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Analysis of flexure in pre-stressed concrete

Flexure causes normal stresses in opposite sign on either side o f neutral axis. An effective distribution of pre-stress can be achieved by adopting optimum CG for the pre-stressing steel in the tensile zone. Analysis of flexure can be done using any of the three elastic theory approaches – Theory of flexure approach / Line of thrust approach / Load balancing (free body) approach.

Theory of Flexure Approach Line of Thrust Approach Load Balancing (Free body) Approach

Flexural stresses indicated by the BMs caused by pre-stressing force, DL and LL are computed and considered with the direct stresses caused by the pre-stressing.

Normal stresses at any cross-section are evaluated as line of thrust at determined location and the whole section is evaluated as an eccentrically loaded section. (Typical approach used by conventional RCC design).

Member is considered to be a self straining system – pre-stressing steel and concrete subjected to forces of equal magnitude but in opposite direction. First the free body of the pre-stressing steel is considered and from that the forces on the free body of the concrete is evaluated. This approach is quite useful in statically indeterminate structures.

Initial Stage:

In a SSB, sagging BM results and hence the centroid of the cable is placed below the centroid of the section for the major portion of the member. This produces direct compression and the hogging bending moment in the pre-stressed beam.

Pre-stressing stress is maximum at the time of transfer and then over time gradually reduces by way of losses of pre-stress.

DL, LL and other loads cause their BMs at various stages of erection, loading and usage. Pattern of load distribution, and consequently the elastic strains in the member are effected varyingly and in

Initial stage:

Centre of compression in an un-cracked condition (which is the case here) at any stage is given by the equation;

eC=−M R

C=

−M R

P=eP−

MP

In a SS beam, the dad load produces sagging BM MD. Hence, for the initial stage, the eccentricity of the centre of compression is expressed as

eC=e P−M D

Pi

Earlier we saw that in an externally unloaded

Because of tension in the pre-stressing cables, they tend to straighten and tend to move up. The surrounding concrete restarain this pressure. This result in entire beam is thrusting upward (upward pressure – taut) which shall balance entire transverse load acting on the beam.

In the unloaded condition, the forces acting on pre-stressing steel and the forces acting on the concrete are equal and opposite. Here, the argument is put forward that the shear and the flexural stiffness of the cables are negligible, they can carry only axial forces.Key assumptions are;

1. Pre-stressing force is uniform throughout the cable.

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continuous manner.Initial and final stages of loading are of concern to us. In the initial stages Pi is maximum and the self weight and some placement loading and permanent loads would be present. In the final stages, the effective pre=stress P along with the DL, LL and other permanent loads (SERVICE LOADS) would be present.

Figure below shows a typical cross-section of a flexural member, being subjected to a pre-stressing force Pi with an eccentricity of ep.

y

yT

C eP

yB

y

f T=−Pi

A+

(−P ieP ) (− yT )I

= −Pi

A [1− eP yT

k2 ]f B=

−Pi

A+

(−P ieP ) ( yB )I

= −Pi

A [1+ eP yB

k2 ]At the bottom fiber, below the centroidal line, direct and the bending stresses have the same sign and at the top fiber (above the centroidal line) theses stress3es will have the opposite signs.

Therefore, eccentric pre-stress produces high compressive strength at the bottom fibre and lower compressive or tensile stress at the top fibre.

member, M = 0 and hence, eC=e P. However, in this case, we see that the term involving dead load

moment MD and Pi, M D

P ishifts the center of

thrust upwards; i.e eC<eP. We shall say that an internal lever arm results between resultant compressive and tensile forces in the cross-section of the beam. This eccentricity of the line of thrust,

represented as j DP=M D

P i

renders the

evaluation of eC, which shall be used in the

equations for f T , f B in the place of eP as the resultant forces’ momentum is with respect to the centroidal line of forces and not with respect to the line of forces.

f T=¿ −Pi

A [1− eC yT

k2 ]f B=¿

−Pi

A [1+ eC yB

k2 ]

2. Slope of the cable line is small that sinθ=tanθ=θ and cosθ=1

3. Resulting transverse load acting throughout the beam does not affect the pre-stressing force P.

If S is the sag of cable line, span of the beam is L, the cable line has got eccentricity of ePA and ePB at ends A (left face) and B(right face), then the slopes at A and B are given by;

θA=4 Sl

+ePB−ePA

l

θB=−4 S

l+

ePB−ePA

l

Equation of the cable line is

eP=e pA+(e pB−e pA ) xl+ 4 S

l2x (l−2 x )

Curvature of cable line is

dθdx

=d2 ep

d x2=−8S

l2

We see that the upward pressure exerted by the cable on the concrete beam is uniform and is given by the equation

w p=8 PS

l2

Special case 1:For a cable symmetrical about the mid-span, we have e pA=epB and hence

eP=e pA+4 S

l2x ( l−2 x )

θA=θB=4 Sl

Special case 2:

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Incorporating the dead load moment MD, the stresses at the top and bottom fibres in the initial stage are stated as

f T=¿ −Pi

A [1− eP yT

k2 ]−M D yT

I

f B=¿ −Pi

A [1+ eP yB

k2 ]+ M D yB

I

For a kink in the cable, there is a sudden change in the direction of the cable δθ, change in the cable angle, then the force at the kink exerted by the cable on the concrete is W p=Pδθ.

If the kink downwards, then the force exerted by the concrete on cable is downward and the force exerted by the cable on concrete is upward. P Wp exerted by concrete on cable

P δθ

Wp exerted by cable on concrete

If the kink upwards, then the force exerted by the concrete on cable is upward and the force exerted by the cable on concrete is downward.

