2003-01-18 - Division of Solid Mechanics

More documents

Recommendations

Info

5. (3 points) ( ) 0 Examination in Damage Mechanics and Life Analysis (TMHL61) LiTH 2003-01-18 y 2 a 2 W P x Part 2 A large thin plate contains a through-thickness crack with initial crack length 2a 0 = 15 mm. The plate is loaded by a force P. Upon loading the plate, symmetric stable crack growth was observed. At load P = 300 kN the plate fractured. During the loading the crack opening δ(0) was recorded, and just prior to the fracture, the crack opening displacement was found to be δ(0) = 0.1 mm. Use the Dugdale model to determine the crack length just before the plate fractured. P Numerical data: Modulus of elasticity E = 210 GPa, yield strength σ Y = 480 MPa, plate width 2W = 0.4 m and plate thickness t = 0.002 m (i.e. the plate is so thin that the conditions at the crack tip may be regarded as plane stress). Solution: The load P = 300 kN gives stress σ ∞ = P/2Wt = 375 MPa in the plate. Since plane stress is at hand, the Dugdale model may be used. It gives δ(0) = 8 σ Y a 1 + sin(πσ E π ln⎧ ∞ /2σ Y ) ⎫ ⎨ cos(πσ ∞ /2σ Y ) Enter the given numerical values and solve for the unknown crack (half-)length a = a crit . One then obtains a crit = δ(0) E π ⎛ ⎜⎝ ln 1 + sin(πσ ∞/2σ Y ) ⎞ ⎟ 8 σ Y cos(πσ ∞ /2σ Y ) ⎠ − 1 210 000 ⋅ π ⎛ 1 + sin(π ⋅ 375/2 ⋅ 480) ⎞ = 0, 0001 ⎜ln ⎟ = 0, 0098 m 8 ⋅ 480 ⎝ cos(π ⋅ 375/2 ⋅ 480) ⎠ Thus, the total crack length at failure is 2a crit = 19.6 mm (= 1.31⋅2a 0 ). − 1 ⎬ ⎭ 8 2003-01-23/TD
6. 2 h Examination in Damage Mechanics and Life Analysis (TMHL61) LiTH 2003-01-18 a a a Q P P Q M b 2h 2h b Part 2 Determine, by use of superposition of stress intensity factors, the stress intensity factor K I for the split beam with a long crack, see figure. The crack length is a, beam thickness is b and beam height is 2h, where a >> b and a >> h. The material is linear elastic with modulus of elasticity E. The beam is loaded symmetrically with two opposite forces P at the beam ends. The sliding boundary condition to the right is such that the slope of the beam end is always zero. Assume plane stress conditions. The following results, taken from Problems 5/2 and 5/3 in the textbook can be used: Double cantilever beam with stress intensity factor K I = 2 Qa / bh Double cantilever beam with energy release rate G = 12M 2 / Eb 2 h 3 Solution: First, remove the boundary condition to the right and add a bending moment M at the beam end. The bending moment M should be such that the slope at the free end of the cantilever beam is zero. One obtains, for slope Θ, a Q M M √⎺3 which gives M = Qa / 2, and here Q = P Problem 5/3 gives, for plane stress, that K M I = = 2 M / bh Now superposition can be used to calculate the stress intensity factor in the case asked for. One obtains, using M = Qa / 2 and Q = P, K I = K I Q − K I M = 2 √⎺3 Qa bh√⎺h √⎺h Θ = Q a 2 2EI − M a = 0 EI √⎺⎺⎺EG √⎺3 √⎺h − 2 √⎺3 M bh√⎺h = 2 √⎺3 Qa bh√⎺h − 2 √⎺3 Qa 2bh√⎺h = √⎺ 3 Qa bh√⎺h = √⎺ 3 Pa bh√⎺h 9 2003-01-23/TD
Page 1 and 2: Avd Hållfasthetslära, IKP, Linkö
Page 3: Examination in Damage Mechanics and
Page 7: Examination in Damage Mechanics and

2003-01-18 - Division of Solid Mechanics

Create successful ePaper yourself

Delete template?

Save as template?