Two wooden members of 80 × 120-mm uniform rectangular cross section are joined by the simple glued scarf splice shown, Knowing that β = 22° and that the maximum allowable stresses in the joint are, respectively 400 kPa in tension (perpendicular to the splice) and 600 kPa in shear (parallel to the splice), determine the largest centric load P that can be applied.

Question-AnswerCategory: Strength of MaterialsTwo wooden members of 80 × 120-mm uniform rectangular cross section are joined by the simple glued scarf splice shown, Knowing that β = 22° and that the maximum allowable stresses in the joint are, respectively 400 kPa in tension (perpendicular to the splice) and 600 kPa in shear (parallel to the splice), determine the largest centric load P that can be applied.
Beer Johnston asked 12 months ago

Two wooden members of 80 × 120-mm uniform rectangular cross section are joined by the simple glued scarf splice shown, Knowing that β = 22° and that the maximum allowable stresses in the joint are, respectively 400 kPa in tension (perpendicular to the splice) and 600 kPa in shear (parallel to the splice), determine the largest centric load P that can be applied.

Fig.14. 80

Mazurek Gravity answered 12 months ago

Step No: 1

Draw the force distribution of forces on the splice.

Step No: 2

Calculate the area of the inclined plane by using the equation:

Here,  is the area of the inclined plane,  is the width of the rectangular cross-section, is the depth of the rectangular cross-section and  is the angle of inclination of the plane.
Substitute  for  and  for .

Step No: 3

Apply the equilibrium equation along the inclined plane.

Substitute  for .

Here,  is the force along the inclined plane.

Step No: 4

Calculate the Shear stress by using the equation:

Here,  is the Shear stress.
Substitute  for  for and  for .

Step No: 5

Apply the equilibrium perpendicular to the inclined plane.

Here,  is the force perpendicular to the inclined plane.

Step No: 6

Calculate the normal stress by using the equation:

Here,  is the normal stress.
Substitute  for  for  and  for .

Therefore, as per design the allowable value of P is the smaller value:
.