Chapter 13
Q. 13.5
Solve the above example using principal axes of inertia and Eq. (13.9) of flexural stress.
\sigma_{x x}=-\frac{M_{\xi} \eta}{\bar{I}_{\xi \xi}}+\frac{M_\eta \xi}{\bar{I}_{\eta \eta}} (13.9)
Step-by-Step
Verified Solution
We obtain first extreme values of area moment of inertia using Mohr’s circle as shown in Figure 13.18:
Clearly,
\theta_{ p }=22.15^{\circ} ⦪, \quad \bar{I}_{\max }=(24.6+20.62)\left(10^6\right) mm ^4=45.22 \times 10^6 mm ^4
and \bar{I}_{\min }=(24.6-20.62)\left(10^6\right) mm ^4=3.98 mm ^4
Thus, we redraw the axes and resolve our moments along η and ξ axes {}^2. Clearly,
M_\eta=1.83 \times 10^6 N mm \text { and } M_{\xi}=-27.94 \times 10^6 N mm
Thus, flexure stress at any point (η, ξ ) on the beam section is given by Eq. (13.9) as
\sigma_{x x}=-\frac{M_{\xi} \eta}{\bar{I}_{\xi \xi}}+\frac{M_\eta \xi}{\bar{I}_{\eta \eta}}
The angle made by the NA with ξ-axis is given by
\tan \alpha=\frac{\eta}{\xi}, \quad \text { from } \sigma_{x x}=0
or \tan \alpha=\frac{M_\eta}{M_{\xi}} \cdot \frac{\bar{I}_{\xi \xi}}{\bar{I}_{\eta \eta}}=\frac{1.83}{-27.94} \cdot \frac{45.22}{3.98}=-0.744
So, α = 143.34°. We mark the NA in Figure 13.19. Thus, the total angle made by NA with z-axis is 165.5° as before. Now invoking coordinate transformation equation as shown in the inset of Figure 13.8, we get (η, ξ ) coordinates of point A (or point B) as
\begin{aligned} &\eta_{ A }=(100) \cos 22.15^{\circ}-(10) \sin 22.15^{\circ}=88.85 mm \\ &\xi_{ A }=(100) \sin 22.15^{\circ}-(10) \cos 22.15^{\circ}=46.97 mm \end{aligned}
Therefore,
\left.\sigma_{x x}\right|_{ A }=-\frac{(-27.94)(88.85)}{45.22}+\frac{(1.83)(46.97)}{3.98} N / mm ^2
or \left.\sigma_{x x}\right|_{ A }=76.5 MPa \text { (as before) }
{}^2While rotating the axes, we first locate the h-axis by turning the y-axis by an angle 22.15° in the clock-wise direction and then set the ξ-axis in such a way that η-ξ-x axes (x-axis is perpendicular to the plane of the figure) form a right-hand triad. Observe Figure 13.19 very carefully.
