reinforced concrete design (ii) dr. hussain a. jabir · reinforced concrete design (ii) dr. hussain...

REINFORCED CONCRETE DESIGN (II) Dr. Hussain A. Jabir

1

PRESTRESS CONCRETE

Definition of Prestress:

Pre-stress is defined as a method of applying pre-compression to control the stresses

resulting due to external loads below the neutral axis of the beam tension developed

due to external load which is more than the permissible limits of the plain concrete.

The pre-compression applied (may be axial or eccentric) will induce the compressive

stress below the neutral axis or as a whole of the beam c/s. Resulting either no

tension or compression.

Terminology:

1. Tendon: A stretched element used in a concrete member of structure to impart

prestress to the concrete.

2. Anchorage: A device generally used to enable the tendon to impart and maintain

prestress in concrete.

3. Pretensioning: A method of prestressing concrete in which the tendons are

tensioned before the concrete is placed. In this method, the concrete is introduced

by bond between steel & concrete.

4. Post-tensioning: A method of prestressing concrete by tensioning the tendons

against hardened concrete. In this method, the prestress is imparted to concrete by

bearing.

Materials for prestress concrete members:

1. Cement: The cement used should be any of the following

(a) Ordinary Portland cement.

(b) Rapid hardening Portland cement.

(c) High strength ordinary Portland cement.


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2. Concrete: Prestress concrete requires concrete, which has a high compressive

strength reasonably early age with comparatively higher tensile strength than

ordinary concrete. Minimum cement content of 300 to 360 kg/m3

is prescribed for

the durability requirement.

The water content should be as low as possible.

3. Steel: High tensile steel, tendons, strands or cables.

High strength steel contains:

0.7 to 0.8% carbons,

0.6% manganese,

0.1% silica

Durability, Fire Resistance & Cover Requirements For P.S.C Members:

20 mm cover for pretensioned members

30 mm or size of the cable which ever is bigger for post tensioned members.

If the prestress members are exposed to an aggressive environment, these covers

are increased by another 10 mm.

Necessity of high grade of concrete & steel:

Higher the grade of concrete higher the bond strength which is vital in

pretensioned concrete, Also higher bearing strength which is vital in post-

tensioned concrete. Further creep & shrinkage losses are minimum with high-

grade concrete.

Generally minimum C30 grade concrete is used for post-tensioned & C40

grade concrete is used for pretensioned members.

The losses in prestress members due to various reasons are generally in the range of

250 N/mm2

to 400 N/mm2

. Hence high tensile steel wires are used which varies from

1600 to 2000 N/mm2

.


3

Classifications and Types:

Prestressed concrete structures can be classified in a number of ways depending

upon the feature of designs and constructions.

1. Pre-tensioning: In which the tendons are tensioned before the concrete is

placed, tendons are temporarily anchored and tensioned and the prestress is

transferred to the concrete after it is hardened.

2. Post-tensioning: In which the tendon is tensioned after concrete has hardened.

Tendons are placed in sheathing at suitable places in the member before casting

and later after hardening of concrete.

Prestressing Systems:

1. Pretensioning system:

In the pre-tensioning systems, the tendons are first tensioned between rigid

anchor-blocks cast on the ground or in a column or unit –mould types

pretensioning bed, prior to the casting of concrete in the mould. The tendons

comprising individual wires or strands are stretched with constant eccentricity

or a variable eccentricity with tendon anchorage at one end and jacks at the

other. With the forms in place, the concrete is cast around the stressed tendon.

The system is shown in Fig.1 below.

Fig. 1 Prestressing system


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2. Post-tensioned system:

In post-tensioning the concrete unit are first cast by incorporating ducts or

grooves to house the tendons. When the concrete attains sufficient strength, the

high-tensile wires are tensioned by means of jack bearing on the end of the face

of the member and anchored by wedge or nuts. The forces are transmitted to the

concrete by means of end anchorage and, when the cable is curved, through the

radial pressure between the cable and the duct. The space between the tendons

and the duct is generally grouted after the tensioning operation.

Comparative Study: Pretension Vs Post-tensioned Member

Pretension member Post-tensioned member

1. In pretensioned prestress concrete,

steel is tensioned prior to that of concrete.

It is released once the concrete is placed

and hardened. The stresses are transferred

all along the wire by means of bond.

1. Concreting is done first then wires are

tensioned and anchored at ends. The

stress transfer is by end bearing not by

bond.

2. Suitable for short span and precast

products like sleepers, electric poles on

mass production.

2. Suitable for long span bridges

3. In pretensioning the cables are

basically straight and horizontal. Placing

them in curved or inclined position is

difficult. However the wire’s can be kept

with eccentrically. Since cables can not

be aligned similar to B.M.D. structural

advantages are less compare to that of

post-tensioned.

3. The post tensioning cables can be

aligned in any manner to suit the B.M.D

due to external load system. Therefore it

is more economical particularly for long

span bridges. The curved or inclined

cables can have vertical component at

ends. These components will reduce the

design shear force. Hence post-tensioned

beams are superior to pretensioned beams

both from flexural and shear resistances

point.

4. Prestress losses are more compare to

that of post-tensioned concrete.

4. Losses are less compare to pre-

tensioned concrete


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Basic Concept of Pre-stressing

It’s well known that the concrete tensile strength varies from (8 to 14) percent of its

compressive strength. Due to this fact, flexural cracks develop at early stage of

loading. To solve such problem, the concept of pre-stress concrete was developed.

