Challenge 2
Solution:
Aim:
- To determine the design loads on structure per Working Stress Method (WSM) and Limit State Method (LSM) of design.
- To determine the moment of resistance of the section per m width of slab and to classify the section as an under-reinforced, over-reinforced, or balanced section.
Introduction:
It is given that; a structure is subjected to the following loads.
DL-100kN
SIDL-50kN
LL-200kN
WL- ±150kN
EL- ±200KN
A slab has the following data.
Slab depth –150mm
Dia. of bar-8mm
Spacing of bar-125mm
Clear Cover -20mm
Fy- 415MPa
Fck- 30MPa
Effective depth (d) = 150-20-(8/2) =126 mm
Calculations and Explanations:
LSD Method:
Load Combinations for Design Loads:
- 5 (DL+WL) = 1.5 (100+50+150) =450kN
- 9 DL+1.5 WL = 0.9 x 150 + 1.5 x 150 =360 kN
- 2 (DL+LL+WL) = 1.2 (150+200+150) = 600 kN
- 5 (DL+EL) = 1.5 (150+200) = 525 kN
- 9 DL + 1.5 EL =0.9 x 150 +1.5 x 200 = 435 kN
- 2 (DL+LL+EL) = 1.2 (150+200+200) = 660 kN
- 5 (DL+LL) = 1.5 (150+200) = 525 kN
Design Load = Characteristic load x Partial safety factor
WD = 660 kN x 1.5 =990 kN
Characteristics load on serviceability:
- 0 (DL+LL) = 150+200=350 kN
- 0 (DL+QL) = 150+200 =350 kN
- 0 DL + 0.8 (LL+QL) = 150 + 0.8 (200+200) =470kN
Characteristic Load = 470 kN
Moment of Resistance determination:
Given data:
- Slab depth –150mm
- of bar-8mm
- Spacing of bar-125mm
- Clear Cover -20mm
- Fy- 415MPa
- Fck- 30MPa
- Effective depth (d) = 150-20-(8/2) =126 mm
Steel area in Strip, Ast = (1000/125)x (3.14/4)x8x8)=402.123 mm2
Neutral axis depth, Xu = (0.87 x 415x 402.123)/(0.36 x 30x 1000) = 13.443mm
Moment of Resistance Mr= 0.87 fy Ast (d –(1.015 fy Ast/1000 fck))
= 0.87 x 415x 402.123 (126 – (1.015 x415x402.123/1000 x 30))
Mr = 17.47 kNm
P = (100/d) ((3.14/4)xdxd) = (100/126)(0.785x8x8/125)=0.3189
Pt, lim=41.38x(Xu max/d)xfck/fy
Pt, lim =41.38x0.4791x30/415=1.433
P<Pt, lim, hence the section is Under reinforced section.
Results:
Design Load=990KN
Mr = 17.47 kNm, the section is the Under-reinforced section.