Problem 3.6. Timber I beam

Maximum stress, σ^max, [N/mm²]=

To perform checking the maximum normal stress must be compared to the strength of the material.

Step 1. Find the relevant cross section, give the maximum moment.

Check maximum momentHide moment

Step 2. Calculate the section properties.

Check section propertiesHide section properties

Area:

$A=\sum _{i=1}^3{b}_i{t}_i=40.5\times {10}^3 {\mathrm{mm}}^2$

Center of gravity:

$s=\frac{S}{A}=\frac{b_1\times t_1\times {\displaystyle \frac{t_1}{2}}+b_2\times t_2\times \left(t_1+\frac{b_2}{2}\right)+b_3\times t_3\times \left(t_1+b_2+\frac{t_3}{2}\right)}{A}=166.7 \mathrm{mm}$

Moment of inertia of the cross section:

$I=\frac{b_1{t}_1^3}{12}+b_1{t}_1^{}{\left(s–\frac{t_1}{2}\right)}^2+\frac{t_2{b}_2^3}{12}+b_2{t}_2^{}{\left(t_1+\frac{b_2}{2}–s\right)}^2+\frac{b_3{t}_3^3}{12}+b_3{t}_3^{}{\left(t_1+b_2+\frac{t_3}{2}–s\right)}^2=9.825\times {10}^8{\mathrm{mm}}^4$

Step 3. Determine the maximum stress in the cross section. Check the cross section.

Check maximum stressHide stress

The stress at an arbitrary location:

Eq.(3-29)

$\sigma =\frac{M}{I}y$

The maximum stress arises at the bottom of the cross section.

Eq.(3-30)

The maximum stress arise at the bottom of the cross section.

The elastic section modulus is:

$W=\frac{I}{y}=\frac{I}{h–s}=\frac{I}{t_1+b_2+t_3–s}=\frac{9.825\times {10}^8}{460–166.7}=3.35\times {10}^6{\mathrm{mm}}^3$

$$\begin{gathered} {\mathit{\sigma }}^{\mathbf{m}\mathbf{a}\mathbf{x}}=\frac{M}{W}=\frac{50\times {10}^6}{3.35\times {10}^6}\mathbf{\text{=14.92}}\frac{\mathbf{N}}{\mathbf{m}{\mathbf{m}}^{\mathbf{2}}}\mathbf{ }\mathbf{\left(}\mathbf{compression}\mathbf{\right)} \\ {\sigma }^{\max }\text{=14.92}\frac{\mathrm{N}}{{\mathrm{mm}}^2} <f=60\frac{\mathrm{N}}{{\mathrm{mm}}^2} \end{gathered}$$

The cross section (and the whole girder) is safe for bending.

To perform checking the maximum normal stress must be compared to the strength of the material.

First the relevant cross section is chosen. The maximum moment arises above the middle support. The value is given in the internal force diagram: M = -50 kNm.

The section properties are given below.

Area:

$A=\sum _{i=1}^3{b}_i{t}_i=40.5\times {10}^3 {\mathrm{mm}}^2$

Center of gravity:

$s=\frac{S}{A}=\frac{b_1\times t_1\times {\displaystyle \frac{t_1}{2}}+b_2\times t_2\times \left(t_1+\frac{b_2}{2}\right)+b_3\times t_3\times \left(t_1+b_2+\frac{t_3}{2}\right)}{A}=166.7 \mathrm{mm}$

Moment of inertia of the cross section:

$I=\frac{b_1{t}_1^3}{12}+b_1{t}_1^{}{\left(s–\frac{t_1}{2}\right)}^2+\frac{t_2{b}_2^3}{12}+b_2{t}_2^{}{\left(t_1+\frac{b_2}{2}–s\right)}^2+\frac{b_3{t}_3^3}{12}+b_3{t}_3^{}{\left(t_1+b_2+\frac{t_3}{2}–s\right)}^2=9.825\times {10}^8{\mathrm{mm}}^4$

The stress at an arbitrary location:

Eq.(3-29)

$\sigma =\frac{M}{I}y$

The maximum stress arises at the bottom of the cross section.

Eq.(3-30)

The elastic section modulus is:

$W=\frac{I}{y}=\frac{I}{h–s}=\frac{I}{t_1+b_2+t_3–s}=\frac{9.825\times {10}^8}{460–166.7}=3.35\times {10}^6{\mathrm{mm}}^3$

$$\begin{gathered} {\mathit{\sigma }}^{\mathbf{m}\mathbf{a}\mathbf{x}}=\frac{M}{W}=\frac{50\times {10}^6}{3.35\times {10}^6}\mathbf{\text{=14.92}}\frac{\mathbf{N}}{\mathbf{m}{\mathbf{m}}^{\mathbf{2}}}\mathbf{ }\mathbf{\left(}\mathbf{compression}\mathbf{\right)} \\ {\sigma }^{\max }\text{=14.92}\frac{\mathrm{N}}{{\mathrm{mm}}^2} <f=60\frac{\mathrm{N}}{{\mathrm{mm}}^2} \end{gathered}$$

The cross section (and the whole girder) is safe for bending.