This concept is based on imposing a concentric or an eccentric compressive force in

the longitudinal direction of the structural element. This force will eliminate or

considerably reduce the tensile stresses at the critical sections (mid span and

supports), Figure below. Hence, the cracks will be prevented from developing.

[Nawy, 2003].

Concrete fibber stress distribution in a rectangular girder with straight tendon


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The typical loading history and the corresponding stress distribution across the depth

of the critical section are shown in Figure below;

Load versus deformation (camber or deflection) was, schematically, plotted for

various loading stages due to self-weight effect up to rupture, see below;


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A pre-stressed member, when compared with an equivalent reinforced concrete

member, requires less concrete, and about (1 5/ 𝑡𝑜 1/3) of the amount of reinforcing

steel. The depth of pre-stressed concrete members is usually less than the depth of

the reinforced concrete members for the same span length and loading conditions,

thereby the required casting concrete is less by about (20-25%) than in case of

reinforced concrete but the difference in initial cost is not proportional to the

difference in weight of materials. On the other hand, formwork or mould may be

more expensive, and the additional cost of the pre-stressing operation itself must be

considered. However, there is a little difference in the initial cost of reinforced and

pre-stress concrete, but the indirect savings ( long working live, avoidance of

maintenance, reduction of dead weight imposed on the supporting members … ext)

which accrue from the pre-stressing are often substantial and should be fully taken

into account.

Advantages of Prestressing

Prevents tension cracks under working loads to be used for water and oil tanks

and member subjected to corrosion conditions.

Smaller section will be required for design (the entire section remains

effective for stresses).

Larger span will be possible because of reduction in weight.

Deflection under working loads will be reduced (Cambering of the members).

Prestress reduces diagonal tension stresses at working loads. This led to use of

modified ( I and T sections).

Prestress concrete has more shear resistance, where as shear resistance of

R.C.C is less.


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Analysis of Prestress Member

Basic assumptions

1. Concrete is a homogenous material.

2. Within the range of working stress, both concrete & steel behave elastically,

notwithstanding the small amount of creep, which occurs in both the materials

under the sustained loading.

3. A plane section before bending is assumed to remain plane even after bending,

which implies a linear strain distribution across the depth of the member.

Analysis of prestress member

The stress due to prestressing alone are generally combined stresses due to the action

of direct load bending from an eccentrically applied load. The following notations

and sign conventions are used for the analysis of prestress members.

P: Prestressing force (Negative when compressive). (N)

e: Eccentricity of prestressing force. (mm)

M = P.e : Moment. (N.mm)

A: Cross-sectional area of the concrete member. (mm2)

I: Second moment of area of the section about its centroid. (mm4)

St , Sb: Section modulus of the top & bottom fibre respectively. (mm3)

ftop , fbot: Prestress in concrete developed at the top & bottom fibres. (MPa)

ct, cb: Distance of the top & bottom fibre from the centroid of the section. (mm)

r: Radius of gyration. (mm)


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(i) Concentric tendon

In this case, the load is applied concentrically and a compressive stress of

magnitude (P/A) will act through out the section. Thus the stress will generate in the

section as shown in the figure below.

Stress distribution for Concentric Prestressing

(ii) Eccentric tendon

Direct stress Bending stress Resultant stress

Stress distribution for Eccentric Prestressing

Thus the stresses developed at the top & bottom fibres of the beam can be

written as:

𝑓𝑡𝑜𝑝 = −𝑃

𝐴(1 −

𝑒 𝑐𝑡

𝑟2 ) 𝑓𝑏𝑜𝑡 = −𝑃

𝐴(1 +

𝑒 𝑐𝑏

𝑟2 )

S S


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Flexural Analysis

Pi : Initial prestressing force. (N)

Pe : Effective prestressing force ( after all losses). (N)

f : Stress. (MPa)

Initial Pre-stressed

𝑓𝑖𝑡𝑜𝑝 = −𝑃𝑖

𝐴(1 −

𝑒 𝑐𝑡

𝑟2)

𝑓𝑖𝑏𝑜𝑡 = −𝑃𝑖

𝐴(1 +

𝑒 𝑐𝑏

𝑟2)

Applying Beam Self-weight


𝐴(1 −

𝑒 𝑐𝑡

𝑟2) −

𝑀𝑔 𝑐𝑡

𝐼

𝑓𝑖𝑏𝑜𝑡 = −𝑃𝑖

𝐴(1 +

𝑒 𝑐𝑏

𝑟2) +

𝑀𝑔 𝑐𝑏

𝐼

Applying Full Service Loads

𝑓𝑡𝑜𝑝 = −𝑃𝑒

𝐴(1 −

𝑒 𝑐𝑡

𝑟2) −

𝑀𝑔 𝑐𝑡

𝐼−

𝑀𝑠 𝑐𝑡

𝐼

𝑓𝑏𝑜𝑡 = −𝑃𝑒

𝐴(1 +

𝑒 𝑐𝑏

𝑟2) +

𝑀𝑔 𝑐𝑏

𝐼+

𝑀𝑠 𝑐𝑏

𝐼

Asp

e

ct

cb

-Pi/A +Pi.e ct /I

-Pi.e cb /I

−𝑃

𝐴(1 −

𝑒 𝑐𝑡

𝑟2)

−𝑃

𝐴(1 +

𝑒 𝑐𝑏

𝑟2)

-Mg.ct /I

-Mg.cb /I

-Ms.ct /I

-Ms.cb /I

f top

f bot

+ = + + =

Initially

Self wt. Service load


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ACI-Allowable Stresses:

1. Immediately after Pre-stress Transfer;

(a) Ext. fiber stress in compression ………..................................................0.60fci

(b) Ext. fiber stress in compression at end of simply supported member …. 0.70fci

(c) Ext. fiber stress in Tension .......………...................................................0.25√fci

(d) Ext. fiber stress in Tension at end of simply supported member………...0.5√fci

2. At Service Loads

(a) Ext. fiber stress in compression …............0.45fc (Pre-stress+ Sustained load).

