Problem 4.1. Bending of an inhomogeneous cross section

Problem a)

Maximum stress, σ_max [MPa]=

CHECK

Bending resistance, M_R,e [kNm]=

CHECK

Problem b)

Bending resistance, M_R,p [kNm]=

CHECK

Problem c)

Maximum stress, σ_max [MPa]=

CHECK

Bending resistance, M_R,e [kNm]=

CHECK

Problem a)

Step 1. Calculate the section properties.

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Step 2. Determine stress distribution from moment, M = 250 kNm. Draw stress diagram. Calculate the maximum values.

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Step 3. Calculate bending resistance (when the maximum stress reaches the strength).

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Problem b)

Step 1. Draw the stress distribution along the height of the cross section which arises from M = 250 kNm.

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Step 2. Draw the stress diagram which belongs to plastic failure. Calculate moment resistance.

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Problem c)

Step 1. Replace the inhomogeneous cross section with an equivalent homogeneous one. 

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Step 2. Calculate the section properties of the replacement cross section (area, center of gravity, moment of inertia).

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Step 3. Determine stress distribution from moment, M = 250 kNm. Draw stress diagram. Calculate the maximum values.

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Step 4. Calculate bending resistance (when the maximum stress reaches the strength).

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Problem a)

Since the material properties of the two parts of the cross section are the same, the cross section can be treated as homogeneous. (Ratio of the elastic moduli is α=1.) The section properties are

$$\begin{gathered} I_z=\frac{b{h}^3}{12}=\frac{250\times {450}^3}{12}=1.90\times {10}^9{\mathrm{mm}}^4 \\ {W}_z=\frac{b{h}^2}{6}=\frac{250\times {450}^2}{6}=8.44\times {10}^6{\mathrm{mm}}^3 \end{gathered}$$

Stress diagram from moment, M = 250 kNm and the maximum values ares given below. 

 
${\mathit{\sigma }}_{\mathbf{m}\mathbf{a}\mathbf{x}}^{\mathbf{\pm }}={\sigma }_{\mathrm{t}}={\sigma }_{\mathrm{b}}=\frac{M}{I_z}\frac{h}{2}=\frac{M}{W_z}=\frac{250\times {10}^6}{8.44\times {10}^6}\mathbf{=}\mathbf{29}\mathbf{.}\mathbf{63}\mathbf{ }\frac{\mathbf{N}}{{\mathbf{mm}}^{\mathbf{2}}}>f=25 \mathrm{MPa}$

Thus the material fails.

The bending resistance can be calculated assuming that the maximum stress reaches the strength.

${\mathit{M}}_{\mathbf{R}\mathbf{,}\mathbf{e}}=W_{z}f=8.44\times {10}^6\times 25=210.9\times {10}^6 \mathrm{Nmm}\mathbf{=}\mathbf{210}\mathbf{.}\mathbf{9}\mathbf{ }\mathbf{kNm} < M=250 \mathrm{kNm}$

The elastic bending resistance is not sufficient for this moment.

Problem b)

The cross section is still homogeneous, the stress distribution from moment, M = 250 kNm is shown in the Figure. 

The middle of the cross section is in elastic stage, top and bottom parts are plastic. From the moment equilibrium the height of the elastic part can be calculated.

$$\begin{gathered} M=2f\frac{x^2}{2}\frac{2}{3}b+2\left(\frac{h}{2}–x\right)fb\left(\frac{h}{2}–\frac{{\displaystyle \raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}–x}{2}\right) \\ \to x=\sqrt{\left(\frac{h^2}{8}–\frac{M}{2fb}\right)6}=178.5 \mathrm{mm} \end{gathered}$$

The stress diagram which belongs to plastic failure is shown below. The stress reaches the strength value at each points of the cross section. The moment resistance is calculated as

${\mathit{M}}_{\mathbf{R}\mathbf{,}\mathbf{p}}=f\frac{b{h}^2}{4}=25\frac{250\times {450}^2}{4}\mathbf{=}\mathbf{316}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{ }\mathbf{Nmm}=316.4 \mathrm{kNm} > M=250 \mathrm{kNm}$

The cross section is safe.

Problem c)

The inhomogeneous cross section is replaced with an equivalent homogeneous one.The bottom material is chosen to be the reference material, the top part of the cross section is increased by the ratio of the elastic moduli

$\alpha =\frac{E_a}{E_b}=2$

(Width of the cross section is increased in order not to change the centre of gravity of that part.)

Section properties of the replacement cross section (area, center of gravity, moment of inertia) are

$$\begin{gathered} A_{\mathrm{e}}={\alpha }_{\mathrm{e}}\frac{h}{3}b+\frac{2h}{3}b=2\frac{450}{3}250+\frac{2\times 450}{3}250=1.5\times {10}^5{\mathrm{mm}}^2 \\ {y}_{\mathrm{c}}\mathit{=}\frac{{\alpha }_{e}b{\displaystyle \frac{h}{3}\frac{h}{6}}+b\frac{2h}{3}\frac{2h}{3}}{A_{\mathrm{e}}}\mathit{=}\frac{2\times 250{\displaystyle \frac{450}{3}}+250\frac{4}{9}450}{1.5\times {10}^5}\mathit{=}187.5 \mathrm{mm} \\ {I}_{\mathrm{e},z}=\frac{{\alpha }_{e}b{\left(\frac{h}{\mathit{3}}\right)}^{\mathit{3}}}{\mathit{12}}+{\alpha }_{\mathrm{e}}b\frac{h}{3}{\left(y_{\mathrm{c}}–\frac{h}{6}\right)}^2+\frac{{\alpha }_{\mathrm{e}}b{\left(\frac{2h}{3}\right)}^3}{12}+b\frac{2h}{3}{\left(\frac{2h}{3}–y_{\mathrm{c}}\right)}^2\mathit{=}2.602\times {10}^9{\mathrm{mm}}^4 \end{gathered}$$

Stress distribution from moment, M = 250 kNm is given in the Figure below. The maximum value is also calculated.

${\mathbf{\sigma }}_{\mathbf{max}}^{}={\sigma }_{\mathrm{t}}={\alpha }_{\mathrm{e}}\frac{M}{I_{\mathrm{e},z}}{y}_{\mathrm{c}}=2\frac{250\times {10}^6}{2.06\times {10}^9}187.5\mathbf{=}\mathbf{36}\mathbf{.}\mathbf{04}\mathbf{ }\frac{\mathbf{N}}{{\mathbf{mm}}^{\mathbf{2}}}>f=25 \mathrm{MPa}$

Thus the material fails.

The moment which equilibrates the stress distribution where the maximum stress reaches the strength is the bending resistance.

${\mathit{M}}_{\mathbf{R}\mathbf{,}\mathbf{e}}=\frac{f{I}_{\mathrm{e},z}}{{\alpha }_{\mathrm{e}}{y}_{\mathrm{c}}}=\frac{25\times 2.602\times {10}^9}{2\times 187.5}=173.4\times {10}^6 \mathrm{Nmm}\mathbf{=}\mathbf{173}\mathbf{.}\mathbf{4}\mathbf{ }\mathbf{kNm} > M=250\mathrm{kNm}$

The elastic bending resistance is not sufficient.