Download 3.5 The plastic region of the stress strain curve for a metal is

Review
3.3
State Hooke's law.
E = a constant of proportionality called the modulus of elasticity.
3.5
Define tensile strength of a material.
Answer. The tensile strength is the maximum load experienced during the tensile test divided by
the original area.
3.6
Define yield strength of a material.
Answer. The yield strength is the stress at which the material begins to plastically deform. It is
usually measured as the 0.2% offset value, which is the point where the stress strain curve for the
material intersects a line that is parallel to the straight-line portion of the curve but offset from it by
0.2%.
3.8
What is work hardening?
Answer. Work hardening, also called strain hardening, is the increase in strength that occurs in
metals when they are strained.
Multiple Choice
3.2
Which one of the following is the correct definition of ultimate tensile strength, as derived from
the results of a tensile test on a metal specimen: (a) the stress encountered when the stress strain curve
transforms from elastic to plastic behavior, (b) the maximum load divided by the final area of the
specimen, (c) the maximum load divided by the original area of the specimen, or (d) the stress observed
when the specimen finally fails?
Answer. (c).
3.5
The plastic region of the stress strain curve for a metal is characterized by a proportional
relationship between stress and strain: (a) true or (b) false?
Answer. (b). It is the elastic region that is characterized by a proportional relationship between
stress and strain.
3.9
Which one of the following materials has the highest modulus of elasticity: (a) aluminum, (b)
diamond, (c) steel, (d) titanium, or (e) tungsten?
Answer. (b).
3.1
The shear strength of a metal is usually (a) greater than or (b) less than its tensile strength?
Answer. (b).
Problems
3.1
A tensile test uses a test specimen that has a gage length of 50 mm and an area = 200 mm2. During
the test the specimen yields under a load of 98,000 N. The corresponding gage length = 50.23 mm.
This is the 0.2 percent yield point. The maximum load of 168,000 N is reached at a gage length =
64.2 mm. Determine (a) yield strength, (b) modulus of elasticity, and (c) tensile strength. (d) If
fracture occurs at a gage length of 67.3 mm, determine the percent elongation. (e) If the
specimen necked to an area = 92 mm2, determine the percent reduction in area.
Solution: (a) Y = 98,000/200 = 490 MPa.
(b) s = E e
Subtracting the 0.2% offset, e = (50.23 - 50.0)/50.0 - 0.002 = 0.0026
E = s/e = 490/0.0026 = 188.5 x 103 MPa.
(c) TS = 168,000/200 = 840 MPa.
(d) EL = (67.3 - 50)/50 = 17.3/50 = 0.346 = 34.6%
(e) AR = (200 - 92)/200 = 0.54 = 54%
Online Problems
Online 1: Three methods of solid solution strengthening.
a. Grain size reduction: as we reduce the size of grains we make it harder for defects to move due
to the increased numbers of grain boundaries. However, we bring ourselves closer to the tensile
strength and strain at failure since defects will tend to aggregate at boundaries.
b. Allloying : by adding “impurity” atoms we may distort the regular order of the matrix. These
local matrix distortions make it harder for defects to move thus increasing “Strengh” I nthe form
of yield and tensile strengths. Changing the atomic in attraction through alloying would affect
the modulus of elasticity.
c. Cold Work: by deforming a metal we reshape grains and add energy into the grain structure; the
distorted grains are not easily distorted further hence the “strength” increases in the for of
increased yield strength.
Online 2: Assume a bond
Area = 1 sq. inch
between two materials exists on
a plane at an angle of =60
F
F
degrees with respect to the

horizontal. Assume also that the
bond can withstand a shear stress of 600 psi in the plane. For a bar with a 1 in. by 1 in. cross section,
determine the maximum horizontal force, F, that the piece can withstand.
𝜏𝑅 = σcos(𝜙) cos(𝜆)
n
And
σ=
F
A



where  and are the angles between the force and the normal to the plane, and between the force
and the plane itself, and A is the area of the cross section (not the slanted area which is As).
From geometry,  and there fore
𝜏𝑅 = σcos(𝜃) sin(𝜃)
F
A
𝜏𝑅 = cos(𝜃) sin(𝜃)
or
𝑙𝑏
1𝑖𝑛2 (600 2 )
𝐴𝜏𝑅
𝑖𝑛 ≅ 1386 𝑙𝑏
𝐹=
=
cos(𝜃) sin(𝜃) cos(60) sin(60)