BAB 3 PERENCANAAN PEMBEBANAN KELOMPOK 9 revisi mmmmmmmmmmmmmmmmmmmmmm

(1)

BAB III

PERENCANAAN

PELAT

(2)

III.1. Perencanaan Penulangan Pelat Atap Arah X Beban rencana pelat atap arah x, qu = 5.374 kN/m²

Dari hasil analisa menggunakan software STAAD-PRO diperoleh hasil sebagai berikut:

Mu Tumpuan : 13,574 kNm Mu Lapangan : 14,100 kNm

1) Daerah Penulangan Lapangan Atap Arah Lx

 Mu lapangan = 14,100 kNm

h = 10 cm = 100 mm

b = 100 cm = 1000 mm Fc’ = 25 Mpa

Fy = 240 Mpa

 d rencana = 100 – 30 = 70 mm

 Nilai Rn

�� = ��

�. �. �²

14,100 × 10⁶

= 0,9 .1000 .70² = 3,197 �/��²

 Nilai β1

β1 = 0,85 −

= 0,85 −

= 0,871

��^′− 28

7 × 0,05 25 − 28

7 × 0,05

(3)

 Nilai ρ

0,85 × ��^′ 2 × ��

ρ = �� × (1 − √1 − 0,85× ��^′) 0,85 × 25

= 240 × (1 − √1 − 2 × 3,197 0,85 × 25)

= 0,0145

 Nilai ρb

0,85 × ��′× β1 600

ρb = �� × 600 + ��

0,85 × 25 × 0,87 600

= 240 × 600 + 240

= 0,0550

 Nilai ρmaks

ρmaks = 0,75 × ρb = 0,75 × 0,0550 = 0,0412

 Nilai ρmin

1,4 1,4

ρmin = �� = 240 = 0,00583 Maka :ρmin≤ ρ ≤ ρmaks

0,00583 ≤ 0,0145 ≤ 0,0412 Jadi, yang dipakai adalah ρ = 0,0145

Luas Tegangan Tarik

 �� = � × � × �

= 0,0145 × 1000 × 70

= 1015 ��²

Dipakai tulangan∅14 – 150 (As 1026 mm²)

(4)

�� = 0,0014 × � × ℎ

= 0.0014 × 1000 × 100

= 140 ��²

Jarak Max

 �� = 3 × ℎ

= 3 × 100 �� = 300 ��

Kontrol

 �� = � − �� −^�∅tulangan

14 �

= 100 − 20 − ( )

= 73 �� 2

Syarat Rasio Penulangan : daktual≥ drencana

73 mm ≥ 70 mm

(5)

��²

∅14 − 125

��. �� 1232 ��²× 240 �⁄��²

� = 0,85.��^′. � = 0,85 × 25 �⁄ × 1000 ��

= 13,914 ��

�� = ��. ��. (� − 2)�

= 1232 ��²× 240 �⁄ × (70 �� –13,914 ��

)

��² 2

= 18640554,24 ��

= 18,640 ��

�� = �. �� = 0,9 × 18,640 = 16,776 ��

Syarat Kontrol Penulangan Mr ≥ Mu

16,776 kNm ≥ 14,100 kNm (AMAN)

(6)

h = 10 cm = 100 mm b = 100 cm = 1000 mm fc’ = 25 Mpa

fy = 240 Mpa

 drencana = 100 - 30 = 70 mm

 Nilai Rn

�� = ��

�. �. �²

13,574 × 10⁶

= 0,9 .1000 .70² = 3,078 �/��²

 Nilai β1

β1 = 0,85 −

= 0,85 −

= 0,871

��^′− 28

7 × 0,05 25 − 28

7 × 0,05

 Nilai ρ

0,85 × ��^′ 2 × ��

ρ = �� × (1 − √1 − 0,85× ��^′) 0,85 × 25

= 240 × (1 − √1 − 2 × 3,078 0,85 × 25)

= 0,0139

 Nilai ρb

0,85 × ��′× β1 600

ρb = �� × 600 + ��

0,85 × 25 × 0,87 600

= 240 × 600 + 240

= 0,0550

 Nilai ρmaks

ρmaks = 0,75 × ρb = 0,75 × 0,0550 = 0,0412

 Nilai ρmin

1,4 1,4

ρmin = �� = 240 = 0,00583

(7)

