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Structure Design-I (Moments)
The document discusses structural mechanics and principles of moments. It defines a moment as the product of a force and its perpendicular distance from a fulcrum. Moments can be used to calculate unknown support reactions in beams and other structural systems. Couples, consisting of two equal and opposite parallel forces, produce pure rotation and cannot be balanced by a single force. The degree of static indeterminacy refers to the number of redundant unknown internal forces and reactions that must be specified to fully describe the static behavior of a structure.
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Structure Design-I (Moments)
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- 2. MOMENTS, PRINCIPLE OFTHE LEVER The forces studied so far tend to either distort an object or move bodies from one position to another. Forces can also have rotational effect on bodies if these are hinged at a point. In the figure below the force F would rotate the body around the point O. The point O is generally referred to as a fulcrum. O F F
- 3. Try openinga door by pushing it near the hinge. It will not open till force is applied at a distance from the hinge. Try opening the door by pulling it parallel to the door - again same result. This suggests that forces have a best rotational affect when these are at a distance from the hinge or fulcrum and the perpendicular distance between the hinge and the line of action of the force will determine the total rotational affect. This phenomenon is determined by the principle of “moments”. A moment is thus a product of the magnitude of force and the perpendicular distance of the fulcrum from the line of action of the force. This distance is also referred to as the ‘arm of the moment’.
- 4. Resultant Moments Moment= force x arm Moments can be clockwise or anti clockwise. These can be arithmetically added Find the resultant moment about the point O in the figure above 4kN 6kN 2kN 5kN O30 cm 30 cm 20 cm 40 cm
- 5. Calculating Beam SupportReactions This principle of moment can be used to calculate the reactions at the support points in a structural system. Take the case of the following beam: Assuming the system to be in equilibrium Rl x 5m = (3kN x 1m) + (6kN x 3m) Rr x 5m = (6kN x 2m) + (3kN x 4m) 6kN 3kN 2 m 2 m 1 m Rl Rr
- 6. Beam Aof effective span 8m has a self load of 30N per meter and carries a wall load of 710N per meter. It also carries the load of 3 beams B, C and D with point loads of 2000, 2000 and 4000N respectively. Beams B, C and D are equally spaced at 2 meters each. Find the reactions Rl and Rr. Hint: First draw the force system diagram. WALL B DC A OPENINGRl Rr
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- 8. 2000 2000 4000 UDL Rl 2m2m2m Rr 2m UDL= (30+710) x 8 N Taking moments about Rr: Rl x 8 = (4000x2) + (2000x4) +( (30+70)x8 x4) + (2000x6) Taking moments about Rl: Rr x 8 = (2000x2) + (2000x4) +( (30+70)x8 ) x4 + (4000x6) Crosscheck: Rl + Rr = 2000 + 2000 + 4000 + UDL
- 9. 5kN 4kN10kN Rl 8kN 9m Rr 1m 1.5m2.5m5m1.5m FindRl and Rr in the loading system shown in the diagram above.
