tutorial(30th March- 2nd April)

tutorial(30th March- 2nd April) - Emblem 5.4-4 A cantilever...

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Unformatted text preview: Emblem 5.4-4 A cantilever beam AB is loaded by a couple M0 at its free end (see figure). The length of the beam is L = 1.5 m and the longitudinal normal strain at the top surface is 0.001. The distance from the top surface of the beam to the neutral surface is 75 mm. Calculate the radius of curvature p, the curvature K, and the vertical deflection 8 at the end of the beam. Snlutlnn 5.4-4 Cantilever beam tnhlem 5.5-1ll A railroad tie (or sleeper) is subjected to two rail loads, each of magnitude P = 175 ltN, acting as shown in the figure. The reaction q of the ballast is assumed to be uniformly distributed over the length of the tie. which has cross-sectional dimensions b=300mmandh=250mm Calculate the maximum bending stress am in the tie due to the loads P, assuming the distance L = 1500 mm and the overhang length a = 500 mm. l_.;4 M. ASSume that the deflection curve is nearly flat. Then the distance BC is the same as the length L of the beam. , _£_l.5n‘t= .smB—p —?5m 0.02 8 = arcsin 0.02 = 0.02rad 5 = p(l - cos 9) = (75 m}(l — cos (0.02 rad» =l5.0mm 4- L NOTE: 5 = 100. which confirms that the deflection curve is nearly flat. 3%};1. #aa' Fly—1 -1“: Sululinn 5.5.10 Railroad tie (or sleeper) DATA P=17SkN L=1500mm _ 2P bhl q L+2a s BENDING-MOMENT DIAGRAM Ml 0 A my “2 z 2 P Mean.“ 2 L+2a qL )1 PL l1,1:— —+ _—_. 2 2(2 " 2 — P (5 «f—fl +21 2 2 b=300mm a=500mm s = — = 3.125 x10" m3 h: 250mm Substitute numerical values: M1 = 17.500 N -m Mm M2 = —21,375 N - m = 21.875N-tn MAmIM BENDING STRESS a,“ M 21,875N'm =—m‘““=—"—m~—“—=7.UW 4— s 3.125x twin3 3 (Tension on top: compression on bottom] Problem 5.5-1'9 A beam ABC with an overhang from B to C supports a uniform load of 160 lhfft throughout its length (see figure). The beam is a channel section with dimensions as shown in the figure. The moment of inertia about the z axis (the neuual axis) equals 5.14 111.4 Calculate the maximum tensile su'ess a", and maximum compressive stress a} due to the uniform load. Solution 5.5-19 Beam with an overhang AT CROSS SECTION OF MAXIMUM Pom-NE M] BENDING MOMENT M 13,500 lb: . 2.496 ' . _ 0 _ a: = 1‘72 ___ W = 6560 p51 1._,| ‘7’ 1a 5.14 m. 335 n- M l3 5001b: . 0.674 ' . “1 m= [‘C'=——-—( ’ 53.”. mar/70951 1: = 5.14 in.‘ = ' "" Cl = 0‘6“ “1' 02 = 2'496 111' AT CROSS sac-non 0F MAXIMUM mam RA. = 600 lb R8 = 1300 "3 SENDING MOMENT M = 11251b-fi=135001b-in. . . 1 ' M 24, ll:- . 0.674 . , M2 = 200010-11 = 24,000 10-111. 0', = 10' = = 3,150 1331 I, 5.14 tn. M262 (24.000 lb-in.)(2.496 in.) , 0‘ 11 5.14 'm.‘ pm MAXIMUM 511153555 0" = 6,560 psi 0} = 11,650 psi 4— Pffllllfll'l‘l 5.5-3 A cantilever beam of length L = 6 ft supports a uniform load of intensity q = 200 lblft and a concean load P = 2500 lb (see figure). Calculate the required Section modulus 5 if awn“, = 15,000 psi. Then Select a suitable wide-flange beam (W shape) from Table 5-1, Appendix E. and recalculate 5' taking into account the weight of the beam. Select a new beam size if necessary. Solution 5.6-3 Cantilever beam P=25001b q=2001blft L=6ft TRIALSECT‘ION W8><21 Cam 2 15'0"” Psi 5 = 13.2 in.3 qo = 21 1er 2 REQUIRED SECTION MODULUS Mo = ‘30 L = 373 "3-1-1 = 45351b-1n_ (#3 2 M = + — = . ~ m“ PL 2 15‘000 lb h + 3'6“) !b R Mm = 223.200 + 4.536 = 227.7001b-in. =18,600 lb—fl = 223,2001b-in. R ’ d S _ Mm _ 227,700 “rm _ 15 2 I j s = Mm: = 223v2m1h'in-=14 88 in 3 alum “um... 15.000 psi _ I m- Vflhw '5'090 PSi 15.2 1.1.3 < 13.2 in.3 Beam is satisfactory. Use WSXZI ‘— ...
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This note was uploaded on 03/15/2012 for the course CIVL 000 taught by Professor Kk during the Spring '10 term at HKUST.

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