hw23-10sol - H W 1‘ 13 ’ SDLu‘noNS 9—10. The state...

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Unformatted text preview: H W 1‘ 13 ’ SDLu‘noNS 9—10. The state of stress at a point in a member is shown ' 2 ksi ' on the element. Determine the stress components acting on the inclined plane AB. Solve the problem using the method of equilibrium described in Sec. 9.1. 3 ksi Force Equllibrium: For the sectioned element, '\+2Fyr = 0; ’ AF," — 3(AA sin 30°)sin 60° + 4(A A sin 30°)sin 30° —2(AA cos 30°)sin 30° — 4(AA cos 30°) sin 60° = AFy: = 4.165 AA /'+ EFXI = 0; AFp + 3(AA sin 30°) cos 60° + 4(A A sin 30°)cos 30° —2(AA cos 30°) cos 30° + 4(AA cos 30°) cos 60° = 0 AFX. = —2.714 AA ' ~ acme“ ' . 2,44 toxic ' Normal and Shear Stress: For the inclined plane. AF , q, = limMno A; = —2.71 ksi Ans. - 4.17 ksi Ansr Negative sign indicates that the sense of o’xr, is opposite to that shown on FBD. 9—11. Solve Prob. 9—10 using the stress-transformation equations developed in Sec. 9.2. Show the result on a sketch. Normal and Shear Stress: In accordance with the established sign convention, 9 = +60° a, ‘= -3 ksi 0,, = 2 ksi 1,}, = —-4 ksi 3ksi Stress Transformation Equations: Applying Eqs. 9-1 and 9-2. ax+ay (TX—o", 0-,: = 2 + —2——c0526 + ‘rxy sin 20 _ + _ .— = 3 2 + 3 2 c05120° + (—4 sin 120°) 2 2 = —2.71 ksi Ans. 0',t v cry rxly: = ——--2—- sin 26 + 1,0, cos-20 = — '32— 2 sin120° + (—4c05120°) = 4.17 ksi Ans. ' Negative sign indicates or, is a compressive stress ~9—13. Determine the equivalent state of stress on an element if the element is oriented 60° clockwise from the element shown. Show the result on a sketch. In accordance to the established sign covention, 0 = -60° (Fig. a) 0-,, = 200 psi a'y = —350 psi Txy = 75 psi Applying Eqs 9-1, 9—2 and 9-3, O'x+0'y (TX—'0”. . ox: = -——2—- + ——§—c0529 + 'rxysm20 200 + (—350) + 200 — (—350) 2 2 cos (—120°) + 75 sin (~120") = —277.45 psi = -277 psi Ans. (5+0), O'x—‘O'y a-y, = 2 —- Tcos 20 — 7,0. sin 29 200 + (—350) 200 — (—350) - ___ ___.__ _ ._—-— cps_(—120°) — 75 sm (-120°) 2 2 = 127.45 psi = 127 psi Ans- ax — a". Hy. = —-—?—sin 20 + 7,}, cos 20 200 — —350 = ..___2(___) Sin (—120°) + 75 cos (—120°) = 200.66 psi = 201 psi A“ Negative sign indicates that 0),! is a compressive stress These result, can be represented by the element shown in Fig. b. 7: C (Uiqu - _ ' 09-23. e wood beam is subjected to a load of 12 kN. If a grain of wood in the beam at point A makes an angle of I. 10' with the horizontal as shown, determine the normal and shear stress that act perpendicular and parallel to the grain due to the loading. I = T12- (0.2)(03)3 = o.45(10‘3)m4 Q. = 'y‘A’ = 0.1125(o.2)(0.075) =.1.6875(10'3)m3 M 13.714 103 0.075 ‘ 0,, = A = ~—# = 2.2857MPa (T) 1 - 0.45(10'3) _ g _ 6.875(103)(1.6875)(10'3) _ TA " It ‘ 0.45(1o-3)(o.2) ' 01286 M'Pa 0x = 2.2857 MPa 0, = o 1,, = -0.1286 MPa eéiv (71H? 01-42, “J W 9 07w 1 “7’ 2 £7” + (7)“‘19 W“ 1 W” = “LW‘PK + Q”? N,“ m 44') + (vmmMM Kai“ 40") 1 _ 61': mm m WM + 1m 0' +0” (fiA—(f '. -_—_ {,"Y/ : ELI—Y) _ K TV) 0,3 — V‘XY gnaw MN 119551 Nu _ w 03 40° - (4mm MM ($W40.) 'L 5'7: —_ 0,3500 Vlch TV“) —; - (Vegas.le + m w = — (““iwflwwé + «gamma l "1 rm; 2' .. 0543’“ MY; \ Q1: mac. M04 11"“: oxa'S-N’a/ / I \ M / / /, firm—410° ...
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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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hw23-10sol - H W 1‘ 13 ’ SDLu‘noNS 9—10. The state...

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