Challenge 2
1)
AIM : To design load on the structure for given loads is to be found out as per working stress method as well as Limit state method.
PROCEDURE:
Given: Dead load=100KN, Super imposed load=50KN, Live load=50KN Wind load=+/- 150 KN, Earthquake load=+/- 200KN
Design load: No factor of safety for is used for load in working stress method .
So,
Design Dead load : 100 kN
Design Super imposed load : 50 kN
Design Live load : 50 kN
Design Wind load : +/- 150 kN
Design Earthquake load : +/- 200kN
In this case allowable stress in concrete is reduced by 67% of its characterstic strength.
Thus only 33% of characterstic strength is considered i design
Design loads as per Limit state method for Limit state of collapse . ( Strength case)
Dl LL Super imposed load ( LL) WL EQ
100 200 50 +/- 150 +/- 200
DL+LL 1.5 DL 1.5LL 1.5 SL
DL+QL( QL=WL/EQ) 1.5 DL 1.5 1.5
DL+QL 0.9 DL 1.5 WL 1.5 EQ
DL+LL+QL 1.2 DL 1.2 LL 1.2 SL 1.2WL 1.2 EQ
Design Dead load : 1.5x100 = 150 kN
Design Super imposed load
Case-1 : 1.5( DL+LL+SL) = 1.5( 100+200+50) = 525 kN
Case-2 : 1.5 (DL+ QL) = 1.5( 100+150) = 500kN Or 1.5 ( 100+200) = 600 kN
Case 3 : 0.9 DL+ 1.5 QL = 0.9x 100+1.5x150 = 315 kN Or 0.9x100 + 1.5x200 = 390 kN
Case 4 : 1.2( DL+LL+SL+QL) = 1.2 ( 100+200+50+200) = 660 kN
Design loads as per Limit state method for Limit state of Serviceability.
Case 1: : 1.0 (DL +LL+SL) = 1.0( 100+200+50) = 350 kN
Case 2 : : 1.0( DL+QL) = 1.0 ( 100+ 200) = 300 kN
Case 3 : 1.0 DL + 0.8 ( LL+QL) = 1.0 x200 + 0.8 ( 200+50+ 200) = 560 kN
Design load is maximum of above cases which is 660 kN Apart from above material factor of safety is 1.5 for concrete and 1.15 for steel
RESULT : Thus desired design load have been found out as above
2)
AIM: Moment of resistance of given section is required to be calculated and section is also required to be classified.
PROCEDURE:
Given:
Slab depth = 150 mm,
Yield strength of steel (fy)= 415 MPa
Dia of bar (ϕ)= 20 mm
Strength of concrete (fck) = 30 MPa
Spacing of bar (s) = 125 mm
Clear cover = 20 mm
Effective depth (d) = 150 -20 -8/2 = 126 mm
Area of one bar = (π/4)×ϕ^2
No. of bar in 1 m strip = 1000 mm/ s
Area of steel in strip (Ast) = 401.92mm^2
Percentage of steel (p) (`A_(st)/bd) * 100 = 2512/1000*120 = 0.25 %
Limiting % of steel, p,lim = 41.38x0.4791x30/415 = 1.433 %
p < pt,lim, hence section is under reinforced.
Moment of resistance (Mr) = 0.87fy×d(100×πϕ^2/4s)(1−(1.015fy×πϕ^2/fck×4sd)
= 0.87x 415x 401.92x 126[1) [ 1-(1.015 x 415 x 3.14 x 64/30 x 4 x 125 x 126]
= 18284265.216 [ 0.955]
= 17465349 N mm
= 17.465 kN m
RESULT:
Moment of resistance = 17.465 kN m