(b) Ext. fiber stress in compression …............0.6fc (Pre-stress+ Total load).

(c) Ext. fiber stress in Tension .......………......0.5√fc

3. Permissible Stress in Pre-stressing Tendons

(a) Due to jacking force …………..............................0.94fpy ≤ 0.80fpu

(b) After pre-stress transfer……….............................0.82fpy ≤ 0.74fpu

(c) Post tensioning tendons …………….….................0.7fpu

Example (1): For a simply supported symmetrically I-section beam with the

properties as follow:

I=5×109 mm4 , A=114000mm2 , r2=44000mm2 , h= 610mm , carries uniformly

distributed load (D+L) = 8 kN/m in additional to its own weight over on span length

of 12m, e=130mm , Pi=750kN , Pe=640kN.

Find the flexural stresses at mid-span and at ends due to initial prestress plus self wt.

and due to effective prestress plus full service load.

Solution:

Wg=A.γ = (114000×10-6

).24=2.7 kN/m

Mg=Wg.L2/8

=2.7× (12)2/8=48.6 kN.m

Mid-span stresses due to Initial prestressing and self-weight:


𝐴(1 −

𝑒 𝑐𝑡

𝑟2) −

𝑀𝑔 𝑐𝑡

𝐼

= −750000

114000(1 −

130 × 305

44000) −

48.6 × 106 × 305

5 × 109= −0.65 − 2.96 = −3.61 𝑀𝑃𝑎

𝑓𝑖𝑏𝑜𝑡 = −750000

114000(1 +

130 × 305

44000) +

48.6 × 106 × 305

5 × 109= −12.51 + 2.96 = −9.55𝑀𝑃𝑎


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Mid-span stresses after all losses and applying service loads:

Ms=Ws.L2/8

=8×122/8 = 144 kN.m


𝐴(1 −

𝑒 𝑐𝑡

𝑟2) −

𝑀𝑔 𝑐𝑡

𝐼−

𝑀𝑠 𝑐𝑡

𝐼

= −640000

114000(1 −

130 × 305

44000) −

48.6 × 106 × 305

5 × 109−

144 × 106 × 305

5 × 109

= −0.55 − 2.96 − 8.78 = −12.3 𝑀𝑃𝑎

𝑓𝑏𝑜𝑡 = −640000

114000(1 +

130 × 305

44000) +

48.6 × 106 × 305

5 × 109+

144 × 106 × 305

5 × 109

= −10.68 + 2.96 + 8.78 = 1.03 𝑀𝑃𝑎

At ends:

𝑓𝑖𝑡𝑜𝑝 = −0.65 𝑀𝑃𝑎 𝑓𝑖𝑏𝑜𝑡 = −12.51 𝑀𝑃𝑎

𝑓𝑡𝑜𝑝 = −0.55 𝑀𝑃𝑎 𝑓𝑏𝑜𝑡 = −10.68 𝑀𝑃𝑎

Example (2): For ex. (1) if fci=25MPa, and fc=35MPa. Check the resultant stress

with the ACI permissible stresses.

Solution:

At midspan stresses:

Initially 𝑓𝑖𝑡𝑜𝑝 = −3.61 𝑀𝑃𝑎 (𝐶𝑜𝑚𝑝. ) < 𝐴𝐶𝐼 𝑙𝑖𝑚𝑖𝑡 = 0.6 × 𝑓𝑐𝑖 = −15 𝑀𝑃𝑎 ∴ 𝑂𝐾

𝑓𝑖𝑏𝑜𝑡 = −9.55 𝑀𝑃𝑎 (𝐶𝑜𝑚𝑝. ) < 𝐴𝐶𝐼 𝑙𝑖𝑚𝑖𝑡 = 0.6 × 𝑓𝑐𝑖 = −15 𝑀𝑃𝑎 ∴ 𝑂𝐾

Finally 𝑓𝑡𝑜𝑝 = −12.3 𝑀𝑃𝑎 (𝐶𝑜𝑚𝑝. ) < 𝐴𝐶𝐼 𝑙𝑖𝑚𝑖𝑡 = 0.45 × 𝑓𝑐 = −15.75 𝑀𝑃𝑎 ∴ 𝑂𝐾

𝑓𝑏𝑜𝑡 = 1.03 𝑀𝑃𝑎 (𝑇𝑒𝑛𝑠. ) < 𝐴𝐶𝐼 𝑙𝑖𝑚𝑖𝑡 = 0.5 × √𝑓𝑐 = 2.96 𝑀𝑃𝑎 ∴ 𝑂𝐾

At ends: all stresses are OK with ACI limits.


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Example (3): For ex. (1) calculate stresses at (14⁄ ) span and check them with ACI

limitations.

Example (4): If the tendon profile in ex.(1) is as shown below, calculate stresses at

midspan, at ends and at (14⁄ ) span. Then check the stresses with ACI limitations.

Example (5): For ex. (1), find Pi (Initial prestress force) such that the bottom fiber

stress is finally zero, (15% losses) and check other stresses with ACI limitations.

