# Prestressed Concrete Structures Questions and Answers – Design of Post Tensioned Beams

This set of Prestressed Concrete Structures Multiple Choice Questions & Answers (MCQs) focuses on “Design of Post Tensioned Beams”.

1. Calculate ultimate moment and shear of effective span is 30m, live load is 9kn/m, dead load excluding self weight is 2kn/m, load factors for dead load is 1.4 for live load is 1.6 cube strength of concrete fcu is 50n/mm2 cube strength at transfer is fci is 35n/mm2, tensile strength of concrete Ec is 34kn/mm2 loss ratio ɳ is 0.85 and 8mm diameter high tensile strength fpu is 1500n/mm2 are available for use and the modulus of elasticity of high tensile wires is 200kn/mm2?
a) 340 and 450kn
b) 240 and 340kn
c) 140 and 240kn
d) 100 and 200kn

Explanation: Wmin/Wud = (50x2400x9.81×0.125x25x30/50×106x0.852) = 0.31
Ultimate load excluding the factored selfweight = (1.4×2)+(1.6×9) = 17.2kn/m, Wud = 17.2/1-1.4×0.31 = 30KN/M, Wmin = (0.31×30) = 9.3kn/m, Ultimate moment, Mu = (0.125x30x302) = 3400knm, Ultimate shear, Vu = (0.5x30x30) = 450kn.

2. Find cross-sectional dimensions thickness of web if hf/d ratio is 0.23 and bw/b ratio is 0.25 and b is 0.5d?
a) 100mm
b) 110mm
c) 120mm
d) 30mm

Explanation: hf/d =0.23 and bw/b = 0.25 and b = 0.5d,
Mu = 0.10fcubd2 d = (3400×106/0.10x50x0.5)1/3 = 1130mm, h = (1130/0.85) = 1300, b = 600mm, hf = (0.2×1130) = 250mm, adopt an effective depth, d = 1150mm, thickness of web, bw = (0.6vu/fth) = (0.6x450x103/1.7×1300) = 120mm.

3. Calculate working moment if design working load is 19.8kn/m covered over a span of 30m (actual self weight of girder is 8.8kn/m)?
a) 3000
b) 2000
c) 4340
d) 2230

Explanation: Actual self weight of the beam and the girder = 8.8kn/m, span = 30m
Minimum moment Mmin = 990knm, Design working load = 19.8kn/m,
Working moment Md = (0.125×19.8×302) = 2230knm.

4. Find the Permissible stresses and range of stresses for class 1 structure fcu = 50n/mm2, fck = 35n/mm2 according to BS: 8110 recommendations for fcu = 50n/mm2 and fci = 35n/mm2,fct = 0.5fci = 17.5n/mm2?
a) 16.5n/mm2
b) 12.56n/mm2
c) 13.56n/mm2
d) 12.00n/mm2

Explanation: fcu = 50n/mm2,fck = 35n/mm2 according to BS: 8110 recommendations for fcu = 50n/mm2 and fci = 35n/mm2, fct = 0.5fci = 17.5n/mm2For class 1 structure fu = htw = 0, fbr = (ɳfct-ftw) = (0.85×17.5) = 15n/mm2, fcw = 0.33fcu = (0.33×50) = 16.5n/mm2, fcu = (fcw-ɳfu) = 16.5n/mm2.

5. Find prestressing force if area is 36.75mm2 of eccentricity 580 given finf is 26.5kn/m and zb is 99×106?
a) 405
b) 308
c) 453
d) 206

Explanation: Area = 36.75mm2, e = 580, finf = 26.5kn/m, zb = 99×106
p = (AfinfZb/Zb+Ae) =(367500×26.5x99x106/(99×106)+(367500×580)) = 308x104kn/m2.
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6. Find force in cable using Freyssinet cables 12-8mm diameter and stressed to 1100n/mm2 of eccentricity 50 and the prestressing force is given as 1000n/mm2?
a) 660kn
b) 234kn
c) 300kn
d) 230kn

Explanation: 12 diameter, stress = 1100n/mm2, e = 50, prestressing force = 1000n/mm2
Force in each cable = ((12x50x1100)/1000)) = 660kn.

7. Find ratio for ultimate flexural strength at the centre – span section given that Aps = 3000mm2, d = 1150mm, fcu = 50n/mm2, bw = 150mm, fpu = 1500n/mm2, b = 600mm, ht = 250mm, design ultimate moment mud = 3400knm?
a) 9.5
b) 0.23
c) 6.7
d) 3.4

Explanation: Aps = 3000mm2, d = 1150mm, fcu = 50n/mm2, bw = 150mm, fpu = 1500n/mm2, b = 600mm, ht = 250mm, design ultimate moment mud = 3400knm, according to BS: 8110-1985, Aps = (Apw+Apf) = Apf = 0.45×50(600-150)(250/1500) = 0.45xfcu(b-bw)(hf/fpu) = 1680mm2, Apw = (1300-1680) = 1320mm2, ratio(fpuApw/fcubwd) = (1500×1320/50x150x1150) = 0.23.

8. Calculate the slope of cable section at support uncracked in flexure given that eccentricity is 410, length is 30m and stress induced is 1000?
a) 0.0547
b) 2.456
c) 0.0234
d) 0.0123

Explanation: e = 410, length = 30m, stress induced = 1000
Slope of cable θ = (4e/l) = ((4×410)/(30×1000)) = 0.0547.

9. Calculate the span section cracked in flexure (M=M0) Fcp = 23.4n/mm2, zb is 99×106 and stress induced is 1000?
a) 1200kn
b) 1850kn
c) 2300kn
d) 4300kn

Explanation: Fcp = 23.4n/mm2, zb is 99×106, stress is 1000
m0 = (0.8fcpZb) = (0.8 x 23.4 x (99×106/1000)) = 1850knm.

10. Find resultant maximum long term deflection if ϕ is 2.6, αy is 38.5mm, αg is 46mm, αp is 74.7mm?
a) 95mm
b) 35mm
c) 55mm
d) 20mm

Explanation: Ece = (Ec/1+ϕ) = (Ec/2.6), ϕ = 2.6, αy = 38.5mm, αg = 46mm, αp = 74.7mm, resultant maximum long term deflection = (2.6×46)+38.5-(0.85×74.7) = 95mm which is less than the code limit (span/250) = 120mm, ɳ = 0.85.

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