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.

Analysis of the Problem:
From the provided image, we have two cylindrical logs (Log 1 and Log 2) bonded at an inclined interface. The horizontal axis of loading has a uniform tensile stress applied to it:
${\sigma }_x=4\text{MPa}$
The vertical dotted line represents a plane perpendicular to the longitudinal axis of the logs. The inclined interface makes an angle of:
$\theta =30°$
with this vertical plane.

Therefore, the normal to the inclined interface plane makes the same angle $\theta =30°$ with the axis of loading (the horizontal axis along which the tensile stress is applied).

1. Calculation of Normal Stress (${\sigma }_n$) on the Interface:
The normal stress acting on an inclined plane at an angle $\theta$ to the cross-sectional plane is given by the formula:
${\sigma }_n={\sigma }_x{\cos }^2\left(\theta \right)$
Substituting the given values into the formula:
${\sigma }_n=4\times {\cos }^2(30°)$
${\sigma }_n=4\times {\left(\frac{\sqrt{3}}{2}\right)}^2$
${\sigma }_n=4\times \frac{3}{4}=3\text{MPa}$

2. Calculation of Shear Stress ($\tau$) on the Interface:
The shear stress acting on the inclined plane is given by the formula:
$\tau ={\sigma }_x\sin \left(\theta \right)\cos \left(\theta \right)$
Substituting the given values:
$\tau =4\times \sin (30°)\cos (30°)$
$\tau =4\times \frac{1}{2}\times \frac{\sqrt{3}}{2}$
$\tau =\sqrt{3}\approx 1.732\text{MPa}$

3. Evaluation of Failure Criteria:
We compare the calculated stresses with the given threshold limits at the interface:

• Criterion 1: The interface fails if the normal stress exceeds 2.5 MPa. Since the calculated normal stress is ${\sigma }_n=3\text{MPa}>2.5\text{MPa}$, the interface fails under Criterion 1.
• Criterion 2: The interface fails if the shear stress exceeds 1.5 MPa. Since the calculated shear stress is $\tau =1.732\text{MPa}>1.5\text{MPa}$, the interface also fails under Criterion 2.

Since both failure limits are exceeded, the glued interface fails due to both criteria 1 and 2.