Structure Design -1(Lecture 9 bm and sf solved examples)
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1) The document provides examples of calculating bending moment (BM) diagrams for beams with various loading conditions including point loads, uniformly distributed loads (UDL), and combined loading systems. 2) Key steps discussed include drawing shear force (SF) diagrams to determine the location of maximum BM, and calculating reactions before determining BM values. 3) For a simply supported beam, the maximum BM occurs where the SF is zero, and this point can be found by equating the total load to the left reaction.
Structure Design -1(Lecture 9 bm and sf solved examples)
- 1. Structural Design I Structural Mechanics BM SF Continued Lecture 9
- 3. 6kN/m Calculate the magnitude and location of maximum B in the beam shown above and draw the B M diagram We need to first draw the FBD: 1m 1m 1m 22kN 10kN 2m 22kN10kN 1m 1m 1m2m 6kN/m
- 4. To find the location of max BM we need to draw the SF diagram. For that we need to first calculate the reactions Rl × 5 = (22 × 1) + (18 × 3.5) + (10 × 4) = 22 + 63 + 40 = 125 Rl = 25kN Rr × 5 = (10 × 1) + (18 × 1.5) + (22 ×4) = 10 + 27 + 88 = 125 Rl = 25kN Rl + Rr = 10 + 18 + 22 = 50, Rl + Rr = 25 + 25 = 50 22m10m 1m 1m 1m2m 6kN/m
- 5. 1m 1m 1m2m 6kN/m 25 kN 25 kN 22 kN10 kN Just left of Rr SF = 25 kN, At 1 m from Rr, SF = 25 – 22 = 3 kN At 2 m from Rr, SF = 25 – 22 = 3 kN At 3 m from Rr, SF = 3 – 6 = -3kN At 4 m from Rr, SF = - 3 - 12 = -15 kN At 5 m from Rr, SF = -15 – 10 = -25 Max BM will lie where SF = 0. From above Sf =0 at 2.5 m from Rr Max BM at 2.5 m from Rr = (25 × 2.5) – (22 × 1.5) – (6 ×.5 × .25 ) = 62.5 – 33 – .75 = 28.75kNm
- 6. 1m 1m 1m2m 6kN/m 25 kN 25 kN 10 kN 22 kN BM at 1 m from Rl = (25 × 1) – ( 6 × 1 × .5) = 25 – 3 = 22 kNm BM at 2.5 m from Rl = 28.75 kNm BM at 3 m from Rl = (25 × 3) – (10 × 2) - (6 × 3 × 1.5) = 75 – 20 – 27 = 28kNm
- 7. Cantilever beam with partial UDL and Point Load A cantilever beam is loaded as shown. BM at D = 0 BM at C = 6×4×2=48kNm (-ve) BM at B = (6×4×4) = 96kNm (-ve) BM at A = (6×4×6) + (6×2) = 156kNm (-ve) From D to C BM will be a curve, from C to B it will be a straight sloping line and from B to A it will be a straight sloping line. SF will be 24 kN at C, increase by 6 at B to 30 kN And remain constant till A. It will slope from D to C, be parallel to base till B change vertically at B and be parallel to base till A.
- 8. Beams carrying Complicated Load Systems When a beam is carrying a number of loads both continuous and point loads then these can be drawn separately and then added geometrically to plot the total BM diagram. In the figure, say the BM produced by the UDL is a curve that has vertical values a, b and c and BM by point loads A, B and C at the load points by the point loads then the BM diagram can be plotted by taking values A+a, B+b and C+c at the load points as vertical ordinates.
- 9. Position of Maximum Bending Moment in Simply Supported Beams When only the maximum BM needs to be calculated without drawing the BM diagram the following steps can be adopted: The maximum bending moment occurs at the point where the Shear Force is zero. To find this point, first find the left side reaction Then proceed across from the left end until the load on the beam equals in total value the left end reaction. This where the shear force is equal to zero. Calculate the BM at this point by subtracting the total load moments from the reaction moment around that point.
- 10. RrRl 4kN/m2kN/m 4m2m In the beam shown above Rl×6 = (2×2×5)+(4×6×3) = 20+72 = 92 or Rl = 15.3 Rr×6 = (2×2×1)+(4×6×3) = 4+72 = 76 or Rr = 12.7 Rl + Rr = 28kN Total load = (2×2) + (4×6) = 28kN SF at right end of 2kN load = 15.3 – 12 = 3.3kN SF will be zero at a distance of 3.3÷4 = 0.825m right of this point or 2.825m from left end To cross check: SF will be zero at a distance from 12.7÷4 from right end =3.175m BM at 2.825m from left end = (15.3×2.825) – (2×2×1.825) – (4×2.825×1.412) =43.2 – 7.3 – 16=19.9kNm or 20kNm Determine the magnitude and position of the maximum BM in the beam loaded As shown..
- 11. Find the Maximum BM for the beam loaded as shown above 1kN/m 2kN/m 4kN 5m 18m4m 8m First the reactions Rl×18 = (4×5) + (2×22×11) + (1×8×18) Rl ×18 = 20 + 484 + 144 = 648 or Rl = 36 kN Rr×18 = (4×13) + (2×22×7) = 52 + 308 = 360 or Rr = 20kN Total load = 4 + (2×22) + (1×8) = 4 + 44 + 8 = 56 Rl + Rr = 36 + 20 = 56 Rl Rr
- 12. SF at A =0, at B = -(1×4) – (2×4) = -12 then moves up by Rl or +24 At C SF slopes down by (4×1)+ ( 4×2) = 12 so becomes (24 – 12) = 12 The load to the right of C is 2kN/m So SF will be 0 at 6 m to right of C Max BM will be at 10 m right of Rl Max BM = (36×10) – (8×10) – (2×14×7) = 360 – 80 – 196 = 84kNm