Calculate the position of the shear center of the thin-walled section shown in Fig. 16.12; the thickness of the section is 2mm and is constant throughout.

Question

Accepted Answer

Since we are only asked to find the position of the shear center of the section and not a complete shear flow distribution the argument of Ex. 16.3 applies in that if we refer the position of the shear center to the web/flange junction 2 (or 3) it is only necessary to obtain the shear flow distribution in the flange 34 (or 12). First, we must calculate the section’s properties. Taking moments of area about the top flange

[latex]2(50+50+25) \bar{y}=2 imes 50 imes 25+2 imes 25 imes 50[/latex]

which gives

[latex]\bar{y}=20 mm[/latex]

Now taking moments of area about the vertical web

[latex]2(50+50+25) \bar{x}=2 imes 50 imes 25+2 imes 25 imes 12.5[/latex]

from which

[latex]\bar{x}=12.5 mm[/latex]

The second moments of area are then calculated using the methods of Section 15.4.5.

[latex]I_{x x}=2 imes 50 imes 20^{2}+\left(2 imes 50^{3} / 12 ight)+2 imes 50 imes 5^{2}+2 imes 25 imes 30^{2}[/latex]

i.e.,

[latex]I_{x x}=108,333 mm ^{4}[/latex]

[latex]I_{y y}=\left(2 imes 50^{3} / 12 ight)+2 imes 50 imes 12.5^{2}+2 imes 50 imes 12.5^{2}+\left(2 imes 25^{3} / 12 ight)[/latex]

i.e.,

[latex]I_{y y}=54,689 mm ^{4}[/latex]

[latex]I_{x y}=2 imes 50(+12.5)(+20)+2 imes 50(-12.5)(-5)[/latex]

i.e.,

[latex]I_{x y}=31,250 mm ^{4}[/latex]

Note that the flange 34 makes no contribution to [latex]I_{x y}[/latex] since its centroid coincides with the y axis of the section. Considering the horizontal position of the shear center we apply a vertical shear load, [latex]S_{y}[/latex], through the shear center and determine the shear flow distribution in the flange 34. Since [latex]S_{x}=0,[/latex] Eq. (16.14) reduces to

[latex]q_{s}=-\left(\frac{S_{x} I_{x x}-S_{y} I_{x y}}{I_{x x} I_{y y}-I_{x y}^{2}} ight) \int_{0}^{s} t x d s-\left(\frac{S_{y} I_{y y}-S_{x} I_{x y}}{I_{x x} I_{y y}-I_{x y}^{2}} ight) \int_{0}^{s} t y ds[/latex] (16.14)

[latex]q_{s}=\left[\left(S_{y} I_{x y} ight) /\left(I_{x x} I_{y y}-I_{x y}^{2} ight) ight] \int_{0}^{s} t x d s-\left[\left(S_{y} I_{y y} ight) /\left(I_{x x} I_{y y}-I_{x y}^{2} ight) ight] \int_{0}^{s} t y d s[/latex] (i)

Substituting the values of [latex]I_{x x}[/latex] etc. gives

[latex]q_{s}=1.26 imes 10^{-5} S_{y} \int_{0}^{s} x d s-2.21 imes 10^{-5} S_{y} \int_{0}^{s} y d s[/latex] (ii)

On the flange 34, [latex]x=12.5-s[/latex] and [latex]y=-30 mm ext {. }[/latex] Eq. (ii) then becomes

[latex]q_{43}=S_{y} imes 10^{-5} \int_{0}^{s}(82.05-1.26 s) d s[/latex]

Integrating gives

[latex]q_{43}=S_{y} imes 10^{-5}\left(82.05 s-0.63 s^{2} ight)[/latex] (iii)

Now taking moments about the flange/web junction 2

[latex]S_{y} \xi_{S}=\int_{0}^{25} q_{43} imes 50 d s[/latex]

Substituting for [latex]q_{43}[/latex] from Eq. (iii) and integrating gives

[latex]\xi_{ S }=11.2 mm[/latex]

We now apply [latex]S_{x}[/latex] only through the shear center and carry out the same procedure as above. Then

[latex]q_{43}=S_{x} imes 10^{-5}\left(2.19 s^{2}-92.55 s ight)[/latex] (iv)

Taking moments about the web/flange junction 2

[latex]S_{x} \eta_{ S }=-\int_{0}^{25} q_{43} imes 50 d s[/latex]

Note that in this case the moment of the resultant of the internal shear flows is in the opposite sense to that of the applied force. Substituting for [latex]q_{43}[/latex] from Eq. (iv) and carrying out the integration gives

[latex]\eta_{ S }=7.2 mm[/latex]

Question 16.5: Calculate the position of the shear center of the thin-walle...

Related Answered Questions