Problem 4.13. Failure envelop

Problem b)

Moment resistance from simplified failure envelope, M_R [kNm]=

CHECK

Problem c)

Moment resistance from the equilibrium equations, M_R [kNm]=

CHECK

Problem a)

Step 1. Determine the point of the failure envelope which belongs to the maximum normal force.

Check first characteristic pointHide first characteristic point

Step 2. Determine the point of the failure envelope which belongs to the maximum moment.

Check second characteristic pointHide second characteristic point

Step 3. Determine the point of the failure envelope which belongs to pure bending.

Check third characteristic pointHide third characteristic point

Step 4. Draw failure envelope. Verify the load bearing of the cross section for the given load.

Check envelopeHide envelope

Problem b)

Determine approximate moment resistance which belongs to the normal force, N = 400 kN.

Check approximate moment resistanceHide approximate moment resistance

Step 2. Check whether the steels are in yielding stage.

Show checkingHide checking

Step 2. Calculate moment resistance from the moment equilibrium.

Check resistanceHide resistance

Problem a)

First the characteristic points of the simplified failure envelope are determined.

First point of the failure envelope belongs to the maximum normal force. Stress diagram is given in the Figure.

Values of the normal force and the moment are calculated from the force and moment equilibrium of the cross section. Steel bars are in yielding stage (ε_cu= 3.5×10^-3 >ε_s= 2.18×10^-3 ).

$$\begin{gathered} N_1=bh{f}_{\mathrm{c}}+A_{\mathrm{s}1}{f}_{\mathrm{y}}+A_{\mathrm{s}2}{f}_{\mathrm{y}}=300\times 500\times 16.7+(603+1571)\times 435=3451 \mathrm{kN} \\ {M}_1=A_{s\mathit{2}}{f}_{\mathrm{y}}\left(\frac{h}{\mathit{2}}\mathit{-}{d}_{\mathit{2}}\right)\mathit{-}{A}_{s\mathit{1}}{f}_y\left(d_{\mathit{1}}\mathit{-}\frac{h}{\mathit{2}}\right)= \\ =603\times 435\left(250-45\right)-1571\times 435\left(455-250\right)=-86.32 \mathrm{kNm} \end{gathered}$$

Second point of the failure envelope belongs to the maximum moment. Stress diagram is given in the Figure.

Second point is where the yield stage of the tensile steel bars starts that is the expected location of the maximum moment. In the top extreme fiber of concrete the ultimate strain ε_cu = 3.5 ‰, in the tensile steel the strain is ε_s = 2.18‰. From these strains the neutral axis can be determined:

$$\begin{gathered} x=\frac{d{\epsilon }_{\mathrm{cu}}}{{\epsilon }_{\mathrm{cu}}+{\epsilon }_s}=455\frac{3.5}{3.5+2.18}=280.4 \mathrm{mm} \\ \mathrm{from} \mathrm{which} \mathrm{the} \mathrm{compression} \mathrm{zone} \mathrm{is} \\ {x}_{\mathrm{c}}=0.8x=224.3 \mathrm{mm} \end{gathered}$$               

where multiplicator 0.8 takes into account that the „locking”-plastic-brittle stress-strain diagram is partly filled.        

The bending moment and the normal force of this point can be calculated from the force and moment equilibrium, respectively (also compression steel bars yield, because ε_s2 = ε_cu(x–d₂)/x = 2.94‰ > 2.18 ‰):

$$\begin{gathered} N_2=b{x}_{\mathrm{c}}{f}_{\mathrm{c}}-A_{\mathrm{s}1}{f}_{\mathrm{y}}+A_{\mathrm{s}2}{f}_{\mathrm{y}}=300\times 224.3\times 16.7+(603-1570)\times 435=702.6 \mathrm{kN} \\ {M}_2=b{x}_{\mathrm{c}}{f}_{\mathrm{c}}\left(\frac{h}{2}-x_{\mathrm{c}}\right)+A_{s\mathit{2}}{f}_{\mathrm{y}}\left(\frac{h}{\mathit{2}}\mathit{-}{d}_{\mathit{2}}\right)\mathit{+}{A}_{s\mathit{1}}{f}_y\left(d_{\mathit{1}}\mathit{-}\frac{h}{\mathit{2}}\right)= \\ =300\times 224.3\times 16.7\left(250-224.3\right)+603\times 435\left(250-45\right)+1570\times 435\left(455-250\right)=-348.8 \mathrm{kNm} \end{gathered}$$

The third point of the failure envelope belongs to pure bending. Both steel bars are assumed to yield.  Height of the compressed concrete zone can be calculated from the force equilibrium:

$0=b{x}_{\mathrm{c}}{f}_{\mathrm{c}}-A_{\mathrm{s}1}{f}_{\mathrm{y}}+A_{\mathrm{s}2}{f}_{\mathrm{y}}=300\times 16.7x_{\mathrm{c}}-435(603-1570) \to x_{\mathrm{c}}=84.05 \mathrm{mm}$

