{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

ws 16 - For/6 JMQRQQ C9 MEEN 221 Summer 2002 Worksheet 5 Dr...

Info iconThis preview shows pages 1–16. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 8
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 10
Background image of page 11

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 12
Background image of page 13

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 14
Background image of page 15

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 16
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: For /6 JMQRQQ? C9 MEEN 221 Summer 2002 - Worksheet 5 Dr. A. Palazgolo (1) HWSS Due at Beginning of Next Class C115 (2, 11, 14, 29, 39, 89, 107) (2) Attendance Mandatory for 10:00 am- 12:45 pm , unless excused by instructor. Quizzes may be given at any time during this period. (3) Subjects: Centroids and Distributed Force Systems EXAM I ,, WEDS. a ,JUNE 6:00—8:00 PM ZACH 104B EXTRA TIMEAIAYBE GIVEN (A) . Centers of Mass/Gravity and Volume (a) Center of Mass - CM Suppose "that a rigid body is represented by a collection of particles With interaction and external forces. Each point on the rigid body may have a different acceleratiOn (i.e., 5,4 for some arbitrary point A). How Can Newton's Law 2F: M be written, i e which value of if should be utilized? To answer this sum Newton's Law for all particles representing the system the system _ ZMa =Zf. +.[ ~ (1) The first sum on the RHS is zero by the nature of interaction forces and Newton' 5 3rd Law The second sum on the RHS IS the resultant R of all external forces on the body ('0) Select the point G such" that ‘ ' M230 = R (3) Where M = Z M i = total mass of the body (4) Since the RHS of (l) and (3) are both K it must be true that their LHS's are equal. Thus MEG = 2 Mid. ' (5) Integrating twice in time yields M; =§Mia (6) or ‘- 70 firm:- +royi+erI€ <7) The summation in 1 may be expressed as an integral for a continuous body, with Mi:>dM:>pdV,andFi:>F (8) _ This yields the following result: Define the point G as being the center of mass of the body then the coordinates of G are 1 ' 1 1 rm, =—A7JXpdV, r0, =HJYNV, r02 =il—szdV (9) and it is true that MEG = R = resultant of all external forces on the body (10) Center of Gravity — CG The center of gravity is the location where the entire weight of the body could be concentrated and its moment about 0 would be the same as the resultant of the moments for weights of all the - differential masses that comprise the body. (C) F0. x (— mgé)= (raw? + rm] + r0212)" (— mgIE) = J? x (1W: I(Xf+ fi+Zl€)x (— pgdW) 3 rG.ij—ra.,mf= iXde/j— tide (11) :> r” =l prdV, r67 =i [YpdV (12) m V m V From (9) and (12) it is seen that the X and Y coordinates of the CG and CM are identical. The Z coordinate of the CG is arbitrary although it is usually taken as the Z coordinate of the CM. This result is consistent with the principle of transmissibility. Centroid of Volume CV . The volume centroid is the location of the mass center if the density of the body was replaced by a constant, i.e. if (13) Eqs. 9 gives the centroid coordinates as r“ ;—I];V[XdV, rm, =%V[YdV, rcz =%V[ZdV - , (14) 4 p In analogous manner the centroid of an area A in the XY plane ' has coordinates gfi-lfi Three bodies with masses of 3, 6, and 7 kg are loeated at points 'fl (4, -3, 1). (-1, 3, 2), end (2. 2, -4), respeetively. Locate the 3 mass center of the system if the distances are measured in meters. Solution .§+12* :Locate the centroid of the ’ shaded triangular area shown 5" in Fig. 95—12 if b = 200 mm and h = 300 mm. QC)\/C) i=0 _k—— ‘5f16.i Locate the centroid of the shaded area shown in Fig. P5-16. é!- 5-20 , Locate the centroid of the 'shaded area shown in Fig. P5-20. r Riley 5-2-8 6" Locate the mass centef of the .hemisphere shown in Fig. P5-28 if the density p at any point P is proportional to the distence ‘from the xy-plane to point P. [> restart; [> # Example Eh. Riley 5—28 .