# Solid Mechanic

Topics: Elasticity, Tensile strength, Linear elasticity Pages: 7 (1050 words) Published: May 18, 2013
Chapter Objectives
 Understand how to measure the stress and strain through experiments Correlate the behavior of some engineering materials to the stress-strain diagram.

In-class Activities
1. 2. 3. 4. 5. 6. 7. Reading Quiz Applications Stress-Strain diagram Strength parameters Poisson’s ratio Shear Stress-strain diagram Concept Quiz

TENSION AND COMPRESSION TEST

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)
c)

MPa
GPa

d)

All of them

2) The Poisson’s ratio, v of common engineering materials lies in the range: a) b) c) d) 0≤v≤1 0 ≤ v ≤ 0.5 -1 ≤ v ≤ 1 -0.5 ≤ v ≤ 0.5

APPLICATIONS

APPLICATIONS (cont)

STRESS STRAIN DIAGRAM • Note the critical status for strength specification  proportional limit  elastic limit  yield stress  ultimate stress  fracture stress

STRENGTH PARAMETERS • Modulus of elasticity (Hooke’s Law)

  E
• Modulus of Resistance
1 1  pl ur   pl pl  2 2 E
2

Modulus of Toughness – It measures the entire area under the stress-strain diagram

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.

EXAMPLE 1 (cont)
Solution
• 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,

E

450  75.0 GPa 0.006

• From triangle CBD,

BD 600 106 E   75.0 109 CD CD  CD  0.008 mm/mm

 

 

EXAMPLE 1 (cont)
Solution
• 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,

1 1 ur initial   pl pl  4500.006  1.35 MJ/m3 (Ans) 2 2 1 1 ur  final   pl pl  6000.008  2.40 MJ/m3 (Ans) 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

Please refer to the website for the animation: Poisson’s Ratio

 lat v  long

EXAMPLE 2
A bar made of A-36 steel has the dimensions shown in Fig. 3–22. If an axial force of P = 80 kN 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.

EXAMPLE 2 (cont)
Solution
• The normal stress in the bar is

P 80103  z    16.0106  Pa A 0.10.05 • From the table for A-36 steel, Est = 200 GPa

16.0 106 z    80 106 mm/mm Est 200 106
• The axial elongation of the bar is therefore

z

   

 z   z Lz  80106 1.5  120m (Ans) Copyright © 2011 Pearson Education South Asia Pte Ltd

EXAMPLE 2 (cont)
Solution
• The contraction strains in both the x and y directions are

• The changes in the dimensions of the cross section are
 x   x Lx  25.6106 0.1  2.56m...