Nama : Riza Matholiul Umam NPM : 2240503160 Kelas : TS 04
Tugas 3 Struktur Beton II
1. Suatu balok beton bertulang penampang persegi menahan momen Mw = 146 KNm, fc' = 20 Mpa, fy = 400 MPa, b = 310 mm, h = 510 mm , d’ = 95 mm. Tentukan tulangan yang diperlukan ? Diketahui :
Mw = 146 KNm
fc' = 20 MPa
fy = 400 MPa
b = 310 mm
h = 510 mm
d' = 95 mm
n = 9
Fc = 9 MPa
Fs = 170 MPa
Ditanya : Tulangan yang diperlukan Penyelesaian :
0,8924
1,2960 Mpa
410 mm
67,5348 kN.m
1085,7265
78,4652 kN.m 0,3227
0,0085
1085,7265
(3D25 = 1472,6 mm^2) kN
1465,2701 249,0959 𝑚 = 𝐹𝑐
𝐹𝑐 +𝐹𝑠 𝑛
= 9
9 +170 9
=
j= 1 −13𝑚 = 1 −130,3227 =
𝜌 =1 2
𝑚 𝐹𝑐 𝐹𝑠 =1
2×0,3227× 9
170 =
𝑑 = ℎ − 100 = 510 − 100 =
𝐴𝑠1= 𝜌. 𝑏. 𝑑 =0,0085 (310) (410) =
𝐴𝑠1= 𝑀𝑤
𝐹𝑠. 𝑗. 𝑑 = 67,5348× 106 1700,8924(410)=
𝑀𝑤2= 𝑀𝑤− 𝑀𝑤1= 146 −67,5348=
𝑁𝐷2= 𝑁𝑇2= 𝑀𝑤2
(𝑑 − 𝑑′)=78,4652(10)3 (410 − 95) =
𝐴𝑠2=𝑁𝑇2
𝐹𝑠 =249,0959× 103
170 =
𝑀𝑤1= 𝑘. 𝑏. 𝑑2=1,2960 (310) (410)210−6= k=12𝐹𝑐 𝑚 𝑗 =12× 9 ×0,3227×0,8924=
𝑚𝑚2
𝑚𝑚2
𝑚𝑚2
( Tulangan Tarik ) (4D29 = 2642,1 mm^2)
132,3108 mm
45,68292683 Mpa < 170 mpa (ijin maksimum)
Rasio modulus elastisitas 2n dapat diterima :
Digunakan tulangan 6D36 (6107,2 mm^2)
2.Suatu balok beton bertulang persegi bertulangan rangkap, fc’ = 25 MPa, fy = 350 MPa, b = 320 mm, h = 675 mm, d’ = 95 mm, As’ = 4D25, As = 4D29+2D32. Tentukan Mw ?
Diketahui:
fc' = 25 MPa
fy = 350 MPa
b = 320 mm
h = 675 mm
d' = 95 mm
n = 9
Fc = 11,25 Mpa
Fs = 155 Mpa
As'= 4D25= 1963,5 mm^2
As= 4D29+2D32= 4250,5749 mm^2
d= h-100= 575 mm
Ditanya : Tentukan Mw Penyelesaian :
0,39512
0,86829
0,22231 MPa
0,01434 2550,9966
(Tulangan tekan) Mpa
2,5379
5773,4616 𝐴𝑠= 𝐴𝑠1+𝐴𝑠2 =1085,7265+1465,2701 =
md= 𝑚 𝑑 =0,3227 x410 =
𝑓𝑐1′ =𝐹𝑐(𝑚𝑑 − 𝑑′)
𝑚𝑑 =9(132,3108− 95) 132,3108 =
2. 𝑛. 𝑓𝑐1= 2(9)(2,5379 ) =
𝑚𝑚2
𝐴𝑠′ = 𝑁𝐷2
(2𝑛 − 1)𝑓𝑐1′ = 249,0959(10)3
2 9 − 1 (2,5379)= 𝑚𝑚2
4.1 4. 𝜋. 252=
4.1
4. 𝜋. 292+ 2.1 4. 𝜋. 322=
𝑚 = 𝐹𝑐
𝐹𝑐 +𝐹𝑠 𝑛
= 11,25 11,25 +155
9
=
j = 1 −13𝑚 = 1 −130,395122 =
k =12𝐹𝑐 𝑚 𝑗 =12× 9 ×0,395122×0,868293 =
𝜌 =1
2 𝑚 𝐹𝑐
𝐹𝑠 =1
2×0,395122× 11,25
155 =
Menentukan sumbu berat penampang transformasi:
x= 231,566 mm atau x= -679,28 mm
digunakan x = 231,566 mm Bandingkan 2nfc1 terhadap fs ijin
6,63468136 MPa
119,4242644 MPa 155 MPa Penampang transformasi dapat diterima
Menentukan kopel momen lentur dalam dengan menggunakan cara kopel momen:
kN
kN
kN
kN.m
Pemeriksaan momen lentur dalam dengan menggunakan menggunaka cara penampang transformasi:
mm^4
313,798839 kN.m Atau
323,905936 kN.m 6459118419
658,8391033
416,818458
221,4618285
313,7986564 𝑓𝑐1′ =𝐹𝑐(𝑥 − 𝑑′)
𝑥 =11,25(231,5658− 95)
231,5658 =
2. 𝑛. 𝑓𝑐1′ = 2(9)(6,634681357) = 1
2𝑏𝑥2+ 2𝑛 − 1 𝐴𝑠′ 𝑥 − 𝑑′ = 𝑛(𝐴𝑠)(𝑑 − 𝑥) 1
2320𝑥2 + 17 (1963,5) 𝑥 − 95 = 9(4250,575)(575 − 𝑥) 160𝑥2 + 33379,5 𝑥 − 95 = 38255,175(575 − 𝑥) 160𝑥2 + 33379,5𝑥 − 3171052,5 = 21996725,625 − 38255,175𝑥
160𝑥2+ 71634,675𝑥 − 25167778,125 = 0
<
𝑁𝐷2 = 2𝑛 − 1 × 𝐴𝑠′ × 𝑓𝑐1′ = 33379,5 ×6,634681357× 10−3=
𝑀𝑤= 𝑁𝐷1× 𝑑 −1
3𝑥 + 𝑁𝐷2× (𝑑 − 𝑑′) = ((416,8185× (575 −1
3 231,5658))) +(221,4618× 575 − 95 ) × 10−3= 𝑁𝑇 = 𝐹𝑠 × 𝐴𝑠 = 155 ×4250,575× 10−3=
𝑁𝐷1 =1
2× 𝐹𝑐 × 𝑏 × 𝑥 =1
2× 11,25 × 320 × 231,5658 × 10−3=
𝐼𝑐𝑟 =1
3× 𝑏 × 𝑥3+ 2𝑛 − 1 𝐴𝑠′(𝑥 − 𝑑′)2+ 𝑛 × 𝐴𝑠(𝑑 − 𝑥)2
𝐼𝑐𝑟 =1
3× 320 ×231,56583+ 2 × 9 − 1 × (1963,5) × (231,5658− 95)2+ 9 × (4250,575) × (575 −231,5658)2
𝑀𝑤 =𝐹𝑐(𝐼𝑐𝑟)
𝑥 =11,25 ×6459118419× 10−6
231,5658 =
𝑀𝑤 = 𝐹𝑠(𝐼𝑐𝑟)
𝑛(𝑑 − 𝑥) =155 ×6459118419× 10−6 9 × (575 − 231,5658) = 𝐼𝑐𝑟 =
