Problem 5.4 Shrinkage

Problem a)

Maximum stress in the concrete, σ_c [N/mm²]=

Stress in the steel, σ_s [N/mm²]=

Problem b)

Maximum stress in the concrete, σ_c [N/mm²]=

Stress in the steel, σ_s [N/mm²]=

Problem c)

Maximum stress in the concrete, σ_c [N/mm²]=

Stress in the top steel, σ_s1 [N/mm²]=

Stress in the bottom steel, σ_s2 [N/mm²]=

Problem a)

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

Check section propertiesHide section properties

$$\begin{gathered} A_{\mathrm{c}}=300\times 400=1.2\times {10}^5 {\mathrm{mm}}^2 \\ {A}_{\mathrm{s}}=4\frac{{\varphi }^2\pi }{4}={12}^2\pi =452 {\mathrm{mm}}^2 \\ {A}_{\mathrm{e}}=A_{\mathrm{c}}+\frac{E\mathrm{s}}{E_{\mathrm{c}}}{A}_{\mathrm{s}}=120000+\frac{200}{31}452=1.229\times {10}^5 {\mathrm{mm}}^2 \\ {s}_{\mathrm{e}}=\frac{h}{2}=200 \mathrm{mm} \\ {I}_e=\frac{b{h}^3}{12}+2\frac{E\mathrm{s}}{E_{\mathrm{c}}}\frac{A_s}{2}{\left(d–s_{\mathrm{e}}\right)}^2=\frac{300\times {400}^3}{12}+452\frac{200}{31}{\left(355–200\right)}^2=1.670\times {10}^9 {\mathrm{mm}}^4 \end{gathered}$$

Holes in the concrete and the moment of inertia of the steel bars about their centre of gravity are neglected.

Step 2. Determine the kinematic load and deformations of the beam.

Check deformationsHide deformations

The kinematic load is

$N_{\mathrm{K}}=A_{\mathrm{c}}{E}_{\mathrm{c}}{\epsilon }_{\mathrm{cs}}$

The cross section is symmetric, thus the curvature from the axial load is zero, the elongation of the beam is

Eq.(5-36)

${\epsilon }_{\mathrm{oK}}=–{\epsilon }_{\mathrm{cs}}\frac{A_{\mathrm{c}}}{A_{\mathrm{e}}}=–5\times {10}^{–4}\frac{1.2\times {10}^5}{1.23\times {10}^5}=–4.888\times {10}^{–4}$

Step 3. Give the stresses in the concrete and in the steel.

Check stressesHide stresses

Eq.(5-37)

$$\begin{gathered} {\left({\mathbf{\sigma }}_{\mathbf{K}\mathbf{,}\mathbf{R}}\right)}_{\mathbf{c}}=E_{\mathrm{c}}\left({\epsilon }_{\mathrm{oK}}+{\epsilon }_{\mathrm{cs}}\right)=31\times {10}^3\left(–4.88\times {10}^{–4}+5\times {10}^{–4}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{368}\frac{\mathbf{N}}{{\mathbf{mm}}^{\mathbf{2}}} \\ {\left({\mathbf{\sigma }}_{\mathbf{K}\mathbf{,}\mathbf{R}}\right)}_{\mathbf{s}}=E_{\mathrm{s}}{\epsilon }_{\mathrm{oK}}=200\times {10}^3–4.88\times {10}^{–4}\mathbf{=}\mathbf{–}\mathbf{97}\mathbf{.}\mathbf{6}\frac{\mathbf{N}}{{\mathbf{mm}}^{\mathbf{2}}} \end{gathered}$$

Problem b)

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

Check section propertiesHide section properties

$$\begin{gathered} A_{\mathrm{c}}=300\times 400=1.2\times {10}^5 {\mathrm{mm}}^2 \\ {A}_{\mathrm{s}}=4\frac{{\varphi }^2\pi }{4}={12}^2\pi =452.4 {\mathrm{mm}}^2 \\ {A}_{\mathrm{e}}=A_{\mathrm{c}}+\frac{E\mathrm{s}}{E_{\mathrm{c}}}{A}_{\mathrm{s}}=120000+\frac{200}{31}452=1.229\times {10}^5 {\mathrm{mm}}^2 \\ {s}_e=\frac{b{h}^2/2+\frac{Es}{E_c}{A}_{s}d}{A_e}=\frac{300\times {400}^2/2+{\displaystyle \frac{31}{200}}452\times 355}{1.23\times {10}^5}=203.7 \mathrm{mm} \\ {I}_e=\frac{b{h}^3}{12}+bh{\left(s_e–\frac{h}{2}\right)}^2+\frac{Es}{E_c}{A}_s{\left(d–s_e\right)}^2=1.668\times {10}^9{\mathrm{mm}}^4 \end{gathered}$$

Holes in the concrete and the moment of inertia of the steel bars about their centre of gravity are neglected.

