield Strength fyv = 2400.00 kg/cm2

Main rebar diameter dbm = 1.90 cm

2nd main rebar diameter dbn = 1.90 cm

Stirrups Rebar diameter dbv = 1.00 cm

Min concrete cover, Cover = 4.00 cm

Min rebar clear space, Minclrspc = 2.50 cm

RESULT:

Effective Depth, d = 44.05 cm

Distance, d1 = H - d = 5.95 cmds = 43.32 cm

ds1 = 5.95 cm

Minimum Rebar Ratio, romin = 0.00352 (Using Romin = 200/Fy)

Maximum Rebar Ratio, romax = 0.01874

Flexural Reinforcement: OK

Normal Direction:

Beta Factor, Beta = 0.850

Location of neutral axis, c = 9.6063 cm

Length of compression block, a = 8.1654 cm

Total Rebar, nb = 4/ 6 d19


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Struktur

Beton 1

8

Total Rebar area, Ast = 17.0117 cm2 (ro = 0.01287)

Nominal Moment Capacity, Md = 20963.44 kg.m

Inverted Section Direction


Location of neutral axis, c1 = 7.5919 cm

Length of compression block, a1 = 6.4532 cm

Total Rebar, nb = 6/ 4 d19 Total Rebar area, As1 = 11.3411 cm2 (ro1 = 0.00858)

Nominal Moment Capacity, Md1 = 14471.02 kg.m

Shear Reinforcement: OK

Av = 2*(0.25*Pi*dbv*dbv) = 0.00 cm2

Ag = bw*h = 0.00 cm2

pw = Ast/(bw*d) = 0.0000

x1 = bw-2*cv-2*dbv = 0.00 cm

y1 = h-2*cv-2*dbv = 0.00 cm

Vn = Vu/phi = 0.00 kg

Tn = Tu/phi = 0.00 kg.m

Vc = = 0.00 kg

Tc = = 0.00 kg.m

Vnc = Vn - Vc = 0.00 kg

Tnc = Tn - Tc = 0.00 kg.m

Avs = Vnc/(fy*d) = 0.00 cm2

Ats = Tnc/(zt*x1*y1*fy) = 0.00 cm2Atst = Avs + 2*Ats = 0.00 cm2

Additional Longitudinal Rebar, At = 0.00 cm2

Stirrups Spacing, spc = 2540.01 cm


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Struktur

Beton 1

9

1.b Diketahui : fc’ = 200 kg/cm2

( 20 Mpa )

fy = 4000 kg/cm2

( 400 Mpa )

b = 35 cmh = 50 cm

d = 0.9 x 50 = 45 cm

Mu = 20 tm = 2000000 kg/cm


Jumlah Tulangan (n)

Penyelesaian :


Rn =2^***85.0* ' d b fc

Mu

=2^54*35*200*85.0*8.0

2000000

=9639000

2000000

= 0,20749

2.

Indeks tulangan ( n )

n = 1- )*21( Rn

= 1 - )20749,0*21(

= 1 – 0,76488

= 0,23512


= n fy

fc'*85.0

= 0,235124000

200*85.0

= 0,00999


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Struktur

Beton 1

10

ρ min =14

=14

4000

= 0,00350

ρ max = 0,75 .0,85 .0,85 .fc ′ fy .6000

/ (6000+fy)

=0,75 .0,85 .0,85 .200

4000 .6000 / (6000+4000)

= 0,01626

4. 45Luas Tulangan (As)

As = *b*d

= 0,00999 * 30 * 45

= 15,7383 cm2

5.

