# Secondary Prestress Effects

## Secondary Prestress on Continuous Beams

When pre-stress is applied to a beam it will often be applied at an eccentricity to the neutral axis, giving rise to a moment being applied to the beam,
typically prestress is applied to the bottom of the beam causing a hogging moment. This moment applied to the beam is known as the *primary prestress*, it is the direct
effect of the prestress causing a moment on the beam.

When a simply supported beam is prestressed it will typically hog upwards, as the beam is simply supported then there is no restraint to this movement and the beam simply deflects.
If the beam is part of a continuous span, then the whole length of the beam over multiple spans will want to hog upwards when prestressed, however the movement is restrained by the intermediate supports.
This restraint prevents the deflection of the beam an causes additional moments on the beam, this is called *secondary prestress* can lead to significant additional moments being applied to the beam.

This article aims to explain how to calculate these secondary effects for both a two span and a four span structure.

#### Two Span Continuous Beam

To calculate the additional secondary moments that must be applied to a continuous pre-stressed, beam the reactions at each of the support required to restrain the beam must be calculated.
To do this we need to establish the amount the beam would like to deflect upwards due to the prestress and then from that deflection calculate a reaction that would need to be applied to
prevent the deflection, *I.e. Deflection upwards due to presstress - Deflection downwards due to restrain reaction = 0. *
We start by establishing the deflection function for a beam under uniform moment, M, arising from the prestress times the prestress eccentricity from the neutral axis, M = Pe.

The deflection function is found as follows:

$$EI\frac{d^2v}{dx^2} = \text{Moment Function} = Pe$$ $$EI\frac{dv}{dx} = Pex + A $$ $$EIv = \frac{Pex^2}{2} + Ax + B $$ $$\text{Apply boundary conditions to establish coefficients A and B.}$$ $$\qquad\text{At x = 0, v = 0}$$ $$\qquad\qquad\text{Therefore, B = 0}$$ $$\qquad\text{At x = L/2,} \frac{dv}{dx} = 0$$ $$\qquad\qquad0 = \frac{PeL}{2} + A$$ $$\qquad\qquad\text{Therefore, } A = -\frac{PeL}{2}$$ $$\text{Therefore, the full expression for deflection is:}$$ $$\qquad EIv = \frac{Pex^2}{2} - \frac{PeLx}{2} = \frac{Pe}{2}(x^2 - Lx)$$ $$\text{At } x = L/2$$ $$\qquad EIv = \frac{PeL^2}{8} - \frac{PeL^2}{4} = -\frac{PeL^2}{8}$$

We now need to calculate the deflection function for a beam under a midspan point load. The point load has the same effect as the reaction of the intermediate support. Square brackets in the equation below are Macaulay Brackets, when the value in the square brackets is negative then the bracket term is equal to zero and ignored, when the value in the brackets is positive, the brackets behave as a normal term. The whole bracket acts as the "x" term for purposes of intergration, not the value in front of the bracket.

The deflection function is found as follows:

$$EI\frac{d^2v}{dx^2} = \text{Moment Function} = \frac{Rx}{2} - R\left[x-\frac{L}{2}\right]$$ $$EI\frac{dv}{dx} = \frac{Rx^2}{4} - \frac{R}{2}\left[x-\frac{L}{2}\right]^2 + A $$ $$EIv = \frac{Rx^3}{12} - \frac{R}{6}\left[x-\frac{L}{2}\right]^3 + Ax + B $$ $$\text{Apply boundary conditions to establish coefficients A and B.}$$ $$\qquad\text{At x = 0, v = 0}$$ $$\qquad\qquad\text{Therefore, B = 0}$$ $$\qquad\text{At x = L, v = 0}$$ $$\qquad\qquad0 = \frac{RL^3}{12} - \frac{R}{6}\left[L-\frac{L}{2}\right]^3 + AL$$ $$\qquad\qquad0 = \frac{RL^3}{12} - \frac{R}{6}\left[\frac{L}{2}\right]^3 + AL$$ $$\qquad\qquad0 = \frac{RL^3}{12} - \frac{RL^3}{48} + AL$$ $$\qquad\qquad0 = \frac{4RL^2}{48} - \frac{RL^2}{48} + A$$ $$\qquad\qquad\text{Therefore,} A = -\frac{3RL^2}{48} = -\frac{RL^2}{16}$$ $$\text{Therefore, the full expression for deflection is:}$$ $$\qquad EIv = \frac{Rx^3}{12} - \frac{R}{6}\left[x-\frac{L}{2}\right]^3 -\frac{RL^2x}{16} $$ $$\text{At } x = L/2$$ $$\qquad EIv = \frac{RL^3}{96} - \cancel{\frac{R}{6}\left[\frac{L}{2}-\frac{L}{2}\right]^3} -\frac{RL^3}{32} $$ $$\qquad\text{Note: The struckthrough term above is ignored here as the value in the Macaulay brackets is zero or less} $$ $$\qquad EIv = \frac{RL^3}{96} - \frac{3RL^3}{96} = -\frac{RL^3}{48} $$

