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

1. Design a pretensioned roof pull-in to suit the data F_{cu}, concrete cube strength = 50n/mm^{2}, effective span = 6m, applied load = 5kn/m, dead load = 1.4, live load = 1.6, β = 0.125, k = 7.5, Dc = 2400, and determine ultimate moment and shear?

a) 42 and 27.75

b) 54 and 27.75

c) 34 and 27.75

d) 20 and 28

View Answer

Explanation: W

_{min}/W

_{ud}= KD

_{c}gβ(L/h)L/f

_{cu}(d/h)

^{2}= 7.5×2400 x 9.81 x 0.125×25 x 6/50x 10

^{6}x(0.85)

^{2}= 0.094

F

_{cu}, concrete cube strength = 50n/mm

^{2}, effective span = 6m, applied load = 5kn/m, dead load = 1.4, live load = 1.6, β = 0.125, k = 7.5, D

_{c}= 2400, f

_{cu}= 50n/mm

^{2}, w

_{min}= (0.094)(9.25) = 0.86kn/m, m

_{u}= (0.125×9.25×6

^{2}) = 42knm, v

_{u }= (0.5×9.25×6) = 27.75kn.

2. Design cross sectional dimensions of pretensioned roof pull given that b is 0.5d?

a) 250

b) 260

c) 270

d) 280

View Answer

Explanation: M

_{u}= 0.10f

_{cu}bd

^{2}and if b = 0.5d

D = (42×10

^{6}x2/0.10×50)

^{1/3}= 270mm.

3. Find the approximate thickness of web if b is 0.5d, d is 270mm, d/h ratio is 0.85, h is 315mm, adopt effective depth, d = 275mm overall depth , h is 320mm, width of flange of 160mm and Average thickness of flange is 70mm since sloping flanges are used, increases the flange thickness by 20 percent?

a) 45mm

b) 43mm

c) 41mm

d) 42mm

View Answer

Explanation: b = 0.5d, d = 270mm, d/h = 0.85, h = 315mm, adopt effective depth, d = 275mm overall depth , h is 320mm, width of flange of 160mm and Average thickness of flange is 70mm since sloping flanges are used, increases the flange thickness by 20 percent:

Thickness of flange = (0.2×275) =55mm Approximate thickness of web = (0.85v

_{u}/f

_{t}h) = (0.85×27.75×10

^{3}/1.7×320) = 43mm.

4. Find minimum range of stresses if f_{ct} is 15n/mm^{2}, f_{cw} is 17, f_{tw} is zero, f_{u} is -1n/mm^{2}, ɳ is 0.8?

a) 12 and 18n/mm^{2}

b) 13 and 14n/mm^{2}

c) 12 and 15n/mm^{2}

d) 10 and 16n/mm^{2}

View Answer

Explanation: Range of stress f

_{br}= (ɳf

_{ct}-f

_{cw}) = (0.85×15-0) = 12n/mm

^{2}, f

_{tr}= (f

_{cw }– ɳ

_{fu}) = (17-0.8x(-1)) = 17.8n/mm

^{2}, f

_{ct}= 15n/mm

^{2}, f

_{cw}= 17, f

_{tw}= 0, f

_{u}= -1n/mm

^{2}, ɳ = 0.8.

5. Find minimum section modulus given data is mg is 3.86×10^{6}, m_{q} is 22.50×10^{6}, f_{br} is given as 12 and the loss ratio is 0.8?

a) 134×10^{4}

b) 182×10^{4}

c) 123×10^{4}

d) 120×10^{4}

View Answer

Explanation: m

_{g}= 3.86×10

^{6}, m

_{q}= 22.50×10

^{6}, f

_{br}= 12, loss ratio = 0.8

Z

_{b}> or equal (m

_{q}+(1-ɳ)m

_{g}/f

_{br}) > or equal ((22.50×10

^{6})+(1-0.8)3.86×10

^{6})/12)

Greater than equal to 182x104mm

^{3}.

6. Find the supporting force if given characteristic strength is -1, moment of gravity is 3.86×10^{6}, z_{t} = 230×10^{4}?

a) -2.68n/mm^{2}

b) -3.45n/mm^{2}

c) -1.23n/mm^{2}

d) 13.56n/mm^{2}

View Answer

Explanation: p = (A(f

_{inf}Z

_{b}+f

_{sub}Z

_{t})/Z

_{t}+Z

_{b})

F

_{inf}= ((f

_{tw}/ɳ+(m

_{q}+m

_{g})/ɳz

_{b})) = ( 0+ (26.36×106/0.8x230x104))

F

_{sup}= (f

_{u}– m

_{g}/z

_{t}) = (-1 – (3.86×106)/(230×104)) = -2.68n/mm

^{2}.

7. Check for ultimate flexural strength if given A_{ps} is 154mm^{2}, f_{pu} is 1600n/mm^{2}, b is 160mm, f_{cu} is 50n/mm^{2}and diameter is 265mm?

a) 9.65

b) 0.116

c) 3.442

d) 2.345

View Answer

Explanation: A

_{ps}= (38.5xy) = 154mm

^{2}, f

_{pu}= 1600n/mm

^{2}, b = 160mm, f

_{cu}50n/mm

^{2},, d = 265mm

(A

_{ps}f

_{pu}/bdf

_{cu}) = (154×1600/160x265x50) = 0.116.

8. Find ultimate shear strength (check it for safe against shear failure) if v_{u} is 27.75kn, Loss ratio is 0.8, prestressing force is 182000, area is 31400, breadth is 50 where height is 320, prestressing force is 1.7, f_{cp} = 4.65, f_{t} is 1.7?

a) Safe

b) Unsafe

c) Zero

d) Collapse

View Answer

Explanation: F

_{cp}= (ɳp/A) = (0.8×182000/31400) = 4.65n/mm

^{2}

V

_{cw}= 0.67bh(f

_{1}

^{2}+0.8f

_{cp}f

_{t})

^{1/2}= (0.67x50x320(1.72+0.8×4.65×1.7)1/2/10

^{3}) = 33.2kn

V

_{cw}> V

_{u}hence safe against shear failure.

9. Check for deflection due to prestressing force if given data is Prestressing force is 182×10^{3} eccentricity of cable is 10^{5}, Length of the cable is 1000, elastic modulus of concrete is 34×10^{3}, Moment of inertia is 3200×10^{5}?

a) 9.4

b) 4.5

c) 6.8

d) 9.8

View Answer

Explanation: P = 182×10

^{3}e = 10

^{5}, L = 1000, elastic modulus of concrete = 34×10

^{3}, I = 3200×105

PeL

^{2}/8E

_{c}I = (182×10

^{3}x10

^{5}x6

^{2}x1000

^{2}/8x34x10

^{3}x3200x10

^{5}) = 6.8mm.

10. Find the deflection due to self weight given that ϕ = 1.6, E_{e} = 2.6E_{ce}, elastic modulus of concrete is 34×10^{3}, gravity is given as 6, self weight is 0.76, Length of the cable is 1000, elastic modulus of concrete is 34×10^{3} , Moment of inertia is 3200×10^{5}?

a) 1.66mm

b) 5.3mm

c) 23.4mm

d) 1.02mm

View Answer

Explanation: E

_{ce}= E

_{c}/1+ϕ, ϕ = 1.6, E

_{e}= 2.6E

_{ce}

Deflection due to self weight g = (5gL

^{4}/384E

_{c}I) = (5×0.76×6

^{4}x1000

^{4}/384x34x10

^{3}x3700x10

^{5}) = 1.02mm.

**Sanfoundry Global Education & Learning Series – Prestressed Concrete Structures.**

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