Jumlah tulangan (nb)


Dimana Ab = ¼*3.14*1,92


Ab = ¼*3.14*1,92

= 2,83385 cm2

nb = As/Ab

= 15,7383 cm2

/ 2,83385 cm2



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Struktur

Beton 1

11


DATA:







Concrete Compressive Strength fc1 = 200.00 kg/cm2Concrete Crack Strength fcr = 150.00 kg/cm2









RESULT:


Distance, d1 = H - d = 5.95 cm

ds = 44.05 cmds1 = 5.95 cm




Normal Direction:







12/44

Struktur

Beton 1

12




Location of neutral axis, c1 = 5.6190 cm

Length of compression block, a1 = 4.7761 cm


Total Rebar area, As1 = 5.6706 cm2 (ro1 = 0.00368)Nominal Moment Capacity, Md1 = 7723.30 kg.m


Av = 2*(0.25*Pi*dbv*dbv) = 0.00 cm2

Ag = bw*h = 0.00 cm2

pw = Ast/(bw*d) = 0.0000

x1 = bw-2*cv-2*dbv = 0.00 cm

y1 = h-2*cv-2*dbv = 0.00 cm



Vc = = 0.00 kg

Tc = = 0.00 kg.m

Vnc = Vn - Vc = 0.00 kg



Ats = Tnc/(zt*x1*y1*fy) = 0.00 cm2

Atst = Avs + 2*Ats = 0.00 cm2




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Struktur

Beton 1

13

1.c Diketahui : fc’ = 200 kg/cm2

( 20 Mpa )

fy = 4000 kg/cm2

( 400 Mpa )

b = 30 cm

h = 50 cm

d = 0.9 x 50 = 45 cmMu = 10 tm = 1000000 kg/cm


Jumlah Tulangan (n)

Penyelesaian :


Rn =2^***85.0*8.0 ' d b fc

Mu

=2^45*30*200*85.0*8.0

1000000

=82617316

1000000

= 0,12104


n = 1- )*21( Rn

= 1 - )12104,0*21(

= 1 – 0,87059

= 0,12941


= n fy

fc'*85.0

= 0,129414000

200*85.0

= 0,00550


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Struktur

Beton 1

14

ρ min =14

=14

4000

= 0,00350

ρ max = 0,75 .0,85 .0,85 .fc ′ fy .6000

/ (6000+fy)

=0,75 .0,85 .0,85 .200

4000 .6000 / (6000+4000)

= 0,01626


As = *b*d

= 0,00550 * 30 *45

= 7,42487 cm2

5.



Dimana Ab = ¼*3.14*1,92


Ab = ¼*3.14*1,92

= 2,83385 cm2

nb = As/Ab

= 7,42487 cm2

/ 2,83385 cm2



15/44

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Beton 1

15


DATA:

















RESULT:



ds = 44.05 cm

ds1 = 5.95 cm




Normal Direction:




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Beton 1

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Location of neutral axis, c1 = 5.7023 cmLength of compression block, a1 = 4.8470 cm


Total Rebar area, As1 = 5.6706 cm2 (ro1 = 0.00429)



Av = 2*(0.25*Pi*dbv*dbv) = 0.00 cm2

Ag = bw*h = 0.00 cm2

pw = Ast/(bw*d) = 0.0000

x1 = bw-2*cv-2*dbv = 0.00 cm

y1 = h-2*cv-2*dbv = 0.00 cm



Vc = = 0.00 kg

Tc = = 0.00 kg.m

Vnc = Vn - Vc = 0.00 kg

Tnc = Tn - Tc = 0.00 kg.mAvs = Vnc/(fy*d) = 0.00 cm2






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Struktur

Beton 1

17

1.d Diketahui : fc’ = 200 kg/cm2

( 20 Mpa )

fy = 4000 kg/cm2

( 400 Mpa )

b = 40 cm

h = 70 cm

d = 0,9 x 70 = 63 cmMu = 40 tm = 4000000 kg/cm


Jumlah Tulangan (n)

Penyelesaian :


Rn =2^***85.0*8.0 ' d b fc

Mu

=2^63*40*200*85.0*8.0

4000000

=21591360

4000000

= 0,18525


n = 1- )*21( Rn

= 1 - )18525,0*21(

= 1 – 0,7934

= 0,20660


= n fy

fc'*85,0

= 0,206604000

200*85,0

= 0,00878


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Struktur

Beton 1

18

ρ min =14

=14

4000

= 0,00350

ρ max = 0,75 .0,85 .0,85 .fc ′ fy .6000

/ (6000+fy)

=0,75 .0,85 .0,85 .200

4000 .6000 / (6000+4000)

= 0,01626


As = *b*d

= 0,00878 * 40 *63

= 22,12686 cm2

5.