Example (6): A rectangular concrete beam of cross-section 300 mm deep and 200

mm wide is prestressed by means of 15 wires of 5 mm diameter located 65 mm from

the bottom of the beam and 3 wires of diameter of 5 mm, 25 mm from the top.

Assuming the prestress in the steel Pe =840 MPa, calculate the stresses at the

extreme fibers of the mid-span section when the beam is supporting its own weight

over a span of 6 m. If a uniformly distributed live load of 6kN/m is imposed,

evaluate the maximum working stress in concrete.

centroid tendon

50mm

150mm

12m


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Solution:

Distance of the centroid of prestressing force from the base:

𝒚 = ((15 × 65) + (3 × 275)

18) = 100 𝑚𝑚

Eccentricity, e = 150-100= 50 mm

Prestressing force, P = (840×18×19.7) =3×105 N

Area of concrete section, A = (300 × 200) = 6×104

mm2

Second moment of area, I = 200×3003

/12 = 45×107

mm4

Section modulus (St & S

b) = (45×10

7

/150)=3×106

mm3

Reduce of gyration, r= 86.603 mm, r2=7500 mm

2

Self weight of the beam = (0.2×0.3×24) = 1.44kN/m

Moment due to self weight, Mg = 1.44×(6)2/8 = 6.48 kN.m

Moment due to service load Ms= 6×(6)2/8 = 27 kN.m


𝐴(1 −

𝑒 𝑐𝑡

𝑟2) −

𝑀𝑔 𝑐𝑡

𝐼−

𝑀𝑠 𝑐𝑡

𝐼

= −3 × 105

6 × 104(1 −

50 × 150

7500) −

6.48 × 106 × 150

45 × 107−

27 × 106 × 150

45 × 107

= −0 − 2.16 − 9 = −11.16 𝑀𝑃𝑎 (Comp.)

𝑓𝑏𝑜𝑡 = −𝑃𝑒

𝐴(1 +

𝑒 𝑐𝑡

𝑟2) +

𝑀𝑔 𝑐𝑡

𝐼+

𝑀𝑠 𝑐𝑡

𝐼

= −10 + 2.16 + 9 = 1.16 𝑀𝑃𝑎 (𝑇𝑒𝑛𝑠. )

+ + + = = _

_

+

_

_ _ _

+ + +

5

5

5

5

0

10 2.16

2.16 9

9

11.16

1.16

Prestress Self wt. Service load Resultant

stress


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Losses in Prestress

The initial prestressing concrete undergoes a gradual reduction with time from the

stages of transfer due to various causes. This is generally defined as total “Loss of

Prestress”. The various losses are explained below:

Types of losses in prestress:

Pretensioning

1. Elastic shortening of concrete (ES).

2. Creep of concrete (CR).

3. Shrinkage of concrete (SH).

4. Relaxation of stress in steel (RE).

Post-tensioning

1. No loss due to elastic shortening if all wires are simultaneously tensioned. If

the wires are successively tensioned, there will be loss of prestress due to

elastic shortening of concrete.

2. Creep of concrete.

3. Shrinkage of concrete.

4. Relaxation of stress in steel.

5. Friction.

6. Anchorage slip (ANC).

1. Loss due to elastic shortening of the concrete

As the prestress force is transferred to concrete, the member shortens and the

prestress steel shortens with it. Hence there is a loss of prestress. The loss of

prestress due to deformation of concrete depends on the modular ratio & the average

stress in concrete at the level of steel, then:


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𝑓𝑐𝑖𝑟 =𝑃𝑖

𝐴+

𝑃𝑖 . 𝑒2

𝐼−

𝑀𝑔. 𝑒

𝐼

𝐸𝑆 = 𝐾𝑒𝑠. 𝐸𝑠.𝑓𝑐𝑖𝑟

𝐸𝑐𝑖

Where:

𝑓𝑐𝑖𝑟: 𝑇ℎ𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 𝑖𝑛 𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒 𝑎𝑡 𝑠𝑡𝑒𝑒𝑙 𝑙𝑒𝑣𝑒𝑙.

𝐾𝑒𝑠 = 1.0 𝑓𝑜𝑟 𝑝𝑟𝑒𝑡𝑒𝑛𝑠𝑖𝑜𝑛𝑒𝑑 𝑚𝑒𝑚𝑏𝑒𝑟.

𝐾𝑒𝑠 = 0.5 𝑓𝑜𝑟 𝑝𝑜𝑠𝑡𝑒𝑛𝑠𝑖𝑜𝑛𝑒𝑑 𝑚𝑒𝑚𝑏𝑒𝑟.

𝐸𝑐𝑖: 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝑜𝑓 𝑒𝑙𝑎𝑠𝑡𝑖𝑐𝑖𝑡𝑦 𝑜𝑓 𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒.

𝐸𝑠: 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝑜𝑓 𝑒𝑙𝑎𝑠𝑡𝑖𝑐𝑖𝑡𝑦 𝑜𝑓 𝑠𝑡𝑒𝑒𝑙.

If the initial stress in steel is known, the percentage loss of stress in steel due to

elastic deformation of concrete can be computed.

Example (1): (Elastic deformation)

A pre-stressed concrete beam of 8m length, 100 mm wide and 300 mm deep, is pre-

tensioned by straight, wires carrying an initial force of 150kN at an eccentricity of

50 mm. The modulus of elasticity of steel and concrete are 210000 and 35000

N/mm2

respectively. Estimate the percentage loss of stress in steel due to elastic

deformation of concrete if the area of steel wires is 188 mm2

.