Wp exerted by cable on concrete

δθ P P Wp exerted by concrete on cable

Final Stage:

In the final stage, the forces that act on the section are the effective pre-stress (initial pre-stres less the losses in pre-stress) and full working or service loads (DL, LL, etc.).

Final Stage:

DL and LL at final stage produces sagging bending moment (MD + ML) in SS beam. Eccentricity of the centre of compression at final stage is expressed as

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Taking into account the moment due to service loads ML, the stress levels in the top and bottom fibres can be expressed as

f T=¿ −Pi

A [1− eP yT

k2 ]−(M ¿¿ D+ M L) yT

I¿

f B=¿ −Pi

A [1+ eP yB

k2 ]+(M ¿¿ D+M L) yB

I¿

It shall be seen that at initial stage, compressive stress at the bottom fiber is high and more or less equal to the working stress. Stress at the top fiber is zero or at the most the working stress in tension.

In the final stage, the stress at top fiber is the working stress in compression and in the stress in the bottom fiber is zero or at the most te working stress in tension.

eC=e P−M D+M L

P

Note that the denominator is P, the net force after losses and additions to the initial pre-stress force; and

not Pi, the initial pre-stress force.

As in the initial stages, we shall see that the stresses in the top and bottom fibers are again

f T=¿ −PA [1− eC yT

k2 ]f B=¿

−PA [1+ eC y B

k2 ]Note that the numerator term is P, the net force after losses and additions to the initial pre-stress force; and

not Pi, the initial pre-stress force.

Exercise:A simply supported pre-stressed beam of span 20 m having a cross section as below; yT

yB eP

Top width being 1000 mm and the bottom width is 600 mm and the height is 1200 mm. It carries a self weight of 24 kN/m at initial stage. Determine the prestressing force required to induce a net stress

Exercise:Evaluate and check for the initial and final stresses for the problem given. Use line of

We take the MD, k2 from the previous working.

MD = 1.2 MN.mk2 = 0.42 m2

Using the equations for f T , f B

Exercise:For the section given, the cable is having eccentricities in face A and face B as 0.1 m and 0.2 m. If the pre-stressing force is 5 MN and the uniform load is 0.08 MN/m including the self weight, analyze the stresses in mid-span.

0.1m 0.4m 0.2m

10m 10m

Sectional properties are;A = 0.96 m2

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equal to ZERO and -20 MPa at the top and bottom fibres respectively.

For a trapezoidal cross-section, by convention, a < b a = 0.6 m and b = 1.0 m

Cross-sectional area

A = (a+b )2

h=(0.6+1.0 )

2x1.2 = 0.96 m2

yT=h3 ( 2a+b

a+b )=1.23 x ((2x 0.6)+1.00.6+1.0 )=0.55m

Therefore yB=h− yT

¿1.2−0.55=0.65m

Moment of inertia about the centrodidal axis, (XX direction)

I=h3 (3a+b )12

=1.23 x ( (3 x0.6 )+(1.0 ) )

12=0.4032m4

Radius of gyration about the centroidal axis in XX

direction is k 2= IA

=0.40320.96

=0.42m2

Self weight = 24 kN/m 0.024 MN/m (uniform load over the span) Bending moment at mid span (maximum) due to self weight is given by the equation

M D=w l2

8=0.024 x202

8=1.2MN .m

M L is zeroas apart from self weight, no other load is specified here.

f T=¿ −Pi

A [1− ec yT

k2 ]=

−Pi

0.96 [1− ec x0.55

0.42 ] = 0 (given)

f B=¿ −Pi

A [1+ ec yB

k2 ]¿

−Pi

0.96 [1+ ec x o .65

0.42 ] = - 20 MPa

Solving we get,Pi=8.79MN∧ec=0.764m

In the final stages, we have obtained from the previous approach, eP=0.321m

Substituting the values in the equation below,

eC=e P−M D

P 0.764=0.321−

1.2P

We get the value of P (the net pre-stressing force) as - 2.70 MPa (compression)

yT=0.55myB=0.65m

I=0.4032m4

Given: P = 5 MN

P θA=P[ 4Sl

+e pB−epA

l ]= 5 x [ 4 x0.4

20+ 0.2−0.1

20 ] = 0.425 MN

+ve therefore upward.

P θB=P[ 4 Sl

−e pB−e pA

l ]= 5 x [ 4 x0.4

20−0.2−0.1

20 ] = 0.375 MN

+ve therefore upward.

Uniform load due to pre-stress on the beam is

w p=8 PS

l2 = 8 x5 x0.4

202 = 0.04 MN/m

Now we shall see using the free-body considerations, we deduce that an external load of 0.08 MN/m and the support reactions at A and B as 0.8 MN and 0.8MN magnitude are the resultants.

BM at mid-span is

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Using the equations for f T , f B

f T=¿ −Pi

A [1− eP yT

k2 ]−(M ¿¿ D+ M L) yT

I¿

= −Pi

0.96 [1− eP x0.55

0.42 ]−1.2 x0.550.4032

= 0 (given)

f B=¿ −Pi

A [1+ eP yB

k2 ]+(M ¿¿ D+M L) yB

I¿

¿ −Pi

0.96 [1+ eP x o .65

0.42 ]+1.2 x 0.650.4032

= - 20

MPa

Solving we get,Pi=8.79MN∧eP=0.321m

−P+( RA−P θA ) l2−(W −w p ) l

2l4

= −0.5+ (0.8−0.425 ) x 10−(0.08−0.04 ) x 10x 5

= 1.25 MN.m

Stresses at fibers are as follows;

Stress at top fibre f T=−PA

+ M yT

I

= −50.96

+1.25 x (−0.55)0.4032

= - 7.458 MPa

Stress at bottom fibre

f B=−P

A+ M yT

I

= −50.96

+1.25 x 0.650.4032

= - 3.363 MPa

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