A tall building has the cross section shown in Fig. P.9.23 and is subjected to horizontal wind loads which are resisted by an unsymmetrical arrangement of concrete shear walls. If the height of the building is 60 m determine the maximum direct stress in the cross section produced by a uniform

Question

A tall building has the cross section shown in Fig. P.9.23 and is subjected to horizontal wind loads which are resisted by an unsymmetrical arrangement of concrete shear walls. If the height of the building is 60 m determine the maximum direct stress in the cross section produced by a uniform wind pressure of 750 N/m² acting as shown. Specify the point at which the maximum direct stress acts and assume, for the purposes of calculation, that the shear walls, all of thickness 200 mm, are thin.

Accepted Answer

Referring to Fig. S.9.23 and taking moments of areas about AH

[latex]\begin{aligned}200(2+4+8+4+3) imes 10^{3} \bar{y}=&(2 imes 200 imes 1+4 imes 200 imes 6+8 imes 200 imes 4+4 imes 200 imes 8\&+3 imes 200 imes 1.5) imes 10^{6}\end{aligned}[/latex]

which gives

[latex]\bar{y}=4.5 imes 10^{3} mm[/latex]

Now taking moments of areas about EF

[latex]\begin{aligned}200(2+4+8+4+3) imes 10^{3} \bar{z}=&(-2 imes 200 imes 8-4 imes 200 imes 6+3 imes 200\& imes 8+4 imes 200 imes 6) imes 10^{6}\end{aligned}[/latex]

from which

[latex] \bar{z}=0.4 imes 10^{3} mm[/latex]

Then

[latex]\begin{aligned}I_{z}=& \frac{200 imes\left(2 imes 10^{3} ight)^{3}}{12}+200 imes 2 imes 10^{3} imes\left(3.5 imes 10^{3} ight)^{2}+200 imes 4 imes 10^{3} imes\left(-1.5 imes 10^{3} ight)^{2} \&+\frac{200 imes\left(8 imes 10^{3} ight)^{3}}{12}+200 imes 8 imes 10^{3} imes\left(0.5 imes 10^{3} ight)^{2}+\frac{200 imes\left(3 imes 10^{3} ight)^{3}}{12} \&+200 imes 3 imes 10^{3} imes\left(3 imes 10^{3} ight)^{2}+200 imes 4 imes 10^{3} imes\left(-3.5 imes 10^{3} ight)^{2}\end{aligned}[/latex]

which gives

[latex]\begin{aligned}I_{z} &=3142 imes 10^{10} mm ^{4} \I_{y}=& 200 imes 2 imes 10^{3} imes\left(8.4 imes 10^{3} ight)^{2}+\frac{200 imes\left(4 imes 10^{3} ight)^{3}}{12}+200 imes 4 imes 10^{3} imes\left(6.4 imes 10^{3} ight)^{2} \&+200 imes\left(8 imes 10^{3} ight) imes\left(0.4 imes 10^{3} ight)^{2}+200 imes\left(3 imes 10^{3} ight) imes\left(-7.6 imes 10^{3} ight)^{2} \&+\frac{200 imes\left(4 imes 10^{3} ight)^{3}}{12}+200 imes\left(4 imes 10^{3} ight) imes\left(-5.6 imes 10^{3} ight)^{2}\end{aligned}[/latex]

which gives

[latex]\begin{gathered}I_{y}=12312.5 imes 10^{10} mm ^{4} \I_{z y}=200[2 imes(3.5)(8.4)+4(-1.5)(6.4)+8(0.4)(0.5)+4(-5.6)(-3.5)+3(3)(-7.6)] imes 10^{9}\end{gathered}[/latex]

from which

[latex]I_{z y}=640 imes 10^{10} mm ^{4}[/latex]

Also [latex] M_{z}=-750 imes 16 imes 60 imes 30 imes 10^{3}=-2.16 imes 10^{10} Nmm ext { and } M_{y}=0 [/latex]. Substituting these values in Eq. (9.31) and simplifying

[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_{z}=-0.36 imes 10^{-4} z+6.95 imes 10^{-4} y[/latex] (i)

The maximum value will occur at H where [latex] z=-7.6 imes 10^{3} mm ext { and } y =+4.5 imes 10^{3} mm [/latex] Then, from Eq.(i)

[latex] \sigma_{z}=3.4 N / mm ^{2} ext { (tension) } [/latex]

Question 9.23: A tall building has the cross section shown in Fig. P.9.23 a...

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