Problem 3.11. Shear deflection

Ratio of bending and shear deflection, v_Bo/v_So=

Step 1. Draw the moment diagram. Write and sketch bending and the shear deformation functions.

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Shear deflection is proportional to the bending moment diagram:

Eq.(3-64)

$v_{\mathrm{S}}=\frac{M}{S}+C+Dx$

where constant, C and D are determined from the boundary conditions:

$$\begin{gathered} v\left(0\right)=0=v_B\left(0\right)+v_S\left(0\right)=0 \\ v\left(L\right)=v_B\left(L\right)+v_S\left(L\right)=0 \end{gathered}$$

Here the bending deflection, and the moment are zero at the support, thus constants, C and D are zero. C = D = 0.

Bending deflection is derived by double integration of the moment function. Maximum deflection is given, the shape is shown below.

$v_{\mathrm{B}}={\int }_0^x{\chi }_{y}dx=–\frac{1}{EI}{\int }_0^x{\int }_0^x{M}_{y}dxdx+C_1+C_{2}x$

Step 2. Determine bending and shear stiffnesses of the cross section.

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$$\begin{gathered} EI=E\left(\frac{b{h}^3}{12}–\frac{\left(b–2t\right){\left(h–2t\right)}^3}{12}\right)=1.707\times {10}^{13} {\mathrm{Nmm}}^2=1.707\times {10}^4 {\mathrm{kNm}}^2 \\ \mathrm{S}\mathit{=}\frac{GA}{n}\mathit{\approx }G{A}_{\mathrm{w}}=G\times 2\times t\times \left(h–2t\right)=3.55\times {10}^8 \mathrm{N}=3.55\times {10}^5 \mathrm{kN} \\ \mathrm{where} n=\frac{A}{A_{\mathrm{w}}} \end{gathered}$$

See Figure 73.

Step 3. Calculate bending and shear deflection at the middle. Compare results.

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Shear deflection is proportional to the bending moment diagram:

Eq.(3-64)

where constants, C and D are determined from the boundary conditions:

$$\begin{gathered} v\left(0\right)=0=v_B\left(0\right)+v_S\left(0\right)=0 \\ v\left(L\right)=v_B\left(L\right)+v_S\left(L\right)=0 \end{gathered}$$

Here the bending deflection, and the moment are zero at the support, thus constants, C and D are zero. C = D = 0.

Bending deflection is derived by double integration of the moment function. 

$v_{\mathrm{B}}={\int }_0^x{\chi }_{y}dx=–\frac{1}{EI}{\int }_0^x{\int }_0^x{M}_{y}dxdx+C_1+C_{2}x$

The bending and shear deformation functions are sketched below:

The bending and shear stiffnesses of the cross section are:

$$\begin{gathered} EI=E\left(\frac{b{h}^3}{12}–\frac{\left(b–2t\right){\left(h–2t\right)}^3}{12}\right)=1.707\times {10}^{13} {\mathrm{Nmm}}^2=1.707\times {10}^4 {\mathrm{kNm}}^2 \\ \mathrm{S}\mathit{=}\frac{GA}{n}\mathit{\approx }G{A}_{\mathrm{w}}=G\times 2\times t\times \left(h–2t\right)=3.553\times {10}^8 \mathrm{N}=3.55\times {10}^5 \mathrm{kN} \\ \mathrm{where} n=\frac{A}{A_{\mathrm{w}}} \end{gathered}$$

See Figure 73.

Maximum bending and shear deflection at the middle of the beam are the following:

$$\begin{gathered} {\mathit{v}}_{\mathbf{Bo}}=\frac{Pa}{24EI}\left(3L^2–4a^2\right)=\frac{70\times 2.6/3}{24\times 1.707\times {10}^4}\left(3\times 2.6^2–4\frac{2.6^2}{3^2}\right)=0.00256 \mathrm{m}=\mathbf{2}\mathbf{.}\mathbf{56}\mathbf{ }\mathbf{mm} \\ {\mathit{v}}_{\mathbf{so}}=\frac{Pa}{S}=\frac{70\times 2.6/3}{3.55\times {10}^5}=0.000171 \mathrm{m}=\mathbf{0}\mathbf{.}\mathbf{171}\mathbf{ }\mathbf{mm} \end{gathered}$$

Shear deflection is negligable compared to the bending deflection.