Problem 2.3. Stresses of an I beam

Point A

Maximum stress, ${\sigma }_1 \left[\mathrm{MPa}\right]=$

Minimum stress, ${\sigma }_2$[MPa]=

Point B

Maximum shear stress, ${\tau }^{\max } \left[\mathrm{MPa}\right]=$

Point A

Step 1. Write stress vector of point A in x-y  coordinate system attached to the beam’s axis.

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Step 2. Determine principal stresses of point A.

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see Eq.(2-11)

$$\begin{gathered} {\mathit{\sigma }}_{\mathbf{1}}=\frac{{\sigma }_x+{\sigma }_y}{2}+\sqrt{{\left(\frac{{\sigma }_x–{\sigma }_y}{2}\right)}^2+{\tau }_{xy}^2}={\tau }_{xy}^{}=\mathbf{15}\mathbf{.}\mathbf{4}\mathbf{ }\mathbf{MPa}\mathbf{ }\mathbf{\left(}\mathbf{tension}\mathbf{\right)} \\ {\mathbf{\sigma }}_{\mathbf{2}}=\frac{{\mathrm{\sigma }}_{\mathrm{x}}+{\mathrm{\sigma }}_{\mathrm{y}}}{2}–\sqrt{{\left(\frac{{\mathrm{\sigma }}_{\mathrm{x}}–{\mathrm{\sigma }}_{\mathrm{y}}}{2}\right)}^2+{\mathrm{\tau }}_{\mathrm{xy}}^2}={\mathrm{\tau }}_{\mathrm{xy}}^{}=\mathbf{–}\mathbf{15}\mathbf{.}\mathbf{4}\mathbf{ }\mathbf{MPa}\mathbf{ }\mathbf{\left(}\mathbf{compression}\mathbf{\right)} \end{gathered}$$

Step 3. Calculate and draw the principal directions.

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$\cot 2{\beta }_0=\frac{{\sigma }_x–{\sigma }_y}{2{\tau }_{xy}}=0 \to {\beta }_0=\pm {45}^{\circ }$ 

Principal directions are shown in the Figure.

See Mohr circle in Figure 2.7c

Step 4. Draw the stresses.

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Point B

Step 1. Write stress vector of point B in x-y  coordinate system attached to the beam’s axis.

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Step 2. Calculate the maximum shear stress.

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Eq.(2-13)

${\mathit{\tau }}^{\mathbf{max}}=\sqrt{{\left(\frac{{\mathrm{\sigma }}_{\mathrm{x}}–{\mathrm{\sigma }}_{\mathrm{y}}}{2}\right)}^2+{\mathrm{\tau }}_{\mathrm{xy}}^2}=\frac{{\mathrm{\sigma }}_{\mathrm{x}}}{2}=\frac{167}{2}=\mathbf{83}\mathbf{.}\mathbf{5}\mathbf{ }\mathbf{MPa}$

See also Mohr circle in Figure 2.7a

Point A

According to the approximation the web is subjected to pure shear stage. The stress vector is

$\left\{\begin{array}{c}{\sigma }_x\\ {\sigma }_y\\ {\tau }_{xy}\end{array}\right\}=\left\{\begin{array}{c}0\\ 0\\ 15.4\end{array}\right\}\mathrm{MPa}$

The principal stresses of point A are

see Eq.(2-11)

$$\begin{gathered} {\mathit{\sigma }}_{\mathbf{1}}=\frac{{\sigma }_x+{\sigma }_y}{2}+\sqrt{{\left(\frac{{\sigma }_x–{\sigma }_y}{2}\right)}^2+{\tau }_{xy}^2}={\tau }_{xy}^{}=\mathbf{15}\mathbf{.}\mathbf{4}\mathbf{ }\mathbf{MPa}\mathbf{ }\mathbf{\left(}\mathbf{tension}\mathbf{\right)} \\ {\mathbf{\sigma }}_{\mathbf{2}}=\frac{{\mathrm{\sigma }}_{\mathrm{x}}+{\mathrm{\sigma }}_{\mathrm{y}}}{2}–\sqrt{{\left(\frac{{\mathrm{\sigma }}_{\mathrm{x}}–{\mathrm{\sigma }}_{\mathrm{y}}}{2}\right)}^2+{\mathrm{\tau }}_{\mathrm{xy}}^2}={\mathrm{\tau }}_{\mathrm{xy}}^{}=\mathbf{–}\mathbf{15}\mathbf{.}\mathbf{4}\mathbf{ }\mathbf{MPa}\mathbf{ }\mathbf{\left(}\mathbf{compression}\mathbf{\right)} \end{gathered}$$

The principal directions are shown in the Figure

$\cot 2{\beta }_0=\frac{{\sigma }_x–{\sigma }_y}{2{\tau }_{xy}}=0 \to {\beta }_0=\pm {45}^{\circ }$ 

See Mohr circle in Figure 2.7c

Pure shear is equivalent to the sum of a pure compression and a pure tension in $\pm {45}^{\circ }$directions.

Point B

According to the approximation the bottom flange is in pure tension stage. The stress vector is

$\left\{\begin{array}{c}{\sigma }_x\\ {\sigma }_y\\ {\tau }_{xy}\end{array}\right\}=\left\{\begin{array}{c}{\sigma }_1\\ 0\\ 0\end{array}\right\}=\left\{\begin{array}{c}167\\ 0\\ 0\end{array}\right\}\mathrm{MPa}$.

The x and y directions are the principal directions, the principal stress, ${\sigma }_1$ is equal to the tensile stress, ${\sigma }_{\mathrm{f}}$.

The maximum shear stress is

Eq.(2-13)

${\mathit{\tau }}^{\mathbf{max}}=\sqrt{{\left(\frac{{\mathrm{\sigma }}_{\mathrm{x}}–{\mathrm{\sigma }}_{\mathrm{y}}}{2}\right)}^2+{\mathrm{\tau }}_{\mathrm{xy}}^2}=\frac{{\mathrm{\sigma }}_{\mathrm{x}}}{2}=\frac{167}{2}=\mathbf{83}\mathbf{.}\mathbf{5}\mathbf{ }\mathbf{MPa}$

See also Mohr circle in Figure 2.7a