Unformatted text preview: 2. Find: Determine Young's Modulus. Data: Upper yield point stress of 207 MPa, Yield point
strain of 0.001. Solution: Recall that Young's Modulus is the slope of the
stress/strain curve in the elastic portion of the
stress strain curve. That is: E=Stress/Strain
Substituting values into the above equation
yields: E = 207/0.001 = 207,000 MPa = 207 GPa. 3. FIND: Calculate the straintofail of silicate glasses. GIVEN: The modulus is 107 psi and the strengths of three
samples are 5, 50, and 500 ksi. ASSUMPTIONS: The material behaves in a linear elastic manner
up to failure. SKETCH: The beginnings of the three curves lie on top of
one another. The higher the straintofail the longer the curve extends to the upper right. The x's indicate the
points of.fai1ure. Stress
smm
SOLUTION: Hooke‘s law states:
6 = Es.
Rearranging,
s = a/E. Substitution for each of the samples: a = 5000 psi / 107 psi = 0.05%
s = 50,000 psi / 107 psi = 0.5% s = 500,000 psi / 107 psi — 5%. .
COMMENTS: The smallest value represents that of ordinary window glass. The largest value is characteristic of an optical
fiber. 4. Find: Determine if the stress is above or below the
yield stress. If the stress is below the yield stress
compute Young's Modulus. Data: The yield stress for the mild steel is 207 MPa. A
specimen has a diameter of 0.01m and a length of
0.1m. It is loaded in tension to 1000N and
deflects 6.077 x 104m. Solution: To solve this problem, we must first determine the
applied tensile stress. Recall that the tensile
stress is given by the formula: stress=Force
(P)/Area Normal to force (A). The cross sectional area is: As nDz/4 = 3.1416x(0.01)2/4 = 7.85x10“ The stress is: Stress = 1000/7.85x104 = 12.7MPa
The applied stress is much less than the yield stress. To obtain the Modulus recall the definition: E=o/s. We must thus compute the
strain in order to determine the Modulus: sA1/1o = 6.077x104/0.1 = 6.077x104
Thus E=12.7/6.077x104 = 2.09x105 MPa. 5. Find: Data: Solution: 6. Find: Data:
Solution: 8. Find: Data: Solution: ' 1/2011: * (313:: 01? hrs“ Compute deflection of specimens. Eu=70,460MPa, Em=122,500MPa, Ew=388,080MPa, A=0.01m x 0.01m=10"m2, Length(lo)=lm, Load(P)=5000N. Start with the relationship between stress and strain for linear elastic behavior: 02E; (1) Note that: ‘a=P/A (2) I a=Al/lo (3) Substitute (2) & (3) into (1): Thus P/A.=EAl/1° Rearranging to solve for Al we have:
AlPlolAE=5000 x 1/(104E)=5x107/E (5) Note that for Al to be in meters we require E to
be expressed in Pa. Using (5) and E in Pa we get
the following deflection values: Al==7 . 09x10‘4m, Cu=4 . 08x10'4m, W=1.29x10"m. (4) Deflection at 5000N.
ENylcnsfs=2.08GPa, Load(P)=5000N, 1°=1m, new“i 2.
Using the formula developed in the preceding problem we have: Al°=Plo/(AE). Determine the shear strain at yield.
For a particular steel: v=0.295, E = 205,000MPa,
ow300MPa and 1” = 1/26”.
Recall Hooke's Law in shear: tw= G5,.
stated that I” = 1/20”. Substitution yields: ow/ZG. The problem now is to determine G. Recall that G=E/(2(1+v))=
205,000/(2(1+0.295))=79,151 MPa. Substituting we have: 7,” 300/(2x79,151)=1.895x10'3. Also, it is 15 22. Find: Data: Solution: Compute the energy per unit volume, UN . to deform to a strain of 0.01. [1]
[2} Recall that the energy to deform to a given strain is the area under the stressstrain curve to the
strain of interest. This is equal to the integral with respect to strain of the stressstrain curve. Stressstrain curve for Cu may be represented by: ac“ = 319(MPa) 8°" Stressstrain curve for steel may be represented by: as = 450(MPa) 2°" U 0.0!
__ = o:
[1,)“ a[310 (MPa) .9 d3 0.01 (U) 310 (MP0) 5”
V c, 1.5 M
MN MJ J (9.) = 2m Mpg = 2.07 —T = 2.07 —3 = 2.07::106 —,
V Cr: m m m U 0.01
(—] = [450 (MP0) 5‘” ds
V s o 0.01 9:) = 450 (MPa) 1:” [V s 1.3
( 0.0 g] = 0.87 MPa = 0.87 M? = 0.87 —M—3] = 8.7x10’ is S m m m Find: (a) Explain why normal the tensile test is
used (b) differences in three of four point bend
specimens and to) some limitations on data.
Solution: (a) Due to the brittle nature of ceramics,
there is a very significant risk of failure
outside the gage section of the specimen during gripping. Also, such specimens are difficult and
expensive to machine. (b)The three or four point bend specimens are
subjected to negative (or compressive) loads and
do not require grips for loading. Hence, there is
no risk of premature failure. (c)The stress distribution in a 3—point or 4point
bend specimen is nonuniform along the cross
section of the Specimen. Hence, the strength is
obtained by calculating the outer fiber stress in
the specimen at the time of the failure. The strength values obtained from these tests have
considerable scatter. ...
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 Summer '08
 Tannebaum
 Tensile strength

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