Maka :ρmin≤ ρ ≤ ρmaks

 �� = � × � × �

= 0,0139 × 1000 × 70

= 973 ��²

Luas Tualangan Susut

 �� = 0,0014 × � × ℎ

= 0.0014 × 1000 × 100

= 140 ��²

(8)

��² Jarak Max

 �� = 3 × ℎ

= 3 × 100 �� = 300 ��

Kontrol

 �_{��} = � − �� −^�_�∅tulangan

= 100 − 20 − ( )20

= 76 �� 2

76 mm ≥ 70 mm

��. �� 1005 ��²× 240 �⁄��²

� = 0,85.��^′. � = 0,85 × 25 �⁄ × 1000 ��

= 11,350 ��

�� = ��. ��. (� − 2)�

= 1005 ��²× 240 �⁄ × (70 �� –11,350 ��

)

��² 2

= 15515190 ��

= 15,515 ��

�� = �. �� = 0,9 × 15,515 = 13,963 ��

13,963 kNm ≥ 13,574 kNm (AMAN)

(9)

 Tabel Penulangan Plat Atap Arah X

Rn ρmin ρ ρmaks Tulangan Pokok (Tarik) Tulangan Susut Lapangan 3,197 0,00583 0,0145 0,0412 ∅14 – 150 (1026 ��²) ∅6 − 150 (188 ��²) Tumpuan 3,078 0,00583 0,0139 0,0412 ∅8 − 50 (1005 ��²) ∅6 − 150 (188 ��²)

(10)

III.2. Perencanaan Penulangan Pelat Atap Arah Y Beban rencana pelat atap arah x, qu = 5.374 kN/m²

Mu Tumpuan: 20,725 kNm Mu Lapangan : 14,900 kNm

1) Daerah Penulangan Lapangan Atap Arah Ly

h = 10 cm = 100 mm

b = 100 cm = 1000 mm Fc’ = 25 Mpa

Fy = 240 Mpa

 d rencana = 100 – 30 = 70 mm

 Nilai Rn

�� = �.�.�� ²

 Nilai β1

14,900 × 10⁶

= 0,9 .1000 .70² = 3,378 �/��²

β1 = 0,85 −

= 0,85 −

= 0,871

 Nilai ρ

��^′− 28

7 × 0,05 25 − 28

× 0,05 7

0,85 × ��^′ 2 × ��

ρ = �� × (1 − √1 − 0,85× ��^′)

(11)

0,85 × 25

= 240 × (1 − √1 − 2 × 3,378 0,85 × 25)

= 0,0154

 Nilai ρb

0,85 × ��′× β1 600

ρb = �� × 600 + ��

0,85 × 25 × 0,87 600

= 240 × 600 + 240

= 0,0550

 Nilai ρmaks

ρmaks = 0,75 × ρb = 0,75 × 0,0550 = 0,0412

 Nilai ρmin

1,4 1,4

 �� = � × � × �

= 0,0154 × 1000 × 70

= 1078 ��²

(12)

Luas Tulangan Susut

 �� = 0,0014 × � × ℎ

= 0.0014 × 1000 × 100

= 140 ��²

Jarak Max

 �� = 3 × ℎ

= 3 × 100 �� = 300 ��

(13)

��² Kontrol

 �_{��} = � − �� −^�_�∅tulangan

= 100 − 20 − ( )19

= 70,5 �� 2

70,5 mm ≥ 70 mm

��. �� 1134 ��²× 240 �⁄��²

� = 0,85.��^′. � = 0,85 × 25 �⁄ × 1000 ��

= 12,807 ��

�� = ��. ��. (� − 2)�

= 1134 ��²× 240 �⁄ × (70,5 �� –12,807 ��

)

��² 2

=17444503,44 ��

= 17,444 ��

�� = �. �� = 0,9 × 17,444 = 19,275 ��

15,699 kNm ≥ 14,900 kNm (AMAN)

(14)

