Engineering science N4 (FET College)
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The document discusses the principles of shear force and bending moment diagrams for a loaded beam, presenting key equations governing vertical forces and moments. It details the calculations of reaction forces, shear force values, and bending moments at various points along the beam. The final results include graphical representations of both the shear force diagram and the bending moment diagram, highlighting the behavior of the beam under applied loads.
![X ;Y coordinates Calculation
Point
A
Y= 14.8125KN;
X = 0 meters
Reaction
Support A
acts
Upwards
Point
C
Y = 5.8125KN
X = 3 m
14.8125KN
– (3KN/m
X 3m)]
Point
D
Y = 5.8125KN
X = 5m](https://image.slidesharecdn.com/engineeringsciencen4-150730053936-lva1-app6891/85/Engineering-science-N4-FET-College-23-320.jpg)
Engineering science N4 (FET College)
- 1. SHEAR FORCE DIAGRAMS & BENDING MOMENT DIAGRAMS +27 73 090 2954 topstudentz017@gmail.com
- 2. Consider the beam as shown below. 3kN/m 30KN 10KN A 3m C 2m D 3m B 3m E RA RB
- 3. The two equations that govern a loaded beam are: a) Total vertical Forces Acting Downward = Total Vertical forces acting upward
- 4. b) Total Clockwise Moment About a Turning Point = Total Anticlockwise Moment about the same turning point.
- 5. Based on the two equations above, we are going to write two equations. a) Total Vertical Forces acting downward = (3KN/m X3m) + 30KN + 10KN
- 6. Total vertical forces acting upward= RA + RB The symbol for our equation a) is usually written ΣFV=0: 49KN = RA + RB (1)
- 7. The symbol for our equation b) is usually written ΣMA=0 This symbol means summation of moments about the turning point (support A) is zero.
- 8. The symbol for our equation b) is usually written ΣMA=0 This symbol means summation of moments about the turning point (support A) is zero.
- 9. The following diagram illustrates clockwise and anticlockwise moments about a support turning point.
- 11. Based on the above diagram: Let the weight of the boy on the left hand side be A KN and the weight of the boy on the right hand side be B KN. The weight of a body acts downwards.
- 12. Anticlockwise moment: AKilonewton exerts an anticlockwise moment of A Kilonewtons X x meters about the support . Assuming that the perpendicular distance of the weight of A from the turning point is x meters.
- 13. Clockwise moment: BKN exerts an clockwise moment of B Kilonewtons X w meters about the support . Assuming that the perpendicular distance of the weight of B from the turning point is w meters.
- 14. Based on the above definitions, we are going to write an equation of ΣMA=0 for the loaded beam. Clockwise moments about RA (3kN/m X 3m X (3/2)m) + (30KN X 5m) + (10KN X 11m) = 273.5KNm
- 15. (3kN/m X 3m X (3/2)m) For the moment shown above: 3kN/m X 3m = magnitude of the Uniformly distributed load. (3/2)m represents the perpendicular distance between the turning point A and the point through which the udl acts.
- 16. In order words, this moment is : 3 represents the span on which the udl acts. The magnitude of the udl , 3kN/m X 3m , acts through the middle of the span.
- 17. (30KN X 5m) This represents the moment of the point load, 30KN, which acts at a perpendicular distance of 5m from the turning point, A.
- 18. For Anticlockwise moments: RB X (3m + 2m + 3m) RB represents the support reaction acting vertically upwards. (3m + 2m + 3m) represents the distance between the support RB and the turning point.
- 19. 273.5KNm = RB X (3m + 2m + 3m) RB = 34.1875KN From the equation RA + RB = 39 RA = (49 - 34.1875)KN = 14.8125KN
- 20. We shall now proceed to draw the SFD. Shear Force Diagram is a graph of the vertical forces plotted on the y axes and the horizontal perpendicular distance of the force on the x axis. The point A is zero meters
- 21. C = 3m ; D = 5m etc A C D B E SHEARFORCEAXIS(KN) DISTANCE , X. (METRES)
- 22. SFD KEY POINTS 1. The final vertical force at the section E = 0 2. The sum of the upward vertical forces = The sum of the downward vertical forces as shown by the SFD. 3. ΣFV=0.
- 23. X ;Y coordinates Calculation Point A Y= 14.8125KN; X = 0 meters Reaction Support A acts Upwards Point C Y = 5.8125KN X = 3 m 14.8125KN – (3KN/m X 3m)] Point D Y = 5.8125KN X = 5m
- 24. X ;Y coordinates Calculation D Y = -24.1875KN X = 5m (5.8125 -30 )KN 30KN acts downward B Y = -24.1875KN X = 8 m B Y = 10KN X = 8 m (34.1875 – 24.1875) KN E Y;X = 10KN ; 11m E Y = OKN X = 11m (10 -10)KN
- 26. BENDING MOMENT DIAGRAM Bending Moment Diagram or Graph is the graph of the bending moment on the y axis plotted against the distance, x metres, along the beam (x axis).
- 27. BENDING MOMENT DIAGRAM The values of the Bending moment on the y axis can be evaluated by calculating the areas under the Shear Force diagram curve section by section progressing from left to right.
- 28. Section Area Bending Moment A – C Trapezium 1/2(14.8125 + 5.8125) 3m = 30.9325KNm MC = 30.9325KNm C – D Rectangle 2m X 5.8125KN = 11.625KNm MD = 30.9325 + 11.625 = 42.5525KNm D – B Rectangle 3m X -24.1875KN = -72.5625KNm MB= (42.5525 – 72.5625) KNm= -30.01KNm B – E Rectangle 3m X 10KN = 30KNm ME = (30 - 30.01) = 0
- 29. BMD KEYPOINTS 1. The final bending moment at the section E = 0 2. The total areas (+ve bending moment) above the x axis = The total areas (-ve bending moment) below the x axis. 3. ΣMA=0
- 30. BENDINGMOMENTAXIS(KNM) DISTANCE , X. (METRES) A C D B E MC= (3m, 30.9325KNm) ; ME = (11m, 0KNm) MD = (5m,42. 5525KNm) MB = = ( 8m, 30.01KNm)
- 31. BMD KEYPOINTS 4. The section of the Beam that carries a uniformly distributed load is represented as a curve on the BMD. 5. The section of the Beam that carries point loads are represented by slant straight lines.
Editor's Notes
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