To find the reaction that occurs from the prestressing moment, we need to set the deflection due to the prestress moment to be equal to the deflection from the restraint reaction and solve to find the reaction, R for a given prestress moment, Pe.

At the intermediate pier point of restraint:

$$v_{prestress} + v_{reaction} = 0 $$ $$v_{prestress} = -v_{reaction} $$ $$-\frac{PeL^2}{8} = \frac{RL^3}{48} $$ $$-\frac{6Pe}{48} = \frac{RL}{48}$$ $$R = -\frac{6Pe}{L} $$

This reaction can now be used to find the bending moment diagram that arises due to the restraint of the beam over the intermediate support. This is finally added to the primary prestressing moment to get the total moment effects on a continuous beam.

An example of how these results are used is shown below:

$$\text{Assume we have a beam with the following properties:}$$ $$\text{Span, L = 50m, Pe = 100kNm.}$$ $$\qquad R = -\frac{6Pe}{L} = -\frac{6 \times 100}{50}$$ $$\qquad R = 12kN$$ $$\text{To satisfy vertical equilibrium end support reactions must be developed to resist the midspan induced reaction:}$$ $$\qquad R_a = R_b = \frac{R}{2}= \frac{12}{2} = 6kN$$ $$\text{Primary Prestress} = Pe = 100kNm$$ $$\text{Secondary Prestress} = -1.5Pe = -150kNm$$ $$\text{Peak Hogging} = Pe = 100kNm$$ $$\text{Peak Sagging} = Pe-1.5Pe = -50kNm$$

The deflection of the beam due to prestress, the deflection caused by the restraint reaction and the resulting total deflections are shown below. Note the total deflection is plotted on the right hand axis as the total deflection is a much lower magnitude than the applied and restraint deflections:

#### Four Span Continuous Beam

When additional supports are added in to a continuous beam, more restraint reactions are required to ensure these additional points of restraint also have zero deflection. If only the middle support restraint reaction was applied then the deflections at the midspan support would be zero, however the midspan restraint reaction would not be sufficient to ensure zero deflection at other intermediate supports. However, if restraint reactions are only applied to the new intermediate supports then this will lead to excessive deflections at the midspan support, this excess deflection will need to be counteracted by an opposite reaction at midspan, which will in turn effect the deflection at the intermediate support as well as the deflection at midspan. It is possible to see that there is a balancing act between the reactions that are needed at the new intermediate supports and the reaction needed at the midspan support.

Using a similar approach to the two span beam it is possible to calculate the reactions at each support for a given applied moment. The main difference between the two span beam and the four span beam is that a new deflection function needs to be calculated for the new supports. This is calculated below:

