## Subscribe \$4.99/month

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

## 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. 23.4

A two-cell beam has singly symmetrical cross-sections 1.2 m apart and tapers symmetrically in the y direction about a longitudinal axis (Fig. 23.12). The beam supports loads which produce a shear force $S_y$ = 10 kN and a bending moment $M_x$ = 1.65 kN m at the larger cross-section; the shear load is applied in the plane of the internal spar web. If booms 1 and 6 lie in a plane which is parallel to the $y_z$ plane calculate the forces in the booms and the shear flow distribution in the walls at the larger cross-section. The booms are assumed to resist all the direct stresses while the walls are effective only in shear. The shear modulus is constant throughout, the vertical webs are all 1.0 mm thick while the remaining walls are all 0.8 mm thick:

Boom areas : $B_1=B_3=B_4=B_6=600 \mathrm{~mm}^2 \quad B_2=B_5=900 \mathrm{~mm}^2$

## Verified Solution

At the larger cross-section

$I_{x x}=4 \times 600 \times 90^2+2 \times 900 \times 90^2=34.02 \times 10^6 \mathrm{~mm}^4$

The direct stress in a boom is given by Eq. (16.18) in which $I_{x y}=0 \text { and } M_y=0$, i.e.

$\sigma_z=\left(\frac{M_y I_{x x}-M_x I_{x y}}{I_{x x} I_{y y}-I_{x y}^2}\right) x+\left(\frac{M_x I_{y y}-M_y I_{x y}}{I_{x x} I_{y y}-I_{x y}^2}\right) y$  (16.18)

$\sigma_{z, r}=\frac{M_x y_r}{I_{x x}}$

whence

$P_{z, r}=\frac{M_x y_r}{I_{x x}} B_r$

or

$P_{z, r}=\frac{1.65 \times 10^6 y_r B_r}{34.02 \times 10^6}=0.08 y_r B_r$  (i)

The value of $P_{z,r}$ is calculated from Eq. (i) in column ② of Table 23.2;$P_{x,r}$ and $P_{y,r}$ follow from Eqs (21.10) and (21.9), respectively in columns ⑤ and ⑥. The axial load $P_r$ is given by [②² + ⑤² + ⑥²]$^{1/2}$ in column ⑦ and has the same sign as $P_{z,r}$(see Eq. (21.12)). The moments of $P_{x,r}$ and $P_{y,r}$, columns ⑩ and ⑪, are calculated for a moment centre at the mid-point of the internal web taking anticlockwise moments as positive.

$P_{y, r}=P_{z, r} \frac{\delta y_r}{\delta z}$  (21.9)

$P_{x, r}=P_{z, r} \frac{\delta x_r}{\delta z}$  (21.10)

$P_r=P_{z, r} \frac{\left(\delta x_r^2+\delta y_r^2+\delta z^2\right)^{1 / 2}}{\delta z}$  (21.12)

From column ⑤

$\sum_{r=1}^6 P_{x, r}=0$

(as would be expected from symmetry).
From column ⑥

$\sum_{r=1}^6 P_{y, r}=764.4 \mathrm{~N}$

From column ⑩

$\sum_{r=1}^6 P_{x, r} \eta_r=-117846 \mathrm{~N} \mathrm{~mm}$

Table 23.2

 ① ② ③ ④ ⑤ ⑥ ⑦ ⑧ ⑨ ⑩ ⑪ $P_{z,r}$ $\frac{\delta_{x_r}}{\delta_z}$ $\frac{\delta_{y_r}}{\delta_z}$ $P_{x,r}$ $P_{y,r}$ $P_{r}$ $\xi_r$ $\eta_r$ $P_{x,r}\eta_r$ $P_{y,r}\xi_r$ Boom (N) (N) (N) (N) (mm) (mm) (N mm) (N mm) 1 2619.0 0 0.0417 0 109.2 2621.3 400 90 0 43680 2 3928.6 0.0833 0.0417 327.3 163.8 3945.6 0 90 −29457 0 3 2619.0 0.1250 0.0417 327.4 109.2 2641.6 200 90 −29466 21840 4 −2619.0 0.1250 −0.0417 −327.4 109.2 −2641.6 200 90 −29466 21840 5 −3928.6 0.0833 −0.0417 −327.3 163.8 −3945.6 0 90 −29457 0 6 −2619.0 0 −0.0417 0 109.2 −2621.3 400 90 0 −43680

