Exam 2 Review Oct 13, 2014

Dr. Koen – 3D Rigid Body, Trusses

5 - 76 The boom AC is supported at A by a ball-and-socket joint and by two cables BDC and CE. Cable BDC is continuous and passes over a pulley at D. Calculate the tension in the cables and the x,y,z components of reaction at A if a crate has a weight of 80 lb.

Thought process…

3d Rigid Body Equilibrium Problem

Draw FBD to identify unknowns

Ball & Socket at A

Two cables (always pull)

Assume all unknowns as positive

Decide on equations that can be used

Express each cable as a Cartesian Vector

Apply equations

Ax Ay

Az

CD

BD

CE

CD = CD { -.231i - .923j + .308k} r CD = < -3, -12, 4 > CE = CE { .218i - .873j + .436k} r CE = < 3, -12, 6 > BD = BD { -.468i - .625j + .625k} r BD = < -3, -4, 4 > ƩFx = 0; Ax - .231CD + .218CE - .468BD = 0 ƩFy = 0; Ay - .923CD - .873CE - .625BD = 0 ƩFz = 0; Az + .308CD + .436CE+ .625 BD – 80 = 0 ƩMx = 0; .625BD(4) – 80(12) + .436CE(12) + .308CD(12) = 0 ƩMz = 0; -.218(CE12) + .231(CD)12) +.468BD(4)=0 Also… BD = CD Using Solver…. Ax = 19.4 lb Ay = 192 lb Az = -25.8 lb CD = BD = 62lb CE = 110 lb

5 – 80 The bent rod is supported at A, B, and C by smooth journal bearings. Determine the magnitude of F2 which will cause the reaction Cy at the bearing C to equal to zero. The bearings are in proper alignment and exert only force reactions on the rod. Set F1 = 300 lb

Thought process…

3d Rigid Body Equilibrium Problem

Draw FBD to identify unknowns

Proper Alignment means no couple moment reactions at supports… just forces

Assume all unknowns as positive

6 unknowns, so need all 6 equations

Express F1 and F2 as Cartesian Vectors

Apply all six equations

Ax

Bx

Ay

Cy=0 Bz

Cz

F1 = {0i – 300 cos45j - 300sin45k} = {0 i – 212j – 212k} lb F2 = {F2cos45sin30i + F2cos45cos30j - F2sin45k} = {.354(F2) i + .612F2j – .707(F2)k} lb ƩFx = 0; Ax + Bx + .354F2 = 0 ƩFy = 0; Ay – 212 + .612F2 = 0 ƩFz = 0; Bz + Cz – 212 - .707F2 = 0 ƩMx = 0; -Ay(4) – Bz(3) + 212(5) + 212(5) = 0 ƩMy = 0; Ax(4) + Cz(5)= 0 ƩMz = 0; Ax(5) + Bx(3)= 0 Using Solver…. Ax = 357 lb Ay = -200 lb Bx = -595 lb Bz = 974 lb Cz = -286 lb F2= 674 lb

Prob 6 - 9 Determine the force in each member of the truss and state if the members are in tension or compression. Hint: The vertical component of force at C must equal zero.

6 kN

AE

AB 3

4

DB AB

3

4

EB

BC

Thought process…

Truss by Joints problem

Find starting point – without supports?

Draw FBD, can assume tension for all, but THINK!

Decide on path.

Find by inspection.

Joint A

Joint B

6 kN

AE

AB 3

4

Joint A ƩFy = 0; 4/5 AB - 6 = 0 AB = 7.5 kN (T) ƩFx = 0; 3/5 (7.5) - AE = 0 AE = 4.5 kN (C)

DB AB

3

4

EB

BC

Joint B

ƩFy = 0; -4/5 (7.5) – 8 + DBsin45 = 0 DB = 19.8 kN (C) ƩFx = 0; -3/5 (7.5) + BC – DB cos 45 = 0 BC = 18.5 kN (T)

AE

EB = 8kN (T)

ED = 4.5 kN(C)

Joint E

8 kN By inspection….

Prob 6 – 42

Determine the force in members LK, LC, and BC of the truss, and state if the members are in tension or compression.

Thought process…

Truss by Section

Decide where to cut, which side to use

Find support reactions

Draw FBD, assume tension for all

Look for points to sum moments about

LK

BC

LC

7500 lb

ƩMA = 0; 3000(4) + (3/5)LC(8)= 0

LC = -2500 lb….. 2500 lb (C)

ƩMC = 0; 7500(8) + (4/5) LK(6) – 3000(4) = 0

LK = -10,000 lb…. = 10,000 (C)

ƩML = 0; 7500(4) – BC(3)= 0

BC = 10,000 lb (T)

Clockwise moments called positive…..

Ms. Campbell

2D Rigid Body Equilibrium

Frames & Machines