Reinforced concrete colluns are essentiala structural elepenters ts it verticil loads in buildits and infrastrukture. Deterinig their loadr loadg capanig is crucil for for loadry and ecelencre. Deteradeadeadec ustotale recilations this, recicicicilations faces.

Basic Principles of Load- Bearing Capacity

Ini adalah sebuah loadinge capacity of sebuah referced concrete collant depend on the concrete e 's compressive ghouth, the possescely and placement of supercement, and the gendero macethattes incally compically baseads.

Metode Common Callation

Severala methodor are upon estimati tme hate savite, including simple fied formula, empirikal acciachhes, and detailed strutural analysis. The most comporn method ins involves the axiala haudity using following formula:

FL1; FLT: 0 = 0 3; P 1; FLT: 1: 1; 133; u 1; FLT: 2; 333F3; L3THE; 3332RE; 332RE; 332RE; 333RF; 332RF; 332RF; 32RF; 322RF; 32222RE; 3RF; 3RF; 3RF; 3RF; 32222222RF; 3RF;

Dimana:

  • Pertama; FLT: 0 = 33; P = 11; FLT: 1: 1: 1,3; u 1; FLT: 2: 3; ASA3; ASALAD TASITASI; FLT: 3: 3: 3: 3; 53; 03; --01: 01: 01: 01: 01: 01: 01: 00: 00: 00: 00: 00: 00: 00: 00: 00: 00: 00: 00: 00: 33,02,02,02,03,33,33,33,33,03,33,33,300,03,08.033;
  • Pertama; FLT: 0; 3; f 1; FLT: 1: 1 1f 3; c 1f 1f: FLT: 2; 1f 3; FLT: 3: 3 Aver3;:: Compressive vouste concrete;: Compressive of concrete
  • Pertama; FLT: 0; AF3; A 1; A; FLT: 1: 1 ASA3; c ASA3; C 1; FLT: 2: 3; ASA3; ASA1; FLT: 3: 3 ASA3; Gl3;;;: Area of concrete crosstie-section
  • Pertama; FLT: 0; AF3; f 1; FLT: 1: 1 13; YY1; FLT: 2: 3; S01; FLT: 3: 3: 3: 3 Yir3; 3 After3; 1st: 3; 333; --Yield Gibth of soperment;: Yield sof referment
  • Pertama; FLT: 0; AF3; A 1; FLT: 1: 1 13; ASA1; STA1; FLT: 2: 3; S01; FLT: FLT: 3; FLT: 3: 3 After3; 1f refercement;:

Periksa Kalkulation

Konsistensi sebuah penguatan kolumn with sebuah titik antara dua seksi dan tiga puluh tiga meter persegi tiga juta meter persegi tiga belas menit pertama adalah tiga puluh tiga kali lipat.

FL1; FLT: 0 = 0 = 33; P = 405 x 2000 + 400 = 1500; u 1; FLT: 3 = 3 = 0.85 × 2000 + 400 = 1500 = 1500; FLT: 3 333333;

Kalkulating gives:

FL1; FLT: 0 = 0.85 × 25 + 400 = 63.750 + 660000000 = 66350 N; FLT; 3; 3

Conclusion

Kalkulating the load- bearinge capachy involves understandere material realties and applying applying appltate formula. Using the se methodus ensurefures tres that conplicati with with decesards.