From column ⑪

$\sum_{r=1}^6 P_{y, r} \xi_r=-43680 \mathrm{~N} \mathrm{~mm}$

From Eq. (21.15)

$S_{x, w}=S_x-\sum_{r=1}^m P_{z, r} \frac{\delta x_r}{\delta z} \quad S_{y, w}=S_y-\sum_{r=1}^m P_{z, r} \frac{\delta y_r}{\delta z}$  (21.15)

$S_{x, w}=0 \quad S_{y, w}=10 \times 10^3-764.4=9235.6 \mathrm{~N}$

Also, since $Cx$ is an axis of symmetry, $I_{xy}$ = 0 and Eq. (20.6) for the ‘open section’ shear flow reduces to

\begin{aligned}q_s=&-\left(\frac{S_x I_{x x}-S_y I_{x y}}{I_{x x} I_{y y}-I_{x y}^2}\right)\left(\int_0^s t_{\mathrm{D}} x \mathrm{~d} s+\sum_{r=1}^n B_r x_r\right) \\&-\left(\frac{S_y I_{y y}-S_x I_{x y}}{I_{x x} I_{y y}-I_{x y}^2}\right)\left(\int_0^s t_{\mathrm{D}} y \mathrm{~d} s+\sum_{r=1}^n B_r y_r\right)\end{aligned}  (20.6)

$q_{\mathrm{b}}=-\frac{S_{y, w}}{I_{x x}} \sum_{r=1}^n B_r y_r$

or

$q_{\mathrm{b}}=-\frac{9235.6}{34.02 \times 10^6} \sum_{r=1}^n B_r y_r=-2.715 \times 10^{-4} \sum_{r=1}^n B_r y_r$  (ii)

‘Cutting’ the top walls of each cell and using Eq. (ii), we obtain the $q_b$ distribution shown in Fig. 23.13. Evaluating δ for each wall and substituting in Eq. (23.10) gives for cell I

$\frac{\mathrm{d} \theta}{\mathrm{d} z}=\frac{1}{2 A_R G}\left(-q_{s, 0, R-1} \delta_{R-1, R}+q_{s, 0, R} \delta_R-q_{s, 0, R+1} \delta_{R+1, R}+\oint_R q_{\mathrm{b}} \frac{\mathrm{d} s}{t}\right)$  (23.10)

$\frac{\mathrm{d} \theta}{\mathrm{d} z}=\frac{1}{2 \times 36000 G}\left(760 q_{s, 0, \mathrm{I}}-180 q_{s, 0, \mathrm{II}}-1314\right)$  (iii)

for cell II

$\frac{\mathrm{d} \theta}{\mathrm{d} z}=\frac{1}{2 \times 72000 G}\left(-180 q_{s, 0, \mathrm{I}}+1160 q_{s, 0, \mathrm{II}}+1314\right)$  (iv)

Taking moments about the mid-point of web 25 we have, using Eq. (23.13)

$S_x \eta_0-S_y \xi_0=\sum_{R=1}^N \oint_R q_{\mathrm{b}} p_0 \mathrm{~d} s+\sum_{R=1}^N 2 A_R q_{s, 0, R}-\sum_{r=1}^m P_{x, r} \eta_r+\sum_{r=1}^m P_{y, r} \xi_r$  (23.13)

\begin{aligned}0=&-14.7 \times 180 \times 400+14.7 \times 180 \times 200+2 \times 36000 q_{s, 0, \mathrm{I}}+2 \times 72000 q_{s, 0, \mathrm{II}} \\&-117846-43680\end{aligned}

or
$0=-690726+72000 q_{s, 0, \mathrm{I}}+144000 q_{s, 0, \mathrm{II}}$ (v)

Solving Eqs (iii)–(v) gives
$q_{s, 0, \mathrm{I}}=4.6 \mathrm{~N} / \mathrm{mm} \quad q_{s, 0, \mathrm{II}}=2.5 \mathrm{~N} / \mathrm{mm}$
and the resulting shear flow distribution is shown in Fig. 23.14.