2) Daerah Penulangan Tumpuan Atap Arah Ly

 Mu Tumpuan = 20,275 KNm

h = 10 cm = 100 mm

b = 100 cm = 1000 mm fc’ = 25 Mpa

fy = 240 Mpa

 drencana = 100-30 = 70 mm

 Nilai Rn

�� = ��

�. �. �²

20,275 × 10⁶

= 0,9 .1000 .70² = 4,597 �/��²

 Nilai β1

β1 = 0,85 −

= 0,85 −

= 0,871

��^′− 28

7 × 0,05 25 − 28

× 0,05 7

 Nilai ρ

0,85 × ��^′ 2 × ��

ρ = �� × (1 − √1 − 0,85× ��^′) 0,85 × 25

= 240 × (1 − √1 − 2 × 4,597 0,85 × 25)

= 0,0218

 Nilai ρb

0,85 × ��′× β1 600

ρb = �� × 600 + ��

0,85 × 25 × 0,87 600

= 240 × 600 + 240

= 0,0550

 Nilai ρmaks

ρmaks = 0,75 × ρb = 0,75 × 0,0550 = 0,0412

 Nilai ρmin

1,4 1,4

ρmin = �� = 240 = 0,00583

(15)

Maka :ρmin≤ ρ ≤ ρmaks

 �� = � × � × �

= 0,0218 × 1000 × 70

= 1526 ��²

 �� = 0,0014 × � × ℎ

= 0.0014 × 1000 × 100

= 140 ��²

(16)

 �� = 3 × ℎ

= 3 × 100 �� = 300 ��

Kontrol

 �_{��} = � − �� −^�_�∅tulangan

= 100 − 20 − ( )14

= 73 �� 2

73 mm ≥ 70 mm

��. �� 1539 ��²× 240 �⁄��²

� = 0,85.��^′. � = 0,85 × 25 �⁄ × 1000 ��

= 17,831 ��

(17)

�� = ��. ��. (� − 2)�

= 1539 ��²× 240 �⁄ × (73 �� –17,831 ��

)

��² 2

= 23670250,92 ��

= 23,670 ��

�� = �. �� = 0,9 × 23,670 = 21,303 ��

21,303 kNm ≥ 20,275 kNm (AMAN)

 Tabel Penulangan Plat Atap Arah Y

Rn ρmin ρ ρmaks Tulangan Pokok (Tarik) Tulangan Susut Lapangan 3,378 0,00583 0,0154 0,0531 ∅19 − 250 (1134 ��²) ∅6 − 150 (188 ��²) Tumpuan 4,597 0,00583 0,0218 0,0531 ∅14 − 100 (1539 ��²) ∅6 − 150 (188 ��²)

(18)

III.3. Perencanaan Penulangan Pelat Lantai Arah X Beban rencana pelat atap arah x, qu = 12,092 kN/m²

Mu Tumpuan: 30,539 kNm Mu Lapangan : 31,728 kNm

1) Daerah Penulangan Lapangan Atap Arah Lx

h = 12 cm = 120 mm

b = 100 cm = 1000 mm Fc’ = 25 Mpa

Fy = 240 Mpa

 d rencana = 120 – 30 = 90 mm

 Nilai Rn

�� = ��

�. �. �²

31,728 × 10⁶

= 0,9 .1000 .90² = 4,352 �/��²

 Nilai β1

β1 = 0,85 −

= 0,85 −

= 0,871

��^′− 28

7 × 0,05 25 − 28

7 × 0,05

(19)

 Nilai ρ

0,85 × ��^′ 2 × ��

ρ = �� × (1 − √1 − 0,85× ��^′) 0,85 × 25

= 240 × (1 − √1 − 2 × 4,352 0,85 × 25)

= 0,0205

 Nilai ρb

0,85 × ��′× β1 600

ρb = �� × 600 + ��

0,85 × 25 × 0,87 600

= 240 × 600 + 240

= 0,0550

 Nilai ρmaks

ρmaks = 0,75 × ρb = 0,75 × 0,0550 = 0,0412

 Nilai ρmin

1,4 1,4

Luas Tulangan Tarik

 �� = � × � × �

= 0,0205 × 1000 × 90

= 1845 ��²

(20)

 �� = 0,0014 × � × ℎ

= 0.0014 × 1000 × 120

= 168 ��²

(21)

