## Chapter 13

## Q. 13.8

A bending moment M_{ o } is applied as shown in a rolled-steel beam, the section details of which are given in Figure 13.24. The web of the section forms an angle θ with the vertical. If \sigma_{ o } be the maximum bending stress developed in the beam, then calculate the angle θ for which the maximum flexural stress developed in the beam is 2 \sigma_{ o } .

## Step-by-Step

## Verified Solution

Clearly from the given figure, we get

\bar{I}_{\eta \eta}=2.82 \text { (units) }^4 \text { and } \bar{I}_{\xi \xi}=103 \text { (units) }{ }^4

In the given case,

M_\eta=M_{ o } \sin \theta \quad \text { and } \quad M_{\xi}=M_{ o } \cos \theta

Flexural 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}} (13.9)

=-M_{ o }\left[\frac{\eta \cos \theta}{103}-\frac{\xi \sin \theta}{2.82}\right]

If \phi be the angle made by NA with ξ-axis, then

\tan \phi=\frac{\eta}{\xi} \text { for } \sigma_{x x}=0

So \tan \phi=\tan \theta(103 / 2.82) \text {. Hence, } \phi>\theta . The orientation of NA is shown in Figure 13.24 itself.

From section geometry, we conclude points A and A′ have the highest stress. But \eta_{ A }=6 units and \xi_{ A }=-2 units. Therefore,

\sigma_{ A }=-M_{ o }\left[\frac{6 \cos \theta}{103}+\frac{2 \sin \theta}{2.82}\right]=2 \sigma_{ o } \quad \text { (as given) } (1)

Clearly when θ = 0,

\sigma_{ A }=-\frac{6 M_{ o }}{103}=\sigma_{ o } \quad \text { (as given) } (2)

Therefore from Eqs. (1) and (2), by eliminating \sigma_o , we get

\frac{12 M_{ o }}{103}=M_{ o }\left[\frac{6 \cos \theta}{103}+\frac{2 \sin \theta}{2.82}\right]

or 6cosθ + 73sinθ = 12

\Rightarrow \quad \cos (\theta-\alpha)=\frac{12}{\sqrt{6^2+73^2}}=\frac{12}{73.25}=\cos \left(-80.57^{\circ}\right)

where tanα = 73/6 ⇒ α = 85.3°. Therefore,

θ − 85.3° = −80.57°⇒ θ = 85.3° − 80.57° = 4.73°

The required angle of inclination is θ = 4.73°.