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Test: Loss of Prestress Due to Friction - Civil Engineering (CE) MCQ


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10 Questions MCQ Test - Test: Loss of Prestress Due to Friction

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Test: Loss of Prestress Due to Friction - Question 1

The total loss of prestress due to friction is of ______

Detailed Solution for Test: Loss of Prestress Due to Friction - Question 1

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.

Test: Loss of Prestress Due to Friction - Question 2

The loss of prestress due to friction generally occurs in case of ______

Detailed Solution for Test: Loss of Prestress Due to Friction - Question 2

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.

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Test: Loss of Prestress Due to Friction - Question 3

The wobble effect due to loss of stress is also known as ______

Detailed Solution for Test: Loss of Prestress Due to Friction - Question 3

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.

Test: Loss of Prestress Due to Friction - Question 4

The loss of stress due to curvature effect depends upon ______

Detailed Solution for Test: Loss of Prestress Due to Friction - Question 4

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.

Test: Loss of Prestress Due to Friction - Question 5

Friction loss depends on

Detailed Solution for Test: Loss of Prestress Due to Friction - Question 5

In the case of post tensioned members the tendons are housed in ducts preformed in concrete. The ducts are either straight or follow a curved profile depending upon the design requirements.
Consequently on tensioning the curved tendons, loss of stress occurs in the post tensioned members due to friction between the tendons and the surrounding concrete ducts.
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.
Loss of stress due to the wobble effect, which depends upon the local deviations in the alignment of cable. The wobble or wave effect is the result of accidental or unavoidable misalignment.

Frictional losses can be reduced by several methods:
Over tensioning the tendons by an amount equal to the maximum frictional loss
Jacking the tendons from both ends of the beam, generally adopted when the tendons are long or when the angles of bending are large.

Test: Loss of Prestress Due to Friction - Question 6

The wobble effect is the result of _____

Detailed Solution for Test: Loss of Prestress Due to Friction - Question 6

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.

Test: Loss of Prestress Due to Friction - Question 7

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?

Detailed Solution for Test: Loss of Prestress Due to Friction - Question 7

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.

Test: Loss of Prestress Due to Friction - Question 8

The value of ‘μ’ in loss of stress equation depends upon ____

Detailed Solution for Test: Loss of Prestress Due to Friction - Question 8

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.

Test: Loss of Prestress Due to Friction - Question 9

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?

Detailed Solution for Test: Loss of Prestress Due to Friction - Question 9

d = 40m, Es = 210kn/mm2, n = 4points, θ = 90˚,
Length of wires = (π × 40 × 1000/4) = 104πmm,
Extension at the jack = (960/210 × 10× 104π) = 144mm.

Test: Loss of Prestress Due to Friction - Question 10

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?

Detailed Solution for Test: Loss of Prestress Due to Friction - Question 10

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.

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