Mechanics.
- 1. Beams, Shear Force & Bending Moment Diagrams • Shall approach this through examining BEAMs. • There are a number of steps that you MUST go through to get the solution. • Make sure you follow this every time – makes solution a lot easier!
- 2. Bending of Beams Many objects in everyday life can be analysed as beams. If we ignore mass of the beam, the forces on the beam are as shown: Support reaction forces. Load Force We shall normally ignore the beam mass in these lectures. The design and analysis of any structural member requires knowledge of the internal loadings acting within it, not only when it is in place and subjected to service loads, but also when it is being hoisted. In this lecture we will discuss how engineers determine these loadings.
- 3. Bending of Beams Support reaction forces. Load Force In the figure, the force shown downwards is acting on the beam. It is a point load, acting at a single point on the beam.
- 4. Bending of Beams However, in the figure below, the books are exerting a uniformly distributed load (UDL) on the shelf. A UDL is represented as shown in figure: w N/m
- 5. length Total load is = w length w length In calculating Reactions the UDL is considered to act at the mid-point of its length.
- 6. Beam loading: Point loading Uniformly Distributed Loading (UDL) UDL e.g. UDL – 20N/m, 3kN/m Combined loading
- 7. Beams can have a range of different forms of section.
- 8. Example Determine the reactions of a beam of length 4.5 m which is supported at its ends and subject to a point load of 9 kN a distance of 1.5 m from the left-hand end. Neglect the weight of the beam. The reactions at the supports can be found by taking moments about LHS: MA = 0 = RB 4.5 – 9 1.5 RB = (9 1.5)/4.5 = 3kN Fy = 0 = RB + RA – 9 RA = 6kN 9 RB RA
- 9. Example A uniform cantilever of length 3.0m has a uniform weight per metre of 120kN. Determine the reactions at the built-in end. First convert the UDL into a point load: MA = 0 = 360 1.5 + Ma Fy = 0 = Ry – 360 120 3 = 360kN acting at the mid point i.e. 1.5m from either end. 120 3 Ry Rx Ma FBD By inspection Rx = 0 as NO forces acting in the x-axis. Ma = - 360 1.5 = - 540 kNm Therefore, direction of Ma is Ry = 360 kN
- 10. Shear Force & Bending Moments. If I ‘cut’ the beam anywhere and examine what is happening, I shall see the following: x SF BM
- 11. SHEAR FORCE • The algebric sum of the vertical forces on either side of the section of a loaded beam is called Shearing Force
- 12. Bending Moment • The algebric sum of the moments of the forces on either side of the section of a loaded beam is called Bending Moment.
- 13. Example. A uniform cantilever of length 3.0m has a uniform weight per metre of 120kN. Determine the shear force and bending moment at distances of (a)1.0m. (b)2.0 m from the built-in end if no other loads are carried by the beam. Reactions have already been determined – slide 10 FBD for (a) 360 540 V1 1 M1 1201 0.5 You will notice that in tackling SF & BM problems you set up Fy = 0 to determine the SF (i.e. V ) AND M = 0 to determine the BM (i.e. M)
- 14. Fy = 0 = 360 – 120 – V1 V1 = 240 kN Mat 1m = 0 = -540 + (360 1) - (120 0.5) – M1 M1 = - 540 + 360 - 60 = - 240 kNm To determine the shear force at 1m from the LHS: To determine the bending moment at 1m from the LHS:
- 15. FBD for (b) 360 540 V2 2 M2 1202 1.0 Fy = 0 = 360 – 240 – V2 V2 = 120 kN Mat 2m = 0 = -540 + (360 2) - (240 1) – M2 M2 = - 540 + 720 - 240 = - 60 kNm To determine the shear force at 2m from the LHS: To determine the bending moment at 2m from the LHS:
- 16. •If we wanted to determine the SF and BM at any point along the beam, would need to go through this process each time. •You will agree this tedious. •We need an alternative method to enable us to determine SF and BM at any point along the beam. Shear force and bending moment diagrams. •Shear force diagrams and bending moment diagrams are graphs used to show the variations of the shear forces and bending moments along the length of a beam.