 �� = 3 × ℎ

= 3 × 120 �� = 360 ��

Kontrol

 �_{��} = � − �� −^�_�∅tulangan

= 120 − 20 − ( )16

= 92 �� 2

92 mm ≥ 90 mm

��. �� 2011 ��²× 240 �⁄��²

� = 0,85.��^′. � = 0,85 × 25 �⁄ × 1000 ��

= 22,712 ��

�� = ��. ��. (� − )� 2

= 2011 ��²× 240 �⁄ × (92 �� –22,712 ��

��² 2 )

= 38922020,16 ��

= 38,922 ��

�� = �. �� = 0,9 × 38,922 = 35,029 ��

35,029 kNm ≥ 31,728 kNm (AMAN)

(22)

2) Daerah Penulangan Tumpuan Lantai Arah Lx

h = 12 cm = 120 mm

b = 100 cm = 1000 mm fc’ = 25 Mpa

fy = 240 Mpa

 drencana = 120 - 30 = 90 mm

 Nilai Rn

�� = ��

�. �. �²

30,539 × 10⁶

= 0,9 .1000 .90² = 4,189 �/��²

 Nilai β1

β1 = 0,85 −

= 0,85 −

= 0,871

 Nilai ρ

��^′− 28

7 × 0,05 25 − 28

7 × 0,05

0,85 × ��^′ 2 × ��

ρ = �� × (1 − √1 − 0,85× ��^′) 0,85 × 25

= 240 × (1 − √1 − 2 × 4,189 0,85 × 25)

= 0,0196

(23)

 Nilai ρb

0,85 × ��′× β1 600

ρb = �� × 600 + ��

0,85 × 25 × 0,87 600

= 240 × 600 + 240

= 0,0550

 Nilai ρmaks

ρmaks = 0,75 × ρb = 0,75 × 0,0550 = 0,0412

 Nilai ρmin

1,4 1,4

 �� = � × � × �

= 0,0196 × 1000 × 90

= 1764 ��²

(24)

 �� = 0,0014 × � × ℎ

= 0.0014 × 1000 × 120

= 168 ��²

Jarak Max

 �� = 3 × ℎ

= 3 × 120 �� = 360 ��

Kontrol

 �_{��} = � − �� −^�_�∅tulangan

= 120 − 20 − ( )16

= 92 �� 2

(25)

��² Syarat Rasio Penulangan : daktual≥ drencana

92 mm ≥ 90 mm

��. �� 2011 ��²× 240 �⁄��²

� = 0,85.��^′. � = 0,85 × 25 �⁄ × 1000 ��

= 22,712 ��

�� = ��. ��. (� − 2)�

= 2011 ��²× 240 �⁄ × (92 �� –22,712 ��

)

��² 2

= 38922020,16 ��

= 38.922 ��

�� = �. �� = 0,9 × 38,922 = 35,029 ��

38,092 kNm ≥ 30,539 kNm (AMAN)

(26)

 Tabel Penulangan Plat Lantai Arah X

Rn ρmin ρ ρmaks Tulangan Pokok (Tarik) Tulangan Susut Lapangan 4,352 0,00583 0,0205 0,0531 ∅16 – 100 (2011 ��²) ∅6 − 150 (188 ��²) Tumpuan 4,189 0,00583 0,0196 0,0531 ∅16 − 100 (2011 ��²) ∅6 − 150 (188 ��²)

(27)

III.4. Perencanaan Penulangan Pelat Lantai Arah Y Beban rencana pelat atap arah x, qu = 12,092 kN/m²

Mu Tumpuan : 46,630 kNm Mu Lapangan : 33,537 kNm

1) Daerah Penulangan Lapangan Atap Arah Ly

h = 12 cm = 120 mm

b = 100 cm = 1000 mm Fc’ = 25 Mpa

Fy = 240 Mpa

 d rencana = 120 – 30 = 90 mm

 Nilai Rn

�� = ��

�. �. �²

33,537 × 10⁶

= 0,9 .1000 .90² = 4,6 �/��²

 Nilai β1

β1 = 0,85 −

= 0,85 −

= 0,871

��^′− 28

7 × 0,05 25 − 28

7 × 0,05

(28)