Solution:

Wg=A.γ = (0.1×0.3).24=0.72 kN/m

Mg=wg.L2/8

=0.72× (8)2/8=5.76 kN.m

I=100×(300)3/12=225×10

6 mm

4 , 𝑓𝑝𝑖 =

150×103

188= 800 𝑀𝑃𝑎

𝑓𝑐𝑖𝑟 =150000

100 × 300+

150000 × 502

225 × 106−

5.76 × 106 × 50

225 × 106= 5.39 𝑀𝑃𝑎

𝐸𝑆 = 𝐾𝑒𝑠 . 𝐸𝑠 .𝑓𝑐𝑖𝑟

𝐸𝑐𝑖

= 1 × 210000 ×5.39

35000= 32.34𝑀𝑃𝑎

∴ 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑙𝑜𝑠𝑠 𝑜𝑓 𝑠𝑡𝑟𝑒𝑠𝑠 𝑖𝑛 𝑠𝑡𝑒𝑒𝑙 =32.34

800× 100 = 4%


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2. Loss due to creep of concrete

The sustained prestress in the concrete of a prestress member results in creep of

concrete which is effectively reduces the stress in high tensile steel. The loss of

stress in steel due to creep of concrete can be estimated if the magnitude of creep-

coefficient is known, then:

𝐶𝑅 = 𝐾𝑐𝑟 .𝐸𝑠

𝐸𝑐. (𝑓𝑐𝑖𝑟 − 𝑓𝑐𝑑𝑠)

Where:

𝐾𝑐𝑟 = 2.0 𝑓𝑜𝑟 𝑝𝑟𝑒𝑡𝑒𝑛𝑠𝑖𝑜𝑛𝑒𝑑 𝑚𝑒𝑚𝑏𝑒𝑟.

𝐾𝑐𝑟 = 1.6 𝑓𝑜𝑟 𝑝𝑜𝑠𝑡𝑒𝑛𝑠𝑖𝑜𝑛𝑒𝑑 𝑚𝑒𝑚𝑏𝑒𝑟.

𝑓𝑐𝑑𝑠: 𝑆𝑡𝑟𝑒𝑠𝑠 𝑖𝑛 𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒 𝑎𝑡 𝑐. 𝑔. 𝑠. 𝑑𝑢𝑒 𝑡𝑜 𝑠𝑢𝑝𝑒𝑟 − 𝑖𝑚𝑝𝑜𝑠𝑒𝑑 𝑑𝑒𝑎𝑑 𝑙𝑜𝑎𝑑𝑠 𝑎𝑝𝑝𝑙𝑖𝑒𝑑 𝑎𝑓𝑡𝑒𝑟

𝑝𝑟𝑒𝑠𝑡𝑟𝑒𝑠𝑠𝑖𝑛𝑔.

The magnitude of creep coefficient varies depending upon the humidity, concrete

quality, duration of applied loading and the age of concrete when loaded.

Example (2): (Creep)

A post-tension concrete beam, 8m length of a rectangular section, 100 mm wide and

300 mm deep, is pre-stressed by five wires of 7 mm diameter located at an

eccentricity of 50 mm, the initial stress in the wires being 1200 N/mm2

. The

modulus of elasticity of steel and concrete are 210000 and 35000 N/mm2

respectively Estimate the percentage loss of stress in steel due to creep of concrete, if

the beam carrying a dead load of 3kN/m.

Solution:

𝑃𝑖 = 38.5 × 5 × 1200 = 23.1 × 104𝑁

𝑓𝑐𝑖𝑟 =231000

100 × 300+

231000 × 502

225 × 106−

5.76 × 106 × 50

225 × 106= 9 𝑀𝑃𝑎

𝑀𝑑 = 3 ×82

8= 24 𝑘𝑁. 𝑚 𝑓𝑐𝑑𝑠 =

24×106×50

225×106= 5.33 𝑀𝑃𝑎

𝐶𝑅 = 1.6 ×210000

35000× (9 − 5.33) = 35.23 𝑀𝑃𝑎


1200× 100 = 2.94%


18

3. Loss due to shrinkage of concrete

Factors affecting the shrinkage in concrete:

1. The loss due to shrinkage of concrete results in shortening of tensioned wires &

hence contributes to the loss of stress.

2. The shrinkage of concrete is influenced by the type of cement, aggregate & the

method of curing used.

3. Use of high strength concrete with low water cement ratio results in reduction in

shrinkage and consequent loss of prestress.

4. The primary cause of drying shrinkage is the progressive loss of water from

concrete.

5. The rate of shrinkage is higher at the surface of the member.

6. The differential shrinkage between the interior surfaces of large member may

result in strain gradients leading to surface cracking.

