hwsoln19

# hwsoln19 - 10.1. [10.2] For the beam shown in Fig. PJDJ,...

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Unformatted text preview: 10.1. [10.2] For the beam shown in Fig. PJDJ, what is the shear force and bending moment at the following positions? (a). 5 ft from the left end (b). 12 ft from the left and (c). 5 ft from the right end Write the shear force and bending moment as a function of x for the beam. -‘ Iooolb ‘ 50k 3 a salty ln—w' \5‘—.L— \5L'——I at x = 5 ft. 1 0. , “0 737.5 —V - 0 SM 7‘ V = 737.5 lb. (I -(737.5)(5) + M = 0 737-5 M = 3633ft~1b. a = 12 t ><| 737.5 — 1000 -v = 0 ‘ v =~262.5 'Ib. -(737.5)(12) + (1000)(2) + M= o M = 6850 ft-lb. l Z7515 L “1'4 at x=35 737.5 - 1000 -‘V = O u. \l =—262.5 'lb. ~(737.5)(35) + (1000)(25)- 500 + M = D M = 1312.5 ‘Ft-1b. r '1 moon. 737.5 lk 10.6. [10.2] BC Is cantilevared at C and pinned to AB at B (see Fig. P.10.6). Formulate the shear-force and bending- moment equations. C .7: Figure P.10.6. 1‘ «co “IML 6 A “k- VIM F- L B. £3 £5 AG EMR = 0 -RA(12) + (100)(12)(6) = 0 RA = 600 N 1 wow“ Arwb _ f) »—~ 1:. too“ V 0 < x < 22 500 - 100x -\I = 0 V =-100x * 600 N -600(x) + (100)(x2/2) + M = 0 M = 600x - 50x2 N-m .nr-um ooﬁ + Am-xv¢.qmﬁ + ﬁdnwi - ngH. u 6:. uém n : mmVxVMH 900. H *. -nr-p+ Hm-xv¢.¢uﬂ + ﬁq-xvom - Kama- u z a u z + Hm-xve.quH - AwuxVod + xooa .2. «Jam u 9 a u p twine." + 9“ - 00H; MHvam o u: + 3.51qu - mm. m + 32 2.. xml :43 3 a n :Ir :43.“ + Km i can. mvam aszsn sen. _ .n—Iuv nxmd 1 non.“- u E a u z + um m + Iona H .groaﬁi x?» > Dublxmuooﬂu mVan. r. H .n— +.\$H n m o u oaﬁ - Amﬁvﬁm - AmHUHmHHmH + ”MNVHOQHM o n man A: 7% u um o u anHmm - uaﬁ - “HHﬁmuﬁmu + Amwaooﬂ_. Ima&n1 a u as” a . . . . ﬁgmﬁh .. any Enos mnﬂnmﬁuvo toga bnﬁﬁ 2: .52 3259.3. .EEEEEB 3a ﬁaﬁﬁ ﬁance“ 5...: .2: ...
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## hwsoln19 - 10.1. [10.2] For the beam shown in Fig. PJDJ,...

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