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CHAPTER – II

F2. F2. F1. F1. F5. F3. F3. F4. Fig. 1. Fig. 2. CHAPTER – II. RESULTANT OF COPLANAR NON CONCURRENT FORCE SYSTEM. Coplanar Non-concurrent Force System:.

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CHAPTER – II

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  1. F2 F2 F1 F1 F5 F3 F3 F4 Fig. 1 Fig. 2 CHAPTER – II RESULTANT OF COPLANAR NON CONCURRENT FORCE SYSTEM Coplanar Non-concurrent Force System:   This is the force system in which lines of action of individual forces lie in the same plane but act at different point of application.

  2. F2 F2 F1 F1 F5 F3 F3 F4 Fig. 1 Fig. 2 TYPES • Parallel Force System – Lines ofaction of individual forces are parallel to each other. 2. Non-Parallel Force System – Linesof action of the forces are not parallel to each other.

  3. MOMENT OF A FORCE ABOUT AN AXIS The applied force can also tend to rotate the body about an axis in addition to motion. This rotational tendency is known as moment. Definition: Moment is the tendency of a force to make a rigid body to rotate about an axis. This is a vector quantity having both magnitude and direction.

  4. Moment Axis: This is the axis about which rotational tendency is determined. It is perpendicular to the plane comprising moment arm and line of action of the force. A B d F Moment Center: This is the position of axis on co-planar system. (B). Moment Arm: Perpendicular distance from the line of action of the force to moment center. Distance AB = d.

  5. EXAMPLE FOR MOMENT Consider the example of pipe wrench. The force applied which is perpendicular to the handle of the wrench tends to rotate the pipe about its vertical axis. The magnitude of this tendency depends both on the magnitude of the force and the effective length ‘d’ of the wrench handle. Force perpendicular to the handle is more effective.

  6. B A d F MAGNITUDE OF MOMENT It is computed as the product of the of the force and the perpendicular distance from the line of action to the point about which moment is computed. (Moment center). MA = F×d Unit – Unit of Force × Unit of distance kN-m, N-mm etc.

  7. M M SENSE OF MOMENT The sense is obtained by ‘Right Hand Thumb’ rule. ‘If the fingers of the right hand are curled in the direction of rotational tendency of the body, the extended thumb represents the sense of moment vector’. For the purpose of additions, the moment direction may be considered by using a suitable sign convention such as +ve for counterclockwise and –ve for clockwise rotations or vice-versa.

  8. VARIGNON’S THEOREM (PRINCIPLE OF MOMENTS) Y Ry Q R Qy q P r Py  p   X A O Statement: The moment of a force about a moment center or axis is equal to the algebraic sum of the moments of component forces about the same moment center (axis). Proof (by Scalar Formulation): Let ‘R’ be the given force. ‘P’ & ‘Q’ are component forces of ‘R’. ‘O’ is the moment center. p, r and q are moment arms from ‘O’ of P, R and Q respectively. ,  and  are the inclinations of ‘P’, ‘R’ and ‘Q’ respectively w.r.t X – axis.

  9. Y Ry Q R Qy q P r Py  p   X A O We have, Ry = Py + Qy R Sin = P Sin + Q Sin  ----(1) From le AOB, p/AO = Sin  From le AOC, r/AO = Sin  From le AOD, q/AO = Sin  From (1), R ×(r/AO) = P ×(p/AO) + Q ×(q/AO) i.e., R × r = P × p + Q × q Moment of resultant R about O = algebraic sum of moments of component forces P & Q about same moment center ‘O’.

  10. VARIGNON’S THEOREM – PROOF BY VECTOR FORMULATION Consider three forces F1, F2, and F3 concurrent at point ‘A’ as shown in fig. Let r is the position vector from ‘O’ to point ‘A’. The sum of moments about ‘O’ for these three forces by cross-product is, Mo = ∑(r×F) = (r×F1) + (r×F2) + (r×F3). By the property of cross product, Mo = r × (F1+F2+F3) = r × R where, R is the resultant of three original forces.

  11. APPLICATIONS OF VARIGNON’S THEOREM • Simplifies the computation of moments by judiciously selecting the moment center. • Moment can be determined by resolving a force into x & y components because finding x & y distances in many circumstances may be easier than finding perpendicular distance from moment center to line of action (d). 2. Location of resultant - location of line of action of resultant in case of non-concurrent force system which is an additional information required in the resultant problems.

  12. F d F COUPLE Two parallel, non collinear (separated by certain distance) forces that are equal in magnitude and opposite in direction form ‘couple’. The algebraic summation of the two forces forming couple is zero. Hence, couple does not produce any translation and produces only rotation.