Hence, proper curing is essential to prevent cracks due to shrinkage in prestress

members. In the case of pretensioned members, generally moist curing is restored in

order to prevent shrinkage until the time of transfer. Consequently, the total residual

shrinkage strain will be larger in pretensioned members after transfer of prestress in

comparison with post-tensioned members, where a portion of shrinkage will have

already taken place by the time of transfer of stress. The loss of prestress due to

shrinkage of concrete can be obtained as follow:

𝑆𝐻 = 8.2 × 10−6𝐾𝑠ℎ . 𝐸𝑠 . (1 −0.06

25

𝑉

𝑆) . (100 − 𝑅𝐻)

Where : 𝑅𝐻: 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 ℎ𝑢𝑚𝑖𝑑𝑖𝑡𝑦

𝑉

𝑆: 𝑣𝑜𝑙𝑢𝑚𝑒 𝑡𝑜 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑟𝑎𝑡𝑖𝑜

𝐾𝑠ℎ = 1.0 𝑓𝑜𝑟 𝑝𝑟𝑒𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑚𝑒𝑚𝑏𝑒𝑟, 𝑓𝑜𝑟 𝑝𝑜𝑠𝑡𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑚𝑒𝑚𝑏𝑒𝑟 𝑠𝑒𝑒 𝑇𝑎𝑏𝑙𝑒 (1):

𝑻𝒂𝒃𝒍𝒆 (𝟏) 𝑉𝑎𝑙𝑢𝑒𝑠 𝑜𝑓𝐾𝑠ℎ 𝑓𝑜𝑟 𝑝𝑜𝑠𝑡𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑚𝑒𝑚𝑏𝑒𝑟

Time after curing to post-

tensioning (days) 1 3 5 7 10 20 30 60

Kes 0.92 0.85 0.8 0.77 0.73 0.64 0.58 0.45


19

Example (3): (Shrinkage)

A concrete beam of a rectangular section, 100 mm wide and 300 mm deep is pre-

stressed by a cable carrying an initial pre-stressing force of 300kN. The cross-

sectional area of the cable is 300 mm2

. Calculate the percentage loss of stress in the

cable due to shrinkage of concrete, RH=75%. Assuming the beam to be, (a) pre-

tensioned and (b) post-tensioned.

Assume Es =210 kN/mm

2

, age of concrete at transfer force is 14 days after curing.

Solution:

𝑉

𝑆=

𝐴. 𝐿

𝑃𝑒𝑟𝑚. 𝐿=

𝐴

𝑃𝑒𝑟𝑚.=

100 × 300

2 × (100 + 300)= 37.5 𝑚𝑚

For pre-tension beam:

𝑆𝐻 = 8.2 × 10−6𝐾𝑠ℎ . 𝐸𝑠 . (1 −0.06

25

𝑉

𝑆) . (100 − 𝑅𝐻)

𝑆𝐻 = 8.2 × 10−6 × 1 × 210000 (1 −0.06

25× 37.5) × (100 − 75)

= 39.18 𝑀𝑃𝑎

𝑓𝑝𝑖 =300 × 103

300= 1000 𝑁


1000× 100 = 3.92%

For post-tension beam:

Ksh from Table (1) by interpolation=0.694

𝑆𝐻 = 8.2 × 10−6 × 0.694 × 210000 (1 −0.06

25× 37.5) × (100 − 75)

= 27.19 𝑀𝑃𝑎


1000× 100 = 2.72%


20

4. Loss due to relaxation of stress in steel

Relaxation of a prestressing tendon depends upon the stress level in the tendon.

Basic relaxation values Kre for different kinds of steel are shown in Table (2).

However, because of other prestress losses, there is a continual reduction in

relaxation. The reduction in tendon stress due to elastic shortening of concrete

occurs instantaneously. On the other hand, the reduction of concrete due to creep and

shrinkage takes place in a prolonged period of time. The factor J in equation below

is specified to approximate these effects, then:

𝑅𝐸 = [𝐾𝑟𝑒 − 𝐽(𝑆𝐻 + 𝐶𝑅 + 𝐸𝑆)]𝐶

Where: Kre, J and C are taken from Tables 2 &3.


21

5. Loss of stress due to friction

For post-tensioned member friction against the wall of duct result in prestress loss

increases with distance from the jack. The magnitude of loss of stress due to friction

is of following types:

a. Loss due to curvature effect, which depends upon the tendon form or

alignment, which generally follows a curved profile along the length of the

beam.

b. Loss of stress due to wobble effect, which depends upon the local deviations in

the alignment of the cable. The wobble or wave effect is the result of

accidental or unavoidable misalignment, since ducts or sheaths cannot be

perfectly located to follow a predetermined profile throughout the length of

beam.

𝑃𝑥 = 𝑃𝑖. 𝑒−(𝜇𝛼+𝑘𝑥)

or 𝑃𝑥 = 𝑃𝑖. [1 − 𝜇𝛼 − 𝑘𝑥]

where:

𝑃𝑥: 𝑝𝑟𝑒𝑠𝑡𝑟𝑒𝑠𝑠𝑖𝑛𝑔 𝑓𝑜𝑟𝑐𝑒 𝑎𝑡 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑥 𝑓𝑟𝑜𝑚 𝑒𝑛𝑑

𝑃𝑖: 𝑝𝑟𝑒𝑠𝑡𝑟𝑒𝑠𝑠𝑖𝑛𝑔 𝑓𝑜𝑟𝑐𝑒 𝑎𝑡 𝑡ℎ𝑒 𝑗𝑎𝑐𝑘𝑖𝑛𝑔 𝑒𝑛𝑑

𝜇: 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑐𝑎𝑏𝑙𝑒 & 𝑑𝑢𝑐𝑡 (𝑐𝑢𝑟𝑣𝑎𝑡𝑢𝑟𝑒 𝑐𝑜𝑒𝑓𝑓. )𝑇𝑎𝑏𝑙𝑒 (4)

𝑘: 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑓𝑜𝑟 𝑤𝑎𝑣𝑒 𝑒𝑓𝑓𝑒𝑐𝑡 (𝑤𝑜𝑏𝑏𝑙𝑒 𝑐𝑜𝑒𝑓𝑓. )𝑇𝑎𝑏𝑙𝑒 (4)

