mendesain balok beton bertulang berdasarkan momen ultimate

1. CONTOH SOAL BETON
MENGHITUNG MOMEN ULTIMATE
PART 2
SHALEH AFIF HASIBUAN
TEKNIK SIPIL USU 2016
SEMOGA BERMANFAAT 

2. Diketahui sebuah simple beam dengan panjang 7 m , dibebanin dead load sebesar
25Kn/m (belum termasuk berat sendiri) dan live load sebesar 20Kn/m.
desainlah sebuah balok, jika fy = 420, mutu beton fc = 25mpa, Es = 200.000 mpa.

3. Hitung Momen ultimate

4. 𝑚𝑒𝑛𝑔ℎ𝑖𝑡𝑢𝑛𝑔 𝑏𝑒𝑟𝑎𝑡 𝑠𝑒𝑛𝑑𝑖𝑟𝑖, 𝑑𝑖 𝑎𝑠𝑢𝑚𝑠𝑖𝑘𝑎𝑛 𝑏𝑒𝑟𝑎𝑡 𝑠𝑒𝑛𝑑𝑖𝑟𝑖 35% 𝑑𝑎𝑟𝑖 𝑑𝑒𝑎𝑑 𝑙𝑜𝑎𝑑
𝑚𝑎𝑘𝑎 𝑏𝑒𝑟𝑎𝑡 𝑠𝑒𝑛𝑑𝑖𝑟𝑖 = 35% 𝑥 25 𝐾𝑛 𝑚⁄ = 8,75 𝐾𝑛 𝑚⁄
𝑤 = 25 + 8,75 = 33,75 𝐾𝑛 𝑚⁄ , 𝑤 = 20 𝐾𝑛 𝑚⁄
𝑊 = 1,2𝐷 + 1,6𝐿 = 1,2 33,75 + 1,6 20 = 72,5 𝐾𝑛 𝑚⁄
𝑀 =
1
8
𝑞𝑙 =
1
8
72,5 𝑥 7 = 444𝐾𝑛𝑚
𝑀 ≥ 𝑀
∅𝑀 = 0,9𝐴 𝑓 𝑑 −
𝑎
2
; 𝑎 =
𝐴 𝑓
0,85𝑓 𝑏
; 𝜌 =
𝐴
𝑏𝑑
; 𝐴 = 𝜌𝑏𝑑; 𝑎 =
𝜌𝑏𝑑𝑓𝑦
0,85𝑓 𝑏
∅𝑀 = 0,9𝜌𝑏𝑑𝑓 𝑑 −
𝜌𝑑𝑓
2𝑥0,85𝑓
;
∅𝑀
𝑏𝑑
= 𝜌𝑓𝑦 1 −
𝜌𝑓
1,7𝑓
𝜌 𝑑𝑖𝑝𝑖𝑙𝑖ℎ 𝑠𝑒𝑘𝑖𝑡𝑎𝑟 0,25 − 0,5 𝜌 ; 𝜌 =
0,85𝛽𝑓
𝑓
600
600 + 𝑓

5. Reinforced concrete Mechanics and Design
by Wight MacGregor
𝜌 =
0,85𝑥0,85𝑥25
420
600
600 + 420
= 0,025; 𝜌 = 0,0031 − 0,0035
𝑎𝑚𝑏𝑖𝑙 𝜌 = 0,3𝜌 ; 𝜌 = 0,3𝑥0,025 = 0,0075
∅𝑀
𝑏𝑑
= 𝜌𝑓𝑦 1 −
𝜌𝑓
1,7𝑓
0,9𝑥444𝑥10
4𝑏
= 0,0075𝑥420 1 −
0,0075𝑥420
1,7𝑥25
𝑏 = 326,64 𝑚𝑚 ; 𝑏 = 350 𝑚𝑚 ; 𝑑 = 700𝑚𝑚; 𝑏𝑒𝑟𝑎𝑡 𝑠𝑒𝑛𝑑𝑖𝑟𝑖 = 𝛾𝑏ℎ = 23,536𝑥0,35𝑥0,7 = 5,76𝐾𝑛/𝑚 ≤ 8,75(𝑜𝑘)
Biasanya d = (1,5-2)b; ambil d= 2b

6. 𝐴 = 𝜌𝑏𝑑 = 0,0075𝑥350𝑥700 = 1837,5𝑚𝑚
𝑐𝑜𝑏𝑎 𝑝𝑎𝑘𝑎𝑖 4 𝑡𝑢𝑙𝑎𝑛𝑔𝑎𝑛 𝑏𝑎𝑤𝑎ℎ 3 𝑡𝑢𝑙𝑎𝑛𝑔𝑎𝑛 𝑎𝑡𝑎𝑠
n = 4
𝐴 = 4𝑥
1
4
𝑥𝜋𝑥𝐷 ; 𝐷 ≥ 24 𝑚𝑚 𝑎𝑚𝑏𝑖𝑙 𝐷 = 28 𝑚𝑚

7. Reinforced concrete Mechanics and Design
by Wight MacGregor

8. 𝐴 = 4𝑥
1
4
𝑥𝜋𝑥28 = 2463𝑚𝑚
𝐴 = 3𝑥
1
4
𝑥𝜋𝑥14 = 461𝑚𝑚
𝑇 = 𝐴 𝑓 = 2463𝑥420 = 1034460𝑁
𝑐 = 461𝑥 420 − 0,85x25 = 183823N
𝑇 = 𝑐 + 𝑐
1034460 = 0,85𝑥25𝑥350𝑥𝑎 + 183823
𝑎 = 114 𝑚𝑚
𝑀 = 𝑐 𝑑 −
𝑎
2
+ 𝑐 𝑑 − 𝑑
𝑀 = 0,85𝑥25𝑥350𝑥114𝑥 650 −
114
2
+ 183823 650 − 50
𝑀 = 613𝐾𝑛𝑚
∅𝑀 = 0,9𝑥613 = 551𝐾𝑛𝑚 > 444𝐾𝑛𝑚 (𝑜𝑘)