Problem 5.7. Prestressed beam

Problem a)

Compression force, F [kN]=

Problem b)

Pre-tensioning strain, ε_ps =×10^-3

Maximum stress in the steel, σ_s =

Effect of prestress is equivalent to the effect of a negative shrinkage.

Step 1. Calculate section properties of the replacement homogeneous cross section.

Check section propertiesHide section properties

Eqs.(4-4)-(4-8)

$$\begin{gathered} A_{\mathrm{c}}=300\times 400=1.2\times {10}^5 {\mathrm{mm}}^2 \\ {A}_{\mathrm{s}1}=A_{\mathrm{s}2}=6\frac{{\varphi }^2\pi }{4}=6\frac{12.5^2}{4}\pi =736 {\mathrm{mm}}^2 \\ {A}_{\mathrm{e}}=A_{\mathrm{c}}+\frac{E\mathrm{s}}{E_{\mathrm{c}}}\left(A_{\mathrm{s}1}+A_{\mathrm{s}2}\right)=120000+2\frac{200}{31}736=1.295\times {10}^5 {\mathrm{mm}}^2 \\ {s}_{\mathrm{e}}=\frac{h}{2}=200 \mathrm{mm} \\ {I}_{\mathrm{e}}=\frac{b{h}^3}{12}+2\frac{E_{\mathrm{s}}}{E_{\mathrm{c}}}\frac{A_{\mathrm{s}1}}{2}{\left(d–s_{\mathrm{e}}\right)}^2=\frac{300\times {400}^3}{12}+2\times 736\frac{200}{31}{\left(355–200\right)}^2=1.83\times {10}^9 {\mathrm{mm}}^4 \end{gathered}$$

Step 2. Give tensile stress in the extreme fibre of concrete from the given moment, M. Calculate the necessary compressive force at the extreme fibre to obtain σ_c = f_c.

Check compressive forceHide compressive force

Eq.(4-9)

$$\begin{gathered} {\sigma }_{\mathrm{Mc}}=f_{\mathrm{ct}}=\frac{M}{I_{\mathrm{e}}}\frac{h}{2}–\frac{F}{A_{\mathrm{e}}} \\ \to \mathbf{ }\mathit{F}=A_{\mathrm{e}}\left(f_{\mathrm{ct}}–\frac{M}{I_{\mathrm{e}}}\frac{h}{2}\right)=1.295\times {10}^5\left(1.2–\frac{60\times {10}^6}{1.83\times {10}^9}200\right)\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{945}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{N} \end{gathered}$$

Problem b)

Step 1. Apply a central pre-tension force, N _K = F  on the unloaded beam, the value of which is determined in Problem a). Calculate the kinematic deformation of the beam.

Check kinematic deformationHide kinematic deformation

Eq.(5-38)

${\epsilon }_{\mathrm{oK}}=\frac{–N_{\mathrm{K}}}{E_{\mathrm{c}}{A}_{\mathrm{e}}}=\frac{–6.95\times {10}^5}{31\times {10}^3\times 1.295\times {10}^5}=–1.73\times {10}^{–4}$

Step 2. Determine the pre-tensioning strain.

Check pre-tensioning strainHide pre-tensioning strain

Eq.(5-38)

$$\begin{gathered} {\epsilon }_{\mathrm{oK}}=1.73\times {10}^{–4}=–{\epsilon }_{\mathrm{ps}}\frac{E_{\mathrm{s}}}{E_{\mathrm{c}}}\frac{A_{\mathrm{s}}}{A_{\mathrm{e}}} \\ \to {\mathit{\epsilon }}_{\mathbf{ps}}=–{\epsilon }_{\mathrm{oK}}\frac{E_{\mathrm{c}}}{E_{\mathrm{s}}}\frac{A_{\mathrm{e}}}{A_{\mathrm{s}}}=1.73\times {10}^{–4}\frac{31}{200}\frac{1.295\times {10}^5}{2\times 736}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{36}\mathbf{\times }{\mathbf{10}}^{\mathbf{–}\mathbf{3}} \end{gathered}$$

Step 3. Give the stress in the steel from the moment and the pretension force.

Check steel stressHide steel stress

Residual stress in the steel from pre-tensioning:

Eq.(5-39)

${\left({\mathrm{\sigma }}_{\mathrm{K},\mathrm{R}}\right)}_{\mathrm{s}}={\mathrm{\epsilon }}_{\mathrm{ps}}\frac{E_{\mathrm{s}}}{A_{\mathrm{e}}}{A}_{\mathit{c}}=2.36\times {10}^{–3}\frac{200\times {10}^3}{1.295\times {10}^5}1.2\times {10}^5=437.1\frac{\mathrm{N}}{{\mathrm{mm}}^2} \left(\mathrm{tension}\right)$

Stress in the steel from bending moment:

