# t2f03 - Home Introduction Policies Requirements & Grading...

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Home Introduction Policies Requirements & Grading Resources Syllabus Assignments Handouts Discussion Web Links Distance Learners Contact MSE 200

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MAT 200 Test#2 K IC = f( π c) ½, ε '= ε o'exp(-Q/kT), ln(l/lo), (l-lo)/lo, σ =F/Ao, σ =F/A, σ = σ o+K/d 1/2 Q1. A sample of a metal of length 2 inches and thickness 0.5 inches and width 0.5 inches has been subjected to a compressive load by drop forging wherein a heavy hammer is dropped from a height to deform the metal. It is required to reduce the thickness of the sample after deformation to 0.45 by drop forging. The width of the sample is constrained not to change. The load versus elongation curve of the sample under compression is shown below. Calculate the following. 1. The final length of the sample. 2. The plastic strain along the length and thickness directions 3. The total strain. 4. Thickness of the sample when the maximum load is applied 5. The maximum load required to deform the sample. 6. The engineering stress and the true stress in the sample 7. The energy spent in deforming the sample using straight line approximation of the load-elongation graph in the plastic region. 8. If the height of the hammer before dropping is 20 feet, what is the weight in pounds of the hammer that is desired for the deformation. 9. Do you expect the sample to crack because of lack of ductility. Answers 1. Volume remains constant: 2x0.5=lx0.45, l=2.22 inches 2. Plastic strain in the sample in the longitudinal direction: (2.22-2)/2=0.11, Length of the sample when the thickness is (~0.498)=2.008 Plastic strain along the thickness direction=(0.45-0.5)/0.5=-0.1 3 Total strain= plastic +elastic strain Elastic strain in the thickness direction=(0.498-0.5)/0.5=-0.004 Elastic strain in the longitudinal direction=(2.008-2)/2=0.004
Total strain in the thickness direction=-0.104 Total strain in the longitudinal direction=0.114 4. Thickness of the sample including elastic and plastic strain when the load is applied=0.448 inches Length of the sample including elastic and plastic strain=2.228 inches 5. Load required to deform the sample=~7750 lbs 6. Engineering compressive stress =7750/(2.0x0.5) = 7750 psi True compressive stress= 7750/(2.228x0.5)= 6957 psi 7. Energy spent = Area under the load-displacement graph(use straight line approximation) = 0.5x(7000)x(0.5-0.498)+7000x(0.498-0.448)+0.5x(7750-7000)x(0.498-0.448)

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## This note was uploaded on 10/11/2009 for the course MSE 200 taught by Professor Wholedepartment during the Spring '08 term at N.C. State.

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t2f03 - Home Introduction Policies Requirements & Grading...

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