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# HW 12 - E150 “'a.r um I 1"\1.25 PROBLEM egg The...

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Unformatted text preview: E150... “ 'a.r um.- ., I 1" \1 .25 PROBLEM egg The components of plane strain at point p of a bit during ' a drilling operation are a. = 0.00400. 8,, = 000300. and :' coordinate system are 2‘. = 0.00125. 0', = 000025, and . 7'... = 0.00910. What is the angle 6'? " The radius of Mohr's circle is: R = 1((00005 — (41003))1 + (0.003)2 = 0.00461 ‘._ —~ 0.003 ; The an lee is: 6' =tao‘I —-—-—— =40.6° _ 9 ‘ 1 [— 0.003 — 0.0005) 7/ / 2 --;[email protected]=; .. - The components of plane strain at point p are a. = a, = so and 1'... = 0. Use Mohr's circle to show that s'. = 0 and i"; 7‘... = 0 for any value of 0. - §QLUTIQN: ' With e,c = 2,. we see that Mohr's circle becomes a POINT. Therefore: _-._ ANS s = so 1' EV Y value f = 0.00600. and the components referred to the x’y’z' 1 E . SOLUTION: The center of Mohr's circle is at: C. = ﬂit-00;“ = 0.0005 ' ‘4 (0.004. 0.003) 0.20 Using the characteristics of the circle to ﬁnd a": e.’ = 0.0005 + (0.00461) (cos 20) = 0.00125 20 = - 80.0°- (-40.00 = -40° ANS 0 = - 20° f. 331 P .14 Use Mohr's circle to determine the principal strains and the maximum in-plane shear strain to which the Space Shuttle‘s nozzie is subjected in Problem 6.3.7. ﬁ) 5: . 00765. —0.001055) 50.130 E 5. ' §QLUTION: (0.00207. 0.001035) :- The center of Mohr’s circle is at: C! = m = 0.00745 0, = 0 2 The radius of the circle is: R = (0.00727 - 0.00745)2 + [000207] = 0.00105 7 '/ 2 The principal strains on the element are: 01 = 0.00745 + R = 0.00745 + 0.00105 Ii" ." ANS 01 = 0.0005 5 02 = 0.00?45 - R = 0.00745 - 0.00105 ANS g2 = 0.00640 7m = 2 R = 2 (0.00105) AN5 1...... = 0.00210 : PROBLEM §~§.15 i Use Mchr's circle to determine the principal strains and the maximum in-plane shear strain at point p of the MacPherson = strut suspension in Problem 6.2.31. (“0.0088, -0.00‘|6) S'j- i SQLUTION: The center of Mohr's circle is at: C4. =w = -0003; (4,0052. 0) i 2 (0.0024. 0.0018) 1121; CT = U .:I 2 _ The radius of the circle is: R = (— 0.0033 — (00032))2 {“003 6] =0.00588 ' The principal strains on the element are: a, = - 0.0032 + R = - 0.0032 + 0.00588 ANS e. = 0.00268 52 = - 0.0032 — R = - 0.0032 — 0.00530 __j ‘ ANS 02 = - 0.00900 . 7...... = 2 R = 2 (0.00588) ANS 1m, .—. 0.01120 33? 'Using the result for 7,, in Equation (6—46): F LE 4.1 _ If the bearing's housing in Problem 6-2.10 is made of steel with elastic modulus E = 200 GPa and Poisson's ratio v = 0.28. what is the strain component a, at p? SQLQTION: From Equation (6-45): 5, =icrz —-‘1(or, +i:ry)=———i——T(0)———0'§"3——2(—253.5x10ﬁiirtm2 +309x10‘Nim’) E 5 200x10 Nr'm 200x10 Nim ANS §,=-. 7 At a point of a material subjected to a state of plane stress, the strains measured by the strain rosette are e, = 0.006, a, = -0.003. and 8,; = -0.002. The modulus of elasticity and Poisson's ratio of the material are E = 30 GPa and v = 0.33. What is the state of stress at the point? SQLQIIQN: Using Equations (6-10) to determine the normal and shear stresses: c, = a" 00520;. + a, sin2 0, + 7., sin 0., cos 0, = c, cos2 (—63“) + s, sin2 (-68”) + in, sin (00°) cos (68°) = 0.000 ab = 0, cos2 0., + s, sin2 0., + 7,, sin 8;, cos 0., = ex (:052 (0) + s, sin2 (0) + 7,, sin (0) cos (0) = - 0.003 c. = s. coszec + s, sin2 0: + 7*, sin 0.: cos 0,, = 5, cos2 (56°) + s, sin2 (56°) + 7,, sin (56") cos (56“) = — 0.002 Solving the above three equations simultaneously: 0; = - 0.003 21. = 0.0041 7,, = - 0.0084 Using the above values in Equations (6-43) & (6-44): 1 0.33 0' ———— 1 —-——-——o‘ I [1] 30iciii°mm2 aoxiotiwm2 ’ — 0.003 = come—Ls +—._‘_s [2] 30x109Nz'm2 "' 30x109Ni’m1 3" Solving equations [1] and [2] together: ANS ox = — 55.4 MPa 6,. = 104.? MPa 1,, = G 'ny ='(11.2a x 10° Nim’) (— 0.0004) ANS I“, = - 04.0 MPa 347 ...
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HW 12 - E150 “'a.r um I 1"\1.25 PROBLEM egg The...

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