Challenge 2
Q-1 A structure is subjected to Dead load of 100 kN, Super imposed dead load of 50 kN, Live load of 200 kN, Wind load of +/- 150 kN and Earthquake load of +/- 200 kN. Determine the design loads on structure per Working Stress Method(WSM) and Limit State Method(LSM) of design.
AIM AND INTRODUCTION : Design load on the structure for given loads is to be found out as per workinf stress method as well as Limit state method.
PROCEDURE IN DETAIL:
Given-
Dead load : 100 kN
Super imposed load : 50 kN
Live load : 50 kN
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
---------------------------------xxxxxxxxxxxx------------ END OD Q-1 -----------------xxxxxxxxxxxxxxxx---------------------------
Q-2. Consider a 150 mm thick slab with 8mm bottom reinforcement spaced at 125mm. Clear cover to bottom rebar is 20 mm. Grade of concrete is M30 and grade of steel is Fe415. Determine the moment of resistance of the section per m width of slab. Also, classify the section as under-reinforced, over-reinforced or balanced section.
AIM AND INTRODUCTION : Moment of resistance of given section is required to be calculated and section is also required to be classified.
PROCEDIRE IN DETAIL :
Slab depth = 150 mm,
Yield strength of steel `f_y`= 415 MPa
DDia of bar `phi`= 20 mm
Strength of concrete `f_(ck)` = 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 = `pi/4xxphi^2`
No. of bar in 1 m strip = 1000 mm/ s
Area of steel in strip, `A_(st)` = = (1000/125)*`(pi/4* 8^2)`
= 401.92 `mm^2`
Percentage of steel, p (`A_(st)/bd) * 100 = 2512/1000*120 = 0.25 %
Limiting % of steel, p,lim = 41.38*`(x_u,max//d)xxf_(ck)/f_y`
= 41.38x0.4791x30/415
= 1.433 %
p < `p_t,lim`, hence section is under reinforced.
Moment of resistance, M_r = `0.87f_yd(100xxpiccphi^2//4s)(1-(1.015f_yxxpiphi^2//f_(ck)xx4sd)`
= 0.87x415x401.92x 126[1) [ 1-(1.015x415x3.14x64/30x4x125x126]
= 18284265.216 [ 0.955]
= 17465349 N mm
= 17.465 kN m
``