🏗️✨ SITE ENGINEER'S CHALLENGE! ┃ የሳይት መሐንዲስ ፈተና! ✨🏗️

📐 Footing 1.5 m × 1.5 m × 0.45 m with 12 mm Ø @ 150 mm c/c both ways ┃ 1.5 ሜ × 1.5 ሜ × 0.45 ሜ መሠረት፣ በሁለቱም አቅጣጫ 12 ሚሜ ብረት በ150 ሚሜ ልዩነት 🔩

📌 Cover: 50 mm ┃ ሽፋን፡ 50 ሚሜ

📌 Bend-up: 300 mm ┃ የታጠፈ ክፍል፡ 300 ሚሜ

📌 Concrete: M20 ┃ ኮንክሪት፡ M20

🧮 Can you calculate? ┃ ማስላት ትችላላችሁ?

1️⃣ 🧱 Volume of concrete ┃ የኮንክሪት መጠን

2️⃣ 🔢 Number of bars each way ┃ በእያንዳንዱ አቅጣጫ የብረት ብዛት

3️⃣ 📏 Cutting length of one bar ┃ የአንድ ብረት የመቁረጫ ርዝመት

4️⃣ ⚖️ Total weight of steel ┃ የጠቅላላ ብረት ክብደት

⬅️ Drop your answers below 👇 ┃ መልሳችሁን ከታች ጻፉ 👇

✅ Solution coming soon! ┃ መፍትሄው በቅርብ ይመጣል!
🔔 Follow for more civil engineering problems ┃ ለተጨማሪ የሲቪል ምህንድስና ጥያቄዎች ይከተሉን
👷‍♂️📚 Learn • Calculate • Build ┃ ተማር • አስላ • ገንባ 🧱

@ETCONP

4 comments

1.1.013m3
2.11+11=22
3.2m
4.39kg
1.5*1.5*0.5=1.125m3
1.5m/0.15mm+1=11*2(side)=22number of bar
D^2/162.5*L= 12^2/162*12=10.63kg
1,1.5 * 1.5 * 0.45=1.0125 m3
2,nobar 11 2side 11
3,2metres
4,total Wight =39.072kg
1, Volume of concrete =1.5*1.5*.45=1.0125m³
2, Number of bar =(L/cc)+1, L= 1.5-cover =1.5-.05=1.45,
=(1.45/0.15)+1=10.67~~11bars required in each side
3, cutting length = L(excluding cover)+bendup both side
=1.45+(0.3*2)=2.05m
4, Weight of steel = L(total)* Kg factor
1m Kg factor ∅12=(12²/162)=0.89
L(total)=no of bar * cutting length
=22*2.05=45.1m
W=45.1*0.89=40.139kg