Step 2. Determine the kinematic load and deformations of the beam.

Check deformationsHide deformations

The kinematic load is

$N_{\mathrm{K}}=A_{\mathrm{c}}{E}_{\mathrm{c}}{\epsilon }_{\mathrm{cs}}$

The curvature and the elongation of the beam are

Eq.(5-36)

$$\begin{gathered} {\epsilon }_{\mathrm{oK}}=–\frac{N_{\mathrm{K}}}{E_{\mathrm{c}}{A}_{\mathrm{e}}}=–{\epsilon }_{\mathrm{cs}}\frac{A_{\mathrm{c}}}{A_{\mathrm{e}}}=–5\times {10}^{–4}\frac{1.2\times {10}^5}{1.23\times {10}^5}=–4.888\times {10}^{–4} \\ {\kappa }_{\mathrm{K}}=–\frac{N_{\mathrm{K}}\left(s_{\mathrm{c}}–s_{\mathrm{e}}\right)}{E_{\mathrm{c}}{I}_{\mathrm{e}}}=–\frac{A_{\mathrm{c}}{\epsilon }_{\mathrm{cs}}\left(s_{\mathrm{c}}–s_{\mathrm{e}}\right)}{I_{\mathrm{e}}}=–\frac{1.2\times {10}^5\times 5\times {10}^{–5}\left(200–203.7\right)}{1.668\times {10}^9}=–1.324\times {10}^{–7}\frac{1}{\mathrm{mm}} \end{gathered}$$

Step 3. Give the stresses in the concrete and in the steel.

Check stressesHide stresses

Eq.(5-37)

$$\begin{gathered} {\left({\mathrm{\sigma }}_{\mathrm{K},\mathrm{R}}\right)}_{\mathrm{c}}^{\mathrm{t}}=E_{\mathrm{c}}\left({\epsilon }_{\mathrm{oK}}+s_{\mathrm{e}}{\kappa }_{\mathrm{K}}+{\epsilon }_{\mathrm{cs}}\right)=31\times {10}^3\left(–4.88\times {10}^{–4}+203.7\times 1.32\times {10}^{–7}+5\times {10}^{–4}\right)=0.468\frac{\mathrm{N}}{{\mathrm{mm}}^2} \\ {\left({\mathbf{\sigma }}_{\mathbf{K}\mathbf{,}\mathbf{R}}\right)}_{\mathbf{c}}^{\mathbf{b}}=E_{\mathrm{c}}\left({\epsilon }_{\mathrm{oK}}–\left(h–s_{\mathrm{e}}\right){\kappa }_{\mathrm{K}}+{\epsilon }_{\mathrm{cs}}\right)= \\ =31\times {10}^3\left(–4.88\times {10}^{–4}–(400–203.7)\times 1.32\times {10}^{–7}+5\times {10}^{–4}\right)\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{17}\frac{\mathbf{N}}{{\mathbf{mm}}^{\mathbf{2}}} \\ {\left({\mathbf{\sigma }}_{\mathbf{K}\mathbf{,}\mathbf{R}}\right)}_{\mathbf{s}}=E_{\mathrm{s}}\left({\epsilon }_{\mathrm{oK}}–\left(d–s_{\mathrm{e}}\right){\kappa }_{\mathrm{K}}\right)=200\times {10}^3\left(–4.88\times {10}^{–4}–(355–203.7)\times 1.32\times {10}^{–7}\right)\mathbf{=}\mathbf{–}\mathbf{97}\mathbf{.}\mathbf{6}\frac{\mathbf{N}}{{\mathbf{mm}}^{\mathbf{2}}} \end{gathered}$$

Problem c)

