A carpenter glues a pair of cylindrical wooden logs by bonding their end faces at an angle of of 30° as shown in the figure.

The glue used at the interface fails if

     Criterion 1: the maximum normal stress exceeds 2.5 MPa.

     Criterion 2: the maximum shear stress exceeds 1.5 MPa.

Assume that the interface fails before the logs fail. When a uniform tensile stress of 4 MPa is applied, the interface

The correct option is: fails because of both criteria 1 and 2

1. Analysis of the Geometry from the Figure:
From the provided image, we can observe two wooden logs, labeled Log 1 and Log 2, joined at an inclined interface.
A uniform tensile stress of ${\sigma }_0=4\text{MPa}$ is applied along the longitudinal axis (the horizontal dashed line).
The vertical dotted line represents a plane perpendicular to the longitudinal axis (the normal cross-section).
The interface is inclined at an angle $\theta ={30}^{\circ }$ relative to this vertical line. This means that the normal to the inclined interface plane makes an angle of $\theta ={30}^{\circ }$ with the longitudinal axis.

2. Stress Transformation Equations:
To determine the stresses acting directly on the interface, we resolve the axial tensile stress into normal and shear components acting on the inclined plane:
The normal stress (${\sigma }_n$) on the inclined plane is given by:
${\sigma }_n={\sigma }_0{\cos }^2\left(\theta \right)$
The shear stress ($\tau$) along the inclined plane is given by:
$\tau ={\sigma }_0\sin \left(\theta \right)\cos \left(\theta \right)$

3. Calculation of Normal and Shear Stresses:
Substituting the given values, ${\sigma }_0=4\text{MPa}$ and $\theta ={30}^{\circ }$:
For the normal stress:
${\sigma }_n=4\times {\cos }^2\left({30}^{\circ }\right)=4\times {\left(\frac{\sqrt{3}}{2}\right)}^2=4\times \frac{3}{4}=3\text{MPa}$
For the shear stress:
$\tau =4\times \sin \left({30}^{\circ }\right)\cos \left({30}^{\circ }\right)=4\times \frac{1}{2}\times \frac{\sqrt{3}}{2}=\sqrt{3}\approx 1.732\text{MPa}$

4. Evaluating Failure Criteria:
We compare the calculated stresses on the interface with the given limits of the glue:
• Criterion 1: The interface fails if the maximum normal stress exceeds $2.5\text{MPa}$.
Since the calculated normal stress ${\sigma }_n=3\text{MPa}>2.5\text{MPa}$, Criterion 1 is violated.
• Criterion 2: The interface fails if the maximum shear stress exceeds $1.5\text{MPa}$.
Since the calculated shear stress $\tau \approx 1.732\text{MPa}>1.5\text{MPa}$, Criterion 2 is also violated.

Consequently, the interface fails due to both criteria 1 and 2.