Option 4 : Initial stress in steel

**Concept:**

**Loss of stress due to relaxation of steel **

Relaxation is assumed to mean the loss of stress in steel under nearly constant strain at constant temperature.

It is similar to creep of concrete. Loss due to relaxation varies widely for different steels and the steel manufacturers based on test data may supply its magnitude.

This loss is generally of the order of 2 to 8% of the initial stress. This is generally - 1000 hours of referred to at loading at 27˚ C.

**Elastic deformation of concrete: **

When the pre-stress is applied to the concrete, an elastic shortening of concrete takes place. This results in an equal and simultaneous shortening of the pre-stressing steel.

**Friction loss in post tensioned members: **

This loss occurs only in the post tensioned members. There are small frictional losses in the jacking equipment. The friction between tendons and surrounding materials is not small and may be considered partly a length effect (wobble effect) and partly a curvature effect.

**Loss due to anchorage slip: **

In most tensioning systems, when the cable is tensioned and the jack is released to transfer pre-stress to concrete, the friction wedges employed to grip the wires, slip over a small distance before the wires are finally housed between the wedges. The magnitude of slip depends upon the type of wedge.

Option 4 : type of jack and anchorage

__Concept:__

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.

Option 3 : Shrinkage and creep

Losses of prestressing:

Pre Tensioning |
Post Tensioning |
||

1. |
Electric shortening loss |
1. |
No elastic shortening loss in cables if pulled simultaneously, however, losses will be there if cables are pulled are by one. |

2. |
Creep loss |
2. |
Friction loss |

3. |
Shrinkage loss |
3. |
Anchor slip loss |

4. |
Relaxation loss |
4. |
Creep loss |

5. |
Shrinkage & Relaxation |

**Concept-**

**Shrinkage loss-**

Time-dependent strain measured in an unloaded and unrestrained specimen at a constant temperature.

Loss of pre-stress (Δfp)due to shrinkage = E_{P} × ϵ_{sh}

Where Ep is the modulus of pre-stressing steel and

ϵ_{sh} is shrinkage strain

The approximate value of shrinkage strain for design shall be assumed as follows (IS 1383):

For pre-tensioning = 0.0003

For post-tensioning \(=\dfrac{0.002}{log\left(t+2\right)}\)

Where t = age of concrete at transfer in days

The factors responsible for the creep of concrete will have an influence on the shrinkage of concrete also except the loading conditions.

**Creep of Concrete-**

Time-dependent increase of deformation under sustained load.

Due to creep, the pre-stress in tendons decreases with time.

The loss in pre-stress (Δfp ) due to creep is given as follows.

Δfp = Ep x ϵ_{cr} = Ep x ϵ_{el }x θ

Since ϵ_{cr} = θ x ϵ_{el}

Ep is the modulus of the pre-stressing steel,

ϵ_{cr} = Ultimate creep strain,

ϵ_{el} = elastic strain

θ = Creep coefficient

Option 3 : Friction

__Concept:__

- Loss due to Friction: The friction generated at the interface of concrete and steel during the stretching of a curved tendon in a post-tensioned member. The loss due to friction does not occur in pre-tensioned members because there is no concrete during the stretching of the tendons.

__Important Points__

Loss due to Elastic Shortening: When the tendons are cut and the prestressing force is transferred to the member, the concrete undergoes immediate shortening due to the prestress.

Loss due to Relaxation: Relaxation is assumed to mean the loss of stress in steel under nearly constant strain at a constant temperature. It is similar to the creep of concrete. This loss is generally of the order of 2 to 8% of the initial stress.

Loss due to Anchorage Slip: In a post-tensioned member, the is loss of prestress due to the consequent reduction in the length of the tendon. The loss due to anchorage does not occur in pre-tensioned members.

An ordinary mild steel bar has been prestressed to a working stress of 200 MPa. If the Young’s modulus of steel is 200 GPa and permanent negative strain due to shrinkage and creep is 0.0008, what is the effective stress left in the steel?

Option 4 : 40 MPa

__Concept:__

Loss due to shrinkage of concrete:

(i) It is well known that concrete shrinks with age.

(ii) This shrinkage of concrete in a pre-stressed concrete member results in the shortening of prestressed wires thereby resulting in the loss of pre-stress.

