Question 3.8: Shear stresses in a beam The section shown in Figure 3.87 is...

In the top flange
Consider a position in the top flange, vertical distance y from the centroid, G, as shown in Figure 3.88(a). Using the discrete form of the shear formula,

\tau = \frac{S}{Iz}A\bar{\gamma } = \frac{S}{2.31×10^{6} \cdot  80}\cdot  (80  \cdot (34 – \gamma ))  \cdot \frac{(34  +  \gamma )}{2}

= \frac{S}{4.62×10^{6}}(1156  –  \gamma ^{2})

= 0.0108(1156  –  \gamma^{2})

At position A, γ = 34   ∴    \tau = 0
At position B, γ = 14     ∴    \tau = 10.4  N/mm² (MPa)

In the lower section
Consider a position in the lower section, again vertical distance y from the centroid, G, as shown in Figure 3.88(b).
At position B, there is a step change in the shear stress given by the ratio of the section widths at this point:

\tau = \frac{10.4  \cdot  80}{40} = 20.8 N/mm^{2} (MPa)

At position G, i.e. the neutral axis, we can use the discrete formula for the shear stress. In this case, to simplify calculation, the relevant area can be regarded as the area below the neutral axis:

\tau = \frac{S}{lz}A\bar{\gamma } = \frac{S}{2.31×10^{6}  \cdot  40} \cdot (46  \cdot  40) \cdot  23

= 22.91 N/mm² (MPa)

At position C, \tau = 0 i.e. a free surface.

A sketch of the variation of the shear stress down the vertical centre line is now given in Figure 3.89.

Shear centre
For the thin-walled semicircular cross-section shown in Figure 3.90, determine the position of the shear centre (assume bending about the axis of symmetry X-X)

Shear stress distribution
To solve this problem it is necessary to change from a rectangular coordinate system (x-γ) to a polar coordinate system (r-θ). Referring to Figure 3.90 and using the integral form of the shear stress formula, we obtain a general expression for the shear stress distribution parallel to the wall of the section. Thus,

\tau = \frac{S}{Iz} \int_{A}^{}{ \gamma dA}

with γ = R \cos\phi

dA = d_{s} ⋅ t = Rtd\phi

z = t

giving,

\tau = \frac{S}{Iz} \int_{0}^{\theta }{ R\cos \phi  \cdot  Rtd\phi }

= \frac{SR^{2}}{I} \int_{0}^{\theta }{ \cos \phi d\phi }

∴                                                                \tau = \frac{SR^{2}\sin \theta }{I}                             (3.35)

Now,

I = \int_{A}^{}{\gamma ^{2}dA} = \int_{0}^{\pi }{ (R \cos \theta )^{2}  \cdot  Rd\theta  \cdot  t}

= \int_{0}^{\pi }{ R^{3} t\cos ^{2}\theta  d\theta }

= \int_{0}^{\pi }{ \frac{R^{3}t}{2}(1 + \cos 2\theta ) d\theta }

\frac{R^{3}t}{2} \left[\theta + \frac{\sin 2\theta }{2} \right]_{0}^{\pi }

∴                                                             I = \frac{\pi R^{3}t}{2}                                (3.36)

From (3.35) and (3.36),

\tau = \frac{SR^{2} \sin \theta  \cdot  2}{\pi R^{3}t}

∴                                                            \tau = \frac{2S \sin \theta }{\pi Rt}

The twisting moment (torque) associated with the above shear stress distribution for the whole cross-section is found by taking moments about O:

Torque = \int_{0}^{\pi }{ \tau \cdot (Rd\theta )t \cdot R = \frac{R^{2}t \cdot 2S}{\pi Rt} } \int_{0}^{\pi }{ \sin \theta  d\theta }

= \frac{2 SR}{\pi } [- \cos \theta ]_{0}^{\pi } = \frac{4 SR}{\pi }

To counteract this twisting moment, as shown in Figure 3.91, the shear force, S, must be applied at the shear centre, a distance e, given by,

S(e + R) = \frac{4SR}{\pi }

∴                                                          e = \frac{4R}{\pi }  –  R = 0.273 R