Problem 3.9. Shear flow in a C beam

Maximum value of the shear flow, q^max [N/mm]=

Draw the shear flow diagram. Calculate all the values.

Check figureHide figure

Step 1. Stresses of the thin walled cross section are reduced to its midline. Assume uniform stress distribution through the thickness. Determine the normal stress distribution along the height of the cross section.

Check normal stresses

The moment of inertia of the cross section is

The resultant of the constant normal stresses acts at the midline, the cross section properties are also calculated to the midline.

$$$I=\frac{t{d}^3}{12}+2\frac{t{a}^3}{12}+2at{\left(\frac{d}{2}–\frac{a}{2}\right)}^2+2\frac{b{t}^3}{12}+2bt{\left(\frac{d}{2}\right)}^2=9.95\times {10}^6 {\mathrm{mm}}^4$

The normal stresses in the specific heights are given in the function of the moment:

$$\begin{gathered} {\sigma }_x=\frac{M}{I}y \\ {\sigma }_{x,1}=\frac{M}{I}\left(\frac{d}{2}–a\right)=M\frac{197/2–28.5}{9.95\times {10}^6}=M\times 7.04\times {10}^{–6} \\ {\sigma }_{x,2}=\frac{M}{I}\frac{d}{2}=M\frac{197/2}{9.95\times {10}^6}=M\times 9.90\times {10}^{–6} \end{gathered}$$

Multiplying the stresses with the thickness results in the normal force for unit length, N_x=σ_xt , the distribution of which is given in the Figure.

Step 2. To obtain the shear flow integrate the normal force for unit length, N_x along the midline.

Check normal stresses

See Equation (3-38) and Footnote c in Section 3.1.1.2.

$$\begin{gathered} N_x={\sigma }_{x}t \\ q=–{\int }_0^s\frac{\partial N_x}{\partial x}d\eta =–t{\int }_0^s\frac{\partial \sigma }{\partial x}d\eta =–t{\int }_0^s{\sigma }_x\frac{V}{m}d\eta =–\frac{V}{M}t{\int }_0^s{\sigma }_{x}d\eta \end{gathered}$$

The above integration can be performed by calculating the area of the normal force diagram, N_x = σ_xt given in the Figure above.

$$\begin{gathered} {\mathit{q}}_{\mathbf{1}}=–\frac{V}{M}\frac{N_{x,1}+N_{x,2}}{2}a=–\frac{V}{M}t\frac{{\sigma }_{x,1}+{\sigma }_{x,2}}{2}a=–\frac{V}{M}t\frac{M(7.04+9.90)\times {10}^{–6}}{2}a= \\ =Vta\frac{(7.04+9.9)\times {10}^6}{2}=10000\times 3\times 28.5\frac{(7.04+9.90)\times {10}^{–6}}{2}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{24}\frac{\mathbf{N}}{\mathbf{mm}} \\ {\mathit{q}}_{\mathbf{2}}=q_1+\frac{V}{M}{N}_{x,2}b=q_1+\frac{V}{M}t{\sigma }_{x,2}b=7.24+10000\times 3\times 9.90\times {10}^{–6}\times 117\mathbf{=}\mathbf{42}\mathbf{.}\mathbf{00}\frac{\mathbf{N}}{\mathbf{mm}} \\ {\mathit{q}}_{\mathbf{3}}=q_2+\frac{V}{M}\frac{N_{x,2}}{2}\frac{d}{d}=q_2+\frac{V}{M}\frac{t{\sigma }_{x,2}}{2}\frac{d}{d}=42.00+10000\times 3\times \frac{9.90\times {10}^{–6}}{2}\times \frac{197}{2}\mathbf{=}\mathbf{56}\mathbf{.}\mathbf{63}\frac{\mathbf{N}}{\mathbf{mm}} \end{gathered}$$

 Stresses of the thin walled cross section are reduced to its midline. The approximate moment of inertia of the cross section is

$$$I=\frac{t{d}^3}{12}+2\frac{t{a}^3}{12}+2at{\left(\frac{d}{2}–\frac{a}{2}\right)}^2+2\frac{b{t}^3}{12}+2bt{\left(\frac{d}{2}\right)}^2=9.95\times {10}^6 {\mathrm{mm}}^4$

The resultant of the constant normal stresses acts at the midline, the cross section properties are also calculated to the midline.

The normal stress distribution is given in the function of the moment:

$$\begin{gathered} {\sigma }_x=\frac{M}{I}y \\ {\sigma }_{x,1}=\frac{M}{I}\left(\frac{d}{2}–a\right)=M\frac{197/2–28.5}{9.95\times {10}^6}=M\times 7.04\times {10}^{–6} \\ {\sigma }_{x,2}=\frac{M}{I}\frac{d}{2}=M\frac{197/2}{9.95\times {10}^6}=M\times 9.90\times {10}^{–6} \end{gathered}$$

Multiplying the stresses with the thickness results in the normal force for unit length, N_x=σ_xt , the distribution of which is given in the Figure.

 To obtain the shear flow the normal force per unit length, N_x is integrated along the midline.

See Equation (3-38) and Footnote c in Section 3.1.1.2.

The above integration can be performed by calculating the area of the normal force diagram, N_x=σ_xt given in the Figure above.

$$\begin{gathered} {\mathit{q}}_{\mathbf{1}}=–\frac{V}{M}\frac{N_{x,1}+N_{x,2}}{2}a=–\frac{V}{M}t\frac{{\sigma }_{x,1}+{\sigma }_{x,2}}{2}a=–\frac{V}{M}t\frac{M(7.04+9.90)\times {10}^{–6}}{2}a= \\ =Vta\frac{(7.04+9.9)\times {10}^6}{2}=10000\times 3\times 28.5\frac{(7.04+9.90)\times {10}^{–6}}{2}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{24}\frac{\mathbf{N}}{\mathbf{mm}} \\ {\mathit{q}}_{\mathbf{2}}=q_1+\frac{V}{M}{N}_{x,2}b=q_1+\frac{V}{M}t{\sigma }_{x,2}b=7.24+10000\times 3\times 9.90\times {10}^{–6}\times 117\mathbf{=}\mathbf{42}\mathbf{.}\mathbf{00}\frac{\mathbf{N}}{\mathbf{mm}} \\ {\mathit{q}}_{\mathbf{3}}=q_2+\frac{V}{M}\frac{N_{x,2}}{2}\frac{d}{d}=q_2+\frac{V}{M}\frac{t{\sigma }_{x,2}}{2}\frac{d}{d}=42.00+10000\times 3\times \frac{9.90\times {10}^{–6}}{2}\times \frac{197}{2}\mathbf{=}\mathbf{56}\mathbf{.}\mathbf{63}\frac{\mathbf{N}}{\mathbf{mm}} \end{gathered}$$