Math Problem Statement

A section of an engineering assembly includes three plates A, B and C, with dimensions as shown in Figure 7. All three plates are manufactured from aluminium with a yield strength of 300 MPa and a fracture toughness, KIC, of 20.0 MPa √m. Plates A and B are connected using a single-shear adhesive bonded lap joint with an overlap of 75.0 mm. The shear strength of the adhesive is 1.00 MPa. Plates B and C are welded together at their ends. Visual inspection reveals that there is a 40.0 mm long crack-like flaw in the middle of the weld, The assembly experiences a tensile load of 20.0 kN applied to the ends of the plates, which is greater than the highest expected design load for the assembly. width of plate 75mm Assuming that the flaw in the weld behaves as a sharp centre crack, Calculate the stress intensity factor, KI, for the crack? State your answer to 3 significant figures.

Solution

To calculate the stress intensity factor KIK_I for the crack in the welded plate, we use the following formula for a crack in an infinite plate subjected to a tensile stress:

KI=σπaK_I = \sigma \sqrt{\pi a}

Where:

  • KIK_I is the stress intensity factor,
  • σ\sigma is the applied tensile stress,
  • aa is half the crack length for a center crack,
  • π\pi is a constant (approximately 3.1416).

Step 1: Find the Applied Stress σ\sigma

The applied stress σ\sigma is the tensile load divided by the cross-sectional area of the plate:

σ=PA=Pwt\sigma = \frac{P}{A} = \frac{P}{w t}

Where:

  • P=20.0kN=20,000NP = 20.0 \, \text{kN} = 20,000 \, \text{N} (applied tensile load),
  • w=75.0mmw = 75.0 \, \text{mm} (width of the plate),
  • tt is the thickness of the plate (not provided, but it’s a necessary parameter).

If the thickness tt is not provided, we cannot calculate the exact stress intensity factor KIK_I without knowing it. However, let's continue assuming you will provide this value.

Step 2: Find Half Crack Length aa

For a center crack, the crack length is 2a2a. Given that the crack length is 40.0 mm, half the crack length aa is:

a=40.02=20.0mma = \frac{40.0}{2} = 20.0 \, \text{mm}

Step 3: Calculate the Stress Intensity Factor KIK_I

Once we have the thickness and hence the applied stress, we will plug the values into the formula for KIK_I:

KI=σπaK_I = \sigma \sqrt{\pi a}

If you could provide the thickness tt of the plate, I can proceed with the exact calculation.

Would you like to provide that information or ask any questions about the process? Here are some related questions to explore:

  1. How does the fracture toughness KICK_{\text{IC}} relate to the failure criteria in this case?
  2. What is the significance of adhesive shear strength in the design of this assembly?
  3. How does the thickness of the plate affect the applied stress and ultimately KIK_I?
  4. What happens if KIK_I exceeds KICK_{\text{IC}}, the fracture toughness of aluminium?
  5. Why is the crack assumed to behave as a sharp center crack in this context?

Tip: Stress intensity factor KIK_I is a critical parameter in fracture mechanics to predict the growth of cracks in materials under stress.

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Math Problem Analysis

Mathematical Concepts

Fracture Mechanics
Stress Analysis
Material Science

Formulas

Stress Intensity Factor (K_I) Formula: K_I = σ √(π a)
Applied Stress Formula: σ = P / (w t)

Theorems

Stress Intensity Factor Theorem
Fracture Mechanics

Suitable Grade Level

Undergraduate Engineering (Mechanical/Civil Engineering)