Question

Shown in Fig. P. 14.40 is the cross section of a thin-walled member which is part of an airplane wing. If the allowable shear stress is $5.5 \times 10^7 \mathbf{P a}$, find the allowable shear flow for this section and the stresses away from corners. Sides $A B$ and $C D$ have a thickness of 1 mm and sides $A C$ and $B D$ have a thickness of 2 mm . The mean height of the section is 265 mm and $G$ for the material is $2.8 \times 10^{10} \mathrm{~Pa}$. Assume no buckling. Find the allowable torque transmitted and the rate of twist. Figure P.14.40.

   Shown in Fig. P. 14.40 is the cross section of a thin-walled member which is part of an airplane wing. If the allowable shear stress is $5.5 \times 10^7 \mathbf{P a}$, find the allowable shear flow for this section and the stresses away from corners. Sides $A B$ and $C D$ have a thickness of 1 mm and sides $A C$ and $B D$ have a thickness of 2 mm . The mean height of the section is 265 mm and $G$ for the material is $2.8 \times 10^{10} \mathrm{~Pa}$. Assume no buckling. Find the allowable torque transmitted and the rate of twist.
Figure P.14.40.
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Introduction to Solid Mechanics
Introduction to Solid Mechanics
Irving H. Shames,… 3rd Edition
Chapter 13, Problem 40 ↓

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For sides \( AB \) and \( CD \) (thickness = 1 mm): \[ q_{AB} = \tau \cdot t_{AB} = (5.5 \times 10^7 \, \text{Pa}) \cdot (1 \times 10^{-3} \, \text{m}) = 5.5 \times 10^4 \, \text{N/m} \] For sides \( AC \) and \( BD \) (thickness = 2 mm): \[ q_{AC} =  Show more…

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Shown in Fig. P. 14.40 is the cross section of a thin-walled member which is part of an airplane wing. If the allowable shear stress is $5.5 \times 10^7 \mathbf{P a}$, find the allowable shear flow for this section and the stresses away from corners. Sides $A B$ and $C D$ have a thickness of 1 mm and sides $A C$ and $B D$ have a thickness of 2 mm . The mean height of the section is 265 mm and $G$ for the material is $2.8 \times 10^{10} \mathrm{~Pa}$. Assume no buckling. Find the allowable torque transmitted and the rate of twist. Figure P.14.40.
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