MOM Chapter 03

Chapter Objectives Understand how to measure the stress and strain through experiments  Correlate the behavior of some engineering materials to the stress-strain diagram. Copyright © 2011 Pearson Education South Asia Pte Ltd TENSION AND COMPRESSION TEST Copyright © 2011 Pearson Education South Asia Pte Ltd READING QUIZ 1) The modulus of elasticity E is a measure of the linear relationship between stress and strain. The common unit is: a) kN/mm2 b) MPa c) GPa d) All of them Copyright © 2011 Pearson Education South Asia Pte Ltd READING QUIZ (cont) 2) The Poisson’s ratio, v of common engineering materials lies in the range: a) 0≤v≤1 b) 0 ≤ v ≤ 0.5 c) -1 ≤ v ≤ 1 d) -0.5 ≤ v ≤ 0.5 Copyright © 2011 Pearson Education South Asia Pte Ltd APPLICATIONS Copyright © 2011 Pearson Education South Asia Pte Ltd APPLICATIONS (cont) Copyright © 2011 Pearson Education South Asia Pte Ltd STRESS STRAIN DIAGRAM • Note the critical status for strength specification  proportional limit  elastic limit  yield stress  ultimate stress  fracture stress Copyright © 2011 Pearson Education South Asia Pte Ltd STRENGTH PARAMETERS • Modulus of elasticity (Hooke’s Law)   E • Modulus of Resistance 1  pl 2 1 ur   pl pl  2 2 E • Modulus of Toughness – It measures the enter area under the stress-strain diagram Copyright © 2011 Pearson Education South Asia Pte Ltd EXAMPLE 1 The stress–strain diagram for an aluminum alloy that is used for making aircraft parts is shown in Fig. 3–19. If a specimen of this material is stressed to 600 MPa, determine the permanent strain that remains in the specimen when the load is released. Also, find the modulus of resilience both before and after the load application. Copyright © 2011 Pearson Education South Asia Pte Ltd EXAMPLE 1 (cont) Solutions • When the specimen is subjected to the load, the strain is approximately 0.023 mm/mm. • The slope of line OA is the modulus of elasticity, 450 E  75.0 GPa 0.006 • From triangle CBD, E BD 600 106     75.0 109   CD CD  CD  0.008 mm/mm Copyright © 2011 Pearson Education South Asia Pte Ltd EXAMPLE 1 (cont) Solutions • This strain represents the amount of recovered elastic strain. • The permanent strain is  OC  0.023  0.008  0.0150 mm/mm (Ans) • Computing the modulus of resilience, ur initial   pl pl  4500.006  1.35 MJ/m3 (Ans) 1 1 2 2 ur  final   pl pl  6000.008  2.40 MJ/m3 (Ans) 1 1 2 2 • Note that the SI system of units is measured in joules, where 1 J = 1 N • m. Copyright © 2011 Pearson Education South Asia Pte Ltd POISSON’s RATIO  lat v  long Copyright © 2011 Pearson Education South Asia Pte Ltd EXAMPLE 2 A bar made of A-36 steel has the dimensions shown in Fig. 3–22. If an axial force of P = 80kN is applied to the bar, determine the change in its length and the change in the dimensions of its cross section after applying the load. The material behaves elastically. Copyright © 2011 Pearson Education South Asia Pte Ltd EXAMPLE 2 (cont) Solutions • The normal stress in the bar is P z   80 103    16.0 106 Pa   A 0.10.05 • From the table for A-36 steel, Est = 200 GPa z  z  16.0 106   6   Est 200 10 6  80 10 mm/mm • The axial elongation of the bar is therefore  z   z Lz  80106 1.5  120m (Ans) Copyright © 2011 Pearson Education South Asia Pte Ltd EXAMPLE 2 (cont) Solutions • The contraction strains in both the x and y directions are  x   y  vst z  0.3280106   25.6 m/m • The changes in the dimensions of the cross section are  x   x Lx  25.6106 0.1  2.56m (Ans)  y   y Ly  25.6106 0.  05  1.28m (Ans) Copyright © 2011 Pearson Education South Asia Pte Ltd SHEAR STRESS-STRAIN DIAGRAM • Strength parameter G – Shear modulus of elasticity or the modules of rigidity • G is related to the modulus of elasticity E and Poisson’s ratio v.   G E G 21  v  Copyright © 2011 Pearson Education South Asia Pte Ltd EXAMPLE 3 A specimen of titanium alloy is tested in torsion and the shear stress– strain diagram is shown in Fig. 3–25a. Determine the shear modulus G, the proportional limit, and the ultimate shear stress. Also, determine the maximum distance d that the top of a block of this material, shown in Fig. 3–25b, could be displaced horizontally if the material behaves elastically when acted upon by a shear force V. What is the magnitude of V necessary to cause this displacement? Copyright © 2011 Pearson Education South Asia Pte Ltd EXAMPLE 3 (cont) Solutions • By inspection, the graph ceases to be linear at point A. Thus, the proportional limit is  pl  360 MPa (Ans) • This value represents the maximum shear stress, point B. Thus the ultimate stress is  u  504 MPa (Ans) • Since the angle is small, the top of the will be displaced horizontally by tan0.008 rad   0.008  d  d  0.4 mm 50 mm Copyright © 2011 Pearson Education South Asia Pte Ltd EXAMPLE 3 (cont) Solutions • The shear force V needed to cause the displacement is V V  avg  ; 360 MPa   V  2700 kN (Ans) A 75100 Copyright © 2011 Pearson Education South Asia Pte Ltd CONCEPT QUIZ 1) The head H is connected to the cylinder of a compressor using six steel bolts. If the clamping force in each bolt is 4000N, determine the normal strain in the bolts. Each bolt has a diameter of 5 mm. If σy= 280 MPa and Est = 210GPa, what is the strain in each bolt when the nut is unscrewed so that the clamping force is released? Copyright © 2011 Pearson Education South Asia Pte Ltd CONCEPT QUIZ (cont) a) 0.970 b) 0.203 c) 0.970(10-3) d) Insufficient information to determine because the stress is beyond yield point Copyright © 2011 Pearson Education South Asia Pte Ltd