Prestressed Concrete Structures Multiple Choice Questions on “Loss Due to Anchorage Slip”.
1. The term anchorage slip means ___________
a) Distance moved by friction wedges
b) Radius by friction wedges
c) Rotation by friction wedges
d) Twisting movement by friction wedges
Answer: a
Clarification: Anchorage slip is the distance moved by the friction wedges (in post tensioned members) after releasing the jacks at the ends of the member and before the wires get fixed perfectly in wedges, the loss during anchoring which occurs with wedge type grips is normally allowed for on the site by over-extending the tendon in the prestressing operation by the amount of the draw in before anchoring.
2. The anchorage slip is observed in __________
a) Post tensioned members
b) Pre tensioned members
c) Anchorage members
d) Tensioned members
Answers: b
Clarification: The anchorage slip is observed in post tensioned members at the time of transfer of prestress to the concrete and the friction wedges employed to grip the wires, slip over a small distance before the wires are firmly housed between the wedges.
3. The amount of anchorage slip generally depends upon __________
a) Type of wedge
b) Type of tendon
c) Type of anchor
d) Type of cement
Answer: a
Clarification: The amount of anchorage slip generally depends upon the type of wedge used at the ends and the magnitude of stress in the wires, since the loss of stress is caused by a definite total amount of shortening the percentage loss is higher for short members than for long ones.
4. The anchorage slip is low in members with __________
a) Small spans
b) Large spans
c) Middle spans
d) End spans
Answer: a
Clarification: The anchorage slip involves the definite total amount shortening of concrete and hence it is low in members with small spans compared to the members with large spans while prestressing a short member, due care should be taken to allow for the loss of stresses due to anchorage slip, which forms a major portion of total loss.
5. In case of long line pre tensioning system, anchorage slip is less than __________
a) Magnitude of wires
b) Length of wires
c) Distance of wires
d) Radius of wires
Answer: b
Clarification: In case of long line pre tensioning system, anchorage slip is much less in comparison with the length of the tensioned wires and hence it can be ignored for calculation of stresses, slip of anchorages, length of cable, cross-sectional area of the cable, modulus of elasticity, prestressing force in cable are considered.
6. Which of the following system is similar to loss due to anchorage slip?
a) Freyssinet system
b) Magnel Balton system
c) Leonhardt-Baur system
d) Gifford Udall system
Answer: c
Clarification: The systems in which tendons are lopped around concrete anchorage blocks, as in the case of Leonhardt-Baur system, loss of stress may take place, the Baur Leonhardt system is included in the third type of anchorages which work under the principle of looping tendon wires at the ends of concrete member, in this method double tendons are wrapped around the end block.
7. The loss of stress due to anchorage slip of anchorages is given as __________
a) EsA/L
b) EaA/L
c) EcA/L
d) EwA/L
Answer: a
Clarification: The method used for slip in anchorage by extending tendon is satisfactory provided by overstress does not exceed the prescribed limits of 80-85percent of the ultimate tensile strength of the wire and the magnitude of the loss of stress due to slip in anchorage is computed as
Anchorage slip Δ = PL/AEs, But moment Loss of stress, (Δf)a = P/A,
By considering both equations (Δf)a = EsΔ/l.
8. A concrete beam is post tensioned by a cable carrying an initial stress of 1000n/mm2, the slip at the jacking end was end was observed to be 5mm; modulus of elasticity of steel is 210kn/mm2. Estimate the percentage loss of stress due to anchorage if length of beam is 30m?
a) 3.5%
b) 4.5%
c) 5.5%
d) 8.7%
Answer: a
Clarification: Loss of stress due to anchorage slip = (EsΔ/l),
For a 30m long beam, loss of stress = (210×103×5)/(30×1000) = 35n/mm2,
Loss of stress = 35/1000×100 = 3.5%.
9. A post tensioned cable of beam 10m long is initially tensioned to a stress of 1000n/mm2 at one end, slope is 1 in 24 tendons curved at each end , area is 600mm2, Es is 210kn/mm2, coefficient of friction between duct and cable is 0.55, friction coefficient for wave effect is 0.0015perm. During anchorage, if there is a slip 3mm at the jacking end, calculate final force?
a) 39.8kn
b) 40.2kn
c) 37.8kn
d) 48.8kn
Answer: c
Clarification: Total change of slope from end to end α = (2×1/24) = (1/12),
μα = (0.55×1/12) = 0.046, kx = (0.0015×10) = 0.015,
P˳(μα+kx) = 1000(0.046+0.015) = 61n/mm2, Slip at the jacking end = 3 = (PL/AE)
P = (3×210×103×600/10×1000) = 37800kn = 37.8kn.
10. The total losses of stress that could be encountered under normal conditions of work were recommended by __________
a) Lin
b) Marks
c) Keifer
d) Neville
Answer: a
Clarification: Typical values of the total losses of stress that could be encountered under normal conditions of work were recommended by Lin, long term field studies on the loss of prestress in post tensioned concrete bridge girders have been carried out by marks and Keifer, Neville gave the losses in prestress considering various influencing parameters.
11. A prestressed concrete beam, 200mm wide and 300mm deep is prestressed with wires (area is 320mm2) located at a constant eccentricity of 50mm, initial stress of 1000n/mm2, span is 10m. Calculate loss of stress due to friction and slip anchorage of post tensioned beam?(Es = 210kn/mm2, Ec = 35kn/mm2).
a) 21 and 15
b) 35 and 25
c) 15 and 20
d) 5 and 10
Answer: a
Clarification: A = 320mm2, b = 200mm, d = 300mm, e = 50mm, p = 1000n/mm2, l = 10m, Es = 210kn/mm2, Ec = 35kn/mm2, Slip at anchorage = (1×210×103/10×1000) = 21, Friction effect = (1000×0.0015×10) = 15.