Calculating the strains of the steel bars results in

$$\begin{gathered} {\epsilon }_{\mathrm{s}1}=\frac{{\epsilon }_{\mathrm{cu}}}{x}\left(d_1-x\right)=\frac{3.5\times {10}^{-3}}{1.25\times 84.05}\left(455-1.25\times 84.05\right)=11.65\times {10}^{-3}>2.18\times {10}^{-3} \\ {\epsilon }_{\mathrm{s}2}=\frac{{\epsilon }_{\mathrm{cu}}}{x}\left(x-d_2\right)=\frac{3.5\times {10}^{-3}}{1.25\times 84.05}1.25\times \left(84.05-45\right)=2.001\times {10}^{-3}<2.18\times {10}^{-3} \end{gathered}$$ thus both tensile and compressed steel bars are yielding, the assumptions hold true only for the tensile bars, the compression steel bars are elastic. The stress is the compression bars is

${\sigma }_{\mathrm{s}2}=E_{\mathrm{s}}{\epsilon }_{\mathrm{s}2}=200\times {10}^3\times 2.001\times {10}^{-3}=400.2\frac{\mathrm{N}}{{\mathrm{mm}}^2}$

Moment resistance is determined from the moment equilibrium around the tension steel bars:

$$\begin{gathered} M_3=b{x}_{\mathrm{c}}{f}_{\mathrm{c}}\left(d_1-x_{\mathrm{c}}\right)+A_{s\mathit{2}}{f}_{\mathrm{y}}\left(d_1\mathit{-}{d}_{\mathit{2}}\right)= \\ =300\times 84.05\times 16.7\left(455-84.05\right)+603\times 435\left(455-45\right)=281.5 \mathrm{kNm} \end{gathered}$$

Failure envelope is shown in the Figure below.

The internal forces of the cross section are

$$\begin{gathered} N= 400 \mathrm{kN} \\ M=Ne=400\times 0.5=200 \mathrm{kNm} \end{gathered}$$

The point given by the above internal forces is inside the failure envelop, thus the eccentrically compressed cross section is safe.

Problem b)

The approximate moment resistance which belongs to normal force, N = 400 kN is ${\mathit{M}}_{\mathbf{R}}=\frac{M_2-M_3}{N_2}N+M_3=\frac{348.8-281.4}{702.6}400+281.4\mathbf{=}\mathbf{319}\mathbf{.}\mathbf{8}\mathbf{ }\mathbf{kNm}$

Problem c)

Assume the yielding of all steel bars. Height of the compressed concrete zone can be calculated from the force equilibrium: $$\begin{gathered} N=b{x}_{\mathrm{c}}{f}_{\mathrm{c}}-A_{\mathrm{s}1}{f}_{\mathrm{y}}+A_{\mathrm{s}2}{f}_{\mathrm{y}} \\ \to x_{\mathrm{c}}=\frac{N+A_{\mathrm{s}1}{f}_{\mathrm{y}}-A_{\mathrm{s}2}{f}_{\mathrm{y}}}{b{f}_{\mathrm{c}}}=\frac{400\times {10}^3+435\left(1570-603\right)}{300\times 16.7}=163.9 \mathrm{mm} \end{gathered}$$

Calculating the strains of the steel bars results in

$$\begin{gathered} {\epsilon }_{\mathrm{s}1}=\frac{{\epsilon }_{\mathrm{cu}}}{x}\left(d_1-x\right)=\frac{3.5\times {10}^{-3}}{1.25\times 163.9}\left(455-1.25\times 163.9\right)=4.27\times {10}^{-3}>2.18\times {10}^{-3} \\ {\epsilon }_{\mathrm{s}2}=\frac{{\epsilon }_{\mathrm{cu}}}{x}\left(x-d_2\right)=\frac{3.5\times {10}^{-3}}{1.25\times 163.9}1.25\times \left(163.9-45\right)=2.73\times {10}^{-3}>2.18\times {10}^{-3} \end{gathered}$$ thus both tensile and compressed steel bars are yielding, the assumptions hold true.

Moment resistance is determined from the moment equilibrium around the tension steel bars:

$$\begin{gathered} {\mathit{M}}_{\mathbf{R}}=b{x}_{\mathrm{c}}{f}_{\mathrm{c}}\left(\frac{h}{2}-x_{\mathrm{c}}\right)+A_{s\mathit{2}}{f}_{\mathrm{y}}\left(\frac{h}{\mathit{2}}\mathit{-}{d}_{\mathit{2}}\right)\mathit{+}{A}_{s\mathit{1}}{f}_y\left(d_{\mathit{1}}\mathit{-}\frac{h}{\mathit{2}}\right)= \\ =300\times 163.9\times 16.7\left(250-163.9\right)+603\times 435\left(250-45\right)+1570\times 435\left(455-250\right)\mathbf{=}\mathbf{331}\mathbf{.}\mathbf{9}\mathbf{ }\mathbf{kNm} \end{gathered}$$