test2 - Home Introduction Policies Requirements &...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Home Introduction Policies Requirements & Grading Resources Syllabus Assignments Handouts Discussion Web Links Distance Learners Contact TEST#2 MSE 200 Test#1 , November 8, 2001 , S.S.No: Read the questions carefully and answer on the backside also. A set of formulae are given at the end that may be useful. Q1. A sample of cross sectional area 0.1 sq. inches has been subjected to a tensile test. The load versus elongation curve is shown below. Identify the proportional limit, UTS, and fracture point in the load elongation graph. Calculate, the engineering stress, engineering strain, true stress, and true strain at the point of necking. Calculate the total elongation of the sample and fracture stress. Points:10
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
length at UTS= l=2.14 inch, initial length=lo=2, engineering strain=0.14/2=0.07 Load at UTS=8000, Initial area of cross-section=0.1 sq.in, engineering stress=8000/0.1=80000 psi True strain at UTS=ln(l/lo)=ln(2.14/2)=0.0677 True stress at UTS= Engineering stress x(1+engineering strain) =80000(1+0.07)=85600 psi Total elongation= 2.22+0.01-2.008-(-0.001)=0.223 Total engineering strain =0.111 Fracture stress=(1.111)x7500/(0.1)=82500 psi Q2. A high strength steel has a yield point of 1460 MPa and a critical fracture toughness of 98 MPa(m) 1/2 . Calculate the size of the crack that will lead to catastrophic failure at an applied stress of (1/2)Yield strength.Points:5 σ ys=1460 Mpa, KIC=98 Mpa(m) 1/2 = σ f( π c) 1/2 σ f=1460/2 Mpa, c=0.0014 m Q3. In a laboratory creep experiment at 1000 o C, a steady state creep rate of 1.5x10 -6 sec -1 is observed. When the experiments are performed at 800 o C, the observed creep rate is 1.0x10 -6 sec -1 . Predict the creep rate at a service temperature of 600 o C. Points 8 Strain rate ε '= ε o'exp(-Q/kT), ε 2'/ ε 1'=exp(-Q/kT2+Q/kT1) ln( ε 2'/ ε 1')=(-Q/k){1/T2-1/T1} ln(1.5)=(-Q/k){0.000786-0.00093}=(-Q/k)(-0.000146)
Background image of page 2
Q/k=2777.16,
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 8

test2 - Home Introduction Policies Requirements &...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online