Show that the deflection due to shear in a rectangular section beam supporting a vertical shear load Sy is 20% greater for a shear stress distribution given by the expression τ = -SyA′Ȳ/boIz than for a distribution assumed to be uniform. A rectangular section cantilever beam 200 mm wide by 400 mm

Question

Show that the deflection due to shear in a rectangular section beam supporting a vertical shear load [latex]S_{y}[/latex] is 20% greater for a shear stress distribution given by the expression

[latex] au=-\frac{S_{y} A^{\prime} \bar{y}}{b_{o} I_{z}}[/latex]

than for a distribution assumed to be uniform.
A rectangular section cantilever beam 200 mm wide by 400 mm deep and 2 m long carries a vertically downward load of 500 kN at a distance of 1 m from its free end. Calculate the deflection at the free end taking into account both shear and bending effects. Take [latex]E=200000 N / mm ^{2}[/latex] and [latex]G=70000 N / mm ^{2}[/latex].
Ans. 2.06 mm (downwards).

Accepted Answer

From Eq. (13.22)

[latex]v_{s}=\frac{\beta}{G} \int_{L}\left(\frac{S_{y}}{A} ight) d x[/latex] (13.22)

it can be seen that for a given beam and loading the deflection due to shear depends upon the value of β. Also, from Eq. (13.20)

[latex]\beta=\frac{A}{S_{y}^{2}} \int_{y_{1}}^{y_{2}} au^{2} b_{0} d y[/latex] (13.20)

when a uniform shear stress is assumed, i.e.

[latex]\beta=\frac{A}{S_{y}^{2}} \int_{y, 1}^{y, 2}\left(\frac{S_{y}}{A} ight)^{2} b_{0} d y[/latex]

The integral in the above is equal to A so that

β = 1

For a rectangular section beam in which the shear stress distribution is given by Eq. (10.4),

[latex] au=-\frac{S_{y} A^{\prime} \bar{y}}{b_{0} I_{z}}[/latex] (10.4)

β = 6/5 = 1.2 (see Ex. 13.19) so that the deflection in this case is 20% greater than for a shear stress distribution which is assumed to be uniform.
The deflected shape of the beam due to shear only is shown in Fig. S.13.23(a). Then, from Eq. (13.22)

[latex]v_{s}=\frac{1.2}{70000} \int_{0}^{1000}\left(\frac{500 imes 10^{3}}{200 imes 400} ight) d x=0.107 mm[/latex]

From Eq. (v) of Ex. 13.1

[latex]v_{ ext {tip }}=-\frac{W L^{3}}{3 E I}[/latex] Eq. (v) from Ex. 13.1

the deflection due to bending at B (Fig. S.13.23(b)) is

[latex]v_{ b }=\frac{500 imes 10^{3} imes 1000^{3}}{3 imes 200000\left(\frac{200 imes 400^{3}}{12} ight)}=0.78 mm[/latex]

From Eq. (iii) of Ex. 13.1

[latex]E I \frac{ d v}{ d x}=-W\left(L x-\frac{x^{2}}{2} ight)[/latex] Eq.(iii) from Ex.13.1

the slope of the beam at B is

[latex]\left(\frac{ d v}{ d x} ight)_{ B }=\frac{500 imes 10^{3} imes 1000^{2}}{2 imes 200000\left(\frac{200 imes 400^{3}}{12} ight)}=0.00117[/latex]

Therefore the total deflection at C due to bending is 0.78 + 0.00117 × 1000 = 1.95 mm and the total deflection at C due to bending and shear is [latex]v_{ ext {total }}=0.107+1.95=2.057 mm[/latex].

Question 13.P.23: Show that the deflection due to shear in a rectangular secti...

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