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# HW 13 - 3,113 = M 7-1.10 Determine the internal forces and...

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Unformatted text preview: 3,113 =. M 7-1.10 Determine the internal forces and moment at A for each loading. §QLQTIQN: r—— 4 m (a) Draw the F30 and determine the reactions at the left and (b) right ends of the beam. SRN kNImJ l f r L, A ln—-J I : m—Jv 4m r—l I L, Ry EMR = 0 = (3,000 N) (2 m) — L, (4 m) —> L, = 4,000 N T 2F, = 0 = L, — 8.000 N + R, = 4.000 N — 8,000N + R=y —> R, = 4.000 N T 2F, = 0 = L,( —> L, = 0 Cut the beam through point A and draw the FBD: V A p MA 1, A (F Elm—0 LY EF,=0=L,-—VA=4.000N-VA -£F,=0=PA ANS \_/A=4000N~L EA.=_Q EMA = 0 = - L,('1m)+ MA =- (4.000 N)('1m)+ MA ANS MA = 4,900 N-m (1:) Draw the F80 and determine the reactions at the left and right ends of the beam. — FERN/m Lx I I I I m 1m-—-{p‘ 4m L, Ry ZMR = 0 = (8.000 Nim) (4 m) (2 m) -— L, (4 m) —> L, = 4.000 N T 2F, = 0 = L,— 8.000 N + R, =4,000 N — 8.000N + R=y —> R, = 4.000 N T IF, = 0 = L, —> L, = Cut the beam through point A and draw the FBD: EkN/m p. P Mn l n 101—4 L y EMA = 0 = ~ (4,000 N} (1 m) +[(2,000 Nlm) (1 m)] (0.5m) + MA ANS MA = 3,000 N—m 2F, = 0 = L,— (2.000 Him) (1 m) - VA = 4.000 N - 2.000 N - VA 2F, = 0 = PA ANS \_/A=2,000 NJ. EAEQ 358 PROBLEM7-2.1§ (9991,) E O 0 N IOUN BOON BUUN v' . WN- -400N-m if 9 “lEUONﬁm ' —1600N-m a h" (a : " .L.>! 3-_ The beam in Problem 7-2.13 will safely support shear j forces and bending moments of magnitudes 2 kN and -; _:- ' 6 kN-rn. respectively. Based on this criterion. can it ' safely be subjected to the loads F = 1 kN, C = 1.6 kN-m? W: ' i' 3; Draw the FBD for the beam and determine the reactions : at points A and B. Drawing the shear force and IR N bending moment diagrams: t» 4 m -'+'- 4 m -|-— 8 m '-"E'1-\-r.":'.: .—_- 1:" lkN EMA: 0 = - (1000 N) (16 m)+1600 N-m + e, (0 m) ey=1300NT EF,=0=—1DODN+BY+Ay=-1DODN+1BDON+Ay A,=300Ni . We see from the sear force and bending moment diagrams that, while the shear force is within the given limit, the magnitude of the maximum bending moment exceed the given limit. . ANS Beam gangs}; safely sggggrt the lgagg. _. 379 P 7- . The bar 30 is rigidly ﬁxed to the beam ABC at 8. Draw the shear force and bending moment diagrams for the beam ABC. sgLUTIQu: ' some The angle made by the cable with the horizontal is: (9: tan" a =14.04° 3ft Summing moments about point A: EMA = o = - (600 lb) (10 ft) + T (sin 14.04“) (5 ft) + T (cos 14.04”) (2 n) T = 1546 lb Summing horizontal forces on the bar: 2F, = o = A,‘ — T (cos 14.04“) = Ax - (1546 lb) (cos 14.04") .: Ag=1500 lb.» 1800131410 Summing vertical forces on the bar: 22Fy = o -- - 600 lb + (1546 lb) (sin 14.04°) + A, AV 2 225 lb -180 OF ‘t -llo mm Draw the shear force and bending moment diagrams for beam ABC. i—Em—ﬂhem—ﬂ—Bm—ﬁ Cy 2m M: We consider only the F80 of bar ABC. The effect of the BID-section 1333N “(N of the beam is reduced to an equivalent force and moment at point B. (3,)50 = 4 RN 4 (MB)ED = - (4 kN} (2 m) = - 8 kN-m -1333M Summing moments on bar ABC about point A: V EMA = 0 = - (4.000 N) (2 n1) - (8.000 N-m) + C), (6 m) -EE>E> 7N c, = 2667 N T 10,667N—m Summing vertical forces on bar ABC: 2F,=0=A,+C,- 4.000N=A,+2667N—4,000N 2667N-m AY=1333NT M ...
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HW 13 - 3,113 = M 7-1.10 Determine the internal forces and...

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