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 23

Q. 23.5

Calculate the deflection at the free end of the two-cell beam shown in Fig. 23.15 allowing for both bending and shear effects. The booms carry all the direct stresses while the skin panels, of constant thickness throughout, are effective only in shear.

Take E = 69 000 N/mm² and G = 25 900 N/mm²
Boom areas: B_1=B_3=B_4=B_6=650 \mathrm{~mm}^2 \quad B_2=B_5=1300 \mathrm{~mm}^2

Screenshot 2022-10-10 084035

Step-by-Step

Verified Solution

The beam cross-section is symmetrical about a horizontal axis and carries a vertical load at its free end through the shear centre. The deflection Δ at the free end is then, from Eqs (20.17) and (20.19)

\Delta_M=\frac{1}{E} \int_L\left(\frac{M_{y, 1} M_{y, 0}}{I_{y y}}+\frac{M_{x, 1} M_{x, 0}}{I_{x x}}\right) \mathrm{d} z  (20.17)

\Delta_S=\int_L\left(\int_{\text {sect }} \frac{q_0 q_1}{G t} \mathrm{~d} s\right) \mathrm{d} z  (20.19)

\Delta=\int_0^{2000} \frac{M_{x, 0} M_{x, 1}}{E I_{x x}} \mathrm{~d} z+\int_0^{2000}\left(\int_{\text {section }} \frac{q_0 q_1}{G t} \mathrm{~d} s\right) \mathrm{d} z  (i)

where
M_{x, 0}=-44.5 \times 10^3(2000-z) \quad M_{x, 1}=-(2000-z)
and
I_{x x}=4 \times 650 \times 125^2+2 \times 1300 \times 125^2=81.3 \times 10^6 \mathrm{~mm}^4
also
S_{y, 0}=44.5 \times 10^3 \mathrm{~N} \quad S_{y, 1}=1

The q_0 \text { and } q_1 shear flow distributions are obtained as previously described (note dθ/dz = 0 for a shear load through the shear centre) and are

\begin{aligned}&q_{0,12}=9.6 \mathrm{~N} / \mathrm{mm} \quad q_{0,23}=-5.8 \mathrm{~N} / \mathrm{mm} \quad q_{0,43}=50.3 \mathrm{~N} / \mathrm{mm} \\&q_{0,45}=-5.8 \mathrm{~N} / \mathrm{mm} \quad q_{0,56}=9.6 \mathrm{~N} / \mathrm{mm} \quad q_{0,61}=54.1 \mathrm{~N} / \mathrm{mm} \\&q_{0,52}=73.6 \mathrm{~N} / \mathrm{mm} \text { at all sections of the beam }\end{aligned}

The q_1 shear flows in this case are given by q_0 / 44.5 \times 10^3. Thus

\begin{aligned}\int_{\text {section }} \frac{q_0 q_1}{G t} \mathrm{~d} s=& \frac{1}{25900 \times 2 \times 44.5 \times 10^3}\left(9.6^2 \times 250 \times 2+5.8^2 \times 500 \times 2\right.\\&\left.+50.3^2 \times 250+54.1^2 \times 250+73.6^2 \times 250\right) \\=& 1.22 \times 10^{-3}\end{aligned}

Hence, from Eq. (i)

\Delta=\int_0^{2000} \frac{44.5 \times 10^3(2000-z)^2}{69000 \times 81.3 \times 10^6} \mathrm{~d} z+\int_0^{2000} 1.22 \times 10^{-3} \mathrm{~d} z

giving
Δ = 23.5 mm