Dimana Ab = ¼*3.14*1,92


Ab = ¼*3.14*1,92

= 2,83385 cm2

nb = As/Ab

= 22,12686 cm2

/ 2,83385 cm2



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Struktur

Beton 1

19


DATA:





Factored Normal, Pu = 0.00 kgFactored Torsion, Tu = 0.00 kg.m











RESULT:



ds = 63.50 cm

ds1 = 5.95 cm




Normal Direction:




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Struktur

Beton 1

20







Location of neutral axis, c1 = 5.9656 cmLength of compression block, a1 = 5.0708 cm


Total Rebar area, As1 = 5.6706 cm2 (ro1 = 0.00221)



Av = 2*(0.25*Pi*dbv*dbv) = 0.00 cm2

Ag = bw*h = 0.00 cm2

pw = Ast/(bw*d) = 0.0000

x1 = bw-2*cv-2*dbv = 0.00 cm

y1 = h-2*cv-2*dbv = 0.00 cm



Vc = = 0.00 kg

Tc = = 0.00 kg.m

Vnc = Vn - Vc = 0.00 kg

Tnc = Tn - Tc = 0.00 kg.mAvs = Vnc/(fy*d) = 0.00 cm2






21/44

Struktur

Beton 1

21

2. Balok Penahan Lintang

2.a. Diketahui : Vu = 20 ton

fc’

= 250 kg/cm2

fyv = 2200 kg/cm2

ds = ∅12 cm b = 35 cm

h = 65 cm

= 0.9 * 65

= 59 cm

Ditanya : Tegangan Geser (Vn)

Kuat Geser (Vc)

Penyelesaian :

a. Tegangan Geser ( Vn ) : Vn =Vu

ɸ.b.d=

20.000

0,75.35.59 =

20.000

1548,75= 12,91 kg/cm2

b. Kuat Geser ( Vc ) : vc = 0,53 √ fcI = 0,53√ 250 = 8,38 kg/cm2

c. Kontrol (Vc

Vn) :

Vn

vc

=12,91

8,38

= 1,5

: 3 <Vn

vc≤ 1

: s =Asv .fyv

b (Vn−vc) =

2.0,25.3,14.1,22 .2200

35 (12,91−8,38) =

4973,76

158,55 =

: 31,4 cmJadi s = 30 cm

∅12 – 300 mm


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Struktur

Beton 1

22

Gambar penulangan geser untuk setengah batang

Gambar detail


23/44

Struktur

Beton 1

23

2.b Diketahui : Vu = 35 ton

fc’

= 200 kg/cm2

fyv = 2100 kg/cm2

ds = ∅14 cm b = 30 cm

h = 55 cm

= 0.9 * 55

= 50 cm


Kuat Geser (Vc)

Penyelesaian :


ɸ.b.d=

35.000

0,75.30.50 =

35.000

1125= 31,1 kg/cm2


c.

Kontrol (VcVn ) : Vn

vc = 31.1

7.5 = 4.1

: 5 <Vn

vc≤ 3

: s =Asv .fyv

b (Vn−vc) =

2.0,25.3,14.1,22 .2100

35 (31.1−7.5) =

6462.12

708 =

: 9.1 cm

Jadi s = 8.5 cm

∅14 – 85 mm


24/44

Struktur

Beton 1

24


Gambar detail


25/44

Struktur

Beton 1

25

2.c Diketahui : Vu = 20 ton

fc’

= 345 kg/cm2

fyv = 5000 kg/cm2

ds = ∅12 cm b = 20 cm

h = 40 cm

= 0.9 * 40

= 36 cm


Kuat Geser (Vc)

Penyelesaian :


ɸ.b.d=

20.000

0,75.20.36 =

20.000

540= 37 kg/cm

2


c.