0,85 × �� 2 × ��

ρ = �� × (1 − √1 − 0,85× ��^′) 0,85 × 25

= 240 × (1 − √1 − 2 × 4,6 0,85 × 25)

= 0,0218

 Nilai ρb

0,85 × ��′× β1 600

ρb = �� × 600 + ��

0,85 × 25 × 0,87 600

= 240 × 600 + 240

= 0,0550

 Nilai ρmaks

ρmaks = 0,75 × ρb = 0,75 × 0,0550 = 0,0412

 Nilai ρmin

1,4 1,4

 �� = � × � × �

= 0,0218 × 1000 × 90

= 1962 ��²

Dipakai tulangan∅14 – 75 (As 2053mm²)

(29)

 �� = 0,0014 × � × ℎ

= 0.0014 × 1000 × 120

= 168 ��²

Jarak Max

 �� = 3 × ℎ

= 3 × 120 �� = 360 ��

(30)

��²

 �_{��} = � − �� −^�_�∅tulangan

= 120 − 20 − ( )14

= 93 �� 2

93 mm ≥ 90 mm

��. �� 2053 ��²× 240 �⁄��²

� = 0,85.��^′. � = 0,85 × 25 �⁄ × 1000 ��

= 23,186 ��

�� = ��. ��. (� − 2)�

= 2053 ��²× 240 �⁄ × (93 �� –23,186 ��

)

��² 2

= 40110857,04 ��

= 40,110 ��

�� = �. �� = 0,9 × 40,110 = 36,099 ��

36,099 kNm ≥ 33,537 kNm (AMAN)

(31)

2) Daerah Penulangan Tumpuan Lantai Arah Ly

h = 12 cm = 120 mm

b = 100 cm = 1000 mm fc’ = 25 Mpa

fy = 240 Mpa

 drencana = 120 - 30 = 90 mm

 Nilai Rn

�� = ��

�. �. �²

46,630 × 10⁶

= 0,9 .1000 .90² = 6,396 �/��²

 Nilai β1

β1 = 0,85 −

= 0,85 −

= 0,871

 Nilai ρ

��^′− 28

7 × 0,05 25 − 28

7 × 0,05

0,85 × ��^′ 2 × ��

ρ = �� × (1 − √1 − 0,85× ��^′) 0,85 × 25

= 240 × (1 − √1 − 2 × 6,396 0,85 × 25)

= 0,0326

 Nilai ρb

0,85 × ��′× β1 600

ρb = �� × 600 + ��

0,85 × 25 × 0,87 600

= 240 × 600 + 240

= 0,0550

 Nilai ρmaks

ρmaks= 0,75 × ρb = 0,75 × 0,0550 = 0,0412

 Nilai ρmin

1,4 1,4

0,00583 ≤ 0,0326 ≤ 0,0412

(32)

 �� = � × � × �

= 0,0326 × 1000 × 90

= 2934 ��²

 �� = 0,0014 × � × ℎ

= 0.0014 × 1000 × 120

= 168 ��²

(33)

 �� = 3 × ℎ

= 3 × 120 �� = 360 ��

Kontrol

 �_{��} = � − �� −^�_�∅tulangan

= 120 − 20 − ( )14

= 93 �� 2

93 mm ≥ 90 mm

��. �� 3079 ��²× 240 �⁄��²

� = 0,85.��^′. � = 0,85 × 25 �⁄ × 1000 ��

= 30,774 ��

�� = ��. ��. (� − 2)�

= 3079 ��²× 240 �⁄ × (93 �� –30,774 ��

��² 2 )

= 57352902,48 ��

= 57,352 ��

�� = �. �� = 0,9 × 57,352 = 51,616 ��

51,616 kNm ≥ 46,630 kNm (AMAN)

(34)

 Tabel Penulangan Plat Lantai Arah Y

Rn ρmin ρ ρmaks Tulangan Pokok (Tarik) Tulangan Susut Lapangan 4,6 0,00583 0,0218 0,0531 ∅14 − 75 (2053 ��²) ∅6 − 150 (188 ��²) Tumpuan 6,396 0,00583 0,0181 0,0326 ∅14 − 50 (3079 ��²) ∅6 − 150 (188 ��²)