Subscribe $4.99/month

Un-lock Verified Step-by-Step Experts Answers.

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

Our Website is free to use.
To help us grow, you can support our team with a Small Tip.

All the data tables that you may search for.

Need Help? We got you covered.

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Products

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Need Help? We got you covered.

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.

13.18
13.19