Kontrol (

Vc

Vn) :

Vn

vc

=37

9.8

= 3,8

: 5 <Vn

vc≤ 3

: s =Asv .fyv

b (Vn−vc ) =

2.0,25.3,14.1,22 .5000

20 (37−9.8) =

11304

544 =

: 20.8 cm

Jadi s = 19 cm

∅12 – 190 mm


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Struktur

Beton 1

26


Gambar detail


27/44

Struktur

Beton 1

27

2.d Diketahui : Vu = 39 ton

fc’

= 300 kg/cm2

fyv = 3000 kg/cm2

ds = ∅16 cm b = 25 cm

h = 55 cm

= 0,9 * 55

= 50 cm


Kuat Geser (Vc)

Penyelesaian :


ɸ.b.d=

39.000

0,75.25.50 =

39.000

937,5= 41,6 kg/cm2


c. Kontrol (Vc

Vn) :

Vn

vc =

41,1

9,2 = 4,5

: 5 <Vn

vc≤ 3

: s =Asv .fyv

b (Vn−vc) =

2.0,25.3,14.1,22 .3000

25 (41,6−9,2) =

12057,6

810 =

: 14,8 cm

Jadi s = 13 cm

∅16 – 130 mm


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Beton 1

28


Gambar detail


29/44

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Beton 1

29

3. Balok Penahan Kolom

3.a Diketahui :b = 50 cm

h = 50 cm

fc’

= 200 kg/cm2

fy = 4000 kg/cm2

Mu = 40 tm

Pu = 100 t

DiameterTulangan = 19 mm

Ditanya :Perencanaan Tulangan Kolom

Penyelesaian :

Sumbu x = Muɸ.b.ℎ2 .0,85.f ′

= 40x105

0,65.50.502 .0,85.200 = 4.000.000

13.812.500 = 0,29

Sumbu y =Pu

ɸ.b.ℎ .0,85.f ′ =

100x103

0,65.50.50.0,85.200 =

100.000

176.250 = 0,57

Dari Grafik: fc’ = 200 kg/cm2

= 20 Mpa = 0,80 r = 0,0422

t = r. = 0,0422 x 0,80

= 0,034

= 0,01

Ast = t x b x h

= 0,034 x 50 x 50

= 85 cm

db = ¼ . 3,14 . 1,92 = 2,83 cm

2

n =Ast

db

=85

2,83

= 30 ~32 D 19 mm


30/44

Struktur

Beton 1

30

Gambar Penampang

Diagram Interaksi


31/44

Struktur

Beton 1

31


DATA:

Column Width, Bw = 50.00 cm

Column Height, Ht = 50.00 cm















RESULT:


Balanced neutral axis, cbal = 23.6102 cm

Balanced Compr. block, abal = 20.0686 cm

Balanced Moment Capacity, Mnb = 70844.83 kg.m

Balanced Normal Capacity, Pnb = 130202.40 kg

Balanced Eccentricity, ebal = 54.41 cm



Total Rebar (incl. side bar) nbt = 32 d19

Total Side Rebar nbs = 0 d19

Total Rebar area, Ast = 90.7292 cm2 = 3.63%


Nominal Normal Capacity, Pd = 100009.10 kg


Av = 2*(0.25*Pi*dbv*dbv) = 0.00 cm2

Ag = bw*h = 0.00 cm2

pw = Ast/(bw*d) = 0.0000

x1 = bw-2*cv-2*dbv = 0.00 cm

y1 = h-2*cv-2*dbv = 0.00 cm



Vc = = 0.00 kg

Tc = = 0.00 kg.m

Vnc = Vn - Vc = 0.00 kg





Additional Longitudinal Rebar, At = 0.00 cm2Stirrups Spacing, spc = 30.40 cm


32/44

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Beton 1

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33/44

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Beton 1

33

3.b Diketahui :b = 55 cm

h = 55 cm

fc’

= 250 kg/cm2

fy = 4000 kg/cm2

Mu = 28 tm

Pu = 200 t



Penyelesaian :

Sumbu x =

Mu

ɸ.b.ℎ2 .0,85.f =

28x105

0,65.55.552 .0,85.250 =

2.800.000

22.980.547 = 0,12

Sumbu y =Pu

ɸ.b.ℎ .0,85.f =

200x103

0,65.55.55.0,85.250 =

200.000

417.828,13 = 0,48


= 20 Mpa = 1,0 r = 0,01

t = r.