  13. F d a O F + ∑ Mo = F × ( a + d) – F × a = F× d Moment of a Couple: Consider two equal and opposite forces separated by a distance ‘d’. Let ‘O’ be the moment center at a distance ‘a’ from one of the forces. The sum of moments of two forces about the point ‘O’ is, Thus, the moment of the couple about ‘O’ is independent of the location, as it is independent of ‘a’. The moment of a couple about any point is constant and is equal to the product of one of the forces and the perpendicular distance between them.

  14. F F F F F d Q Q M=F × d = = P P F Fig. (b) Fig. (a) Fig. (c) RESOLUTION OF A FORCE INTO A FORCE-COUPLE SYSTEM A given force F applied at a point can be replaced by an equal force applied at another point Q together with a couple which is equivalent to the original system. Two equal and opposite forces of same magnitude F and parallel to the force F at P are introduced at Q.

  15. F F F F M=F × d Q Q d = = P P Fig. (a) F Fig. (b) Fig. (c) Of these three forces, two forces i.e., one at P and the other oppositely directed at Q form a couple. Moment of this couple, M = F × d. Third force at Q is acting in the same direction as that at P. The system in Fig. ( c ) is equivalent to the system in Fig. ( a ).

  16. Thus, the force F acting at a point such as P in a rigid body can be moved to any other given point Q, by adding a couple M. The moment of the couple is equal to moment of the force in its original position about Q.

  17. A 240 mm F 60º O PROBLEM - 1 A 100N vertical force is applied to the end of a lever at ‘A’, which is attached to the shaft at ‘O’ as shown in the figure.

  18. Determine, a) The moment of 100N force about ‘O’. b) Magnitude of the horizontal force applied at ‘A’, which develops same moment about ‘O’. c) The smallest force at ‘A’, which develops same effect about ‘O’. d) How far from the shaft a 240N vertical force must act to develop the same effect? e) Whether any of the above systems is equivalent to the original?

  19. A 240 mm F 60º O d SOLUTION Case a) The moment of 100N force about ‘O’. Perpendicular distance from the line of action of force F to moment center ‘O’ = d d = 240 cos 60º =120 mm. Moment about ‘O’ = F × d = 100 × 120 = 12,000 N-mm (Clockwise)

  20. F 240mm d 60º A O Case b) Magnitude of the horizontal force applied at ‘A’, which develops same moment about ‘O’. Perpendicular distance between the line of action of horizontal force F at A to moment center ‘O’, d = 240 sin 60º = 207.85 mm. Moment about ‘O’ = F × d = F × 207.85 = 12,000 N-mm (Clockwise) Therefore, F = 12,000/207.85 = 57.73 N

  21. A d = 240 mm F 60º O Case c) The smallest force at ‘A’, which develops same effect about ‘O’. F = M/d Force is smallest when the perpendicular distance is maximum so as to produce same M. i.e., d = 240 mm. Therefore, Fmin = 12,000/240 = 40N.

  22. A d 60º X F O Case d) How far from the shaft a 240N vertical force must act to develop the same effect? Distance along x-axis, X = M/F = 12,000/240 = 50 mm. Distance along the shaft axis d = (X / cos 60) = 50/cos 60 =100 mm

  23. Case e) Observations: I) None of the above force system is equivalent to the original even though all of them produce same moment. II) Shaft rotates in the same manner but the pulling effect on the shaft is different in different cases.

  24. F=50 kN C 4m D 60º 3m B A PROBLEM - 2 A 50kN force acts on the corner of a 4m x 3m box as shown in the Fig. Compute the moment of this force about A by a) Definition of Moment b) Resolving the force into components along CA and CB.

  25. F=50 kN C 4m D 60º 36.87º 3m 60º AC = CAD = tan-1(3/4) = 36.87º B A d E ACE = 60º – 36.87º = 23.13º SOLUTION a) By Definition of Moment: To determine ‘d’: ECD = 60º From ∆le ACE, d = AC × sin ( ACE) = 5 × sin 23.13º = 1.96 m. Moment about A = 50 × 1.96 = 98.20 kNm.

  26. Fy F=50kN 4m D C 60º Fx 3m A B + ΣMA = - Fx × 3 + Fy × 4 b) By Components: Fx = 50 × cos 60 = 25kN. Fy = 50 × sin 60 = 43.30kN. = - 25 × 3 + 43.3 × 4 = + 98.20kNm.

  27. 50kN B 60kN 60º 30kN 30º A D 200mm 80kN PROBLEM - 3 An equilateral triangle of sides 200mm is acted upon by 4 forces as shown in the figure. Determine magnitude and direction of the resultant and its position from point ‘D’.