- 17. A steel beam 8m long is pin-jointed at the left-hand end and simply supported 4m from the right-hand end. The beam is loaded as shown. For the beam: (a) Determine the reactions at A and C. (b) Derive equations for the shear force and bending moment as a function of distance ‘x’ (horizontal displacement from the left-hand end). (c) Draw the shear force and bending moment diagrams for the beam, and determine the position of counteraflexure (should one exist). 2 2 1 3 15kN 2kN/m E D C B A
- 18. (a) Draw the FBD. 15kN E D C B A 2 3 = 6kN 1 Rax Ray Rc By inspection Rax = 0 Have TWO unknowns: Ray and Rc. Therefore, require TWO equations: Fy = 0 & M = 0 Fy = 0 = Ray – 15 + Rc – 6 Ray = 21 - Rc (1) MA = 0 = (15 2) – (Rc 4) + (6 6.5) 30 - 4Rc + 39 = 0 Rc = 69/4 = 17.25kN 2
- 19. Ray = 21 - 17.25 = 3.75 kN Sub for Rc into equation (1) (b) To derive the equations, split the beam into a number of spans. Start from the LHS and whenever you come across a force OR bending moment, you get a span for which an equation has to be derived. 15kN E D C B A 17.25 3.75 Span AB: 0 x 2 Span BC: 2 x 4 Span CD: 4 x 5 Span DE: 5 x 8 X=0 X=2 X=4 X=5 X=8
- 20. Span AB: 0 x 2 3.75 x Vab Mab Fy = 0 = 3.75 - Vab Vab = 3.75 kN MAB = 0 = 3.75x - Mab Mab = 3.75x kNm A check here is useful. M = 3.75x M/x = 3.75 = Vab
- 21. 3.75 Vbc Mbc x 15 Span BC: 2 x 4 Fy = 0 = 3.75 – 15 - Vcb Vbc = -11.25 kN MBC = 0 = 3.75x – 15(x – 2) - Mbc Mbc = -11.25x + 30 kNm x - 2 2 M = -11.25x + 30 M/x = -11.25 = Vbc CHECK: 0 = 3.75x – 15x + 30 - Mbc
- 22. 3.75 Vcd Mcd x 15 Span CD: 4 x 5 Fy = 0 = 3.75 – 15 +17.25 - Vcd Vcd = 6 kN MCD = 0 = 3.75x – 15(x – 2) + 17.25(x – 4) - Mcd Mcd = 6x - 39 kNm x - 2 2 M = 6x - 39 M/x = 6 = Vcd CHECK: 17.25 4 x - 4 0 = 3.75x – 15x + 30 + 17.25x – 69 - Mcd
- 23. 3.75 Vde Mde x 15 Span DE: 5 x 8 Fy = 0 = 3.75 – 15 +17.25 – (2x – 10) - Vde Vde = -2x + 16 kN MDE = 0 = 3.75x – 15(x – 2) + 17.5(x – 4) – (2x – 10)(0.5x – 2.5) - Mde Mde = -x2 + 16 - 64 kNm 17.25 4 x - 4 0 = 3.75x – 15x + 30 + 17.5x – 69 – x2 + 10x - 25 - Mde x - 5 2(x – 5) (x – 5)/2 0 = 3.75 – 15 + 17.25 – 2x + 10 - Vde
- 24. M = -x2 + 16x -64 M/x = -2x + 16 = Vde CHECK:
- 25. (c) To plot the shear force and bending moment diagrams is now a straight forward process of putting in the limits positions of ‘x’ in each of the equations and plotting the values on a graph.
- 26. 15kN E D C B A 3.75 6.0 6.0 -11.25 3.75 -11.25 -9.0 -15.0 7.5 Shear force diagram Bending moment diagram 0 0 0 0
- 27. Point of counterflexure is a point where the bending moment graph crosses the axis. There are situations where there is NO point of counterflexure, i.e. the bending moment graph DOES NOT cross the axis. In this question we do have point of counterflexure. This occurs over span BC, .i.e. when 2 x 4 At the point of counterflexure, the bending moment is EQUAL TO ZERO. Mbc = -11.25x + 30 = 0 x = 30/11.25 = 2.67m The point of counterflexure is when x = 2.67m