A beam possessing the thick section shown in Fig. P.9.18 is subjected to a bending moment of 12 kNm applied in a plane inclined at 30° to the left of vertical and in a sense such that its components Mz and My are negative and positive, respectively. Calculate the magnitude and position of the

Question

A beam possessing the thick section shown in Fig. P.9.18 is subjected to a bending moment of 12 kNm applied in a plane inclined at 30° to the left of vertical and in a sense such that its components [latex] M_{z} ext { and } M_{y} [/latex] are negative and positive, respectively. Calculate the magnitude and position of the maximum direct stress in the beam cross section.

Accepted Answer

Referring to Fig. S.9.18 the components of the bending moment are

[latex] M_{z}=-12 \cos 30^{\circ}=-10.4 kNm , \quad M_{y}=-12 \sin 30^{\circ}=-6 kNm [/latex]

The beam section is unsymmetrical so that the position of the centroid of area must be found before the second moments of area can be calculated. Suppose that the centroid of area is a distance [latex] \bar{y} [/latex] from the top of the top flange and a distance [latex] \bar{z} [/latex] from the left hand face of the vertical leg. The area of cross section is

[latex]A=140 imes 10+200 imes 10+40 imes 10=3800 mm ^{2}[/latex]

Taking moments of areas about the top of the top flange

[latex] 3800 \bar{y}=140 imes 10 imes 5+200 imes 10 imes 100+40 imes 10 imes 195 [/latex]

i.e.

[latex] \bar{y}=75 mm [/latex]

Now taking moments about the left hand face of the vertical web

[latex]3800 \bar{z}=140 imes 10 imes 80+200 imes 10 imes 5+40 imes 10 imes 30[/latex]

from which

[latex] \bar{z}=35.3 mm [/latex]

The second moments of area are then

[latex]\begin{aligned}I_{z}=& \frac{140 imes 10^{3}}{12}+140 imes 10 imes 70^{2}+\frac{10 imes 200^{3}}{12}+10 imes 200 imes 25^{2} \&+\frac{40 imes 10^{3}}{12}+40 imes 10 imes 120^{2}\end{aligned}[/latex]

i.e.

[latex]\begin{gathered}I_{z}=20.6 imes 10^{6} mm ^{4} \I_{y}=\frac{10 imes 140^{3}}{12}+140 imes 10 imes 44.7^{2}+\frac{200 imes 10^{3}}{12}+200 imes 10 imes 30.3^{2} \+\frac{10 imes 40^{3}}{12}+10 imes 40 imes 5.3^{2}\end{gathered}[/latex]

i.e.

[latex]\begin{gathered}I_{y}=7.0 imes 10^{6} mm ^{4} \I_{z y}=140 imes 10(-44.7)(70)+200 imes 10(30.3)(-25)+40 imes 10(5.3)(-120)\end{gathered}[/latex]

i.e.

[latex]I_{z y}=-6.2 imes 10^{6} mm ^{4}[/latex]

The denominator in Eq. (9.31) is

[latex]\sigma_{x}=-\left(\frac{M_{y} I_{z}-M_{z} I_{z y}}{I_{z} I_{y}-I_{z y}^{2}} ight) z-\left(\frac{M_{z} I_{y}-M_{y} I_{z y}}{I_{z} I_{y}-I_{z y}^{2}} ight) y[/latex] (9.31)

[latex]I_{z} I_{y}-I_{z y}^{2}=\left(20.6 imes 7.0-6.2^{2} ight) imes 10^{12}=105.8 imes 10^{12}[/latex]

Now substituting in Eq. (9.31) for [latex]M_{z}, M_{y}[/latex], etc.

[latex]\sigma_{x}=-\left(\frac{-6 imes 20.6-10.4 imes 6.2}{105.8} ight) z-\left(\frac{-10.4 imes 7.0-6 imes 6.2}{105.8} ight) y[/latex]

i.e.

[latex]\sigma_{x}=1.78 z+1.04 y[/latex] (i)

It is not clear from Eq. (i) at which point the maximum will occur so that the stress at different points in the cross section must be calculated.

At J, [latex] z=-114.7 mm , y=65 mm ext { and } \sigma_{x}=-136.6 N / mm ^{2} [/latex] (this will be greater than the stress at A).

At B, [latex] z=35.3 mm , y=75 mm ext { and } \sigma_{x}=140.8 N / mm ^{2} [/latex] (this will be greater than at H).
At D, [latex] z=-14.7 mm , y=-125 mm ext { and } \sigma_{x}=-156.2 N / mm ^{2}[/latex] (this will be greater than at C, F or E).
Therefore the maximum direct stress is -156.2 N/mm² (compression) and it occurs at D.

Question 9.18: A beam possessing the thick section shown in Fig. P.9.18 is ...

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