  28. 50 sin 60 50kN B 60 sin30 50 cos 60 60kN 60 cos30 + ΣFx = Rx 60º 30kN 30º A + ΣFy = Ry D 200mm R = 80kN SOLUTION Resultant & its inclination: Resolving forces we have, = +30 + 60 cos30º – 50 cos60º + 0. = +56.96kN = -80 + 60 sin30º + 50 sin60º + 0 = -6.69kN. Inclination w.r.t horizontal = θR = tan-1(Ry/Rx) = tan-1(6.69/56.96) = 6.7º

  29. 50 sin 60 50kN B 60 sin30 50 cos 60 60kN 60 cos30 60º 30kN 30º A D 200mm 80kN b) Position w.r.to D: Moment of the component forces about D: + MD = -50 sin60 × 100 + 50 cos60 × (sin60 × 200) - 60 × 100 + 80 × 100 + 30 × 0 = 2000kNmm. = R × d = 57.35 × d. where ‘d’ = perpendicular distance from point D to the line of action of R. d = 2000/57.35 = 34.87mm.

  30. 50kN ΣFx B 60kN x R ΣFy 60º 30kN 30º A D 200mm 80kN Location of the resultant along x-axis from the point D M= R × d M= (ΣFy × x – ΣFx × y)--------- (1) If we resolve the resultant at a point on the x-axis, then the moment of force ΣFx about the point D, becomes zero, and M= (ΣFy × x) x=M/ ΣFy

  31. ΣFx R 50kN ΣFy B 60kN 60º 30kN 30º A D 200mm 80kN Location of the resultant along x-axis from the point D Y

  32. Location of the resultant along x-axis from the point D M= (ΣFy × x) Location of the resultant along y-axis from the point D M= - (ΣFx × y)

  33. F1=2500N F2=500N F5=2000N 1 Ө4 Ө2 1 1m F3=1000N Ө5 O 1m F4=1500N PROBLEM - 4 Find the resultant and its position w.r.t ‘O’ of the non-concurrent system of forces shown in the figure.

  34. F1=2500N F2=500N F5=2000N 1 Ө4 Ө2 1 1m Ө5 F3=1000N O 1m F4=1500N + ΣFx = Rx = F2 cosӨ2 +F3 -F4 cosӨ4-F5 cos Ө5 SOLUTION • To find the resultant – Ө2 = tan-1(1/2) = 26.56° Ө4 = tan-1(3/2) = 56.31° Ө5 = tan-1(1/1) = 45° = 500 × cos26.56 + 1000 –1500 × cos56.31-2000 × cos45 = -799.03N = 799.03N← +↑ΣFy = Ry= F1+F2 sin Ө2-F4 sin Ө4+F5 sin Ө5 = 2500+500 sin26.56-1500 sin56.31+2000 sin45 =2889.70N ↑

  35. Ry R ӨR Rx + Mo =R×d = +2500×2 + 500×sin26.56×5 – 500× cos26.56×3 - 1000×1+1500× cos56.31×0 –1500×sin56.3×1+2000× cos45×1-2000×sin45×0 ∴ Resultant R = = 2998.14N ӨR = tan-1 = tan-1(2889.7/799.03) = 74.54° B) Position of Resultant w.r.t ‘O’: By Varignon’s theorem, Moment of the resultant about ‘O’ is = algebraic sum of the moments of its components about ‘O’. = 2998.14 × d d = 1.43 m from O.

  36. 120 kN 50 kN 30 kN 2 m 3 m 1 m 60º B A 6 m PROBLEM - 5 Determine the resultant of three forces acting on a dam section shown in the figure and locateits intersection with the base. Check whether the resultant passes through the middle one-third of the base.

  37. 120 kN 50 kN 30 kN 2 m 60º 3 m 1 m 60º A B 6 m + ∑Fx = Rx = 50 – 30 × cos 60 = 35 kN = 35 kN + ∑Fy = Ry = -120 – 30 × sin 60 = -145.98 kN = 145.98 kN Resultant, R= SOLUTION

  38. θR= tan-1(Ry/Rx) = tan-1(145.98/35) = 76.52º Location of the resultant w.r.t. B: MB= 30×1 + 120 × (6-2) - 50 × 3 = Ry × X 360 = 145.98 × X Therefore, X = 360/145.98 = 2.47m from B. From A, X = 6 – 2.46 = 3.53 m. Middle 1/3rd distance is between 2m and 4m. 2m<3.53<4m Hence, the resultant passes through the middle 1/3rd of the base.

