Problem 4.7. Effect of compressed steel bars in elastic stage

Cracking moment M_cr [kNm]=

Stress in extreme concrete fiber, σ_c [N/mm²]=

Stress in tensile steel, σ_s1 [N/mm²]=

Stress in compressed steel, σ_s2 [N/mm²]=

Curvature, $\kappa \left[\times {10}^{–7}\frac{1}{\mathrm{mm}}\right]=$

Extend solution of Problem 4.4 with the effect of the compressed steel bars

Follow steps in Example 34

Check uncracked section propertiesHide uncracked section properties

Concrete is chosen to be the reference material. Ratio of elastic moduli of concrete and steel is denoted by

Eq.(4-4)

$\alpha =\frac{E_{\mathrm{s}}}{E_{\mathrm{c}}}=\frac{200}{18.3}=10.93$

The cross sectional properties of the replacement homogeneous cross section are

See Figure 133 and Eqs.(4-6)-(4-8)

$$\begin{gathered} A_{\mathrm{e}}=A_{\mathrm{c}}+\alpha A_{\mathrm{s}1}+\alpha A_{\mathrm{s}2}=bh+\left(\alpha –1\right)\left(A_{\mathrm{s}1}+A_{\mathrm{s}2}\right)=1.75\times {10}^{5 }{\mathrm{mm}}^2 \\ {x}_{\mathrm{c}}=s_{\mathrm{e}}=\frac{S_{\mathrm{e}}}{A_{\mathrm{e}}}=\frac{{\displaystyle \frac{b{h}^2}{2}}+\left(\alpha –1\right)A_{\mathrm{s}}{d}_1+\left(\alpha –1\right)A_{\mathrm{s}}{d}_2}{A_{\mathrm{e}}}= 264.6 \mathrm{mm} \\ {I}_{\mathrm{e},\mathrm{I}}=\frac{b{x}_{\mathrm{c}}^3}{3}+\frac{b{\left(h–x_{\mathrm{c}}\right)}^3}{3}+\left(\alpha –1\right)A_{\mathrm{s}1}{\left(d_1–x_{\mathrm{c}}\right)}^2+\left(\alpha –1\right)A_{\mathrm{s}2}{\left(x_{\mathrm{c}}–d_2\right)}^2=4.14\times {10}^9 {\mathrm{mm}}^4 \end{gathered}$$

The moment of inertias of steel bars about their centroidal axis are neglected

Holes in concrete are often neglected and (α-1) is replaced by α.

Step 2. Calculate the cracking moment.

Check cracking momentHide cracking moment

Concrete cracks when stress in the bottom extreme fiber reaches the tensile strength of concrete.

Eq.(4-9)

${\sigma }_{\mathrm{c}}^{\mathrm{b}}=f_{\mathrm{ct}}=\frac{M_{\mathrm{cr}}}{I_{\mathrm{e},\mathrm{I}}}\left(h–x_{\mathrm{c}}\right) \to {\mathit{M}}_{\mathbf{cr}}=\frac{f_{\mathrm{ct}}{I}_{\mathrm{e},\mathrm{I}}}{\left(h–x_{\mathrm{c}}\right)}=17.57\times {10}^6 \mathrm{Nmm}\mathbf{=}\mathbf{17}\mathbf{.}\mathbf{57}\mathbf{ }\mathbf{kNm}<M\mathit{=} 32 \mathrm{kNm}$

Thus the concrete cracks.

Step 3. Calculate the section properties of the cracked cross section.

See Example 36.

Check cracked section propertiesHide cracked section properties

Step 4. Determine the relevant stress values in the concrete and in the steel bars.

Check stressesHide stresses

We assume that the concrete, the tensile and also the compressed steel bars are in elastic stage.