$$\text{The distance between the ends of the beam and the position of the new intermediate support will be defined as, } a$$ $$\text{The distance to the first of the new intermediate supports will therefore be } a$$ $$\text{The distance to the second of the new intermediate supports will therefore be }(L-a)$$ $$\text{As there are two new intermediate supports } R_{int} \text{ then by inspection the support reactions will each equal } R_{int}$$ $$EI\frac{d^2v}{dx^2} = \text{Moment Function} = R_{int}x - R_{int}[x-a] - R_{int}[x-(L-a)]$$ $$EI\frac{dv}{dx} = \frac{R_{int}x^2}{2} - \frac{R_{int}}{2}[x-a]^2 -\frac{R_{int}}{2}[x-(L-a)]^2 + A $$ $$EIv = \frac{R_{int}x^3}{6} - \frac{R_{int}}{6}[x-a]^3 -\frac{R_{int}}{6}[x-(L-a)]^3 + Ax + B $$ $$\text{Apply boundary conditions to establish coefficients A and B.}$$ $$\qquad\text{At x = 0, v = 0}$$ $$\qquad\qquad\text{Therefore, B = 0}$$ $$\qquad\text{At x = L/2, }\frac{dv}{dx} = 0$$ $$\qquad\qquad0 = \frac{R_{int}L^2}{8} - \frac{R_{int}}{2}\left[\frac{L}{2}-a\right]^2 -\cancel{\frac{R_{int}}{2}\left[\frac{L}{2}-(L-a)\right]^2} + A$$ $$\qquad\qquad\text{Note: The struckthrough term above is ignored here as the value in the Macaulay brackets is zero or less} $$ $$\qquad\qquad\text{Therefore, } A = -\frac{R_{int}L^2}{8} + \frac{R_{int}}{2}\left(\frac{L}{2}-a\right)^2$$ $$\text{Therefore, the full expression for deflection is:}$$ $$\qquad EIv = \frac{R_{int}x^3}{6} - \frac{R_{int}}{6}[x-a]^3 -\frac{R_{int}}{6}[x-(L-a)]^3 -\frac{R_{int}L^2x}{8} + \frac{R_{int}x}{2}\left(\frac{L}{2}-a\right)^2 $$

This gives a total of three different deflection functions, one for the prestress moment, Pe, one for the induced midspan reaction and one for the induced intermediate reactions over the new supports as calculated immediately above. The prestress deflection function and midspan reaction deflection function are quoted from the Two Span beam section above, however to avoid confusion, the midspan reaction has been referred to as \(R_{mid}\)

$$ \text{Prestress Moment Deflection: } EIv = \frac{Pe}{2}(x^2 - Lx)$$ $$ \text{Midspan Support Reaction Deflection: } EIv = \frac{R_{mid}x^3}{12} - \frac{R_{mid}}{6}\left[x-\frac{L}{2}\right]^3 -\frac{R_{mid}L^2x}{16} $$ $$ \text{Intermediate Support Reaction Deflection: } EIv = \frac{R_{int}x^3}{6} - \frac{R_{int}}{6}[x-a]^3 -\frac{R_{int}}{6}[x-(L-a)]^3 -\frac{R_{int}L^2x}{8} + \frac{R_{int}x}{2}\left(\frac{L}{2}-a\right)^2 $$

These equations can be combined to create expressions for deflection at each support.

$$\text{At } x = a$$
$$\qquad v_{prestress} + v_{r,int} + v_{r,mid} = 0$$
$$\qquad v_{prestress} = - v_{r,int} - v_{r,mid}$$
$$\qquad \frac{Pe}{2}(a^2 - La) = -\frac{R_{mid}a^3}{12} + \cancel{\frac{R_{mid}}{6}\left[a-\frac{L}{2}\right]^3} +\frac{R_{mid}L^2a}{16} -\frac{R_{int}a^3}{6} + \cancel{\frac{R_{int}}{6}[a-a]^3} +\cancel{\frac{R_{int}}{6}[a-(L-a)]^3} +\frac{R_{int}L^2a}{8} - \frac{R_{int}a}{2}\left(\frac{L}{2}-a\right)^2 $$
$$\qquad \text{or: }\frac{Pe}{2}(a^2 - La) = -k_1R_{mid} -k_2R_{int} $$

$$\text{At } x = \frac{L}{2}$$
$$\qquad v_{prestress} + v_{r,int} + v_{r,mid} = 0$$
$$\qquad v_{prestress} = - v_{r,int} - v_{r,mid}$$
$$\qquad \frac{Pe}{2}(\frac{L^2}{4} - \frac{L^2}{2}) = -\frac{R_{mid}L^3}{96} + \cancel{\frac{R_{mid}}{6}\left[\frac{L}{2}-\frac{L}{2}\right]^3} +\frac{R_{mid}L^3}{32} -\frac{R_{int}L^3}{48} + \frac{R_{int}}{6}[\frac{L}{2}-a]^3 +\cancel{\frac{R_{int}}{6}[\frac{L}{2}-(L-a)]^3} +\frac{R_{int}L^3}{16} - \frac{R_{int}L}{4}\left(\frac{L}{2}-a\right)^2 $$
$$\qquad \text{or: }\frac{Pe}{2}(\frac{L^2}{4} - \frac{L^2}{2}) = -k_3R_{mid} -k_4R_{int} $$

