HW_ Soln_Ch9-1

# HW_ Soln_Ch9-1 - 2 Find Determine Young's Modulus Data...

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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 strain-to-fail 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 strain-to-fail 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: s-A1/1o = 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: Al-PlolAE=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, ow-300MPa 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 stress-strain curve to the strain of interest. This is equal to the integral with respect to strain of the stress-strain curve. Stress-strain curve for Cu may be represented by: ac“ = 319(MPa) 8°" Stress-strain 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 4-point bend specimen is non-uniform 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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