Relevant stress values are

Eq.(4-9)

in the top extreme concrete fibre:

${\mathit{\sigma }}_{\mathbf{c}}^{\mathbf{t}}=–\frac{M}{I_{\mathrm{e},\mathrm{II}}}{x}_{\mathrm{c}}=–\frac{32\times {10}^6}{2.255\times {10}^9}179.5\mathbf{=}\mathbf{–}\mathbf{2}\mathbf{.}\mathbf{55}\frac{\mathbf{N}}{{\mathbf{mm}}^{\mathbf{2}}}\mathbf{ }\mathbf{\left(}\mathbf{compression}\mathbf{\right)} < f_{\mathrm{c}}=13.33 \frac{\mathrm{N}}{{\mathrm{mm}}^2}$

in the tensile steel bars:${\mathit{\sigma }}_{\mathbf{s}\mathbf{1}}^{\mathbf{b}}=\alpha \frac{M}{I_{\mathrm{e},\mathrm{II}}}\left(d_1–x_{\mathrm{c}}\right)=10.93\frac{32\times {10}^6}{2.255\times {10}^9}(455–179.5)\mathbf{=}\mathbf{42}\mathbf{.}\mathbf{72}\frac{\mathbf{N}}{{\mathbf{mm}}^{\mathbf{2}}}\mathbf{ }\mathbf{\left(}\mathbf{tension}\mathbf{\right)}\mathbf{ }< f_{\mathrm{y}}=435\frac{\mathrm{N}}{{\mathrm{mm}}^2}$

in the compressed steel bars:${\mathit{\sigma }}_{\mathbf{s}\mathbf{2}}^{\mathbf{b}}=–\alpha \frac{M}{I_{\mathrm{e},\mathrm{II}}}\left(d_2–x_{\mathrm{c}}\right)=–10.93\frac{32\times {10}^6}{2.255\times {10}^9}(179.5–45)\mathbf{=}\mathbf{–}\mathbf{20}\mathbf{.}\mathbf{87}\frac{\mathbf{N}}{{\mathbf{mm}}^{\mathbf{2}}}\mathbf{ }\mathbf{\left(}\mathbf{compression}\mathbf{\right)}\mathbf{ }< f_{\mathrm{y}}=435\frac{\mathrm{N}}{{\mathrm{mm}}^2}$

Stresses arising in the cross section are lower than the tensile and compressive strength of the materials, thus the materials of the cross section behave in a linearly elastic manner.

Step 5. Give the curvature from the given moment.

Check curvatureHide curvature

Extend solution of Problem 4.4 with the effect of the compressed steel bars

Follow steps in Example 34

Concrete is chosen to be the reference material. Ratio of elastic moduli of concrete and steel is denoted by

Eq.(4-4)

The cross sectional properties of the replacement homogeneous cross section are

See Figure 133 and Eqs.(4-6)-(4-8)

The moment of inertias of steel bars about their centroidal axis are neglected

Holes in concrete are often neglected and (α-1) is replaced by α.

Concrete cracks when stress in the bottom extreme fiber reaches the tensile strength of concrete. The cracking moment is

Eq.(4-9)

${\sigma }_{\mathrm{c}}^{\mathrm{b}}=f_{\mathrm{ct}}=\frac{M_{\mathrm{cr}}}{I_{\mathrm{e},\mathrm{I}}}\left(h–x_{\mathrm{c}}\right) \to {\mathit{M}}_{\mathbf{cr}}=\frac{f_{\mathrm{ct}}{I}_{\mathrm{e},\mathrm{I}}}{\left(h–x_{\mathrm{c}}\right)}=17.57\times {10}^6 \mathrm{Nmm}\mathbf{=}\mathbf{17}\mathbf{.}\mathbf{57}\mathbf{ }\mathbf{kNm}<M\mathit{=} 32 \mathrm{kNm}$

Thus the concrete cracks.

After cracking of the cross section the tensile stress in the concrete is neglected, both the steel and the compressed concrete zone still behave in a linearly elastic manner. Compressed concrete zone and tensile steel bars are replaced again by an equivalent homogeneous cross section. From the concrete cross section only the compressed concrete zone is taken into consideration (x_cb), where x_c is unknown (see Figure below).

See also Example 36.

