HW 15 - se Eqs(7-4 and(7-5 to solve Problem 7-2 24 sLUTION...

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Unformatted text preview: se Eqs. (7-4) and (7- -5) to solve Problem 7-2. 24. sLUTION: is raw the F30 for the beam and determine the reactions at JomtsAandB. BURN-m 4kN/m EMA = 0 = (6,000N)(6m) + 20,0003: ~ In — {4,000N)(6m)(3m) + By(6m) — [-12-(4,000N {m)(6m):1(8m) 3,. =18.66? N ‘1‘ my = 0 = «5.00010 — (4,0000f xmxam) _ Empoom mam] + Ay + 18,6673! A, = 23.333 N T .- V=—J'wdx=jodx=0+c. @x=0.V=-6000N.so: C1=-6000N *VEANS V(x) =0- BOOODN I M=jm= I— 500033: —6000x + c2 @ x = 0. M = - 20.000 N-m. so: (32 = - 20.000 N-m :th M x = 00 x— 20 00 N-m _"For 6 m < x < 12 rn: ' w— = 4000 me V:—]wdx= 4400020: 4000x+C3 @x= 6V-17333N so 03:41.333N ;ANS Vxl 1=f4Q§19x0+ 41 .§§3 1N ' M: 11%;:- ](—4000x + 41333)dx= —2000x + 41, 333): + C4 @ x = 6m. M = 50.000 N-m. so: = c. = 232.000 N-m :_.ANs M = -20 x2+41 3x—23 00 N- .-For12 rn 4 x < 18 m: w = (666.? Nim) (18 — x) m = (12.000 - 666.7 x) N X .‘I.’ = —dex = — [(12,000 — 666.?x)dx = 42,000): + 333.33:2 +18,667 + C5 @ x = 12 rn. V = - 666.7 N. 50: 5' c5: 39. 301 N . ANS mm = (— 12.000 x + 333.3 x2 + 39.3311 N I 1 M = [de = [(333332 —12,000x + 89.38005: =111.1x3 — 6.000;:2 +108,000x + C6 @ x = 1a m. M = 0. so: [I 0 . ANS Mix} = (111.1 x3 — 6,000 x2 +108,UQQ x + 64§,000l N-m 401 Cs = 648.000 P BLEM - .1 The beam consists of material with modulus of elasticity E = 70 GPa and is subjected to couples M = 250 kN-m - ' - '- - - _ at its ends. (a) What is the resulting radius of curvature l—w—-———l.6m—‘—""i «i Le- ' of the neutral axis? (b) Determine the maximum tensile C 0'51“) " stress due to bending. m“ " §QLQTIONz The moment of inertia for the cross-section is: (a) Using Equation (8.10) to determine the magnitude of the radius of curvature; j. ':_ i=_“i___w__=3.174x10'3m" p a: _ (70x109Nim2)(4.369x10‘4m4) ANS Q .= 1.223 m (b)_The maximum normal stress due to the bending moment is: W. Myw _ (250,000N — m)(o.16m,) 02m 1 4.369x1o"m‘ ANS a...“ = 31.5 MPa W The material of the beam in Problem 8-1.1 will safely support a tensile stress of 180 MPa and a compressive stress of 200 MP3. Based on these criteria, what is the largest couple M to which the beam can be subjected? TIO : The symmetry of the cross~section tells us that: (UthNsiui = (UmkontF-Resswa- The moment of inertia for the cross-section is: 1— g = (0.16m)(0.32m)3 _ = at.369x10"‘m4 12 12 The maximum normal stress due to the bending moment is: _ 0'qu = MyW _) . M = am! = (180x1oimm2)(4.359x10" m“) ‘ ' I yMAX 0.16m - ANS M = 492 kN-m " 403 W ._ The beam is subjected to a uniformly distributed load w = 300 tbiin. Determine the maximum tensile stress due to bending at x = 20 in. if the beam has the cross section (a); if it has the cross (b). (The two cross-sections have approximately the same area.) Lain. T ' I _ -t.4'.l'in. E‘“ Frgg Body Dlggrgm 3001b m D I _ A i— ——————————————— ‘l rmi 'r—J- l l “rm 6m [at {h} 85in “it By SOLUTION: The moments of inertia for the cross-sections in the two cases are: (4.47m)(4.47m)3 , 4 (6m)(6in)3 (4rn)(4m)3 _ t I !fl=-—-——-——-=33.27 I =———=86.67 ____ ' . 12 m b 12 12 m |_ BOUlb/m .. l Summing moments about the cut through the beam at x = 20 in: T 3M M = (12,750 lb) (20 in) + (6,000 lb) (10 in) = 195.000 in—lb BUin—e-{V We see that the maximum tensile occurs at the bottom of each cross-section. {it}, 195.0001» «it: 4-47"! In case (a), the maximum tensile stress is: (afimx = -(—)£A) 33.2713: ANS (film = 1 . k i at the bottom fth In case (b). the maximum tensile stress is: (aflmr = W 86.671114 ANS {film = 6.75 ksi tth fth ross-section 410 ...
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