## Textbooks & Solution Manuals

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

## Tip our Team

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

## Holooly Tables

All the data tables that you may search for.

## Holooly Help Desk

Need Help? We got you covered.

## Holooly Arabia

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

Products

## Textbooks & Solution Manuals

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

## Holooly Arabia

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

## Holooly Help Desk

Need Help? We got you covered.

## 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 $2 \sigma_{ o }$.

## 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°.