# hw24-10sol - Mg ‘14 ’ Sow‘no VS 09—17 Determine the...

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Unformatted text preview: Mg ‘14 ’ Sow‘no VS 09—17. Determine the equivalent state of stress on an element at the same point which represents (a) the principal stress, and (b) the maximum in-plane shear stress and the associated average normal stress. Also, for each case, determine the corresponding orientation of the element with respect to the element shown. Sketch the results on each element. Normal and Shear Stress: 0x = 125 MPa 0, = —75 MPa 7;, = —50 MPa In - Plane Principal Stresses: 03,-0- 0' —a' 2 y x y 012: :l: ( ) +¢xy2 2 2 125 + —75 125 - —75 2 = __g i (A) + HO): 2 2 = 25:]:V12500 01 = 137 MPa 0'2 = ——86.8 MPa Ans. Orientation of Principal Plane: Txy —50 20 = —— = —-—-— = —o.5 tan P (a,r — a,)/2 (125—(—75))/2 0,, = ~13.28° and 76.72° Substitute 6 = —13.28° into (n+0), (Ix—(7y a" = 2 + —2—-cos 20 + TX), sin 20 125 + (—75) 125 — (—75) , = -——-——2—-— + -————3—cos(—26.57°)+(—50) sm(—26.57°) = 137 MPa = 01 Thus, ' (11p)1 = —13.3° and (tap)2 = 76.7° Ans. 125 — (—75)/(—50) The element that represents the state of principal stress is shown in Fig.a. Maximum In - Plane Shear Stress: a" — 0' 2 ’ -' “ 2 T = < y) + 7x 2 = (M) + 252 = 112MPa ADS- ma 2 y 2 Ill-p ne Orientation of the Plane of Maximum In - Plane Shear Stress: TU -50 93 = 31.7° and 122° 9—17. Continued By inspection, 1- has to act in the same sense shown in Fig. b to maintain \$35k“ equilibrium. Average Normal Stress: ax + (Ty _ 125 + (—75) 2 — 2 = 25 MPa Ans. O'avg = . The element that represents the state of maximum in - plane shear stress is shown in Fig. c. 993M?“ 9—30. The cantilevered rectangular bar is subjected to the ‘force of 5 kip. Determine the principal stress at points A and B. = le—(sxé) = 54 in“ A = (6)(3) = 18 in2 A = 2.25(1.5)(3) = 10.125 in3 QB = 2(2)(3) = 121113 Point A: P sz 4 45(1-5) . = — + = — + = .472 k ”A A I 18 54 1 5‘ QA 3(10.125) Z . = = — = . k TA It 54(3) 0 1875 \$1 ox = 1.472 ksi a, = 0 TU = 0.1875 ksi a“: + a". o" — cry 2 . + . — 2 = 1472 0 i (1472 0) + 0.18752 2 2 0'1 = 1.50 ksi 0'2 = —0.0235 ksi Point B: _ P sz _ 4 45(1) _ . 0'3 — A I — 18 54 — 0.6111ksx _ ,zQB__3(12)_ . TB —- It —- 54(3) — 0.2222 k51 0x = —0.6111 ksi a", = 0 TX, = 0.2222 ksi 0' + 0' a — a' 2 X Y X y 01,2 = —2 :1: (—-———2 ) + rxyz = —0.611 + 0 :1: (—0.6111 — 0 2 + . 2 2 2 > 0222 01 = 0.0723 ksi 0'2 = —0.683 ksii ‘— ﬂ'N75 4'5! —{ /- #74 6‘? .——. ~— u-ua r»:- H a. m a.- _. E5 Ans. 9—38. A paper tube is formed by rolling a paper strip in a spiral and then gluing the edges together as shown. Determine the shear stress acting along the seam, which is at 30° from the vertical, when the tube is subjected to an . axial force of 10 N. The paper is 1 mm thick and the tube has an outer diameter of 30 mm. 10N ION 5 10 = ——.—— = 109.76 kP A §(0.032 — 0.0282) a 0,, = 109.76 kPa a, = 0 7,, = 0 e = 30° O‘x—O'y 7,3,: = ~—2— sin 20 + 7,}, cos 20 _106.76 — 0 2 I I #9776 if“ sin 60° + 0 = ~47.5 kPa Ans. 9—39. Solve Prob. 9-38 for the normal stress acting perpendicular to the seam. ION 10N P 10 = — = ————— = 109.76 kP ‘7 A g (0.032 — 0.0282) a (n+0), ax-a'y . 0-,, = ,—-—— + ~——cosZ(9 + TxySII'lza 2 2 109.76 + 0 + 109.76 — 0 2 2 cos (60°) + 0 =_ 82.3 kPa Ans. ...
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