Statics S4 slide 9: Maximum shear stress on an oblique section (ENGR5004)

Statics S4 slide 9: Maximum shear stress on an oblique sectionFrom lectureCurrent spec2:25 marks6 min

ENGR5004 Statics session 04, slide 9 (Stress on oblique cuts of rod under uniaxial load, exercise)

A rod of normal cross-sectional area $A_0$ carries a centric axial load $P$ . On an oblique section whose normal makes angle $\theta$ with the rod axis (area $A_\theta$ , with $A_0 = A_\theta\cos\theta$ ), the internal force resolves into a normal component $F = P\cos\theta$ and a shear component $V = P\sin\theta$ , giving

$\sigma = \frac{P}{A_0}\cos^2\theta \quad (2), \qquad \tau = \frac{P}{A_0}\sin\theta\cos\theta \quad (3)$

Formulas you may need

$\sigma = \dfrac{F}{A_\theta}$ , $\tau = \dfrac{V}{A_\theta}$ , $A_\theta = \dfrac{A_0}{\cos\theta}$ (learn this)
$\sin 2\theta = 2\sin\theta\cos\theta$ , $\cos 2\theta = \cos^2\theta - \sin^2\theta$ (learn this)

(a)
Show that $\tau = 0$ for $\theta = 0^\circ$ and $\theta = 90^\circ$ .
[1]
(b)
Demonstrate that $\tau$ reaches its maximum for $\theta = 45^\circ$ , and find $\tau_{max}$ .
[3]
(c)
Show that on the same $45^\circ$ section the normal stress also equals $P/(2A_0)$ .
[1]