STATIC FORCE ANALYSIS WEEK 1 COUPLE MOMENT OF

STATIC FORCE ANALYSIS WEEK 1

COUPLE MOMENT OF COUPLE Couple: Two equal and opposite forces (F & F’) Moment of Couple: A vector normal to the plane of the couple (M=RBAX F)

Conditions of Equilibrium A system of bodies is in equilibrium, if, and only if: In 2 -D (planar) systems:

Two-Force Members Not in equilibrium F 0, M 0 Not in equilibrium F=0, M 0 In equilibrium F=0, M=0 Condition of equilibrium, for any two-force member with no applied torque: The forces are equal, opposite and have the same line of action.

Three-Force Members O Not in equilibrium M 0 In equilibrium F=0, M=0 Condition of equilibrium, for any three-force member with no applied torque: Forces should be coplanar Four-Force Members: The problem is reduced to one of three-force member. Then the approach above is applied.

Example (Graphical Solution) a) Link 3 is a two-force member C

1 N=0. 225 lb b) Link 4 is a three-force member c )1 st approach: OR c’ ) 2 nd approach: Since link 4 is is a three-force member; lines of action of forces P, F 34 F 14 should intersect at a point. Therefore direction of F 14 is obtained. d) Force triangle is used. F If 1 st approach was used, this triangle gives direction and magnitude of 14 If 2 nd approach was used, this triangle gives magnitudes of F 34 Since, graphically e) Link 3// Action and reaction forces are equal: f) Link 2// F=0 , F 14 and

(Analytical Solution) 4// = 5(cos 68. 4 i+sin 68. 4 j) X 120(-cos 40 i-sin 40 j) +12(cos 68. 4 i+sin 68. 4 j) X F 34 (cos 22. 4 i+sin 22. 4 j) = 0 F 34 = 33. 1 lb 2//

Problem (Analytical Solution) (without friction) 2 O 2 A=75 mm P=0. 9 k. N AB=350 mm M 12 =? 3 4 4// F=0 F 34+P+F 14=0 F 34(cos 11, 95 i-sin 11, 95 j)-900 i+F 14 j=0 i: 0. 978 F 34 = F 34 x= 900 F 34 = 920. 25 N j: 0. 207 F 34 = F 34 y= F 14 = 190. 5 N For the moment balance, all of the force vectors should pass through point B. F 34 = 920, 25 /-11, 95 N = 900 i-190. 5 j N F 14 = 190. 5 j N

3// A two-force member F=0 F 23=-F 43=F 34 2// F=0 F 12=-F 32=F 23=F 34 MO 2=0 M 12 + O 2 A x F 32 = 0 M 12 k + 0. 075(cos 105 i+sin 105 j) x (-900 i+190. 5 j) = 0 M 12 k – 3. 69 k + 65. 2 k = 0 M 12 = -61. 51 N. m M 12 = -61. 51 k N. m

Problem (Analytical Solution) (with friction) 2 =0. 2 (Between piston and cylinder) 3 4 4// 0. 2 N 2// F=0 F 34+P+F 14=0 F 34+P+(0. 2 Ni+Nj)=0 F 34(cos 11. 95 i-sin 11. 95 j)-900 i+0. 2 Ni+Nj=0 i: 0. 978 F 34 -900+0. 2 N = 0 j: 0. 207 F 34 + N = 0. 207 F 34 0. 978 F 34 -900+0. 2(0. 207 F 34)= 0 F 34 = 882. 61 Newton N=182. 7 Newton F 14 = 36. 54 i+182. 7 j Newton F 34 = 882. 61 /-11, 95 Newton = 863. 48 i-182. 75 j Newton F=0 F 12=-F 32=F 23=F 34 MO 2=0 M 12 + O 2 A x F 32 = 0 M 12 k + 0. 075(cos 105 i+sin 105 j) x (-863. 48 i+182. 75 j) = 0 M 12 k – 3. 547 k + 62. 55 k = 0 M 12 = -59. 003 N. m M 12 = -59. 003 k N. m CONCLUSION ?