Prestressed Concrete Structures Questions and Answers – Design of Pretensioned Beams

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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 Fcu, concrete cube strength = 50n/mm2, 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

Answer: a
Explanation: Wmin/Wud = KDcgβ(L/h)L/fcu(d/h)2 = 7.5×2400 x 9.81 x 0.125×25 x 6/50x 106x(0.85)2 = 0.094
Fcu, concrete cube strength = 50n/mm2, effective span = 6m, applied load = 5kn/m, dead load = 1.4, live load = 1.6, β = 0.125, k = 7.5, Dc = 2400, fcu = 50n/mm2, wmin = (0.094)(9.25) = 0.86kn/m, mu = (0.125×9.25×62) = 42knm, vu = (0.5×9.25×6) = 27.75kn.
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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

Answer: c
Explanation: Mu = 0.10fcubd2 and if b = 0.5d
D = (42×106x2/0.10×50)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

Answer: b
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 = (0.85vu/fth) = (0.85×27.75×103/1.7×320) = 43mm.

4. Find minimum range of stresses if fct is 15n/mm2, fcw is 17, ftw is zero, fu is -1n/mm2, ɳ is 0.8?
a) 12 and 18n/mm2
b) 13 and 14n/mm2
c) 12 and 15n/mm2
d) 10 and 16n/mm2
View Answer

Answer: a
Explanation: Range of stress fbr = (ɳfct-fcw) = (0.85×15-0) = 12n/mm2, ftr = (fcw – ɳfu) = (17-0.8x(-1)) = 17.8n/mm2, fct = 15n/mm2, fcw = 17, ftw = 0, fu = -1n/mm2, ɳ = 0.8.

5. Find minimum section modulus given data is mg is 3.86×106, mq is 22.50×106, fbr is given as 12 and the loss ratio is 0.8?
a) 134×104
b) 182×104
c) 123×104
d) 120×104
View Answer

Answer: b
Explanation: mg = 3.86×106, mq = 22.50×106, fbr = 12, loss ratio = 0.8
Zb > or equal (mq+(1-ɳ)mg/fbr) > or equal ((22.50×106)+(1-0.8)3.86×106)/12)
Greater than equal to 182x104mm3.
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6. Find the supporting force if given characteristic strength is -1, moment of gravity is 3.86×106, zt = 230×104?
a) -2.68n/mm2
b) -3.45n/mm2
c) -1.23n/mm2
d) 13.56n/mm2
View Answer

Answer: a
Explanation: p = (A(finfZb+fsubZt)/Zt+Zb)
Finf = ((ftw/ɳ+(mq+mg)/ɳzb)) = (0+ (26.36×106/0.8x230x104))
Fsup = (fu – mg/zt) = (-1 – (3.86×106)/(230×104)) = -2.68n/mm2.

7. Check for ultimate flexural strength if given Aps is 154mm2, fpu is 1600n/mm2, b is 160mm, fcu is 50n/mm2and diameter is 265mm?
a) 9.65
b) 0.116
c) 3.442
d) 2.345
View Answer

Answer: b
Explanation: Aps = (38.5xy) = 154mm2, fpu = 1600n/mm2, b = 160mm, fcu 50n/mm2, d = 265mm
(Apsfpu/bdfcu) = (154×1600/160x265x50) = 0.116.

8. Find ultimate shear strength (check it for safe against shear failure) if vu 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, fcp = 4.65, ft is 1.7?
a) Safe
b) Unsafe
c) Zero
d) Collapse
View Answer

Answer: a
Explanation: Fcp = (ɳp/A) = (0.8×182000/31400) = 4.65n/mm2
Vcw = 0.67bh(f12+0.8fcpft)1/2 = (0.67x50x320(1.72+0.8×4.65×1.7)1/2/103) = 33.2kn
Vcw > Vu hence safe against shear failure.

9. Check for deflection due to prestressing force if given data is Prestressing force is 182×103 eccentricity of cable is 105, Length of the cable is 1000, elastic modulus of concrete is 34×103, Moment of inertia is 3200×105?
a) 9.4
b) 4.5
c) 6.8
d) 9.8
View Answer

Answer: c
Explanation: P = 182×103 e = 105, L = 1000, elastic modulus of concrete = 34×103, I = 3200×105
PeL2/8EcI = (182×103x105x62x10002/8x34x103x3200x105) = 6.8mm.

10. Find the deflection due to self weight given that ϕ = 1.6, Ee = 2.6Ece, elastic modulus of concrete is 34×103, gravity is given as 6, self weight is 0.76, Length of the cable is 1000, elastic modulus of concrete is 34×103 , Moment of inertia is 3200×105?
a) 1.66mm
b) 5.3mm
c) 23.4mm
d) 1.02mm
View Answer

Answer: d
Explanation: Ece = Ec/1+ϕ, ϕ = 1.6, Ee = 2.6Ece
Deflection due to self weight g = (5gL4/384EcI) = (5×0.76×64x10004/384x34x103x3700x105) = 1.02mm.
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Sanfoundry Global Education & Learning Series – Prestressed Concrete Structures.

To practice all areas of Prestressed Concrete Structures, here is complete set of 1000+ Multiple Choice Questions and Answers.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn