Problem 3.8. Overhanging beam

Maximum shear stress, τ^max [N/mm²]=

Step 1. Draw the shear force diagram along the length of the beam. Determine the maximum shear force.

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Step 2. Draw the shear stress distribution along the height of the cross section. Calculate the maximum shear stress.

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Shear stresses are determined by the Zhuravskii formula, results are given in the Figure. The section properties are given in Problem 3.6.

Eq.(3-34)

The maximum shear stress arises at the center of gravity:

${\mathit{\tau }}^{\mathbf{max}}=\frac{VS}{Ib}=\frac{33.33\times {10}^3\times 2.82\times {10}^6}{9.82\times {10}^8\times 50}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{91}\frac{\mathbf{N}}{\mathbf{m}{\mathbf{m}}^{\mathbf{2}}}$

where the moment of area is:

$S=b_1{t}_1\left(s–\frac{t_1}{2}\right)+t_2\frac{{\left(s–t_1\right)}^2}{2}=2.82\times {10}^6{\mathrm{mm}}^3$

Step 3. Perform shear check.

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First the shear force diagram is determined along the length of the beam. The maximum shear force arises at the left side of the middle support.

Shear stresses are determined by the Zhuravskii formula, results are given in the Figure. The section properties are given in Problem 3.6.

Eq.(3-34)

The maximum shear stress arises at the center of gravity:

${\mathit{\tau }}^{\mathbf{max}}=\frac{VS}{Ib}=\frac{33.33\times {10}^3\times 2.82\times {10}^6}{9.82\times {10}^8\times 50}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{91}\frac{\mathbf{N}}{\mathbf{m}{\mathbf{m}}^{\mathbf{2}}}$

where the moment of area is:

$S=b_1{t}_1\left(s–\frac{t_1}{2}\right)+t_2\frac{{\left(s–t_1\right)}^2}{2}=2.82\times {10}^6 {\mathrm{mm}}^3$

The shear check is performed for the maximum shear stress:

${\tau }^{\max }=1.91\frac{\mathrm{N}}{{\mathrm{mm}}^2}< f=2.6\frac{\mathrm{N}}{{\mathrm{mm}}^2}$

The beam is safe for shear.