By substituting in the span length, position of the intermediate supports and the applied moment, it is possible to see that we have two simulataneous equations with two unknowns. These can be solved to find the values for \(R_{int}\) and \(R_{mid}\). An example of this is shown below, using the same overall beam length and prestress moment as the Two Span example, but with two additional supports, one at 10m and another at 40m, i.e. a = 10m.

$$\text{Beam properties:}$$ $$\text{Span, L = 50m, Prestress, Pe = 100kNm, Intermediate Supports, a = 10m}$$ $$\text{Intermediate support deflection equations at a:}$$ $$\frac{Pe}{2}(a^2 - La) = k_1R_{mid} + k_2R_{int} $$ $$\qquad \frac{Pe}{2}(a^2 - La) = -20000$$ $$\qquad k_1 = -\frac{a^3}{12} + \cancel{\frac{1}{6}\left[a-\frac{L}{2}\right]^3} +\frac{L^2a}{16} = 1479.2$$ $$\qquad k_2 = -\frac{a^3}{6} + \cancel{\frac{1}{6}[a-a]^3} +\cancel{\frac{1}{6}[a-(L-a)]^3} +\frac{L^2a}{8} - \frac{a}{2}\left(\frac{L}{2}-a\right)^2 = 1833.3$$ $$\text{Midspan support deflection equations at} \frac{L}{2}$$ $$\frac{Pe}{2}(\frac{L^2}{4} - \frac{L^2}{2}) = k_3R_{mid} +k_4R_{int} $$ $$\qquad \frac{Pe}{2}(\frac{L^2}{4} - \frac{L^2}{2}) = -31250$$ $$\qquad k_3 = -\frac{L^3}{96} + \cancel{\frac{1}{6}\left[\frac{L}{2}-\frac{L}{2}\right]^3} +\frac{L^3}{32} = 2604.2$$ $$\qquad k_4 = -\frac{L^3}{48} + \frac{1}{6}[\frac{L}{2}-a]^3 +\cancel{\frac{1}{6}[\frac{L}{2}-(L-a)]^3} +\frac{L^3}{16} - \frac{L}{4}\left(\frac{L}{2}-a\right)^2 = 2958.3$$ $$\text{These form two simulataneous Equations:}$$ $$\qquad -20000 = 1479.2R_{mid} +1833.3R_{int} $$ $$\qquad -31250 = 2604.2R_{mid} +2958.3R_{int} $$ $$\text{Solving these gives values for} R_{mid} \text{ and } R_{int}$$ $$\qquad R_{mid} = +4.71kN \text{ (Upwards)}$$ $$\qquad R_{int} = -14.71kN \text{ (Downwards)}$$ $$\text{To satisfy vertical equilibrium end support reactions must be developed to resist the midspan induced reaction:}$$ $$\qquad R_a = R_b = -\left[\frac{R_{mid}}{2} + \frac{2R_{int}}{2}\right] = -\left[\frac{4.71}{2} + \frac{2\times-14.71}{2}\right]= +12.35kN \text{ (Upwards)}$$ $$\text{Using the above reactions and the applied prestress, Pe, it is possible to work out the full bending moment diagram.}$$

The deflection of the beam due to prestress, the deflection caused by the restraint reactions and the resulting total deflections are shown below. Note the total deflection is plotted on the right hand axis as the total deflection is a much lower magnitude than the applied and restraint deflections:

#### Other Applications - Shrinkage, Creep and Differential Temperature Effects

Although this article focusses on calculating the effects on secondary prestress, the principles can be applied to calculating the bending moment diagram that arises due to any distributed moment. Typical situations that generate similar distributed moment include:

- Differential shrinkage of a beam with a cast in-situ deck that is cast at a different time to the beam.
- Creep effects of continuous beams
- Differential temperature effects, where the top face of a beam expands or contracts at a different rate to the rest of the beam.