For pre-tensioned members, the total residual shrinkage is 3 × 10-4

For Post-tensioned members, total residual shrinkage strain is given by,

\(\dfrac{{2 × {{10}^{ - 4}}}}{{{{\log }_{10}}(t + 2)}}\)

Where t = age of concrete at the time of transfer in days

Loss of prestress = Residual shrinkage strain × E_{s}

**Calculation:**

Given,

Initial Prestress = 200 MPa

Shrinkage strain = 8 × 10^{-4}

Modulus of elasticity of steel (Es) = 200 GPa = 200 × 10^{3} MPa

Now, loss of prestress = Shrinkage strain × E_{s} = 8 × 10^{-4} × 200 × 10^{3} = 160 MPa

∴ stress left = 200 - 160 = 40 MPa

A concrete beam is post-tensioned by a cable carrying initial stress of 1000 N / mm^{2}, the slip at jacking end was observed to be 5 mm, modulus of steel is 210 kN / mm^{2} and span of beam is 30 m; what is % of loss of stress due to anchorage?

Option 1 : 3.5%

__Concept:__

**Post-Tensioned:**

Post-tensioned concrete is a type of prestressed concrete where the concrete is strengthened via an arrangement of reinforcement held in tension. steel cables, called Post-tensioned tendons. a tendon is **generally made of wires, strands, or bars. Wires and strands can be tensioned in groups,** whereas bars are tensioned one at a time.

\({\rm{Loss\;of\;stress\;}} = {\rm{Strain\;due\;to\;anchorage}} × {\rm{Modulus\;of\;Elasticity\;}}\)

\({\rm{Strain\;due\;to\;anchorage}} = {\rm{\;}}\frac{{{\rm{Anchorage\;slip}}}}{{{\rm{length\;of\;beam}}}}\)

\({\rm{Percentage\;loss\;in\;pre}} {\rm{stress}} = \frac{{{\rm{loss\;in\;pre}} {\rm{stress}}}}{{{\rm{initial\;pre}}{\rm{stress}}}} × 100\)

__Calculation:__

Given,

Anchorage slip = 5 mm; Beam length = 30 m; Modulus of Elasticity of steel = 2.1 × 105 N/mm2,

initial pre-stress = 1000 N/mm2

\({\rm{Strain\;due\;to\;anchorage}} = {\rm{\;}}\frac{5}{{30 × 1000}} = {\rm{\;}}{1.67\ × 10^{ - 4}}\)

Loss of stress = 1.67 × 10-4 × 2.1 × 105 = 35 MPa

\({\rm{Percentage\;loss\;in\;pre}} - {\rm{stress}} = {\rm{\;}}\frac{{35}}{{1000}} × 100\)

∴ Loss in pre-stress = 3.5%

Option 1 : relaxation

__Concept:__

Loss of stress due to relaxation of steel

Relaxation is assumed to mean the loss of stress in steel under nearly constant strain at constant temperature.

It is similar to creep of concrete. Loss due to relaxation varies widely for different steels and the steel manufacturers based on test data may supply its magnitude.

This loss is generally of the order of 2 to 8% of the initial stress. This is generally - 1000 hours of referred to at loading at 27˚ C.

A post-tensioned concrete beam is prestressed by means of three cables each 100 mm^{2} area and stressed to 1100 MPa. All three cables are straight and located at an eccentricity of 50 mm. If modulus modular ratio (m) = 6 and stress in concrete at the level of steel (F_{c}) = 5 MPa, then what is the loss of stress in cables due to elastic shortening if all cables are simultaneously tensioned and anchoring?

Option 4 : 0 MPa

** Concept**:

**Elastic shortening of concrete**:

**Concrete shortens**when a**prestressing force is applied**. As the tendons that are bonded to the adjacent concrete simultaneously shorten, they lose part of prestressing force that they carry.

__Important Points__

**Loss in post-tensioned beams**:

- When all the
**tendons are jacked simultaneously**, the elastic shortening**loss is zero**. - When several
**sequential jacking**steps are used, it is given by the following expression –

**\({\rm{\Delta }}{p_{EC}} = \frac{{{E_s}}}{n}\mathop \sum \limits_{j = 1}^n {\left( {\frac{{{\rm{\Delta }}l}}{l}} \right)_j}\)**

Here, j – denotes the number of jacking operations, ∆l – elastic shortening.

- A tendon that was
**tensioned last does not suffer any losses**due to elastic shortening while the tendon that was**tensioned first suffers the maximum amount of loss**.

__Explanation:__

Since,

All the cables are tensioned simultaneously, the **loss of stress in cables due to elastic shortening will be zero**.

__Additional Information__

**Loss in pre-tensioned beams:**

- For pre-tensioned elements, the compressive force imposed on the beam by the tendon results in the longitudinal shortening of the beam as shown below.

The shortening in concrete

\({{\rm{\Delta }}_{ES}} = \frac{{{P_i}L}}{{{A_c}{E_c}}}\)

Since prestressing tendons suffer the same magnitude of shortening.