𝛼: 𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑖𝑣𝑒 𝑎𝑛𝑔𝑙𝑒 𝑖𝑛 𝑟𝑎𝑑𝑖𝑎𝑛 𝑡ℎ𝑟𝑜𝑢𝑔 𝑡ℎ𝑒 𝑡𝑎𝑛𝑔𝑒𝑛𝑡 𝑡𝑜 𝑡ℎ𝑒 𝑐𝑎𝑏𝑙𝑒 𝑝𝑟𝑜𝑓𝑖𝑙𝑒 ℎ𝑎𝑠

𝑡𝑢𝑟𝑛𝑒𝑑 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑎𝑛𝑦 𝑡𝑤𝑜 𝑝𝑜𝑖𝑛𝑡𝑠 𝑢𝑛𝑑𝑒𝑟 𝑐𝑜𝑛𝑠𝑖𝑑𝑒𝑟𝑎𝑡𝑖𝑜𝑛


22

Table (4) Friction Coefficients for Posttensioning Tendons

Type of Tendon Wobble Coeff. (k) per

meter

Curvature Coeff. (μ)

Tendons in flexible metal

sheathing:

Wire tendons

7-wire strands

High strength bars

0.0033-0.0049

0.0016-0.0066

0.0003-0.0020

0.15-0.25

0.15-0.25

0.08-0.30

Tendons in rigid metal

duct

7-wire strand

0.0007

0.15-0.25

Pregreased tendons

Wire tendon and

7-wire strand

0.001-0.0066

0.05-0.15

Mastic-coated tendons

Wire tendon and

7-wire strand

0.0033-0.0066

0.05-0.15

Example (4): (Friction)

A concrete beam of 10 m span, 100 mm wide and 300 mm deep, is pre-stressed by 3

cables type (7-wire strand in flexible metal sheathing). The area of each cable is 200

mm2

and the initial stress in the cable is 1200 N/mm2

. Cable 1 is parabolic with an

eccentricity of 50 mm above the centroid at the supports and 50 mm below at the

center of span. Cable 2 is also parabolic with zero eccentricity at supports and 50

mm below the centroid at the center of span. Cable 3 is straight with uniform

eccentricity of 50 mm below the centroid. If the cables are tensioned from one end,

estimate the percentage loss of stress in each cable due to friction.

30

0m

m

100mm 10 m

1

2

3

50mm

50mm e=50mm


23

Solution:

Equation of parabola is given by: 𝑦 =4𝑒

𝐿2 (𝐿 − 𝑥)

Slope at ends at (𝑥)=0 𝑑𝑦

𝑑𝑥⁄ =

4𝑒

𝐿2(𝐿 − 2𝑥) =

4𝑒

𝐿

For cable 1

𝑒 = 100 𝑚𝑚

𝑠𝑙𝑜𝑝𝑒 𝑎𝑡 𝑒𝑛𝑑𝑠 =4 × 100

10000= 0.04

∴ 𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑖𝑣𝑒 𝑎𝑛𝑔𝑙𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑠, 𝛼 = 2 × 0.04 = 0.08 𝑟𝑎𝑑

For cable 2

𝑒 = 50 𝑚𝑚

𝑠𝑙𝑜𝑝𝑒 𝑎𝑡 𝑒𝑛𝑑𝑠 =4 × 50

10000= 0.02

∴ 𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑖𝑣𝑒 𝑎𝑛𝑔𝑙𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑠, 𝛼 = 2 × 0.02 = 0.04 𝑟𝑎𝑑

Initial pre-stressing force in each cable 𝑃𝑖 = 1200 × 200 = 240000 𝑁

𝑃𝑥 = 𝑃𝑖. 𝑒−(𝜇𝛼+𝑘𝑥) 𝑓𝑜𝑟 𝑠𝑚𝑎𝑙𝑙 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓(𝜇𝛼 + 𝑘𝑥)𝑤𝑒 𝑐𝑎𝑛 𝑤𝑟𝑖𝑡𝑒:

𝑃𝑥 = 𝑃𝑖. [1 − (𝜇𝛼 + 𝑘𝑥)]

𝜇 = 0.2 , 𝑘 = 0.0041/𝑚 From Table (4)

∴ 𝐿𝑜𝑠𝑠 𝑜𝑓 𝑝𝑟𝑒𝑠𝑡𝑟𝑒𝑠𝑠, 𝑃𝑥 = 𝑃𝑖. [𝜇𝛼 + 𝑘𝑥]

𝐶𝑎𝑏𝑙𝑒 1 = 1200. [0.2 × 0.08 + 0.0041 × 10] = 68.4 𝑀𝑃𝑎 = 5.7%

𝐶𝑎𝑏𝑙𝑒 2 = 1200. [0.2 × 0.04 + 0.0041 × 10] = 58.8 𝑀𝑃𝑎 = 4.9%

𝐶𝑎𝑏𝑙𝑒 3 = 1200. [0 + 0.0041 × 10] = 49.2 𝑀𝑃𝑎 = 4.1%

∴ 𝑇𝑜𝑡𝑎𝑙 𝐿𝑜𝑠𝑠 𝑜𝑓 𝑝𝑟𝑒𝑠𝑡𝑟𝑒𝑠𝑠 𝑑𝑢𝑒 𝑡𝑜 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 = 5.7 + 4.9 + 4.1 = 14.7%


24

6. Loss due to Anchorage slip (ANC)

For the posttensioned member due to the deformation of anchorage fixtures at

transfer of prestress, the magnitude of loss of stress due to the slip in anchorage is

computed as follows:

