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ENGR-1100 Introduction to Engineering Analysis

ENGR-1100 Introduction to Engineering Analysis. Lecture 8. Lecture outline. Moment of a force Principle of moments and Varignon’s law. For concurrent forces R = 0 was sufficient for equilibrium. For non concurrent forces the particle idealization is no longer valid.

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ENGR-1100 Introduction to Engineering Analysis

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  1. ENGR-1100 Introduction to Engineering Analysis Lecture 8

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

  3. 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 For non concurrent forces the particle idealization is no longer valid.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 Non Concurrent Forces

  4. Moment M0 of Force F about AA • Magnitude of Moment: M0= |F|d • Direction of Moment is along axis A-A • That is, moment is a vector along the direction of A-A: M0= |F|d eAA whereeAA is unit vector along axis A-A

  5. M0  O d F • In2D, 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: • M0 = |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 M0 : lb•ft or lb•in in U.S. Customary system; and N•m or kN•m in SI units.

  6. M0  O d F • In2D, we don’t need the complication of treating M0 as a vector. We treat M0 as a scalar (it is understood that it is a vector perpendicular to plane). • The magnitude of moment M0 in the plane is: • M0=|F|d • The sign of M0 is: • (+) when counterclockwise (figure) • (-) when clockwise (opposite to figure)

  7. MA= 135 Nm MB= 90 Nm MC= 180 Nm Example P4-2 C  A  B  MA=|F|dA=225*0.6 = 135 Nm MB=|F| dB=225*0.4 = 90 Nm MC=|F| dC=225*0.8 = 180 Nm

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

  9. MA1=|F1|dA=500*30 = 15,000 lb•in MB1=|F1| dB=500*20 = 10,000 lb•in MA1= 15 kip•in MB1= -10 kip•in MB2= 9 kip•in MB= -7.5 kip•in MB2=|F2| dA=300*30 = 9,000 lb•in MC2=|F2| dC=300*25 = 7,500 lb•in

  10. 600 A  dac C 160mm 600 200 mm B L Cos (600) = 200/L Cos (600) = dac/(L-160) L = 200/ Cos (600) L=160+dac/Cos (600) Example P4-10 dac=200-160 Cos (600)=120 mm

  11. MA= -60 Nm MB= -100 Nm 600 L A  dac C 160mm 600 200 mm B dac=120 mm F MA=|F|dAC=500*0.12 = 60 Nm MB=|F| dB=500*0.2 = 100 Nm

  12. The moment M0 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= F1+ F2+…..+ Fn M0= Rdr=F1d1+ F2d2+…..+ Fndn Principle of moments:

  13. 1) MR= R.d=R(h cos g) 2) MA= A.a=A(h cos a) a a 3) MB= B.b=B(h cos b) g d b • From Geometry: • 4) R cos g = b A cos a + B cos b a From equations 1, 2, 3, and 4: b B Cos b A Cos a MR=MA+ MB R Cos g Application of the principle: Varignon’s Theorem y O  R h A B g x

  14. Example P4-18

  15. Solution Fx=F cos(320) = 250 cos(320) = 212 N Fy Fy=F sin(320) = 250 sin(320) = 132.5 N F Fx Using Varignon’s Theorem MA= Fx *0.25 + Fy *0.21 = 80.8 Nm

  16. Answer: MA1 = 650 ft.lb MA2 = -525 ft.lb 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 F1 and F2 about point A. y F1 = 250 lb 60o x A B 3 ft F2 = 175 lb

  17. Answer: MA=5.47 in.kip MB=2.25 in.kip 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. y 350 lb A 40o 12 in x C B 10 in

  18. 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.

  19. It is often more convenient to use the vector approach to simplify moment calculation. Vector Representation of a Moment  O r d A F a M0=r X F = | r || F |sin a e Where: r is the position vector from point O to a point A on the line if action of the forceF. e – unit vector perpendicular to plain containing r and F.

  20. z F The location of point A A rAB rA= xAi+ yAj+ zA k B rB rA zA zB The location of point B y xB yB rB= xBi+ yBj+ zB k xA yA Since: x rA= rB+ rAB Finding the Position Vector r= rAB = rA-rB= (xAi+ yAj+ zA k ) - (xBi+ yBj+ zB k ) = (xA- xB) i+ (yA- yB) j+ (zA- zB) k

  21. y F A r j O x i i j k rx ry 0 Fx Fy 0 M0= = (rx Fy - ry Fx)k= Mzk (rx Fy - ry Fx)k= Mzk Two Dimensional Case F= Fxi+ Fyj r= rxi+ ryj M0=r X F = (rxi+ ryj) X (Fxi+ Fyj) = rx Fx(ix i ) + rx Fy(ix j ) + ry Fx(jx i ) + ry Fy(jx j ) =

  22. r2 r1 a2 A1 a1 A2 Sin a1= d / r1 d = r1Sin a1 = r2Sin a2 Sin a2=d / r2  O d F

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