When we cut the tendons of the pre-tensioned steel, the steel undergoes compression, therefore the steel should have compressive strain. But when we calculate the residual strain, the strain in steel and the compressive strains in concrete due to shrinkage/creep are considered opposite in nature.Why?


3 Answers 3


Unless I'm misunderstanding your are referring to this process of pre-stressing concrete, where

  1. a steel reinforcement/tendon is put under tensile loading
  2. the concrete is cast
  3. the steel reinforcement/tendon is released.

enter image description here

In step 3, (i.e. when the tendons are released), the tendons which is extended tries to return to his original size (that is what you probably refer to as When we cut the tendons of the pre-tensioned steel, the steel undergoes compression, therefore the steel should have compressive strain. However, the steel tendon is never in compression. Actually, it is always in tension and tries to return to its "un-strained" condition.

However, because now steel is encased in concrete, shear forces from the concrete resist the movement. So the steel under tension tries to relax by shrinking to its original state, but is resisted by the initially relaxed concrete. So what happens is that in the end there is a compromise: a slightly less tensed steel tendon and a slightly compressed concrete.

So the steel reinforcement is still under tensile strain, while concrete is under compressive strain.

enter image description here


  • Step 1: Pre-stressing tendon

Assume a steel reinforcement tendon 20[mm] diameter(Probably a bit too much but for calculation simplicity), with a length of 10 [m]. If you apply a tensile force of 50kN, then a stress will develop. The stress will be: $$\sigma_t=\frac{F}{A}\approx 160[MPa]$$

The corresponding strain $\epsilon$, and extension $\Delta L$

$$\epsilon_t = \frac{\sigma}{E}\approx 0.0008 \qquad \Delta L_t = \epsilon *L\approx 8[mm] $$

  • 2. When the concrete is cast and left to solidify, the steel stresses remain the same, and the stress on the concrete is close to zero (there maybe some residual stresses during shrinking but we'll neglect those).

Lets assume that around that one tendon the unit cell is a rectangular concrete crosssection of 100 [mm]x100[mm]x 10[m], with $E_c = 50 [GPa]$.

  • 3. Cutting of the tendons:

When the tendon is cut then the steel is trying to release all the elastic energy it accumulated, by shrinking by 8[mm]. However concrete is resisting. Immediately before the tendon is cut the force in the steel is 50[kN]. After the tendon is cut and while it shrinks then force is reduced.

Let's assume that the steel extension is halved (to 4[mm]), then the force will be 25[kN]. (this is not exactly accurate, because forces are distributed along the cable, and you can have end effects, but it's easier to explain this way). That means that the steel is trying to shrink and therefore it is exerting a compressive force of 25kN to the concrete.

Now, lets assume that the steel shrinkage is a[m]. At that point, the concrete will be compressed (roughly) by a [m], so its strain will be

$$\epsilon_c=\frac{\Delta L}{L}=\frac{a m}{10[m]}=a/10$$

Therefore, the stress on the concrete will be:

$$ \sigma_c = E_c \cdot\epsilon_c$$

And the Force developed by the concrete will be: $$F_c = \sigma_c\cdot A_c = \frac{a E_c A_c}{10}$$

At the same time the force on the steel will be reduced by $\frac{a E_s A_s}{10}$, and the total force from the steel will be equal to $$F_s = 50kN - \frac{a E_s A_s}{10}$$

At that point, because the forces of steel and concrete will need to be equal and opposite you'd get:

$$F_s + F_c =0 $$

$$ 50kN - \frac{a E_s A_s}{10} - \frac{a E_c A_c}{10}=0$$ $$ \frac{a }{10}(E_s A_s - E_c A_c)=50kN$$ $$ a =\frac{50kN}{(E_s A_s + E_c A_c)} *10[m]$$ $$ a =0.9139[mm]$$

So, if the length of the steel cable reduces by 0.914[mm] (from the initial 8), then the steel tendon will be compressing the concrete with a force of approximattely $44259 N$, and the concrete will be resisting by an equal and opposite force.

The steel tendon will still be extended by approximately (8-0.914)[mm], while the concrete compressed by 0.914[mm].

  • $\begingroup$ Can you please shed some more light on the development and nature of stresses and strains in the pre-stressed concrete? $\endgroup$ Commented Jan 15, 2021 at 17:01
  • $\begingroup$ I've tried giving a simplified numerical example. $\endgroup$
    – NMech
    Commented Jan 15, 2021 at 18:09

When a tendon is cut is releases some of the tension near the cut . It does not create any compression in the steel.

  • $\begingroup$ That helped a lot. Thanks $\endgroup$ Commented Jan 15, 2021 at 18:23

Because the steel area in RC beams is much smaller than that of concrete. The neutral axis is unnecessarily too high and a lot of concrete below it is useless, just sitting there adding to the mass. This is due to the fact that the low rebar has to expand for it to take the stress shifting the neutral axis up.

Straight prestress bars

They put pre-tensioned bars in the beam before concrete is poured, held tight by jacks calibrated. These bars once released from the forms are cut from the jigs and try to shrink and will pull in the concrete with them, causing the beam to camber up a bit and make the neutral axis shift down bringing a large portion of the upper section of the concrete ready to take moment compression loads. But nevertheless, they still keep much of their tensile prestress, or they would be useless. They do this by two means. Either by bound stress or by the high strength jacks holding them at the ends.

Let's say the design calls for 65ksi tendon tension after the release. Depending on the type of concrete and crosssection of the beam, it is easy to calculate how much extra tensin the jack needs to impart on the tendon before the release, say 80ksi.

How the prestress works

The tension left in pre or post-tension bars will bring the neutral axis much lower. This does two things.

  • Saving much weight thus allowing wider spans. because a lot of useless concrete is eliminated, especially important in earthquake resistance.

  • Because the beam deflects much lass, hence rendering the structure stiffer and impede the propagation of hairline cracks and fatigue.

  • $\begingroup$ Much appreciated. $\endgroup$ Commented Jan 16, 2021 at 13:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.