Fracture Project Report (2)

FRACTURE MECHANICS IN ENGINEERING DESIGN (Course: ME 5620)

Project Report

Department of Mechanical Engineering

Project by –

Kad Abhishek (fw8663)

Kshire Sagar (fw8276)

Patil Harshada (fx4157)

Under the guidance of Professor Dr. Raveendra Satha

In partial fulfillment of the requirements of ME 5620: FRACTURE MECHANICS IN ENGINEERING DESIGN

Introduction:

The following project work helps us understand how introduction of crack along elliptical, radial, and circular features have an effect on the rectangular bar subjected to a tensile force. This approach in fracture mechanics helps us find out critical section of any geometry with cracks at different crack locations as well as which type of crack could affect the part in the worse manner. It also gives us an overview of how plasticity affects the originally generated crack by extending it in a very small amount. Finally, J integral is also considered to compute the SIF.

Problem Statement:

Calculating nominal stress at various crack locations and estimating the SIF for a given rectangular filleted bar under tension.

t = 6 mm, c = 18 mm, r = 4.5 mm, H = 54 mm, h = 36 mm, L = 60 mm

Material Properties

• Young’s modulus, E = 200 GPa • Poisson’s ratio, ν = 0.3 • Density, = 7500 kg/m3

Group Crack A (1&2) Crack B (3&4) Crack C (5&6) 5 2mm 3mm 2mm

Summary Results and Verification:

Table 1.1

Parameter By Computational – FE Analysis By Analytical Solution

KI (Stress Intensity Factor)

For Shoulder

(MPa√ )

For Circular Hole

(MPa√ )

For Elliptical

(MPa√ )

For Shoulder (MPa√ )

For Circular Hole

(MPa√ )

For Elliptical Hole(MPa

√ )

68.22 204.5 220 72.0225 255.0206 267.7716

% Difference in comparison with the Computational FE Analysis

%6 %24.6 %21.7

Table 1.2

Nominal Stress @ Crack Location, σ0 For Shoulder (MPa) For Circular Hole (MPa) For Elliptical

Hole (MPa) 22.7083 45.4166 45.4166

Calculations:

[A] Analytical Solution:

For Shoulder, a (crack length) = 3 mm

= ( )

( )=

× .= 22.7083

SCF (Stress Concentration Factor) = 1.85, Using Standard Graph Values

We Know That, = × = 1.85×22.7083 = 42.0103

From the figure the crack is at an angle, therefore considering the angle as -1350 and using the transformation we have,

=

2 +

2(2 ) =

2+ 0 =

42.01032

= 21.0051

Now, SIF is given by,

= 1.12 ( × ×2

) = 1.12×42.0103× ( ×3××3

2 ∗ 36)

= 72.0225MPa√

Therefore, = 72.0225 MPa√

For Circular Hole, a (crack length) = 2 mm

= ( )

( )=

( )×=

× .

( × . )× .= 45.4166

SCF (Stress Concentration Factor) = 2, Using Standard Graph Values

We Know That, = × = 2×45.4166 = 90.8333

Now, SIF is given by,

= 1.12 ( × ×2

) = 1.12×90.8333× ( ×2××2

2 ∗ 36= 255.0206MPa√ =

Therefore, = 255.0206 MPa√

For Elliptical Hole, a (crack length) = 2 mm

= ( )

( )=

( )×=

× .

( × . )× .= 45.4166

SCF (Stress Concentration Factor) = 2.1, Using Standard Graph Values

We Know That, = × = 2.1×45.4166 = 95.375

Now, SIF is given by,

= 1.12 ( × ×2

) = 1.12×95.375× ( ×2××2

2 ∗ 36= 267.7716MPa√ =

Therefore, = 267.7716 MPa√

[B] Computational FE Analysis:

Fig. 1 Rectangular Filleted bar geometry with dimensions

Fig. 2 Meshing of Rectangular bar fillet and crack modelling

Fig. 3 Geometry after applying Boundary Condition

Stress plots:

Fig. 4 Maximum Stresses at the crack locations

Fig. 5 Comparison SIF for different element sizes

J Integral:

Crack J Integral SIF MPa√ Crack 1 0.2423 220.1 Crack 2 0.2424 220.2 Crack 3 2.566E-2 68.23 Crack 4 2.57E-2 68.25 Crack 5 0.2093 204.7 Crack 6 0.2094 204.7

Fig. 6 J integral for different contours

Displacement:-

Fig7: Displacement for crack 1

Similarly obtaining values for remaining cracks.

Crack Crack 1&2 Crack3&4 Crack5&6 Displacement (mm) 1.6e-3 3.7e-4 1.06e-3

Deformed crack distance

Fig8: Displacement for crack 1

Similarly obtaining values for remaining cracks.

Crack extension = deformed crack value – original value = 2.61-2.0 Crack extension = 0.61 mm

Conclusion:

1. Stresses at elliptical hole are greater than the actual stresses at fillet and circular hole, the elliptical hole represents critical region of this part. Hence, crack at elliptical hole represents the critical crack.

2. Analyzing on the ABAQUS software, we could conclude that SIF of an Elliptical hole has a higher value. So, it can be concluded that this section might fail the rectangular bar.

3. Different mesh sizes were taken into consideration until SIF values converges to the analytical solution values.

4. We can also infer that for finer mesh, stress distribution being uniform, we get more precise values of SIF.

Crack Crack 1&2 Crack3&4 Crack5&6 Deformed distance

(mm) 2.61 3.15 2.515

Crack extension (mm)

0.61 0.15 0.515