  39. W kN/m L W kN/m L TYPES OF LOADS ON BEAMS 1. Concentrated Loads – This is the load acting for very small length of the beam. 2. Uniformly distributed load – This is the load acting for a considerable length of the beam with same intensity of W kN/m throughout its spread. Total intensity = W × L (acts at L/2 from one end of the spread) 3. Uniformly varying load – This load acts for a considerable length of the beam with intensity varying linearly from ‘0’ at one end to W kN/m to the other representing a triangular distribution. Total intensity of load = area of triangular = 1/2× W × L. (acts at 2×L/3 from ‘Zero’ load end)

  40. 20kN/m 30kN 40kN-m B A 45° 2m 1m 2m 3m + ΣFx = Rx = -30 cos45 = - 21.21kN = 21.21kN← + ΣFy = Ry = -30 sin45 – 20x2 = -61.21kN = 61.21kN↓ Rx R = θR Ry R PROBLEM - 6 Locatethe resultant w. r. to point A. SOLUTION: θR=tan-1(Ry/Rx) = tan-1(61.21/21.21) = 70.89º

  41. + MA = -40 – 30×cos45×0 – 30×sin45×(3+2) – 20×2× (3+2+1+2/2) = -426.07 = 426.07 kNm Position w.r.to A: For this clockwise moment, the line of action must be onto the right of A. The perpendicular distance to the line of action of R from A d = M/R = 426.0/64.78 = 6.58m The X-distance from A along the beam, X = M/Ry = 426.07/61.21 = 6.96m

  42. 10kN 10kN 1m 5kN/m 30° 2kN/m 1m B A 2m 2m 1m 2m + ΣFx = Rx = -10 cos30 = - 8.66 kN = 8.66kN← + ΣFy = Ry = -1/2×2×2-10-10×sin30-5×2 = -27kN =27kN↓ R = PROBLEM - 7 Locatethe resultant w. r. t point A. SOLUTION:

  43. Rx θR R + MA = - (1/2×2×2)×2/3×2 – 10×(2+2) + 10× cos30×1- 10 sin30×(2+2+1+1) – 5×2×(2+2+1+1/2×2) = -124.006kNm = 124.006kNm Ry θR=tan-1(Ry/Rx) = tan-1(27/8.66) = 72.22º Position w.r.t A: For this clockwise moment, the line of action must be onto the right of A. The perpendicular distance to the line of action of R from A, d = M/R = 124.06/28.35 = 4.38m. The X-distance from A along the beam, X= M/Ry = 124.06/27 = 4.59m

  44. B 30 mm A 50 mm 30º 100 mm 50N PROBLEM 8 A 50 N force is applied to the corner of a plate as shown in the fig. Determine an equivalent force - couple system at A. Also determine an equivalent force system consisting of a 150 N force at B and another force at A.

  45. B 30 mm Fx = 50 ×cos 60= 25 N. A 50 mm Fx=50 sin 30 Fy = 50 × sin60= 43.3 N 30º 100 mm 50N Fy=50 cos 30 Fx=50 sin 30 A + ∑MA= Fx×50-Fy×100 50 mm Ma=3080N-mm Fy=50 cos 30 = -3080 N-mm. = 3080 N-mm 100 mm SOLUTION A) Force – Couple System at A: These forces can be moved to A by adding the couple. Moment of the couple about A = 25×50 - 43.3×100

  46. B 30 mm Ma=3080N-mm A A Fx=50 cos60 50 mm Fx=50 sin 30 Fy=50 sin 60 30º 100 mm 50N 100 mm Fy=50 cos 30 SOLUTION

  47. 150 N θ B 30 mm A 50 mm FA 100 mm ∑ FAY = -50×sin60 - 150×sin46.8 => FAY = -152.65N = 152.65N SOLUTION B) Forces at A and B : The couple MA is because of two equal and opposite forces at A and B. i.e., MA = 150 × cosθ × 30 = 3080 Therefore, θ = 46.8º. The force at A is calculated as follows: ∑ FAX = 50×cos60 - 150×cos46.8 => FAX = -77.68N = 77.68N ←

  48. 150 N B 30 mm θ A B 50 mm 30 mm A 63.03 30º 100 mm 171.28N 50N 100 mm SOLUTION = FA = 171.28N θA=tan-1(Fy/Fx) = tan-1(152.65/77.68) = 63.03º

  49. 600 N o 30º 60º 1000 N 2000 N 10º 60º 400 N PROBLEMS FOR PRACTICE 1. Determine the resultant of the parallel coplanar force system shown in fig. (Ans. R=800N towards left, d=627.5mm)

  50. 2. Four forces of magnitudes 10N, 20N, 30N and 40N acting respectively along the four sides of a square ABCD as shown in the figure. Determine the magnitude, direction and position of resultant w.r.t. A. (Ans:R=28.28N, θ=45º, x=1.77a) 20N D C 30N a A 10N a B 40N

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