Challenge 2
Q-01: 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.
Answer:
Given That,
Dead Load (D.L.1) = 100 KN
Super Imposed Dead Load (D.L.2) = 50 KN
Total Dead Load (D.L) = D.L.1 + D.L.2
= 100 + 50
= 150 KN
Live Load (L.L) = 200 KN
Wind Load (W.L) = +/- 150 KN
Earthquake Load (E.L) = +/- 200 KN
Solution,
AS PER WORKING STRESS METHOD (WSM) :-
1- D.L + L.L = 150 + 200
= 350 KN
2.a- D.L + 0.8 * L.L + 0.8 * W.L = 150 + 0.8 * 200 + 0.8 * 150
= 430 KN
2.b- D.L + 0.8 * L.L - 0.8 * W.L = 150 + 0.8 * 200 - 0.8 * 150
= 190 KN
3.a- D.L + W.L = 150 + 150
= 300 KN
3.b- D.L - W.L = 150 – 150
= 0 KN
4.a- D.L + 0.8 * L.L + 0.8 * E.L = 150 + 0.8 * 200 + 0.8 * 200
= 470 KN
4.b- D.L + 0.8 * L.L - 0.8 * E.L = 150 + 0.8 * 200 - 0.8 * 200
= 150 KN
5.a- D.L + E.L = 150 + 200
= 350 KN
5.b- D.L - E.L = 150 – 200
= -50 KN
# NOTE: Negative sign indicates that the force acting on opposite direction.
AS PER LIMIT STATE METHOD (LSM) :-
LIMIT STATE OF COLLAPSE:
1- 1.5 * (D.L + L.L) = 1.5 * (150 + 200)
= 525 KN
2.a- 1.5 * (D.L + E.L) = 1.5 * (150 + 200)
= 525 KN
2.b- 1.5 * (D.L - E.L) = 1.5 * (150 - 200)
= -75 KN
3.a- (0.9 * D.L + 1.5 * E.L) = (0.9 * 150 + 1.5 * 200)
= 435 KN
3.b- (0.9 * D.L - 1.5 * E.L) = (0.9 * 150 - 1.5 * 200)
= -165 KN
4.a- 1.2 * (D.L + L.L + E.L) = 1.2 * (150 + 200 + 200)
= 660 KN
4.b- - 1.2 * (D.L + L.L - E.L) = 1.2 * (150 + 200 - 200)
= 180 KN
LIMIT STATE OF SERVEACIBILITY:
1- D.L + L.L = 150 + 200
= 350 KN
2.a- D.L + E.L = 150 + 200
= 350 KN
2.b- D.L - E.L = 150 – 200
= -50 KN
3.a- D.L + 0.8 * L.L + 0.8 * E.L = 150 + 0.8 * 200 + 0.8 * 200
= 470 KN
3.b- D.L + 0.8 * L.L - 0.8 * E.L = 150 + 0.8 * 200 - 0.8 * 200
= 150 KN
# NOTE: Negative sign indicates that the force acting on the opposite direction.
Q-02: 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.
Answer:
Given That,
D = 150 mm, ɵ = 8 mm, s = 125 mm, c.c = 20 mm, Fck = 30 KN, Fy = 415 KN
Effective Depth (d) = D – Clear Cover (c.c) - ɵ/2
d = 150 – 20 – 8/2
d = 126 mm
Area of Steel (Ast) = [(1000 / s) * {(Π / 4 * (ɵ^2)}]
Ast = [(1000 / 125) * {(Π / 4 * (8 ^ 2)}]
Ast = 402.12 mm^2
Here,
Xu, Lim = K * d
= 0.48 * 126
= 60.48 mm
Xu = (0.87 * Fy * Ast) / (0.36 * Fck * b)
= (0.87 * 415 * 402.12) / (0.36 * 30 * 1000)
= 13.44 mm
Therefore,
Xu, Lim > Xu ; Section is Under Reinforced
Now,
Moment of Resistance (M.O.R) = 0.87 * Fy * Ast * (d – 0.42 * Xu)
= 0.87 * 415 * 402.12 (126 – 0.42 * 13.44)
= 17.47 * 10^6 N-mm
= 17.47 KN-m (Answer)