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.