- Mass Center ef a Hemisphere ' >‘26 := int(k*pi/m*z‘2*(2*R*z-z*2),z=0..R); 3 knRs M - 10m A > m := int(k*pi*z*(2*R*z—z*2),z=0..R); 2f E mz=jiknR4 é: ; 12 ® {> 26; I 18 25 _> #ENGR221 Centroids > restart; _ > :> :> # ' > AY:=J.n ,y=0..h*xlb); A:=int(AY,x=O..b); H -2 ' b A —l-hb _ “2 ‘ > XC1:=int(x,y=0..h*x/b); Kc:= int(Xc1,x=0..b)/A; x2}: Xc1:=—— b X 'Zb _ C' 3 r > Ycl:=int(y,y=0..h*xlb); Yc:= int(!'c1,x=0..b) IA; 1th2 Yc1.=‘ 2 b 1 Yc:="h 3 lllllfll V V V V # AY:=J.n ,y=0 . .2.0*x-x‘2/25.0) ; A:=int (AY,x=O. .50.); AY := 2.0 x — .04000000000 1:2 A := 8333333333 > Kc1:=int(x,y=0..2.0*x-x*2/25.0); Kc:= int(Xc1,x=0. .50.)/A; XcI := x (2.0 x — .04ooooooooo x2) Xc := 25 .00000000 > Yc1:=int(y,y=0 . .2.0*x—x*2/25.0) ,- ‘Ic:= int(Yc1,x=O. .50.)/A; z YcI := .5000000000 (2.0 x — .04000000000 x2) Y6 := 1000000000 :> :> :> # ’> Ax:=1nt( ,x=0..sqrt(40.*y)); A:=int(Ax,y=25.0..90.0); AX := 6324555320 J; L A -.= 3072953723 > Xc1:=int(x,x=0. .sqrt(40.*y)); XC:= int(Xc1,y=25.0..90.0)/A; X6] := 2000000000 y [ Xc := 2432513039 [> Yc1:=int(y,x=0. .sqrt(40.*y)),- Yc:= int(Yc1,y=25.0. .90.0)/A; [ E (3/2) YcI := 6.324555320 y Yc -.= 6068894055 > > Cg/foe 010w OF compowre HREA‘J‘ fill I—a éO/e 1 1 =,_— fl: —— a! */d d - Xe fldg"; 4c é’fX4/4ax/4v‘éx/4flfxd/4f =—- M Xe +4 >42 +443-4Xcflf 2) 4c genera! C‘W‘E Id: 2/" (3’) K6 {-Z/Kc @J yd=/Z_:—Z/4€ yet. (3] .3 a Y‘n'u '_ ,. W-ahfpfjéw. -Sem1ciiciilar'éré‘a_ \ dr'a‘ntof a parabola 255; ~ - ‘ iii? 5-38 Locate the centroid of 4‘ k- r ' the shaded area shown T in Fig. P5—38. 1 f q> W '25 mm Fig. [5-38 “—45 Locate the centroid of the shaded area. shown in Fig. p5—45. (B) Distributed Loads on Beams g ‘ Mir, l IIIIII-lllll H Agimgmm ‘ ' ‘ ' ' ' .. Let P = pressure acting on the beam then WOO = line density ofload = P(X)*t = force per length The two load systems shown above are equivalent, in the sense that they will have an equivalent effect on the other loads in the structure if R = L]'W(x)dx = total load on beam 0 :and d R My =fodF=foW(x)dx=Rd z fl’ 4; 23 0 ‘|L :> d=— W dx p mix (30 4 + 4. ‘1 4? 4 4’: J’ u/dx .-.- J wdxfz‘ndx +fwa€x '1 o 0 4’1va X, I? X2. ,(7 3 A7 x = x=x¢ x-IX ’ " 3 I 2 ’ 3:4, k‘“ L z, ‘2 L: “2 A9:- (fo’wdag-fofmexz +01; wdxg —- 4%M3 L 47“; L :z- [éflxi’dx hfxwdx +waqf},X,=X =9 afx=a€><l 4! 4,441 X4: X-il =b 4X= 06¢? 43 —_ ifdyg + f(4,+XAJW¢QXZ +0! (41.4.; ,Lx3)wd>(3§ = %[‘f*'€z % (way/a, +64%; “we; r» F—— L —el :6 ,er. gag-6)“ 5-74 Determine the resultant i of the system of distributed loads and locate its line of action with respect to the left support for the beam shown in Fig. P5-74. Fig. P5-74 Forces on Submer ed Surfaces Consider the column of liquid shown below. W} F2, 0 T / (P w» d I 5? .19 ‘_J;‘ 6;?111drfjr 4TH ,0 The pressure P must balance the weight of the column plus PM by Newton's Law. So ‘ PA=P0A+Apgd :> P—Po =pgd Thus pressure increases linearly with depth. 5-81* If the dam shown in Fig. P5-81 is 200 ft wide, determine the magnitude of the resultant force fl exerted on the dag by the water (1 = 62.4 lb/ft ) ‘ pressure. 5-87* The width of the rectangular g!“ gate shown in Fig. P5—87 is 4 7 ft. Determine the magnitude of the resultant force fi exerted on the gatg by the water (7: 62. 4 lb/ft ) pressure and the location of the center of pressure with respect to the hinge at the bottom of the gate. Fig. P5-87 :’[> #Example Pb. 5 Riley (5—875 ~-[> dt := 3. 8: L := 6.: rhog :=62. 4 : b:= 4.: ) [> F :=rhog*b*int(dt+. 8666*x, x=0. .L); RI F:= 9584. 340480 - D:= rhog*b/F*int( (L—x)*(dt+. 8666*x), x=0. .6) use the derivative symbol I Error, attempting to assign to #note error. Do amt D which is protected [2» mm: rhog*b/F*int( (L-x)*(<:1t+. 8666*x) ,x= .-6) , mfl/ DD := 25937685559 _' ...
View Full Document

{[ snackBarMessage ]}