The stress-strain diagram of the material of a column can be approximated as shown. Plot P / A vs. K L / r for the column.
Tangent Moduli. From the stress – strain diagram,

Question

The stress-strain diagram of the material of a column can be approximated as shown. Plot P / A vs. K L / r for the column.
Tangent Moduli. From the stress - strain diagram,

Accepted Answer

[latex]\left(E_{t} ight)_{1}=\frac{200\left(10^{6} ight)}{0.001}=200 GPa \quad 0 \leq \sigma<200 MPa \\left(E_{t} ight)_{2}=\frac{(350-200)\left(10^{6} ight)}{0.004-0.001}=50 GPa \quad 200 MPa <\sigma \leq 350 MPa[/latex]

Critical Stress. Applying Engesser's equation,

[latex]\sigma_{c r}=\frac{P}{A}=\frac{\pi^{2} E_{t}}{\left(\frac{K L}{r} ight)^{2}} \quad(1)[/latex]

If [latex]E_{t}=\left(E_{t} ight)_{1}=200 GPa[/latex], Eq. (1) becomes

[latex]\frac{P}{A}=\frac{\pi^{2}\left[200\left(10^{9} ight) ight]}{\left(\frac{K L}{r} ight)^{2}}=\frac{1.974\left(10^{6} ight)}{\left(\frac{K L}{r} ight)^{2}} MPa[/latex]

When [latex]\sigma_{ cr }=\frac{P}{A}=\sigma_{Y}=200 MPa[/latex], this equation becomes

[latex]200\left(10^{6} ight)=\frac{\pi^{2}\left[200\left(10^{9} ight) ight]}{\left(\frac{K L}{r} ight)^{2}} \\frac{K L}{r}=99.346=99.3[/latex]

If [latex]E_{t}=\left(E_{t} ight)_{2}=50 GPa[/latex], Eq. (1) becomes

[latex]\frac{P}{A}=\frac{\pi^{2}\left[50\left(10^{9} ight) ight]}{\left(\frac{K L}{r} ight)^{2}}=\frac{0.4935\left(10^{6} ight)}{\left(\frac{K L}{r} ight)^{2}} MPa[/latex]

when [latex]\sigma_{ cr }=\frac{P}{A}=\sigma_{ Y }=200 MPa[/latex], this equation gives

[latex]200\left(10^{6} ight)=\frac{\pi^{2}\left[50\left(10^{9} ight) ight]}{\left(\frac{K L}{r} ight)^{2}} \\frac{K L}{r}=49.67=49.7[/latex]
Using these results, the graphs of [latex]\frac{P}{A} ext { vs. } \frac{K L}{r} ext { as shown in Fig. } a ext { can be plotted. }[/latex]

Question 13.73: The stress-strain diagram of the material of a column can be...

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