Problem 5.6. Creep

Problem a)

Change in stress [%]:

CHECK

Problem b)

Change in stress [%]:

CHECK

Problem c)

Change in stress [%]:

CHECK

Problem a)

Step 1. Calculate the section properties with the aid of the effective Young modulus.

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Step 2. Determine stress at t = ∞.

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Problem b)

Step 1. Calculate the section properties. Modify the effective Young modulus according to Trost’s model .

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Step 2. Determine stress at t = ∞.

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Problem c)

Step 1. Determine change in stress according to Dischinger model.

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Problem a)

The effective Young modulus is

$E_{\mathrm{eff}}=\frac{E}{1+\phi }=\frac{31\times {10}^3}{1+2}=10.33\times {10}^3 \mathrm{MPa}$

The area of the homogeneous cross section at t = 0 is

$A_{\mathrm{eo}}=A_{\mathrm{c}}+\frac{E\mathrm{s}}{E_{\mathrm{c}}}{A}_{\mathrm{s}}=120000+\frac{200}{31}452=1.229\times {10}^5 {\mathrm{mm}}^2$

The area of the homogeneous cross section at t = ∞ becomes

$A_{\mathrm{e}}=A_{\mathrm{c}}+\frac{E\mathrm{s}}{E_{\mathrm{eff}}}{A}_{\mathrm{s}}=120000+\frac{200}{10.33}452=1.288\times {10}^5 {\mathrm{mm}}^2$

Stress at t = ∞ results in

${\mathit{\sigma }}_{\mathbf{c}}={\sigma }_{\mathrm{co}}\left(1–\phi \frac{E_{\mathrm{s}}}{E_{\mathrm{c}}}\frac{A_{\mathrm{s}}}{A_{\mathrm{e}}}\right) \to \frac{{\sigma }_{\mathrm{c}}}{{\sigma }_{\mathrm{co}}}=\left(1–\phi \frac{E_{\mathrm{s}}}{E_{\mathrm{c}}}\frac{A_{\mathrm{s}}}{A_{\mathrm{e}}}\right)=\left(1–2\frac{200}{31}\frac{452}{1.29\times {10}^5}\right)=0.9547\mathbf{=}\mathbf{95}\mathbf{.}\mathbf{47}\mathbf{\%}$

Problem b)

In the calculation of the section properties the effective Young modulus is modified according to Trost’s model:

$E_{\mathrm{eff}}=\frac{E}{1+\chi \phi }=\frac{31\times {10}^3}{1+0.8\times 2}=11.92\times {10}^3 \mathrm{MPa}$

The area of the homogeneous cross section at t = 0 is unchanged. The area of the homogeneous cross section at t = ∞ becomes

$A_{\mathrm{e\chi }}=A_{\mathrm{c}}+\frac{E\mathrm{s}}{E_{\mathrm{eff}}}{A}_{\mathrm{s}}=120000+\frac{200}{11.92}452=1.276\times {10}^5 {\mathrm{mm}}^2$

Stress at t = ∞ results in:

${\mathit{\sigma }}_{\mathbf{c}}={\sigma }_{\mathrm{co}}\left(1–\phi \frac{E_{\mathrm{s}}}{E_{\mathrm{c}}}\frac{A_{\mathrm{s}}}{A_{\mathrm{e\chi }}}\right) \to \frac{{\sigma }_{\mathrm{c}}}{{\sigma }_{\mathrm{co}}}=\left(1–\phi \frac{E_{\mathrm{s}}}{E_{\mathrm{c}}}\frac{A_{\mathrm{s}}}{A_{e\chi }}\right)=\left(1–2\frac{200}{31}\frac{452}{1.276\times {10}^5}\right)=0.9542\mathbf{=}\mathbf{95}\mathbf{.}\mathbf{42}\mathbf{\%}$

Problem c)

Change in stress according to Dischinger model is

${\mathit{\sigma }}_{\mathbf{c}}={\sigma }_{\mathrm{co}}{e}^{–\left(1–A_{\mathrm{c}}/A_{\mathrm{eo}}\right)\phi } \to \frac{{\sigma }_{\mathrm{c}}}{{\sigma }_{\mathrm{co}}}=e^{–\left(1–A_{\mathrm{c}}/A_{\mathrm{eo}}\right)\phi }=e^{–\left(1–1.2/1.23\right)2}=0.9536\mathbf{=}\mathbf{95}\mathbf{.}\mathbf{36}\mathbf{\%}$