Problem 4.8. Effect of compressed steel bars on plasic resistance

Moment resistance, M_Rp [kNm]=

Follow steps of Example 4.4.

Step 1. Draw the stress and strain diagrams which belongs to plastic failure.

Check diagramsHide diagrams

Step 2. Write normal force equilibrium of the stress resultants. Determine the height of the compressed concrete zone.

Check compressed zone's heightHide compressed zone height

Assume that both tensile and compressed steel bars are in yielding stage. From the force equilibrium the height of the compressed zone, x_c can be expressed:

See also Figure 4.27.

$$\begin{gathered} N_{\mathrm{c}}+N_{\mathrm{s}2}=N_{\mathrm{s}1} \\ {f}_{\mathrm{c}}{x}_{\mathrm{c}}b+A_{\mathrm{s}2}{f}_{\mathrm{y}}=A_{\mathrm{s}1}{f}_{\mathrm{y}} \to x_{\mathrm{c}}=\frac{\left(A_{\mathrm{s}1}–A_{\mathrm{s}2}\right)f_{\mathrm{y}} }{f_{\mathrm{c}}b}=\frac{(1257–628)\times 435}{16.7\times 300}=109.1 \mathrm{mm} \end{gathered}$$

Step 3. Check whether steels are in yielding stage.

Check yielding of steelHide checking

Step 4. Determine plastic moment resistance of the cross section.

Check moment resistanceHide moment resistance

Writing the moment resistance about the centre of gravity of the tensile steel bars, the moment resistance is obtained:

See Figure 4.27.

$$\begin{gathered} {\mathit{M}}_{\mathbf{R}\mathbf{,}\mathbf{p}}=f_{\mathrm{c}}{x}_{\mathrm{c}}b\left(d–\frac{x_{\mathrm{c}}}{2}\right)+A_{\mathrm{s}2}{f}_{\mathrm{y}}\left(d_1–d_2\right)= \\ =16.7\times 109.1\times 300\left(455–\frac{109.1}{2}\right)+628\times 435\times \left(455–45\right)\mathbf{=}\mathbf{331}\mathbf{ }\mathbf{kNm} \end{gathered}$$

Follow steps of Example 4.4.

Stress and strain diagrams which belongs to plastic failure is given in the Figure below.

Assume that both tensile and compressed steel bars are in yielding stage. From the force equilibrium the height of the compressed zone, x_c can be expressed:

See also Figure 4.27.

$$\begin{gathered} N_{\mathrm{c}}+N_{\mathrm{s}2}=N_{\mathrm{s}1} \\ {f}_{\mathrm{c}}{x}_{\mathrm{c}}b+A_{\mathrm{s}2}{f}_{\mathrm{y}}=A_{\mathrm{s}1}{f}_{\mathrm{y}} \to x_{\mathrm{c}}=\frac{\left(A_{\mathrm{s}1}–A_{\mathrm{s}2}\right)f_{\mathrm{y}} }{f_{\mathrm{c}}b}=\frac{(1257–628)\times 435}{16.7\times 300}=109.1 \mathrm{mm} \end{gathered}$$

In the calculation we assumed that steel bars are yielding. The assumption can be checked  by comparing the strain of steel bars to the yield strain. Crush strain of concrete, ε_cu = 3.5 ‰ and yield strain of steel, ε_s = 2.1 ‰ are given.

Checking is performed using the strain diagram of the cross section shown in the Figure above:

$$\begin{gathered} {\epsilon }_{\mathrm{s}1}={\epsilon }_{\mathrm{cu}}\left(\frac{d_1}{x}–1\right)=3.5\times {10}^{–3}\left(\frac{455}{1.25\times 109.1}–1\right)=8.17\times {10}^{–3}>{\epsilon }_s=2.1\times {10}^{–3} \\ {\epsilon }_{s2}={\epsilon }_{\mathrm{cu}}\left(1–\frac{d_2}{x}\right)=3.5\times {10}^{–3}\left(1–\frac{45}{1.25\times 109.1}\right)=2.34\times {10}^{–3}>{\epsilon }_s=2.1\times {10}^{–3} \\ \mathrm{where} x=1.25x_{\mathrm{c}} \end{gathered}$$

thus the assumption is valid, both tensile and compressed steel bars are yielding.

Writing the moment resistance about the centre of gravity of the tensile steel bars, the moment resistance is obtained:

See Figure 4.27.

$$\begin{gathered} {\mathit{M}}_{\mathbf{R}\mathbf{,}\mathbf{p}}=f_{\mathrm{c}}{x}_{\mathrm{c}}b\left(d–\frac{x_{\mathrm{c}}}{2}\right)+A_{\mathrm{s}2}{f}_{\mathrm{y}}\left(d_1–d_2\right)= \\ =16.7\times 109.1\times 300\left(455–\frac{109.1}{2}\right)+628\times 435\times \left(455–45\right)\mathbf{=}\mathbf{331}\mathbf{ }\mathbf{kNm} \end{gathered}$$