Loss of prestress,

\({\rm{\Delta }}{p_{EC}} = \frac{{{E_s}{{\rm{\Delta }}_{ES}}}}{L} = \frac{{{P_i}}}{{{A_c}}} \times \left( {\frac{{{E_s}}}{{{E_c}}}} \right) = m{p_i}\)

Here,

pi – initial prestress, A_{c} – the area of concrete, E_{c} – modulus of elasticity of concrete, E_{s} – modulus of elasticity of steel.

Option 3 : Permanent loads and pre-stressing force

**Concept:**

As per **IS 1343(1980)**: Code of Practice for Prestressed Concrete **19.5.2.1**

**Loss of prestress due to creep of concrete:**

- The loss of prestress due to the creep of concrete under load shall be determined for all the
**permanently applied**loads including the**prestress**. - The creep loss due to
**live load stresses**,**erection stresses**, and other stresses of short duration may be**ignored**. - The loss of prestress due to creep of concrete is obtained as the product of the modulus of elasticity of the prestressing steel and the ultimate creep strain of the concrete fiber integrated along the
**line of center of gravity**of the prestressing tendon at the section at which creep loss is being calculated for bonded tendons. - In the case of unbonded tendons, the creep loss is a product of modulus of elasticity of prestressing steel and creep strain calculated by integrating and averaging creep stress along the line of center of gravity between the anchorage points.
- The total creep strain during any specific period shall be assumed for all practical purposes, to be the creep strain due to sustained stress equal to the average of the stresses at the beginning and end of the period.

Option 2 : 0.1 radians

__Concept:__

Loss due to friction is given by,

PL = Po (kx + μα)

Where, Po = Prestressing force at the jacking end

K = Wobble correction factor

α = Cumulative angle in radians through which tangent to the cable profile has turned between any two points under consideration

If Jacking is from one end, then α = 2θ

If Jacking is from both end, then α = θ

μ = Coefficient of friction

\(θ =\dfrac{4e}{L}\)

e = e1 + e2

Cumulative angle (α):

\(α =θ +θ =2θ =\dfrac{2\times 4\left ( e_{1}+e_{2} \right )}{L}\)

Calculation:

\(α =θ +θ =2θ =\dfrac{2\times 4\left ( e_{1}+e_{2} \right )}{L}\)

\(α =\dfrac{2\times 4\left ( 15+45 \right )}{4.8\times 10^{3}}\)

= 0.1 radians

Option 2 : 13 MPa

__Concept:__

The loss of stress due to creep in concrete is given as

F_{c} = θmf_{cs}

Where,

θ is the ultimate creep coefficient

m is the modular ratio and it is given as

\({\rm{m}} = \frac{{{\rm{Modulus\;of\;Elasticity\;of\;steel}}}}{{{\rm{Modulus\;of\;Elasticity\;of\;Concrete\;}}}}\)

f_{cs} is the stress in concrete at the level of steel.

The loss of stress due to shrinkage in concrete is given as

F_{s} = E_{s} × ϵ

Where,

E_{s} is the Modulus of elasticity of steel

ϵ is the shrinkage strain in concrete

__Calculation:__

Given,

θ = 1.6, ϵ = 200 × 10^{-6} ; f_{1} = 3f_{2}; E_{s} = 200 GPa and E_{c} = 35.35 GPa

m = 200/35.35 = 5.66

f_{1} = 1.6 × 5.66 × f_{cs} = 9.05f_{cs} N/mm^{2}

f_{2} = 200 × 1000 × 200 × 10^{-6} = 40 N/mm^{2}

It is given that

f_{1} = 3f_{2}

9.05f_{cs} = 3 × 40

**⇒**** f _{cs} = 13.26 ≈ 13 N/mm^{2}**

Option 3 : Sequence- stressing loss

**Explanation:**

Loss due to elastic shortening in the post-tensioned beam

**Case 1:**- If there is
**only one wire**to be tensioned**or**There are**multiple wires**but**all tension at the same time**, in that case, Elastic shortening of the concrete takes place at the time of tensioning of wires, so there will be no further elastic shortening of concrete after anchoring the wires. So there will be**no loss of stress in steel due to the elastic shortening**of concrete.

- If there is
**Case 2:****If**there are multiple wires and wires are tensioned one after another then in that case losses occur due to elastic shortening.**Tensioned**the**wire one after the other**is**known**as**Subsequent tensioning**and losses are known as**Sequence stressing**losses

__Important Points__

Losses of prestress:

Losses in pre-tensioned member |
Losses in post-tensioned member |

Loss due to elastic deformation of concrete |
If wires are tensioned simultaneously, then no loss due to elastic deformation of concrete but however if the wires are tensioned successively, then loss due to elastic deformation occurs |

Loss due to stress relaxation in steel | Loss due to stress relaxation in steel |

Loss due to creep and shrinkage of concrete | Loss due to creep and shrinkage of concrete |

No loss due to anchorage slip and friction | Loss due to anchorage slip and friction |