Problem 4.2. Shear of an inhomogeneous cross section

Problem a)

Maximum shear stress, τ_max [MPa]=

Problem b)

Maximum shear stress, τ_max [MPa]=

Problem c)

Maximum shear stress, τ_max [MPa]=

Problem a)

Step 1. Draw the shear stress distribution along the height of the cross section.

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Shear stress is given by the Zhuravskii formula. It is proportional to the area of the the normal stress diagram, which is given in the previous problem (Problem 4.1 a). The distribution is shown in the Figure below.

Eq.(4-10)

${\tau }_{xy}=\frac{V}{bM}{\int }_A^{}{\sigma }_{x}dA=\frac{VS}{I_{z}b}$

Step 2. Determine the section properties.

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Step 3. Calculate the maximum shear stress.

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Problem b)

Step 1. Draw the shear stress distribution along the height of the cross section.

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Shear stress is proportional to the area of the the normal stress diagram, which is given in the previous problem (Problem 4.1 b). Integration of the piecewise linear normal stress function results in linear shear stress distribution at the top and at the bottom, and second order distribution at the middle of the cross section. Shear stress diagram is shown in the Figure below. 

Eq.(4-10)

${\tau }_{xy}=\frac{V}{bM}{\int }_A^{}{\sigma }_{x}dA$

Step 2. Determine shear stress function. 

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Step 3. Calculate the maximum shear stress.

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Problem c)

Step 1. Draw the shear stress distribution along the height of the cross section.

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Shear stress is determined by the integration of the normal stress diagram, which is calculated in the previous problem (Problem 4.1 c) defining an equivalent cross section. The distribution is shown in the Figure below.

Eq.(4-10)

${\tau }_{xy}=\frac{V}{bM}{\int }_A^{}{\sigma }_{x}dA$

Step 2. Calculate the maximum shear stress.

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Problem a)

Shear stress is given by the Zhuravskii formula. It is proportional to the area of the the normal stress diagram, which is given in the previous problem (Problem 4.1 a). The distribution is shown in the Figure below.

Eq.(4-10)

The cross section is homogeneous, α = 1. Moment of inertia is given in the previous problem (Problem 4.1 a), the moment of area function is

$S=b{\int }_y^{\frac{h}{2}}y\mathrm{d}y=b{\left[\frac{y^2}{2}\right]}_y^{\frac{h}{2}}=\frac{b{h}^2}{8}–\frac{b{y}^2}{2}$

The shear stress function reaches it maximum at the centre of gravity of the cross section, at y = 0:

$$\begin{gathered} S(y=0)=\frac{b{h}^2}{8}=\frac{250\times {450}^2}{8}=6.33\times {10}^6 {\mathrm{mm}}^3 \\ {\mathbf{\tau }}_{\mathbf{max}}=\frac{VS}{I_{z}b}=\frac{30\times {10}^3\times 6.33\times {10}^6}{1.90\times {10}^9\times 250}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{400}\frac{\mathbf{N}}{{\mathbf{mm}}^{\mathbf{2}}} \end{gathered}$$

Problem b)

Shear stress is proportional to the area of the the normal stress diagram, which is given in the previous problem (Problem 4.1 b). Integration of the piecewise linear normal stress function results in linear shear stress distribution at the top and at the bottom, and second order distribution at the middle of the cross section. Shear stress diagram is shown in the Figure below. 

${\tau }_{xy}=\frac{V}{bM}{\int }_A^{}{\sigma }_{x}dA$

The cross section is homogeneous, α = 1.

At the top (and at the bottom) of the cross section the shear stress function is linear:

$\tau =\frac{V}{bM}fb\left(\frac{h}{2}–y\right)=\frac{V}{M}f\left(\frac{h}{2}–y\right)$

At the middle part:

$\tau =\frac{V}{M}f\left(\frac{h}{2}–x\right)+\frac{V}{M}\frac{f}{2}\left(1+\frac{y}{x}\right)\left(x–y\right)$

where x is the height of the elastic part, which is determined in Problem 4.1 b).

The shear stress function reaches it maximum at the centre of gravity of the cross section, at y = 0:

$\mathit{\tau }=\frac{V}{M}f\left(\frac{h}{2}–x\right)+\frac{V}{M}\frac{f}{2}x=\frac{30\times {10}^3}{250\times {10}^6}25\left(\frac{450}{2}–\frac{179}{2}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{407}\mathbf{ }\frac{\mathbf{N}}{{\mathbf{mm}}^{\mathbf{2}}}$

Problem c)

Shear stress is determined by the integration of the normal stress diagram, which is calculated in the previous problem (Problem 4.1 c) defining an equivalent cross section. The distribution is shown in the Figure below.

${\tau }_{xy}=\frac{V}{bM}{\int }_A^{}{\sigma }_{x}dA$

The shear stress function reaches it maximum at the centre of gravity of the inhomogeneous cross section, at y  = y_c (See Problem 4.1.c):

$$\begin{gathered} {\sigma }_{\mathrm{b}}=\frac{M}{I_{\mathrm{e}}}\left(h–y_{\mathrm{c}}\right)=\frac{250\times {10}^6}{2.06\times {10}^9}(450–187.5)=25.22\frac{\mathrm{N}}{{\mathrm{mm}}^2} \\ {\mathbf{\tau }}_{\mathbf{max}}=\frac{V}{bM}{\int }_{h–y_{\mathrm{c}}}^{0}b{\sigma }_{x}dy=\frac{V}{bM}b\frac{{\sigma }_x\left(h–y_{\mathrm{c}}\right)}{2}=\frac{30\times {10}^3\times 25.22\times \left(450–187.5\right)}{250\times {10}^6}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{397}\frac{\mathbf{N}}{{\mathbf{mm}}^{\mathbf{2}}} \end{gathered}$$