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

Check section propertiesHide section properties

 

$$\begin{gathered} A_{\mathrm{c}}=300\times 400=1.2\times {10}^5 {\mathrm{mm}}^2 \\ {A}_{\mathrm{s}1}=4\frac{{\varphi }^2\pi }{4}={12}^2\pi =452 {\mathrm{mm}}^2, A_{\mathrm{s}2}=2\frac{{\varphi }^2\pi }{4}=\frac{1}{2}{12}^2\pi =226 {\mathrm{mm}}^2 \\ {A}_{\mathrm{e}}=A_{\mathrm{c}}+\frac{E\mathrm{s}}{E_{\mathrm{c}}}\left(A_{\mathrm{s}1}+A_{\mathrm{s}2}\right)=1.244\times {10}^5 {\mathrm{mm}}^2 \\ {s}_e=\frac{b{h}^2/2+\frac{Es}{E_c}{A}_{s1}{d}_1+\frac{Es}{E_c}{A}_{s2}{d}_2}{A_e}=201.8 \mathrm{mm} \\ {I}_e=\frac{b{h}^3}{12}+bh{\left(s_e–\frac{h}{2}\right)}^2+\frac{Es}{E_c}{A}_{s1}{\left(d_1–s_e\right)}^2+\frac{Es}{E_c}{A}_{s2}{\left(s_e–d_2\right)}^2=1.705\times {10}^9{\mathrm{mm}}^4 \end{gathered}$$

Holes in the concrete and the moment of inertia of the steel bars about their centre of gravity are neglected.

Step 2. Determine the kinematic load and deformations of the beam.

Check deformationsHide deformations

The kinematic load is

$N_{\mathrm{K}}=A_{\mathrm{c}}{E}_{\mathrm{c}}{\epsilon }_{\mathrm{cs}}$

The curvature and the elongation of the beam are

Eq.(5-36)

$$\begin{gathered} {\epsilon }_{\mathrm{oK}}=–\frac{N_{\mathrm{K}}}{E_{\mathrm{c}}{A}_{\mathrm{e}}}=–{\epsilon }_{\mathrm{cs}}\frac{A_{\mathrm{c}}}{A_{\mathrm{e}}}=–5\times {10}^{–4}\frac{1.2\times {10}^5}{1.24\times {10}^5}=–4.824\times {10}^{–4} \\ {\kappa }_{\mathrm{K}}=–\frac{N_{\mathrm{K}}\left(s_{\mathrm{c}}–s_{\mathrm{e}}\right)}{E_{\mathrm{c}}{I}_{\mathrm{e}}}=–\frac{A_{\mathrm{c}}{\epsilon }_{\mathrm{cs}}\left(s_{\mathrm{c}}–s_{\mathrm{e}}\right)}{I_{\mathrm{e}}}=–\frac{1.2\times {10}^5\times 5\times {10}^{–5}\left(200–201.8\right)}{1.704\times {10}^9}=–6.401\times {10}^{–8}\frac{1}{\mathrm{mm}} \end{gathered}$$

Step 3. Give the stresses in the concrete and in the steel.

Check stressesHide stresses

Eq.(5-37)

Problem a)

The inhomogeneous cross section is replaced by a homogeneous one. Properties of the replacement homogeneous cross section are

$$\begin{gathered} A_{\mathrm{c}}=300\times 400=1.2\times {10}^5 {\mathrm{mm}}^2 \\ {A}_{\mathrm{s}}=4\frac{{\varphi }^2\pi }{4}={12}^2\pi =452 {\mathrm{mm}}^2 \\ {A}_{\mathrm{e}}=A_{\mathrm{c}}+\frac{E\mathrm{s}}{E_{\mathrm{c}}}{A}_{\mathrm{s}}=120000+\frac{200}{31}452=1.229\times {10}^5 {\mathrm{mm}}^2 \\ {s}_{\mathrm{e}}=\frac{h}{2}=200 \mathrm{mm} \\ {I}_e=\frac{b{h}^3}{12}+2\frac{E\mathrm{s}}{E_{\mathrm{c}}}\frac{A_s}{2}{\left(d–s_{\mathrm{e}}\right)}^2=\frac{300\times {400}^3}{12}+452\frac{200}{31}{\left(355–200\right)}^2=1.670\times {10}^9 {\mathrm{mm}}^4 \end{gathered}$$

Holes in the concrete and the moment of inertia of the steel bars about their centre of gravity are neglected.

