hw25-10sol - H'W " So’Lo‘noug Stress Elements ' '...

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Unformatted text preview: H'W " So’Lo‘noug Stress Elements ' ' Maximum Principal Stress _ ~ —‘n_.. - W. The figure below shows four stress elements normal and shear stress values as shown. Rank these situations, from greatest to least, en the basis of the maximum principal (stress. 5p51 =01! . 5ps1 ‘ 032.13”; I (GUY-1 "of? a -1 7psi =01 10 psi . 3 psi Greatest 1 A 2 B 3 C 4 Least- Or, the maximum principal stress is the same for every stress element. Please carefully explain your reasoning. . ' ’ (13“(7 Q':—G" 1 1_ i 6V“ 7—7 i [17) “W ‘ 0’ . \ 0’1 v ’E'B\x2 1. - V 4 \ ' ‘ ' .— . Q 1 :fkihw fo. 0 have -5?” ' . $5 ‘ lyre/W a . "I —-\ . L, ’ V} t \1‘oipffin 0b (>54- '—1 +1) 7 a: -r 4?): a. ,L 4. We H (9 SD _—5-3 “L Amara] "molars; /:L , How sure are you of your ranln'ng? (Circle one) Basically Guessed Sure Very Sure 1 ' 2 ' 3 4 I 5 6 7 ‘8 9 10 ~ 115 Stress Elements Maximum Shear Stress 57‘, The figure below shows four stress elements with normal and shear stress values as shown. The out-of-plane principal stress is zero. ' Rankthese situations, from greatest to least, on the basis of the maximum shear stress. 5 psi =6"!- Greatest 1 I 2 A C .4 37 ‘ Least Or, the maximum shear stress is the same for every stress element. Please carefully explain your reasoning. ' an” =-= (rm—my a \‘A-(AMF. 7, + T)”, 5 “T; 1 2' :: O M L?) + ()0) 1 I” - " "I La”“'f“7”“i - f" r, - 12)] \j LIE-J) J, (-m — ,wa mates 0 {L332 + W = W 1 ‘W e4 ’L m J “WT w = M? ” W s V L ' How sure areyouof your rankm' g?. (Circle one) ' ‘ Basically Guessed ' Sure Very Sure 7.} 1 2 . 3 4 - 5 y '6 7 8 9 10 116 9—42. The drill pipe has an outer diameter of 3 in., a wall thickness of 0.25 in., and a weight of 50 lb/ft. If it is subjected to a torque and axial load as shown, determine ' (a) the principal stress and (b) the maximum in-plane shear stress at a point on its surface at section a. Internal Forces and Torque: As shown on FBD(a). Section Properties; 7T . A = 1(32 — 2.52) = 0.68751rm2 7r _ J = — 1.54 — 1.254 = 4.1172 in‘ 2 ( ) I $00 ’3 Normal Stress: ' p ‘ Boa lb}: _ N _ —2500 _ _ . , Wklmlb " ' A ’ 0.687511 ’ 115751351 Shear Stress: Applying the torsion formula. ._ Tc _ 800(12)(1.5) __ _ mam“, 7 — J — 4.1172 — 3497.5p51 v ¢ 7:5“ 1th a) In - Plane Principal Stresses: a, = 0, a, = —1157.5 psi and 7x, = 3497.5 psi for any point on the shaft’s surface. Applying Eq. 9-5. (n+0, a'x—ay2 ' H ' a”: 2 d“(2)1”5 ' l I 0+ —1157.5 / 0— —1157.5 2 i ' ' —————-( , )i (———-——< )) + (3497.5)2 . “’5’” _ 2 2_ “5155‘ = —578.75 4.; 3545.08 0’; = 2966 psi = 2.97 ksi EE 0; = -4124 psi = -4.12 ksi b) Maximum In - Plane Shear Stress: Applying Eq. 9-7 a' - a' 2 T x y ‘ in—p e = ( 2 ) + 7%! = (o — (-1157.5) '2 2 > + (3497.5)2 = 3545 psi = 3.55 ksi ’ Ans. *9—56. Solve Prob. 9—11 using Mohr’s circle. Construction of the Circle: In accordance with the sign conVention, 0-,, = -3 ksi, 0-, = 2 ksi, and 1-,, = —4 ksi. Hence, ' ‘ a'x+a'y —3+2 . ‘2 _ —. 2 —-0.500k51 3 ksi am =. The coordinates for reference point A and C are ‘ A(-3, —4) C(—o.soo, 0) The radius of circle is »R = V(3 — 0.5)2 + 42 = 4.717 ksj Stress on the Inclined Plane: The normal and shear stress components (0,, and r xv ,1) are represented by the coordinates of point P on the circle. 03,: = -0.500 - 4.717 cos 62.01° = -2.71 ksi 11,", = 4.717 sin 62.01° = 4.17 ksi ...
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This note was uploaded on 02/09/2011 for the course CIVIL ENGI 215 taught by Professor Dr.pollock during the Fall '10 term at Washington State University .

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hw25-10sol - H'W " So’Lo‘noug Stress Elements ' '...

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