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# hw26-10sol - “W a 2‘0 — Sownows 9—67 Determine the...

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Unformatted text preview: “W a 2‘0 — Sownows 9—67. Determine the principal stress, the maximum in-plane shear stress, and average normal stress Specify the orientation of the element in each case. Construction of the Circle: In accordance with the sign convention, a" = 350 MPa, 0". = —200 MPa, and 1x, = 500 MPa. Hence, a + a 350 + —200 aavg‘ = x 2 y = -—2(———) = 75.0 MPa Ans. The coordinates for reference point A and C are A(350, 500) C(7S.0, 0) The radius of the circle is R = m = 570.64 MPa at) In - Plane Principal Stresses:,'Ihe coordinate of points B and D represent 01 and 0'2 respectively. 0'1 = 75.0 + 570.64 = 646 MPa Ans. 02 = 75.0 — 570.64 = —496 MPa Ans. Orientaion of Principal Plane: From the circle - 500 tan 201:1 — m - 1.82 OH = 30.6° (Counterclockwise) Ans. b) Maximum In - Plane Shear Stress: Represented by the coordinates of point E on the circle. (“we = R = 571 MPa Ans. Orientation of the Plane for Maximum In - Plane Shear Stress: From the circle tan 29. = 331\$ = 0.55 0s = 14.4° (Clockwise) Ans. 200 MP3 500 MPa 350 MPa *9—68. Draw Mohr’s circle that describes each of the following states of stress. a) Here, a" = 600 psi, 0",. = 700 psi and 7,, = 0.Thus, 0+0- xz y=600:700=650psi 0.an = Thus, the coordinate of reference point A and center of circle are A(600, 0) C (650, 0) Then the radius of the circle is R = CA = 650 — 600 =50psi The Mohr’s circle represents this state of stress is shown in Fig. a. b) Here, 0', = 0,0, = 4 ksi and 7x, = 0.Thus, ax+ay 0+4 aavg=——= 2 2 = 2ksi Thus, the coordinate of reference point A and center of circle are A(0, 0) C(2, 0) Then the radius of the circle is R=CA=2~0=2psi c) Here, crx = 03 = 0 and 1-,, = —-40 MPa.Thus, 0',‘+rr‘v 2 —0 0'an = Thus, the coordinate of reference point A and the center of circle are A(O, ~40) ' C (0, 0) Then, the radius of the circle is R=CA=40MPa The Mohr’s circle represents this state of stress shown in Fig. c (a) (b) 9—68. Continued 09-73. The cantilevered rectangular bar is subjected to the force of 5 kip. Determine the principal stress at point A. Internal Forces and Moment: As shown on FBD. Section Properties: A = 3(6) = 18.0 in2 I = —1—(3)(63) = 540m“ 12 ' QA = 524’ = 2.25(1.5)(3) = 101251113 Normal Stress: \ My ”7*? . 45.0 1.5 400 + ( ) 0,4 = W 540 = 1.4722 kSI . VQ Shear Stress: Applying the shear formula 1- = T __ 3.00(10.125) _ 01875 k , 7" ‘ 54.0(3) ‘ ' 8‘ Construction of the Circle: In accordance with the sign convention, a" = 1.4722 ksi, 0y = 0, and rxy = 40.1875 ksi. Hence, - ’rm: J 0' +0- " 2 ’ =lf4—722;0=0.7361ksi 0'an = The coordinates for reference points A and C are A(1.4722, f0.1875) C(0.73\61, 0) The radius of the circle is R = V(1.4722 — 0.7361)2 + 018752 = 0.7596 ksi In - Plane Principal Stress: The coordinates of point B and D represent 01 and 02, respectively. 01 = 0.7361 + 0.7596 = 1.50 ksi Ans. 0'2 = 0.7361 —- 0.7596 = —0.0235 ksi Ans. My“ 3550C) ”Yuma WW (Wm W1? = L; ML Jr: EM: 0 man 9 - 45 ° v: \ 15M SoLunw LE 1 :3; 1 C63:5 > \0 0k ¥ U» :1er X» mum {W 0' 2 «3: _. bMaQasg .. 500 MP“ ML, w + 0.0\‘5-. VWWMM 1 g : 0:219 - 6001M?“ _ 150 Max Tgoowm :7 SW” “W 78 M sum WW»: (”\$95) D > 150Wa=01 Xv +1» 9.4%» 5- dd» Nuﬁmlwmk 3 [*1 0“ : 07*“, = 500 Ma 3 150 ”pa 0-1, 1 0:; vomm M1 Moulx Cm ; fr WA W I; (IRCIX‘ Fﬂy : SW00 VP‘L’V’LSD Mp“ : 315 ”Pa t‘O—AVW I L 1 > 115 MP“ 31‘3“?“ «mun = 600 Rpm/3'15 MR: 2 \15 MP“ \ 7, K / a} 9 \$45“: A1, = Tm :- wax/mug =L15Mpa \ H 6:;va v {HSMPK / Y AW / \4 /\ m M?“ ...
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