Prestressed Concrete Structures Questions and Answers – Loss Due to Friction

This set of Prestressed Concrete Structures Multiple Choice Questions & Answers (MCQs) focuses on “Loss Due to Friction”.

1. The loss of prestress due to friction generally occurs in case of ____________
a) Post tensioned members
b) Pre tensioned members
c) Chemical members
d) Electrical members

Explanation: Loss of prestress due to friction occurs in the case of post tensioned members, the tendons are housed in the ducts performed in concrete and the ducts are either straight or follow a curved profile depending upon the design requirements.

2. The total loss of prestress due to friction is of ____________
a) 4 types
b) 2 types
c) 8 types
d) 3 types

Explanation: The total loss due to friction is divided into two types:
Loss of prestress due to effect of curvature, Loss of prestress due to wobble effect and frictional losses can be reduced by over tensioning the tendons by an amount equal to the maximum frictional loss and jacking the tendons from both ends of the beam adopted generally, when the tendons are long or when the angles of bearing are large.

3. The loss of stress due to curvature effect depends upon ____________
a) Alignment
b) Midpoint
c) Centerline
d) Exterior point

Explanation: The loss of stress due to the curvature effect, which depends upon the tendon form or alignment which generally follows a curved profile along the length of the beam, curvature coefficient is expressed as μ and wobble coefficient is expressed as k/m.

4. The wobble effect due to loss of stress is also known as ____________
a) Wave effect
b) Ray effect
c) Bubble effect
d) Bulb effect

Explanation: Loss of stress due to wobble effect, which depends upon the local deviations in the alignment of the cable and the wobble effect is also known as wave effect, the friction coefficient values for wave effect k are 0.15 per 100m for normal conditions, 1.5 per 100m for thin walled ducts where heavy vibrations are encountered and in other adverse conditions.

5. The wobble effect is the result of ____________
a) Misalignment
b) Extreme alignment
c) Tensile alignment
d) Anchorage alignment

Explanation: The wobble or wave effect is the result of accidental or unavoidable misalignment since ducts cannot be perfectly located to follow a predetermined profile throughout the length of the beam, the coefficient due to wobble effect may be reduced to zero where the clearance between the duct and cable is sufficiently large to eliminate wave effect so as the sheath is made up of heavy gauge steel tube with water tight joints, where a deformation of duct profile is prevented during the vibration of concrete.
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6. The loss of stress due to friction is given as ____________
a) 1-(μα+Kx)
b) 1-(μα+Kx)
c) 1-(μα+Kx)
d) 1-(μα+Kx)

Explanation: The prestressing force at section is given by:
Px = p˳e-(μα+kx), Loss of stress (Δf)f = 1-(μα+Kx),
Px = Prestressing force at section x, P˳ = initial prestressing force, μ = coefficient of friction between the cable and concrete, k = wobble effect, d = cumulative angle.

7. The value of ‘μ’ in loss of stress equation depends upon ____________
a) Type of curing
b) Type of concrete
c) Type of steel
d) Type of aggregates

Explanation: The values of ‘μ’ (coefficient of curvature effect) depend upon the type of steel and concrete used in construction and are given in Indian standard codes of practice, coefficient of friction can be considerably reduced by variety of lubricants, particularly greases, oil, graphite mixtures, paraffin, the use of paraffin wax gives by far the coefficient of friction especially with high contact pressure.

8. A post tensioned concrete beam 200mm wide and 450mm deep, of span 10m, initial stress of 840n/mm2 is available in the un jacked end immediately after the anchoring. Find the angle between tangents to the cable at supports?
a) 0.13
b) 0.12
c) 0.10
d) 0.23

Explanation: b = 200mm, d = 450mm, l = 10m, r = 84m, d = 5m,
Angle between the horizontal tangent drawn to the cable at support sinα = (5/84) = 0.06radians,
Cumulative angle between tangents to the cable at supports = (2×0.06) = 0.12radians.

9. A concrete tank if has a minimum stress in wires 600n/mm2 immediately after tensioning and the coefficient of friction is 0.5. Calculate the maximum stress to be applied to the wires at the jack?
a) 900n/mm2
b) 960n/mm2
c) 850n/mm2
d) 800n/mm2

Explanation: Px = 600n/mm2, e = 2.7183, μ = 0.5
Px = P˳e-μα,
600 = P˳e-(0.5×π/2),
P˳ = (600) × (2.71830.79) = 1320n/mm2, Average stress in wires = (1320+600/2) = 960n/mm2.

10. A cylindrical concrete tank, 40m external diameter is to be prestressed circumferentially by means of a high strength steel wire (Es = 210kn/mm2) jacked at 4 points and 90 degrees apart. Find the expected extension at the jack?
a) 150mm
b) 130mm
c) 144mm
d) 133mm

Explanation: d = 40m, Es = 210kn/mm2, n = 4points, θ = 90˚,
Length of wires = (π×40×1000/4) = 104πmm,
Extension at the jack = (960/210×103×104π) = 144mm.

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