Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
04-BS-6: Mechanics of Materials — National Exams, December 2017. Closed-book, 3 hours;
any FIVE of the eight questions constitute a complete paper (all eight are answered below as a
complete study resource). Aid sheet permitted; geometric properties of
W-shape sections were supplied as an appendix (not required for any of the eight questions solved
here).
Find. (a) $\sigma_{n},\tau$ on the 15° plane; (b) the maximum
in-plane shear stress, its associated normal stress, and the orientation of those planes.
Mohr's circle for the given plane-stress state (points X and Y are diametrically opposite).
Approach. Plot the two stress points $X(\sigma_x,\tau_{xy})$ and
$Y(\sigma_y,-\tau_{xy})$ on the $\sigma$–$\tau$ plane;
the line XY is a diameter of Mohr's circle. Read the centre and radius directly off the circle's
geometry, then rotate by $2\theta$ (twice the physical angle) to reach the plane of
interest. (The transformation equations below are the trigonometric bookkeeping of exactly this
circle — used here only to confirm the numbers read from the circle, per the question's own
allowance.)
Centre and radius of the circle.
$$C=\frac{\sigma_x+\sigma_y}{2}=\frac{80+(-100)}{2}=-10\text{ MPa}$$
$$R=\sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}=\sqrt{90^2+120^2}=150\text{ MPa}$$
Principal stresses (where the circle crosses the $\sigma$ axis).
$$\boxed{\sigma_1=C+R=140\text{ MPa},\qquad \sigma_2=C-R=-160\text{ MPa}}$$
occurring at $\theta_p=\tfrac12\tan^{-1}\!\left(\dfrac{2\tau_{xy}}{\sigma_x-\sigma_y}\right)=26.57^\circ$
(measured CCW from the x-face on the physical element, i.e. half the angle swept on the circle).
Stresses on the 15° inclined plane (part a). Rotating the radius to X by
$2(15^\circ)=30^\circ$ around the circle (same sense as the physical rotation) and
reading off the new coordinates:
$$\boxed{\sigma_{n}=127.9\text{ MPa (tension)}, \qquad \tau = 58.9\text{ MPa}}$$
with the complementary face carrying $\sigma_{n}'=2C-\sigma_n=-147.9$ MPa, satisfying
$\sigma_n+\sigma_n'=2C$ as it must for any pair of perpendicular planes.
Maximum in-plane shear stress (part b). This is simply the radius of the
circle, occurring 90° around the circle from the principal points, i.e. at
$\theta_s=\theta_p+45^\circ=71.57^\circ$ physically:
$$\boxed{\tau_{max}=R=150\text{ MPa}, \qquad \sigma_{avg}=C=-10\text{ MPa (on both faces of that element)}}$$
Check against the transformation equations.
$$\sigma_{x'}=\frac{\sigma_x+\sigma_y}{2}+\frac{\sigma_x-\sigma_y}{2}\cos2\theta+\tau_{xy}\sin2\theta,\qquad
\tau_{x'y'}=-\frac{\sigma_x-\sigma_y}{2}\sin2\theta+\tau_{xy}\cos2\theta$$
Substituting $\theta=15^\circ$ reproduces $\sigma_n=127.9$ MPa,
$\tau=58.9$ MPa exactly, confirming the circle reading.
