Mechanics of Materials, 7th Edition

Published by McGraw-Hill Education
ISBN 10: 0073398233
ISBN 13: 978-0-07339-823-5

Chapter 1 - Problems: 1.29

Answer

$$\tau_{JOINT} = 488.9 \space kPa$$ $$\sigma_{JOINT} = 488.9 \space kPa$$

Work Step by Step

Solution: This problem involves a slanted ($45^{\circ}$) scarf joint and will require decomposing both the internal forces and the cross sectional area into components. 1) Area: Area of the scarf joint will be larger than the cross sectional area as computed perpendicular to the main axis of the beam: $A_{45JOINT} = \frac{A_{CROSS}}{cos(45^\circ)} = \frac{150 * 75}{cos(45^\circ)} = 15909.9 \space mm^2 = 1.59099 *10^{-2} \space m^2$ 2)Internal Forces The only internal force in the element is the axial force $P = 11 \space kN$ along the main axis of the element. This force can be decomposed into components parallel and perpendicular to the scarf joint. $V_{JOINT} = P_{||} = 11 * cos(45^\circ) = 7.7782 \space kN$ (parallel to Joint) $N_{JOINT} = P_{\perp} = 11 * sin(45^\circ) = 7.7782 \space kN$ (perpendicular to Joint) 3)Shear and Axial Stresses on Joint: $\bullet Shear \space stress:$ $\tau_{JOINT} = \frac{V_{JOINT}}{A_{JOINT}} = \frac{7.7782 * 10^3}{1.59099*10^{-2}} = 488900\space Pa = \boxed{488.9 \space kPa} \space \leftarrow\space\space ANS1$ $\bullet Axial\space stress:$ $\sigma_{JOINT} = \frac{N_{JOINT}}{A_{JOINT}} = \frac{7.7782 * 10^3}{1.59099*10^{-2}} = 488900\space Pa = \boxed{488.9 \space kPa} \space \leftarrow\space\space ANS2$ *NOTE: In this case the axial and shear stresses came out to be equal. This is a special case, due to the fact that $sin(45^\circ) = cos(45^\circ)$ , thus the internal axial force P was decomposed into components of equal magnitude.
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