202hw21_soln - '1’ P PROBLEM 7.30 i, l , A I. '...

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Unformatted text preview: '1’ P PROBLEM 7.30 i, l , A I. ' ‘h'.’_*""'_' "‘7 9' .;T,| I; For the beam and loading shown, (a) draw the shear and bending-moment —; )l diagrams, (b) determine the maximum absolute values of the shear and ‘4—(1 «- u —>\ bending moment. I,-—————> SOLUTION FBD Beam: —~ )ZFt = 0: x = 0, by symmetry Ay = By = P Tr)“ l2Fy=0: P—V=O x A. P K V — P ’“ Qw,=0: M—xP=0 M = Px p l213,=0; P—P—V=0 M £1 K V = 0 ,4 5 v {EMA/=0: M—xP+(x~a)P=0 A M = Pa Along CD: l2}; =0: P—P—P—V=0 E E H V = —P at L-u. 5" x L (XML=O: M—xP+(x—a)P f X, +(x ~ L + a)P = 0 M = P(L — x) Note: Symmetry in M diag. follows symmetry of FBD (17) mm“ = P along AB and CD { lMme = Pa along BC4 PROBLEM 7.31 For the beam and loading shown, (a) draw the shear and bending-moment I B diagrams, ([7) determine the maximum absolute values of the shear and bending moment. SOLUTION h); FBD Section: A ’10 Z I a Q, 24,1: T213}: —V—%x[w0%)=0 a, M I | J— ) (a) I I x ’3 V = ~lfixz v «\\ V 2 L 1 \ *m m V(L) —le a 0 | 2 M \ ' \\ _ ' l _ l x _ \ JULL CEMJ—O. M+§x|:§x[w02]]— _ _ 1mg 6 L M(L) = — éwoLz (b) 1 = E M’OL at B ‘ 1 2 t B 4 | ‘mflx — g WOL a PROBLEM 7.37 _ _';. _ I ' For the beam and loading shown, (a) draw the shear and bending—moment C1 .1. ‘3‘" diagrams, (b) determine the maximum absolute values of the shear and r 1.4 111.1 A: L 1 Lx bending moment. 1.6 111—>+<*l 111 SOLUTION (a) FBD Beam: Q EMA = 0: —(1.3 m)[(1.8 kN/m)(2.6 m)] — (1.6 m)(4 kN) + (4 m)B = 0 AIM/m 13:3.121 kNT 4,, — (1.8 kN/m)(2.6 m) — 4 kN + 3.121 kN = 0 Ay = 5.559 kN I 575317 Along AC: A fAM/M 1.677 I 1 V II! I) 5. (w) 132/ ‘ x ' 17M V .— '3,lU lZF =0; 5.5591d\I—(1.8kN/m)x—V=0 y 45,57 V = 5.559 kN — (1.8 kN/m)x 51.37 (' 2M, = 0: M + %[(1.8 km)x] — x(5.559 kN) = 0 \ (MM; “:0” M = (5.559 kN)x — (0.9 kN/m)x2 Along CD: "MW" (I.le 3 £1- ~-—~—--——— —— K /16M l ‘1/ 5755‘? M yxw A, 94—» A 2F. =0: 5.559kN—x(1.8kN/m)—4kN—V=0 )’ V = (1.559 kN) * (1.8 kN/m)x Q EMK = 0; M + (x — 1.6 m)(4 kN) + %[(1.8 kN/m)x] — x(5.559 kN) = 0 M = 6.4 kN-m+ (1,559 kN)x — (0.9 kN/m)x2 __l Along DB: x; 3.121llN Tsz=0: V+3.121kN=0 V=—3.121kN {XML = 0: —M + xl(3.121kN) = 0 M = (3.121 kN)xl b ( ) [Vimax = 5.56 kN atA 4 IMlm = 6.59 kN-m at C4 PROBLEM 7.72 For the beam and loading shown, (a) draw the shear and bending—moment diagrams, (b) determine the location and magnitude of the maximum bending moment. 1_ _l SOLUTION (a) ( EMA = 0: (3 m) By — (2.1 m)(2.5 kN/m)(4.2 m) = 0 B}, = 7.35 kN l at SF}, = 0: A}, — (2.5 kN/m)(4.2 m) + 7.35 kN = o A}, = 3.15 kN I Shear Diag: V has slope gig = —2.5 kN/m throughout, and jumps at A and B equal to x the forces there. V = 3.15 kN — (2.5 kN/m)(3 m) = —4.35 V =~4.35kN+7.35kN=3kN VC = 3 kN — (2.5 kN/m)(l.2 m) = 0 Note, V = 0 where 3.15 kN — (2.5 kN/m)x = 0, x =1.26m Moment Diag: Mfr dM AtA, M = 0 and 71x— : 3.15 RM The slope decreases to zero at {Ia/Z“) x = 1.26 m and to ~4.35 kN at B, jumps to 3.0 kN and decreases to 0 at C. '//f 1 MD = E(3.15 kN)(1.26 m) = 1.9845 kN-m MB = 1.9845 kN-m — %(4.35 kN)(3 m —1.26 m) = —1.80 kN»m MC = —1.80 kN-m + %(3 kN)(l.2 m) = 0 (1)) From the diagrams, lVlmax = 4.35 kN atB (Mlm = 1.985 kN atD (1.26 m from A) 4 ...
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202hw21_soln - '1’ P PROBLEM 7.30 i, l , A I. '...

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