The cross section of a beam/floor slab arrangement is shown in Fig. P.9.19. The complete section is simply supported over a span of 10 m and, in addition to its self-weight, carries a concentrated load of 25 kN acting vertically downwards at mid-span. If the density of concrete is 2000 kg/m^3,

Question

The cross section of a beam/floor slab arrangement is shown in Fig. P.9.19. The complete section is simply supported over a span of 10 m and, in addition to its self-weight, carries a concentrated load of 25 kN acting vertically downwards at mid-span. If the density of concrete is 2000 kg/m³, calculate the maximum direct stress at the point A in its cross section.

Accepted Answer

The area of cross section is

[latex]A=1 imes 0.15+0.6 imes 0.3=0.33 m ^{2}[/latex]

The self-weight of the beam is then

[latex]SW =0.33 imes 2000 imes 9.81 imes 10^{-3}=6.5 kN / m[/latex]

The maximum bending moment occurs at mid-span and is given by

[latex] M_{\max }=\frac{6.5 imes 10^{2}}{8}+\frac{25 imes 10}{4}=143.75 kNm =M_{z} [/latex]

Referring to Fig. S.9.19

[latex] 0.33 imes 10^{6} \bar{y}=(1 imes 0.15 imes 0.075+0.6 imes 0.3 imes 0.45) imes 10^{9} [/latex]

i.e.

[latex]\begin{gathered}\bar{y}=279.5 mm \0.33 imes 10^{6} \bar{z}=(1 imes 0.15 imes 0.5+0.6 imes 0.3 imes 0.15) imes 10^{9}\end{gathered}[/latex]

i.e.

[latex] \bar{z}=309.1 mm [/latex]

The second moments of area are then

[latex] I_{z}=\frac{1000 imes 150^{3}}{12}+1000 imes 150 imes 204.5^{2}+\frac{300 imes 600^{3}}{12}+300 imes 600 imes 170.5^{2} [/latex]

i.e.

[latex] \begin{gathered}I_{z}=1.72 imes 10^{10} mm ^{4} \I_{y}=\frac{150 imes 1000^{3}}{12}+150 imes 1000 imes 190.9^{2}+\frac{600 imes 300^{3}}{12}+600 imes 300 imes 159.1^{2}\end{gathered}[/latex]

i.e.

[latex]\begin{gathered}I_{y}=2.38 imes 10^{10} mm ^{4} \I_{z y}=150 imes 1000(-190.9)(204.5)+600 imes 300(159.1)(-170.5)\end{gathered}[/latex]

i.e.

[latex]I_{z y}=-1.07 imes 10^{10} mm ^{4}[/latex]

The denominator in Eq. (9.31) is then

[latex] I_{z} I_{y}-I_{z y}^{2}=1.72 imes 10^{10} imes 2.38 imes 10^{10}-\left(-1.07 imes 10^{10} ight)^{2}=2.95 imes 10^{20} [/latex]

The direct stress is then, from Eq. (9.31)

[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]\sigma_{x}=-\left(\frac{143.75 imes 1.07}{2.95} ight) imes 10^{-4} z-\left(\frac{143.75 imes 2.38}{2.95} ight) imes 10^{-4} y[/latex]

i.e.

[latex]\sigma_{x}=-0.0046 z-0.0116 y[/latex] (i)

The coordinates of A are z=9.1 mm, y=-470.5 mm, therefore

[latex] \sigma_{x}( A )=5.4 N / mm ^{2} ext { (tension). }[/latex]

Question 9.19: The cross section of a beam/floor slab arrangement is shown ...

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