- 10. To find thereactions: There are two unknown values. Rl and Rr Taking the system to be in equilibrium ∑M = 0 And taking moments about one of the reaction points, the moment created by the reaction force at that point becomes zero and there is hence only one unknown value. Now, taking moment about Rr we get:
- 11. Taking moments aboutRr: 5kN force is creating a anticlockwise moment Rl force is creating a clockwise moment 10kN force is creating a anticlockwise moment 8kN force is creating a anticlockwise moment 4kN force is creating a clockwise moment. So (Rl x 9) + (4 x 1.5) = (8 x 2.5) + (10 x 7.5) + (5 x 10) 9Rl + 6 = 20 + 75 + 50 Rl = 139 / 9 kN
- 12. Taking moments aboutRl Force 4kn is creating a clockwise moment Force Rr is creating a anticlockwise moment Force 8kn is creating a clockwise moment Force 10kn is creating a clockwise moment Force Rl is creating a zero moment Force 5kn is creating a anticlockwise moment So (Rr x 9) + (5 x 1 ) = ( 10 x 1.5) + (8 x 6.5) + (4 x 10.5) 9Rr + 5 = 15 + 52 + 42 9Rr = 104 or Rr = 104 / 9
- 13. To crosscheck; When systemis in equilibrium then ∑V= 0, ∑H= 0, ∑M= 0 There is no H (horizontal force) To check that ∑V= 0 Upward forces are Rl + Rr = 139/9 + 104/9 = 243/9 = 27 kN Downward forces are 5 + 10 + 8 + 4 = 27 kN
- 14. Parallel Force Systems These can be divided in two classes: Like parallel systems Unlike parallel systems
- 15. 10m 20kN 60kN 2m4m4m 50kN 40kN 10kN Find theresultant of the unlike forces in the diagram above. Magnitude of resultant = -50-60+20+40+10 = -40kN Direction = Downward Position= suppose x m distance from A Taking moments about A 40 x x = (60 x 10) - (20 x 4) - ( 40 x 8) - (10 x 12) x = 2m R A 2m
- 16. COUPLES In theprevious examples the force system was of parallel forces that were not in equilibrium because the downward forces were more than the upward forces. Supposing the sum of upward and downward forces were equal, then two possibilities would be there: If the resultant of the upward and downward forces were in the same line of action the system would be in equilibrium If these were not in a straight line the effect would of pure rotation – which is known as a ‘couple’. Hence, a couple consists of two equal unlike parallel forces acting out of line.
- 17. Anti clock wisecouple Clock wise couple Practical examples of couples can be seen within a beam that is loaded. Within the beam the couple that counters the loading couple is known as the “moment of resistance”
- 18. Properties of acouple: A couple cannot be balanced by a single force. It requires another couple of equal and opposite moment for a state of equilibrium. A couple has the same moment about any point in a plane of forces which is equal to the value of the couple itself.
- 19. 1.2m 20N 20N 20N 20N 20N 20N 0.2m 0.2m 0.2m 0.2m Find the momentof the couple that is generated by the system of forces shown in the figure by: i) finding the force and arm of the couple and ii) taking the moments of the given forces about any convenient point.
- 20. 1.2m A 20N B 20N D20N C 20N E 20N F 20N 0.2m 0.2m 0.4m 0.2m 0.4 + 0.2 + 0.2m i) finding the force and arm of the couple: Moment of resultant = Resultant moment So, find the resultant of the two set of Forces Forces to the right: Value of resultant = sum of all = 60N Position: all the forces are equal and equally spaced So, resultant will be in line with the middle force So, it will be in line with the middle force So, the arm = 0.2 + 0.4 +0 .2 = 0.8 m So the couple is 60 x 0.8 = 48 Nm The couple will be acting CW ii Finding the value of the couple by taking moments about a convenient point : Take moment about the topmost force Moment by B = ACW = 20 x 0.2 = 4 Nm Moment by C = ACW = 20 x 0.4 = 8 Nm Moment by D = CW = 20 x 0.8 = 16 Nm Moment by E = CW = 20 x 1.0 = 20Nm Moment by F = CW = 20 x 1.2m = 24Nm So, the couple will be acting CW Value of couple = (16 + 20 + 24) – (8 + 4) = 60 – 12 = 48 Nm 60N 60N 0.2m
- 21. Degree of staticindeterminacy Degree of static indeterminacy is the number of (independent) unknown static quantities (e.g. internal member forces and support reactions) that must be prescribed in addition to the use of equilibrium equations to completely describe the static state of the structure. The static state of the structure is known if and only if all support reactions and the internal member forces at any locations within the structure can be determined. Those (independent) unknown static quantities are termed as redundant unknowns. The structure is said to be statically determinate if and only if the degree of static indeterminacy is equal to zero; in the contrary, the structure is said to be statically indeterminate if and only if the degree of