The kinematic load from the shrinkage is

$N_{\mathrm{K}}=A_{\mathrm{c}}{E}_{\mathrm{c}}{\epsilon }_{\mathrm{cs}}$

The cross section is symmetric, thus the curvature from the axial load is zero, the elongation of the beam is

Eq.(5-36)

The stresses in the concrete and in the steel are

Eq.(5-37)

$$\begin{gathered} {\left({\mathbf{\sigma }}_{\mathbf{K}\mathbf{,}\mathbf{R}}\right)}_{\mathbf{c}}=E_{\mathrm{c}}\left({\epsilon }_{\mathrm{oK}}+{\epsilon }_{\mathrm{cs}}\right)=31\times {10}^3\left(–4.88\times {10}^{–4}+5\times {10}^{–4}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{368}\frac{\mathbf{N}}{{\mathbf{mm}}^{\mathbf{2}}} \\ {\left({\mathbf{\sigma }}_{\mathbf{K}\mathbf{,}\mathbf{R}}\right)}_{\mathbf{s}}=E_{\mathrm{s}}{\epsilon }_{\mathrm{oK}}=200\times {10}^3–4.88\times {10}^{–4}\mathbf{=}\mathbf{–}\mathbf{97}\mathbf{.}\mathbf{6}\frac{\mathbf{N}}{{\mathbf{mm}}^{\mathbf{2}}} \end{gathered}$$

Problem b)

The inhomogeneous cross section is replaced by a homogeneous one. Properties of the replacement homogeneous cross section are

$$\begin{gathered} A_{\mathrm{c}}=300\times 400=1.2\times {10}^5 {\mathrm{mm}}^2 \\ {A}_{\mathrm{s}}=4\frac{{\varphi }^2\pi }{4}={12}^2\pi =452.4 {\mathrm{mm}}^2 \\ {A}_{\mathrm{e}}=A_{\mathrm{c}}+\frac{E\mathrm{s}}{E_{\mathrm{c}}}{A}_{\mathrm{s}}=120000+\frac{200}{31}452=1.229\times {10}^5 {\mathrm{mm}}^2 \\ {s}_e=\frac{b{h}^2/2+\frac{Es}{E_c}{A}_{s}d}{A_e}=\frac{300\times {400}^2/2+{\displaystyle \frac{31}{200}}452\times 355}{1.23\times {10}^5}=203.7 \mathrm{mm} \\ {I}_e=\frac{b{h}^3}{12}+bh{\left(s_e–\frac{h}{2}\right)}^2+\frac{Es}{E_c}{A}_s{\left(d–s_e\right)}^2=1.668\times {10}^9{\mathrm{mm}}^4 \end{gathered}$$

Effect of shrinkage is taken into account as a kinematic load:

$N_{\mathrm{K}}=A_{\mathrm{c}}{E}_{\mathrm{c}}{\epsilon }_{\mathrm{cs}}$

The curvature and the elongation of the beam are

Eq.(5-36)

The stresses in the concrete and in the steel become

Eq.(5-37)

$$\begin{gathered} {\left({\mathrm{\sigma }}_{\mathrm{K},\mathrm{R}}\right)}_{\mathrm{c}}^{\mathrm{t}}=E_{\mathrm{c}}\left({\epsilon }_{\mathrm{oK}}+s_{\mathrm{e}}{\kappa }_{\mathrm{K}}+{\epsilon }_{\mathrm{cs}}\right)=31\times {10}^3\left(–4.88\times {10}^{–4}+203.7\times 1.32\times {10}^{–7}+5\times {10}^{–4}\right)=0.468\frac{\mathrm{N}}{{\mathrm{mm}}^2} \\ {\left({\mathbf{\sigma }}_{\mathbf{K}\mathbf{,}\mathbf{R}}\right)}_{\mathbf{c}}^{\mathbf{b}}=E_{\mathrm{c}}\left({\epsilon }_{\mathrm{oK}}–\left(h–s_{\mathrm{e}}\right){\kappa }_{\mathrm{K}}+{\epsilon }_{\mathrm{cs}}\right)= \\ =31\times {10}^3\left(–4.88\times {10}^{–4}–(400–203.7)\times 1.32\times {10}^{–7}+5\times {10}^{–4}\right)\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{17}\frac{\mathbf{N}}{{\mathbf{mm}}^{\mathbf{2}}} \\ {\left({\mathbf{\sigma }}_{\mathbf{K}\mathbf{,}\mathbf{R}}\right)}_{\mathbf{s}}=E_{\mathrm{s}}\left({\epsilon }_{\mathrm{oK}}–\left(d–s_{\mathrm{e}}\right){\kappa }_{\mathrm{K}}\right)=200\times {10}^3\left(–4.88\times {10}^{–4}–(355–203.7)\times 1.32\times {10}^{–7}\right)\mathbf{=}\mathbf{–}\mathbf{97}\mathbf{.}\mathbf{6}\frac{\mathbf{N}}{{\mathbf{mm}}^{\mathbf{2}}} \end{gathered}$$

Problem c)

