1. Lecture 8 ENGR-1100 Introduction to Engineering Analysis

2. Lecture outline Moment of a force Principle of moments and Varignon’s law

3. Non Concurrent Forces For concurrent forces R= 0 was sufficient for equilibrium. For non concurrent forces the particle idealization is no longer valid. For this case the condition R= 0, while still necessary for equilibrium, is not sufficient. A second restriction is related to tendency of a force to produce rotation (angular acceleration) of a body. This gives rise to the concept of moment of a force: F F

5. Moment M 0 of Force F about AA Magnitude of Moment: M 0 = |F|d Direction of Moment is along axis A-A That is, moment is a vector along the direction of A-A: M 0 = |F|d e AA where e AA is unit vector along axis A-A

6.  F d  M0M0 In 2D, we work in the plane perpendicular to the axis that contains F. Point O is intersection of axis A-A with that plane. The magnitude of the moment is: M 0 = |F|d where: F = force; and d = perpendicular distance from the line of action of the force to the axis (point O). Units of Moment M 0 : lbft or lbin in U.S. Customary system; and Nm or kNm in SI units.

7.  F d  M0M0 In 2D, we don’t need the complication of treating M 0 as a vector. We treat M 0 as a scalar (it is understood that it is a vector perpendicular to plane). The magnitude of moment M 0 in the plane is: M 0 =|F|d The sign of M 0 is: (+) when counterclockwise (figure) (-) when clockwise (opposite to figure)

8. Example P4-2 M A =|F|d A =225*0.6 = 135 Nm M B =|F| d B =225*0.4 = 90 Nm M C =|F| d C =225*0.8 = 180 Nm M A = 135 Nm M B = 90 Nm M C = 180 Nm  CC BB

9. Class Assignment: Exercise set 4-3 please submit to TA at the end of the lecture

10. M A1 =|F 1 |d A =500*30 = 15,000 lbin M B1 =|F 1 | d B =500*20 = 10,000 lbin M A1 = 15 kipin M B1 = -10 kipin M B2 =|F 2 | d A =300*30 = 9,000 lbin M C2 =|F 2 | d C =300*25 = 7,500 lbin M B2 = 9 kipin M B = -7.5 kipin

11. Example P4-10 A  CC BB L   200 mm 160mm d ac Cos (60 0 ) = 200/L Cos (60 0 ) = d ac /(L-160) L = 200/ Cos (60 0 ) L=160+d ac /Cos (60 0 ) d ac =200-160 Cos (60 0 )=120 mm

12. A  CC BB L   200 mm 160mm d ac d ac =120 mm M A =|F|d AC =500*0.12 = 60 Nm M B =|F| d B =500*0.2 = 100 Nm F M A = -60 Nm M B = -100 Nm

13. Principle of moments: The moment M 0 of the resultant R of a system of forces with respect to any axis or point is equal to the vector sum of the moment of the individual forces of the system with respect to the same axis or point. R= F 1 + F 2 +…..+ F n M 0 = Rd r =F 1 d 1 + F 2 d 2 +…..+ F n d n

14. Application of the principle: Varignon’s Theorem A y B x  R  From Geometry: 4) R cos  = R Cos   A Cos  A cos   B Cos  + B cos  M R =M A + M B From equations 1, 2, 3, and 4: h 1) M R = R. d=R(h cos  )  d  2) M A = A. a=A(h cos  a  3) M B = B. b=B(h cos  b

15. Example P4-18

16. F FxFx FyFy F x =F cos(32 0 ) = 250 cos(32 0 ) = 212 N F y =F sin(32 0 ) = 250 sin(32 0 ) = 132.5 N Solution Using Varignon’s Theorem M A = F x *0.25 + F y *0.21 = 80.8 Nm

17. Class Assignment: Exercise set 4-7 please submit to TA at the end of the lecture Two forces are applied to a beam as shown in the figure. Determine the moments of forces F 1 and F 2 about point A. F 2 = 175 lb F 1 = 250 lb 60 o x y 3 ft A B Answer: M A1 = 650 ft. lb M A2 = -525 ft. lb

18. Class Assignment: Exercise set 4-11 please submit to TA at the end of the lecture Determine the moment of the 350 lb force shown in the figure about points A and B. 350 lb 40 o x y 10 in A C Answer: M A =5.47 in. kip M B =2.25 in. kip B 12 in

19. Class Assignment: Exercise set 4-19 please submit to TA at the end of the lecture Determine the moment of the 750-lb force shown in Fig. P4-19 about point O.

20. Vector Representation of a Moment It is often more convenient to use the vector approach to simplify moment calculation. M 0 =r X F = | r || F |sin  e Where: r is the position vector from point O to a point A on the line if action of the force F. e – unit vector perpendicular to plain containing r and F.  F d  r  

21. Finding the Position Vector r= r AB = r A -r B = (x A i + y A j + z A k ) - (x B i + y B j + z B k )  (x A - x B ) i + (y A - y B ) j + (z A - z B ) k r A = r B + r AB Since: x y z F   xAxA xBxB yAyA yByB zAzA zBzB r AB r B = x B i + y B j + z B k The location of point B rBrB r A = x A i + y A j + z A k The location of point A rArA

22. Two Dimensional Case F= F x i + F y j r= r x i + r y j x y r F O A j i M 0 =r X F = (r x i + r y j) X (F x i + F y j) = r x F x ( i x i ) + r x F y ( i x j ) + r y F x ( j x i ) + r y F y ( j x j ) = (r x F y - r y F x ) k= M z k M 0 = = (r x F y - r y F x ) k= M z k i j k r x r y 0 F x F y 0

23.  F d    r1r1 Sin  1 = d / r 1   r2r2 Sin  2 = d / r 2 d = r 1 Sin  1 = r 2 Sin  2