Structural Design Calculations for a Concrete Beam and T-Bar

Q1)
a) Taking summation of moment at the right end to be 0
∑ M B = 0
RA*12 –(70*9 + 70*7+85*4) = 0
RA = 121.67 kN
Upward forces = 70 + 70 + 85 = 225 kN
RB = 225 – 121.67
= 103.33 kN
Mc = 121.67 *3 = 365.01 kNm
Mc = 121.67 *3 = 365.01 kNm
MD = 121.67 *5 – 70*2 = 468.35 kNm
ME = 103.33 *4 = 413.32 kNm
b) Permissible stress = 165 N/m
Zxx = 468.35∗106
165 = 2.8385 * 106 mm4
= 2838.48 cm3
Try 610 *229 *113 kg
c) Web = d/tw
= 543.1/11.2 = 48.49
ε = √ ( 235
355 ) = 0.8136

d/tw ≤ 72 ε ≤ 58.58
c = 228.2
tf = 17.3
9ε ≥ c
tf =228.2
17.3 =13.19
= 9*0.81367.3224
It fails tosatify the criteria in class 1, 2 and 3, therefore it belongs to class 4
d) Let E = 210 kN/mm2
Permissible difflection = 1/360 * 12000 = 33.33
Actual deflection = 5 w L3
384 EI = 5∗468.35∗12000
384∗87260∗210∗104
= 57.5067 > 33.33
Therefore not ok
Q2)
a)
Ixx = 14307 cm4
Iyy = 4849 cm4
A = 114 cm2
Zxx = 1099 cm3
Ryy = 6.52 cm
Rxx = 11.2 cm
D/T = 15.1
l/ryy = 3000/65.2 = 46
pc = 131.8 N/mm2
pbc = 165 N/mm2
Safe load = PC*AS
= 114∗102∗131.8
103 = 1502.52 kN
Checking combined bending and axial load
Fc/pc + fbc/pbc
= 386.42
131.8 + 241.91
165 = 4.398
From the checking the steel section chosen is not satisfactory
b) Yes, but now the direction of resistance will change to direct oppose the both actions
simultaneously.

Q3)
Thickness of slab = 150 mm – 30 – 10 = 110 mm
Span = 4000 -250 = 3750 mm
Concrete uni weight = 24 kN/m3
Self load = 0.15 * 24 = 3.6 kN/m2
D.L = 1.5 kN/m2
D.L = 5.1 kN/m2
L.L = 4 kN/m2
Total D.L = 5.1 * 3.75 = 19.125 kN/m
Total L.L = 4 * 3.75 = 15 kN/m
Wu = 1.2*19.125 + 1.6* 15 = 46.95 kN/m
+ve moment = + 46.95∗3.752
16 +41.265 kNm
-Ve moment = −46.95∗3.752
11 +60.02 kNm
b)
Applied span/d = 3750/110 = 34.09 mm
c) Shear capacity check
Vu = 1.15∗46.95 8 3.75
2 = 101.235
Vc = 2*√fy∗bd = 14.832