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 f_{cu} is 50n/mm^{2} cube strength at transfer is f_{ci} is 35n/mm^{2}, tensile strength of concrete E_{c} is 34kn/mm^{2} loss ratio ɳ is 0.85 and 8mm diameter high tensile strength f_{pu} is 1500n/mm^{2} are available for use and the modulus of elasticity of high tensile wires is 200kn/mm^{2}?

a) 340 and 450kn

b) 240 and 340kn

c) 140 and 240kn

d) 100 and 200kn

View Answer

Explanation: W

_{min}/W

_{ud }= (50x2400x9.81×0.125x25x30/50×10

^{6}x0.85

^{2}) = 0.31

Ultimate load excluding the factored selfweight = (1.4×2)+(1.6×9) = 17.2kn/m, W

_{ud}= 17.2/1-1.4×0.31) = 30KN/M, W

_{min}= (0.31×30) = 9.3kn/m, Ultimate moment , M

_{u}= (0.125x30x30

^{2}) = 3400knm, Ultimate shear, V

_{u }= (0.5x30x30) = 450kn.

2. Find cross-sectional dimensions thickness of web if h_{f}/d ratio is 0.23 and b_{w}/b ratio is 0.25 and b is 0.5d?

a) 100mm

b) 110mm

c) 120mm

d) 30mm

View Answer

Explanation: h

_{f}/d =0.23 and b

_{w}/b = 0.25 and b = 0.5d,

M

_{u}= 0.10f

_{cu}bd

^{2}d = (3400×10

^{6}/0.10x50x0.5)

^{1/3}= 1130mm, h = (1130/0.85) = 1300, b = 600mm, h

_{f}= (0.2×1130) = 250mm, adopt an effective depth, d = 1150mm, thickness of web, b

_{w}= (0.6vu/fth) = (0.6x450x10

^{3}/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

View Answer

Explanation: Actual self weight of the beam and the girder = 8.8kn/m, span = 30m

Minimum moment M

_{min}= 990knm, Design working load = 19.8kn/m,

Working moment M

_{d}= (0.125×19.8×30

^{2}) = 2230knm.

4. Find the Permissible stresses and range of stresses for class 1 structure f_{cu} = 50n/mm^{2},f_{ck} = 35n/mm^{2} according to BS: 8110 recommendations for f_{cu} = 50n/mm^{2} and f_{ci} = 35n/mm^{2},f_{ct} = 0.5f_{ci}= 17.5n/mm^{2}?

a) 16.5n/mm^{2}

b) 12.56n/mm^{2}

c) 13.56n/mm^{2}

d) 12.00n/mm^{2}

View Answer

Explanation: f

_{cu}= 50n/mm

^{2},f

_{ck}= 35n/mm

^{2}according to BS: 8110 recommendations for f

_{cu}= 50n/mm

^{2}and f

_{ci}= 35n/mm

^{2}, f

_{ct}= 0.5f

_{ci}= 17.5n/mm

^{2}For class 1 structure f

_{u}= h

_{tw}= 0, f

_{br}= (ɳf

_{ct}-ftw) = (0.85×17.5) = 15n/mm

^{2}, f

_{cw}= 0.33f

_{cu}= (0.33×50) = 16.5n/mm

^{2}, f

_{cu}= (fcw-ɳfu) = 16.5n/mm

^{2}.

5. Find prestressing force if area is 36.75mm^{2} of eccentricity 580given f_{inf} is 26.5kn/m and z_{b} is 99×10^{6}?

a) 405

b) 308

c) 453

d) 206

View Answer

Explanation: Area = 36.75mm

^{2}, e = 580, f

_{inf}= 26.5kn/m, z

_{b}= 99×10

^{6}

p =(Af

_{inf}Z

_{b}/Z

_{b}+A

_{e}) =(367500×26.5x99x10

^{6}/(99×10

^{6})+(367500×580)) = 308x104kn/m

^{2}.

6. Find force in cable using Freyssinet cables 12-8mm diameter and stressed to 1100n/mm^{2} of eccentricity 50 and the prestressing force is given as 1000n/mm^{2}?

a) 660kn

b) 234kn

c) 300kn

d) 230kn

View Answer

Explanation: 12 diameter, stress = 1100n/mm

^{2}, e = 50, prestressing force =1000n/mm

^{2}

Force in each cable = ( (12x50x1100)/1000)) = 660kn.

7. Find ratio for ultimate flexural strength at the centre – span section given that A_{ps} = 3000mm^{2}, d= 1150mm, f_{cu} = 50n/mm^{2}, b_{w} = 150mm, f_{pu} = 1500n/mm^{2}, b = 600mm, h_{t} = 250mm, design ultimate moment m_{ud} = 3400knm?

a) 9.5

b) 0.23

c) 6.7

d) 3.4

View Answer

Explanation: A

_{ps}= 3000mm

^{2}, d= 1150mm, f

_{cu}= 50n/mm

^{2}, b

_{w}= 150mm, f

_{pu }= 1500n/mm

^{2}, b = 600mm, h

_{t}= 250mm, design ultimate moment m

_{ud}= 3400knm, according to BS: 8110-1985,

_{Aps}= (A

_{pw}+A

_{pf}) =

A

_{pf}= 0.45×50(600-150)(250/1500) = 0.45xf

_{cu}(b-b

_{w})(h

_{f}/f

_{pu}) = 1680mm

^{2}, A

_{pw }= (1300-1680) = 1320mm

^{2}, ratio(f

_{pu}A

_{pw}/f

_{cu}b

_{wd}) = (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

View Answer

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) F_{cp} = 23.4n/mm^{2}, z_{b} is 99×106 and stress induced is 1000?

a) 1200kn

b) 1850kn

c) 2300kn

d) 4300kn

View Answer

Explanation: F

_{cp}= 23.4n/mm

^{2}, zb is 99×10

^{6}, stress is 1000

m

_{0}= (0.8f

_{cp}Z

_{b}) = (0.8 x 23.4 x (99×10

^{6}/1000)) = 1850knm.

10. Find resultant maximum long term deflection if ϕ is 2.6, α_{y} is 38.5mm, α_{g} is 46mm, α_{p} is 74.7mm?

a) 95mm

b) 35mm

c) 55mm

d) 20mm

View Answer

Explanation: E

_{ce}= (E

_{c}/1+ϕ) = (E

_{c}/2.6), ϕ = 2.6, α

_{y}= 38.5mm, α

_{g}= 46mm, α

_{p}= 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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