${\sigma }_{\mathrm{Ms}}=\pm \frac{E_{\mathrm{s}}}{E_{\mathrm{c}}}\frac{M}{I_{\mathrm{e}}}\left(d–\frac{h}{2}\right)=\pm 32.82\frac{\mathrm{N}}{{\mathrm{mm}}^2}$

Total stress in the top and in the bottom steel bars:

$$\begin{gathered} {\mathit{\sigma }}_{\mathbf{s}\mathbf{1}}={\left({\sigma }_{\mathrm{K},\mathrm{R}}\right)}_{\mathrm{s}}+{\sigma }_{\mathrm{Ms}}=437.1+32.8\mathbf{=}\mathbf{469}\mathbf{.}\mathbf{9}\frac{\mathbf{N}}{{\mathbf{mm}}^{\mathbf{2}}} \\ {\sigma }_{\mathrm{s}2}={\left({\sigma }_{\mathrm{K},\mathrm{R}}\right)}_{\mathrm{s}}–{\sigma }_{\mathrm{Ms}}=437.1–32.8=404.25\frac{\mathrm{N}}{{\mathrm{mm}}^2} \end{gathered}$$

Step 4. Check the stress in the concrete.

Check concrete stressHide concrete stress

Eq.(5-39)

$$\begin{gathered} {\left({\mathrm{\sigma }}_{\mathrm{K},\mathrm{R}}\right)}_{\mathrm{c}}=–{\mathrm{\epsilon }}_{\mathrm{ps}}\frac{E_{\mathrm{s}}}{A_{\mathrm{e}}}{A}_{\mathrm{s}}=2.36\times {10}^{–3}\frac{200\times {10}^3}{1.295\times {10}^5}2\times 736=–5.36\frac{\mathrm{N}}{{\mathrm{mm}}^2} \left(\mathrm{compression}\right) \\ {\sigma }_{\mathrm{Mc}}=\pm \frac{M}{I_{\mathrm{e}}}\frac{h}{2}=\pm \frac{60\times {10}^6}{1.83\times {10}^9}200=\pm 6.56\frac{\mathrm{N}}{{\mathrm{mm}}^2} (\mathrm{tension} \mathrm{at} \mathrm{the} \mathrm{bottom}, \mathrm{compression} \mathrm{at} \mathrm{the} \mathrm{top}) \\ \mathrm{The} \mathrm{concrete} \mathrm{stress} \mathrm{at} \mathrm{the} \mathrm{bottom} \mathrm{is}: \\ {\sigma }_{\mathrm{ct}}={\left({\mathrm{\sigma }}_{\mathrm{K},\mathrm{R}}\right)}_{\mathrm{c}}+{\sigma }_{\mathrm{Mc}}=1.2\frac{\mathrm{N}}{{\mathrm{mm}}^2}=f_{\mathrm{ct}} \left(\mathrm{tension}\right) \\ \mathrm{The} \mathrm{concrete} \mathrm{stress} \mathrm{at} \mathrm{the} \mathrm{top} \mathrm{is}: \\ {\sigma }_{\mathrm{cc}}={\left({\mathrm{\sigma }}_{\mathrm{K},\mathrm{R}}\right)}_{\mathrm{c}}–{\sigma }_{\mathrm{Mc}}=–11.92\frac{\mathrm{N}}{{\mathrm{mm}}^2} \left(\mathrm{compression}\right) \end{gathered}$$

Applying pre-tensioning the concrete remains uncracked for the given moment.

 

Problem a)

Effect of prestress is equivalent to the effect of a negative shrinkage.

First the section properties of the replacement homogeneous cross section are calcultated.

Eqs.(4-4)-(4-8)

Next we calculate the tensile stress in the extreme fibre of concrete from the given moment, M. Then the necessary compressive force is determined at the extreme fibre to obtain σ_c = f_c.

Eq.(4-9)

$$\begin{gathered} {\sigma }_{\mathrm{Mc}}=f_{\mathrm{ct}}=\frac{M}{I_{\mathrm{e}}}\frac{h}{2}–\frac{F}{A_{\mathrm{e}}} \\ \to \mathbf{ }\mathit{F}=A_{\mathrm{e}}\left(f_{\mathrm{ct}}–\frac{M}{I_{\mathrm{e}}}\frac{h}{2}\right)=1.295\times {10}^5\left(1.2–\frac{60\times {10}^6}{1.83\times {10}^9}200\right)\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{945}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{N} \end{gathered}$$

Problem b)

A central pre-tension force, N _K = F  is applied on the unloaded beam, the value of which is determined in Problem a). The kinematic deformation of the beam is

Eq.(5-38)

${\epsilon }_{\mathrm{oK}}=\frac{–N_{\mathrm{K}}}{E_{\mathrm{c}}{A}_{\mathrm{e}}}=\frac{–6.95\times {10}^5}{31\times {10}^3\times 1.295\times {10}^5}=–1.73\times {10}^{–4}$