The cross sectional properties of the equivalent homogeneous cross section are

$$\begin{gathered} A_{\mathrm{e}}=b{x}_{\mathrm{c}}+\alpha A_{\mathrm{s}1}+\left(\alpha –1\right)A_{\mathrm{s}2}=300\times x_{\mathrm{c}}+10.93\times 1257++9.93\times 628 \\ {x}_{\mathrm{c}}=s_{\mathrm{e}}=\frac{S_{\mathrm{e}}}{A_{\mathrm{e}}}=\frac{{\displaystyle \frac{b{x}_{\mathrm{c}}^2}{2}+\alpha A_{\mathrm{s}1}{d}_1+\alpha A_{\mathrm{s}2}{d}_2}}{b{x}_{\mathrm{c}}+\mathrm{\alpha }{A}_{\mathrm{s}1}+\left(\mathrm{\alpha }–1\right)A_{\mathrm{s}2}} \to x_{\mathrm{c}}=179.5 \mathrm{mm} \\ {I}_{\mathrm{e},\mathrm{II}}=\frac{b{x}_{\mathrm{c}}^3}{3}+\alpha A_{\mathrm{s}1}{\left(d_1–x_{\mathrm{c}}\right)}^2+\left(\mathrm{\alpha }–1\right)A_{\mathrm{s}2}{\left(x_{\mathrm{c}}–d_2\right)}^2=2.255\times {10}^9 {\mathrm{mm}}^4 \end{gathered}$$

We assume that the concrete, the tensile and also the compressed steel bars are in elastic stage.

Relevant stress values are

Eq.(4-9)

in the top extreme concrete fibre:

${\mathit{\sigma }}_{\mathbf{c}}^{\mathbf{t}}=–\frac{M}{I_{\mathrm{e},\mathrm{II}}}{x}_{\mathrm{c}}=–\frac{32\times {10}^6}{2.255\times {10}^9}179.5\mathbf{=}\mathbf{–}\mathbf{2}\mathbf{.}\mathbf{55}\frac{\mathbf{N}}{{\mathbf{mm}}^{\mathbf{2}}}\mathbf{ }\mathbf{\left(}\mathbf{compression}\mathbf{\right)} < f_{\mathrm{c}}=13.33 \frac{\mathrm{N}}{{\mathrm{mm}}^2}$

in the tensile steel bars:${\mathit{\sigma }}_{\mathbf{s}\mathbf{1}}^{\mathbf{b}}=\alpha \frac{M}{I_{\mathrm{e},\mathrm{II}}}\left(d_1–x_{\mathrm{c}}\right)=10.93\frac{32\times {10}^6}{2.255\times {10}^9}(455–179.5)\mathbf{=}\mathbf{42}\mathbf{.}\mathbf{72}\frac{\mathbf{N}}{{\mathbf{mm}}^{\mathbf{2}}}\mathbf{ }\mathbf{\left(}\mathbf{tension}\mathbf{\right)}\mathbf{ }< f_{\mathrm{y}}=435\frac{\mathrm{N}}{{\mathrm{mm}}^2}$

in the compressed steel bars:${\mathit{\sigma }}_{\mathbf{s}\mathbf{2}}^{\mathbf{b}}=–\alpha \frac{M}{I_{\mathrm{e},\mathrm{II}}}\left(d_2–x_{\mathrm{c}}\right)=–10.93\frac{32\times {10}^6}{2.255\times {10}^9}(179.5–45)\mathbf{=}\mathbf{–}\mathbf{20}\mathbf{.}\mathbf{87}\frac{\mathbf{N}}{{\mathbf{mm}}^{\mathbf{2}}}\mathbf{ }\mathbf{\left(}\mathbf{compression}\mathbf{\right)}\mathbf{ }< f_{\mathrm{y}}=435\frac{\mathrm{N}}{{\mathrm{mm}}^2}$

Stresses arising in the cross section are lower than the tensile and compressive strength of the materials, thus the materials of the cross section behave in a linearly elastic manner.

The curvature from the given moment is

${\mathbf{\kappa }}_{\mathbf{II}}=\frac{M}{E_{\mathrm{c}}{I}_{\mathrm{e},\mathrm{II}}}=\frac{32\times {10}^6}{18.3\times {10}^3\times 2.255\times {10}^9}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{76}\mathbf{\times }{\mathbf{10}}^{\mathbf{–}\mathbf{7}}\frac{\mathbf{1}}{\mathbf{mm}}$

Compare result to that of Problem 4.4. Tensile stiffness increase stiffness and reduce displacements.