= 0,01 x 1,0

= 0,01

= 0,01

Ast = t x b x h

= 0,01 x 55 x 55

= 30,25 cm

db = ¼ . 3,14 . 1,92 = 2,83 cm

2

n =Ast

db =

30,25

2,83 = 10,6 ~12 D 19 mm


34/44

Struktur

Beton 1

34

Gambar Penampang

Gambar Diagram Interaksi


35/44

Struktur

Beton 1

35


DATA:

















RESULT:















Av = 2*(0.25*Pi*dbv*dbv) = 0.00 cm2

Ag = bw*h = 0.00 cm2

pw = Ast/(bw*d) = 0.0000

x1 = bw-2*cv-2*dbv = 0.00 cm

y1 = h-2*cv-2*dbv = 0.00 cm



Vc = = 0.00 kg

Tc = = 0.00 kg.m

Vnc = Vn - Vc = 0.00 kg





Additional Longitudinal Rebar, At = 0.00 cm2Stirrups Spacing, spc = 30.40 cm


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Beton 1

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37/44

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Beton 1

37

3.c Diketahui :b = 60 cm

h = 60 cm

fc ' = 150 kg/cm2

fy = 4000 kg/cm2

Mu = 10 tm

Pu = 250 t



Penyelesaian :

Sumbu x =

Mu

ɸ.b.ℎ2 .0,85.f =

10x105

0,65.60.602 .0,85.150 =

1.000.000

17.901.000 = 0,05

Sumbu y =Pu

ɸ.b.ℎ .0,85.f =

250x103

0,65.60.60.0,85.150 =

250.000

298.350 = 0,84


= 15 Mpa = 0,6 r = 0,01

t = r.

= 0,01 x 0,6

= 0,006

= 0,01

Ast = t x b x h

= 0,01 x 60 x 60

= 21,6 cm

db = ¼ . 3,14 . 1,92 = 2,83 cm

2

n =Ast

db =

21,6

2,83 = 7,6~8 D 19 mm


38/44

Struktur

Beton 1

38

Gambar Penampang



39/44

Struktur

Beton 1

39


DATA:

















RESULT:


CONCENTRIC LOADING CONDITION







Av = 2*(0.25*Pi*dbv*dbv) = 0.00 cm2

Ag = bw*h = 0.00 cm2

pw = Ast/(bw*d) = 0.0000

x1 = bw-2*cv-2*dbv = 0.00 cm

y1 = h-2*cv-2*dbv = 0.00 cmVn = Vu/phi = 0.00 kg


Vc = = 0.00 kg

Tc = = 0.00 kg.m

Vnc = Vn - Vc = 0.00 kg








40/44

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Beton 1

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Beton 1

41

3.d Diketahui :b = 30 cm

h = 30 cm

fcI

= 150 kg/cm2

fy = 4000kg/cm

2

Mu = 8 tm

Pu = 40 t



Penyelesaian :

Sumbu x =Mu

ɸ.b.ℎ2 .0,85.f =

8x105

0,65.30.30.0,85.150 =

800.000

2.237.625 = 0,36

Sumbu y =Pu

ɸ.b.ℎ .0,85.f =

40x103

0,65.30.30.0,85.150 =

40.000

74.587,5 = 0,54


= 15 Mpa = 0,6 r = 0,0568

t = r.

= 0,0568 x 0,6

= 0,034 = 0,01

Ast = t x b x h

= 0,034 x 30 x 30

= 30,6 cm

db = ¼ . 3,14 . 1,92 = 2,83 cm

2

n =Ast

db =

30,6

2,83 = 10,8~12 D 19 mm


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Gambar Penampang



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DATA:















Min concrete cover, Cover = 4.00 cmMin rebar clear space, Minclrspc = 2.50 cm

RESULT:















Av = 2*(0.25*Pi*dbv*dbv) = 0.00 cm2

Ag = bw*h = 0.00 cm2

pw = Ast/(bw*d) = 0.0000

x1 = bw-2*cv-2*dbv = 0.00 cm

y1 = h-2*cv-2*dbv = 0.00 cm



Vc = = 0.00 kg

Tc = = 0.00 kg.m

Vnc = Vn - Vc = 0.00 kg








44/44

Struktur

Beton 1

laporan tugas ujian mid beton 1

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