Cross sectional properties of the replacement homogeneous cross section are the following

$$\begin{gathered} A_{\mathrm{c}}=300\times 400=1.2\times {10}^5 {\mathrm{mm}}^2 \\ {A}_{\mathrm{s}1}=4\frac{{\varphi }^2\pi }{4}={12}^2\pi =452 {\mathrm{mm}}^2, A_{\mathrm{s}2}=2\frac{{\varphi }^2\pi }{4}=\frac{1}{2}{12}^2\pi =226 {\mathrm{mm}}^2 \\ {A}_{\mathrm{e}}=A_{\mathrm{c}}+\frac{E\mathrm{s}}{E_{\mathrm{c}}}\left(A_{\mathrm{s}1}+A_{\mathrm{s}2}\right)=1.244\times {10}^5 {\mathrm{mm}}^2 \\ {s}_e=\frac{b{h}^2/2+\frac{Es}{E_c}{A}_{s1}{d}_1+\frac{Es}{E_c}{A}_{s2}{d}_2}{A_e}=201.8 \mathrm{mm} \\ {I}_e=\frac{b{h}^3}{12}+bh{\left(s_e–\frac{h}{2}\right)}^2+\frac{Es}{E_c}{A}_{s1}{\left(d_1–s_e\right)}^2+\frac{Es}{E_c}{A}_{s2}{\left(s_e–d_2\right)}^2=1.705\times {10}^9{\mathrm{mm}}^4 \end{gathered}$$

Effect of shrinkage is taken into account as a kinematic load:

$N_{\mathrm{K}}=A_{\mathrm{c}}{E}_{\mathrm{c}}{\epsilon }_{\mathrm{cs}}$

The curvature and the elongation of the beam are

Eq.(5-36)

The stresses in the concrete and in the steel are:

Eq.(5-37)

$$\begin{gathered} {\left({\mathrm{\sigma }}_{\mathrm{K},\mathrm{R}}\right)}_{\mathrm{c}}^{\mathrm{t}}=E_{\mathrm{c}}\left({\epsilon }_{\mathrm{oK}}+s_{\mathrm{e}}{\kappa }_{\mathrm{K}}+{\epsilon }_{\mathrm{cs}}\right)=31\times {10}^3\left(–4.88\times {10}^{–4}+201.86.40\times 1.32\times {10}^{–8}+5\times {10}^{–4}\right)=0.145\frac{\mathrm{N}}{{\mathrm{mm}}^2} \\ {\left({\mathbf{\sigma }}_{\mathbf{K}\mathbf{,}\mathbf{R}}\right)}_{\mathbf{c}}^{\mathbf{b}}=E_{\mathrm{c}}\left({\epsilon }_{\mathrm{oK}}–\left(h–s_{\mathrm{e}}\right){\kappa }_{\mathrm{K}}+{\epsilon }_{\mathrm{cs}}\right)= \\ =31\times {10}^3\left(–4.88\times {10}^{–4}–(400–201.8)\times 6.40\times {10}^{–8}+5\times {10}^{–4}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{939}\frac{\mathbf{N}}{{\mathbf{mm}}^{\mathbf{2}}} \\ {\left({\mathbf{\sigma }}_{\mathbf{K}\mathbf{,}\mathbf{R}}\right)}_{\mathbf{s}\mathbf{1}}=E_{\mathrm{s}}\left({\epsilon }_{\mathrm{oK}}–\left(d_1–s_{\mathrm{e}}\right){\kappa }_{\mathrm{K}}\right)=200\times {10}^3\left(–4.88\times {10}^{–4}–(355–201.8\right)\times 6.40\times {10}^{–8})\mathbf{=}\mathbf{–}\mathbf{94}\mathbf{.}\mathbf{5}\frac{\mathbf{N}}{{\mathbf{mm}}^{\mathbf{2}}} \\ {\left({\mathbf{\sigma }}_{\mathbf{K}\mathbf{,}\mathbf{R}}\right)}_{\mathbf{s}\mathbf{2}}=E_{\mathrm{s}}\left({\epsilon }_{\mathrm{oK}}+\left(s_{\mathrm{e}}–d_2\right){\kappa }_{\mathrm{K}}\right)=200\times {10}^3\left(–4.88\times {10}^{–4}+(201.8–45\right)\times 6.40\times {10}^{–8})\mathbf{=}\mathbf{–}\mathbf{98}\mathbf{.}\mathbf{5}\frac{\mathbf{N}}{{\mathbf{mm}}^{\mathbf{2}}} \end{gathered}$$