∆=𝑃𝐿

𝐴𝐸𝑠

Where:

∆: 𝑠𝑙𝑖𝑝 𝑖𝑛 𝑎𝑛𝑐ℎ𝑜𝑟𝑎𝑔𝑒, 𝑚𝑚

𝑃: 𝑃𝑟𝑒𝑠𝑡𝑟𝑒𝑠𝑠 𝑓𝑜𝑟𝑐𝑒 𝑖𝑛 𝑡ℎ𝑒 𝑐𝑎𝑏𝑙𝑒, 𝑁

𝐿: 𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑐𝑎𝑏𝑙𝑒, 𝑚𝑚

𝐴: 𝐶𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑎𝑏𝑙𝑒, 𝑚𝑚2

𝐸𝑠: 𝑀𝑜𝑑𝑢𝑙𝑢𝑠𝑒 𝑜𝑓 𝑒𝑙𝑎𝑠𝑡𝑖𝑐𝑖𝑡𝑦 𝑜𝑓 𝑠𝑡𝑒𝑒𝑙, 𝑁/𝑚𝑚2

Hence, Loss of stress due to anchorage slip =𝑃

𝐴=

𝐸𝑠∆

𝐿

Example (5): (Anchorage slip)

A concrete beam is post-tensioned by a cable carrying an initial stress of 1000

N/mm2

. The slip at the jacking end was observed to be 5 mm. The modulus of

elasticity of steel is 210 kN/mm2

. Estimate the percentage loss of stress due to

anchorage slip if the length of the beam is 30 m.

Solution:

∴ 𝑙𝑜𝑠𝑠 𝑜𝑓 𝑠𝑡𝑟𝑒𝑠𝑠 𝑑𝑢𝑒 𝑡𝑜 𝑎𝑛𝑐ℎ𝑜𝑟𝑜𝑔𝑒 𝑠𝑙𝑖𝑝 =𝐸𝑠 . ∆

𝐿

=210000 × 5

30000= 35 𝑀𝑃𝑎

∴ 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑙𝑜𝑠𝑠 𝑜𝑓 𝑠𝑡𝑟𝑒𝑠𝑠 =35

1000100 = 3.5 %


25

Limiting Maximum Loss

ACI- ACSE Committee recommended the following max. loss estimates

Type of strand Max. loss (MPa)

Stress-relived strand 345

Low-relaxation strand 275

Example (6): For the pre-tensioned beam over a simply supported condition of 20m

length as shown in the figure below and its carrying a service dead load of 7.6 kN/m

and a service live load of 7 kN/m applied at 30 days, fpi=0.75fpu , where

fpu=1860MPa, fci=31 MPa, fc=41MPa, r.h=75%, 12.7mm dia of stress-relived

strands. Es=200000 MPa. Estimate all pre-stress losses.

Solution:

A=286×103 mm

2

I=3.39×1010

mm4

wg=286×103×10

-6×24=6.86 kN/m Mg=6.86×20

2/8=343 kN.m

(1) 𝐸𝑆 = 𝐾𝑒𝑠. 𝐸𝑠.𝑓𝑐𝑖𝑟

𝐸𝑐𝑖

Pi=0.75×1860×1980=2.762×106 N

𝑓𝑐𝑖𝑟 =2.762×106

286×103+

2.762×106×3552

3.39×1010−

343×106×355

3.39×1010= 16.34 𝑀𝑃𝑎

𝐸𝑆 = 1 × 200000 ×16.34

4700√31= 124.9 𝑀𝑃𝑎

650 mm 150 mm

145 mm

175 mm

550 mm

175 mm

Aps=1980 mm2


26

(2) 𝐶𝑅 = 𝐾𝑐𝑟 .𝐸𝑠

𝐸𝑐. (𝑓𝑐𝑖𝑟 − 𝑓𝑐𝑑𝑠)

𝑀𝑑 = 7.6 ×202

8= 380𝑘𝑁. 𝑚 𝑓𝑐𝑑𝑠 =

380×106×355

3.39×1010= 3.98 𝑀𝑃𝑎

𝐶𝑅 = 2.0 ×.200000

4700√41× (16.34 − 3.98) = 164.3 𝑀𝑃𝑎

(3) 𝑆𝐻 = 8.2 × 10−6𝐾𝑠ℎ . 𝐸𝑠 . (1 −0.06

25

𝑉

𝑆) . (100 − 𝑅𝐻)

= 8.2 × 10−61 × 200000 × (1 −0.06

25×

286000

3900) . (100 − 75)

= 33.8 𝑀𝑃𝑎

(4) 𝑅𝐸 = [𝐾𝑟𝑒 − 𝐽(𝑆𝐻 + 𝐶𝑅 + 𝐸𝑆)]𝐶

Kre=138, J=0.15 and C=1.45 from Tables 2&3

𝑅𝐸 = [138 − 0.15(33.8 + 164.3 + 124.9)]1.45

𝑅𝐸 = 129.8 𝑀𝑃𝑎

∴ 𝑇𝑜𝑡𝑎𝑙 𝑙𝑜𝑠𝑠𝑒𝑠 = 124.9 + 164.3 + 33.8 + 129.8 = 452.8 𝑀𝑃𝑎 > 345𝑀𝑃𝑎 ∴ 𝑁𝑜𝑡 𝑂𝐾

reinforced concrete design (ii) dr. hussain a. jabir · reinforced concrete design (ii) dr. hussain...

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