HW 14 - -.. 4-“;-...

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Unformatted text preview: -.. 4-“;- H-_—;._w.it-u;_._.'.....';:;;.,..,.---. .. .'...'.. _ u g w.-uui-a.—-..n ......n FIST—iii;- gfi; i; 'w '_-;_-‘ M'A-l—r-l - PROBLEM 7-3.2 Determine V and M as functions of x (a) by drawing freebody diagrams and using the equilibrium equations; (b) by using Eqs. (7-4) and (7-5). HW A SOLUTI N: Draw a FBD for an arbitrary length of the left-hand L , l portion of the beam. y W0 (1— X / L) |- H .— -_ ‘- _-- 5" --_ I - i: TV 2M i——x—~i The function which describes the distributed load is: w = wo — ?x = wo[1 e Sum vertical forces to determine the shear force. 2 ANS V = —w0[x — x—J 2L Summing moments about the point of the cut through the beam to W=°=-wot1~~it%i-iétwo*wotsfliiitéitr PROBLEM 7-3.3 Determine Vand M as functions of x by using Eqs. (7-4) and (7-5). SOLUTION: Using equation (7.4) to determine the shear funciiOn: I I ' I 3 x V =—J'wdx= —I3[l—x—]dx=l— yeti] 25 5 0 t] D x] ANS V = —3x + — 25 Using equation (7.5) to determine the bending moment function: .1.‘ I 3 M=Ide= —3x+x—dx 25 0 0 4 ANS M =—--3-x2 +"_ 2 too 389 determine M: w = 3 (1 —x2r25) kNim EROBLEE 73;,fi Use Eqs. (7-4) and (7-5) to determine the internal forces and moment as functions of x for the beam in Problem ?-2.15. Free Body Diagram: F D Y 600th SOLUTION: The angle made by the cable with the horizontal is: 9 = mafia] = 14.o4° 8ft Summing moments about point A: EMA = = - (600 lb) (10 ft) + T (sin 14.04”) (8 ft) + T (cos 14.04") (2 fl) T = 1546 lb Summing horizontal forces on the bar: EFX = 0 = A, — T (cos 14.04”) = Ax — (1546 lb) (cos 14.04°) A, = 1500 lb —> Summing vertical forces on the bar: ER,r = 0 = - 600 lb + (1546 lb) (sin 1404") + A, A5, = 225 lb For0<x<8ft1 w=0 V=—J'wax=—_[0dx=0+q @x=0,V=225lb,so: 0 0 ANS Vix1= 22§ [D X X M=Jde=J225dx=225x+C2 @x=0,M=O,so: 0 0 ANS MIX} = 225 x ForBft-cx-cioft: w=0 V = —]wdx = wiffldx= —(o + C3) 0 o ANS V(x1= 6901b M=Ide=1600dx=600x+C4 @x=10ft.M=0,so: 0 0 ANS M x = 600 x- 6000 ft-Ib 393 @x=10ft.V=-600|b.so: C1 = 225 lb C; = 600 lb C4 = - 6000 ft—lb ; PR .14 E'- -"- Use Eqs. (7-4) and (7-5) to determine the internal forces and moment as functions of x for the beam in Problem 7-2.19. Free Body Diagram: 4 k N E l-(N i. we F—em~+—3m——|T—| W: m Summing vertical forces to determine the magnitude of the distributed load: ZF,=0=-4000 N-2000N +w0(6m) we = 1000 N/rn 1': For0<x<2m: ' w=-1000N!m V=_dex=—J'—1ooodx=1000x+c. @x=0,V=0,so: 01:0 0 0 ANS V(x)=1000x M = [no : [(100000 = 500;:2 + c2 @ x = 0. M = 0, so: 02 = 0 D D ANS M(x}= 000 x2 For2m<x<5m: w=-00me V=—Jw¢ir=—I{—1000)dx=1000x+C3 @x=2m.V=2000.so: Ca=-4000N 0 0 ANS Vix1= HOBOX — 4Q00l N M = Ide = J'uooox F 4000)dx = 500x2 — 4000:: + C4 @ x = 2. M = 2000 him, so: ct = - 3000 N-m 0 0 For5m<x<6mz = -1000 Nfrn ._ V=—J'wdx~ [01.00000=1000x+c5 @x=6m.V=0.so: C5=-6000N 0 U ' ANS mm = (1000 x — 6000) N M = [m = [(10001 — 6000)dx = 500x2 — 0000; + C6 @ x = 6 m, M = 0. so: 06 = 10,000 N-m 0 D ANS M): = 500x2—6000x+18000 N-m .1 399 ...
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This note was uploaded on 11/18/2010 for the course ENGR 213 taught by Professor Keil during the Spring '06 term at Oregon State.

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HW 14 - -.. 4-“;-...

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