The pre-tensioning strain becomes

Eq.(5-38)

$$\begin{gathered} {\epsilon }_{\mathrm{oK}}=1.73\times {10}^{–4}=–{\epsilon }_{\mathrm{ps}}\frac{E_{\mathrm{s}}}{E_{\mathrm{c}}}\frac{A_{\mathrm{s}}}{A_{\mathrm{e}}} \\ \to {\mathit{\epsilon }}_{\mathbf{ps}}=–{\epsilon }_{\mathrm{oK}}\frac{E_{\mathrm{c}}}{E_{\mathrm{s}}}\frac{A_{\mathrm{e}}}{A_{\mathrm{s}}}=1.73\times {10}^{–4}\frac{31}{200}\frac{1.295\times {10}^5}{2\times 736}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{36}\mathbf{\times }{\mathbf{10}}^{\mathbf{–}\mathbf{3}} \end{gathered}$$

Residual stress in the steel from pre-tensioning is

Eq.(5-39)

${\left({\mathrm{\sigma }}_{\mathrm{K},\mathrm{R}}\right)}_{\mathrm{s}}={\mathrm{\epsilon }}_{\mathrm{ps}}\frac{E_{\mathrm{s}}}{A_{\mathrm{e}}}{A}_{\mathit{c}}=2.36\times {10}^{–3}\frac{200\times {10}^3}{1.295\times {10}^5}1.2\times {10}^5=437.1\frac{\mathrm{N}}{{\mathrm{mm}}^2} \left(\mathrm{tension}\right)$

Stress in the steel from bending moment:

${\sigma }_{\mathrm{Ms}}=\pm \frac{E_{\mathrm{s}}}{E_{\mathrm{c}}}\frac{M}{I_{\mathrm{e}}}\left(d–\frac{h}{2}\right)=\pm 32.82\frac{\mathrm{N}}{{\mathrm{mm}}^2}$

Total stress in the top and in the bottom steel bars:

$$\begin{gathered} {\mathit{\sigma }}_{\mathbf{s}\mathbf{1}}={\left({\sigma }_{\mathrm{K},\mathrm{R}}\right)}_{\mathrm{s}}+{\sigma }_{\mathrm{Ms}}=437.1+32.8\mathbf{=}\mathbf{469}\mathbf{.}\mathbf{9}\frac{\mathbf{N}}{{\mathbf{mm}}^{\mathbf{2}}} \\ {\sigma }_{\mathrm{s}2}={\left({\sigma }_{\mathrm{K},\mathrm{R}}\right)}_{\mathrm{s}}–{\sigma }_{\mathrm{Ms}}=437.1–32.8=404.25\frac{\mathrm{N}}{{\mathrm{mm}}^2} \end{gathered}$$

Stress in the concrete:

Eq.(5-39)

$$\begin{gathered} {\left({\mathrm{\sigma }}_{\mathrm{K},\mathrm{R}}\right)}_{\mathrm{c}}=–{\mathrm{\epsilon }}_{\mathrm{ps}}\frac{E_{\mathrm{s}}}{A_{\mathrm{e}}}{A}_{\mathrm{s}}=2.36\times {10}^{–3}\frac{200\times {10}^3}{1.295\times {10}^5}2\times 736=–5.36\frac{\mathrm{N}}{{\mathrm{mm}}^2} \left(\mathrm{compression}\right) \\ {\sigma }_{\mathrm{Mc}}=\pm \frac{M}{I_{\mathrm{e}}}\frac{h}{2}=\pm \frac{60\times {10}^6}{1.83\times {10}^9}200=\pm 6.56\frac{\mathrm{N}}{{\mathrm{mm}}^2} (\mathrm{tension} \mathrm{at} \mathrm{the} \mathrm{bottom}, \mathrm{compression} \mathrm{at} \mathrm{the} \mathrm{top}) \\ \mathrm{The} \mathrm{concrete} \mathrm{stress} \mathrm{at} \mathrm{the} \mathrm{bottom} \mathrm{is}: \\ {\sigma }_{\mathrm{ct}}={\left({\mathrm{\sigma }}_{\mathrm{K},\mathrm{R}}\right)}_{\mathrm{c}}+{\sigma }_{\mathrm{Mc}}=1.2\frac{\mathrm{N}}{{\mathrm{mm}}^2}=f_{\mathrm{ct}} \left(\mathrm{tension}\right) \\ \mathrm{The} \mathrm{concrete} \mathrm{stress} \mathrm{at} \mathrm{the} \mathrm{top} \mathrm{is}: \\ {\sigma }_{\mathrm{cc}}={\left({\mathrm{\sigma }}_{\mathrm{K},\mathrm{R}}\right)}_{\mathrm{c}}–{\sigma }_{\mathrm{Mc}}=–11.92\frac{\mathrm{N}}{{\mathrm{mm}}^2} \left(\mathrm{compression}\right) \end{gathered}$$

Thus applying pre-tensioning the concrete remains uncracked for the given moment.