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Analysis of Beam Reactions and Drawing Bending Moment and Shear Force Diagrams

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This document outlines the step-by-step analysis of a beam structure to determine its stability, reactions, bending moment diagram (BMD), and shear force diagram (SFD). It includes the calculation of unknown reactions and the verification of stability through equilibrium equations. The analysis encompasses various sections of the beam, considering the effects of applied loads and reactions. Additionally, conditions for intermediate pins are discussed, leading to the final determination of key forces and moments in the structure.

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Analysis of Beam Reactions and Drawing Bending Moment and Shear Force Diagrams

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  1. Xb Xa 2m 2m 2m 2m Yb Ya Yc Draw the B.M.D. & the S.F.D. 2 t 2 t / m 1m Assumed reactions Step 1: Stability Check No. of Unknown Reactions? 5 No. of Equilibrium Equations: 3 No. of Extra Conditions: 2 (two intermediate pins) Since the unknowns5 = Equations (3 + 2) Then, Structure is Stable & Statically Determinate Tarek Hegazy, Univ. of Waterloo tarek@uwaterloo.ca

  2.  X + +  Y • Sign Convention • =0 • =0 • =0  M  M + + Since (e) is an intermediate pin, then at (e) right side only = 0 Any point For the whole structure considering all forces including reactions 2 t e Xb 2 t / m 1m Xa d Yb 2m 2m 2m 2m Ya Yc Step 2: Reactions Start writing an equation with least number of unknowns: Tarek Hegazy, Univ. of Waterloo tarek@uwaterloo.ca

  3. 2 t  M + Yb Sign Distance Force at (e), right side only = 0 e Xb 2m 2m +Yb .4 -2 .2 +Xb . 0 = 0 Solve, Yb = +1i.e., correct direction Tarek Hegazy, Univ. of Waterloo tarek@uwaterloo.ca

  4.  M + Since (d) is an intermediate pin, then at (d) right side only = 0 2 t Xb 2 t / m 1m Xa d 1 2m 2m 2m 2m Ya Yc Let’s write another equation: Tarek Hegazy, Univ. of Waterloo tarek@uwaterloo.ca

  5. 1m 2 t Xb 1  M + Sign Distance Force at (d), right side only = 0 2m 2m d +1 .4 -2 .2 +Xb .1= 0 Solve, Xb = 0 Tarek Hegazy, Univ. of Waterloo tarek@uwaterloo.ca

  6. 0 0 Xa  X + Now, = 0 , +Xa - 0 = 0, or Xa = 0  M + Since (d) is an intermediate pin, then at (d) left side only = 0 2 t 2 t / m 1m d 1 2m 2m 2m 2m Ya Yc Let’s now write two equations to get the last two unknowns: Tarek Hegazy, Univ. of Waterloo tarek@uwaterloo.ca

  7. Equivalent = 2 x 4 = 8  M + Sign Distance Force at (d), left side only = 0 2 t / m d 2m 2m Ya Yc - Yc.2 - Ya.4 + 8 .2 = 0 Or, 4 Ya + 2 Yc = 16 . . . . . (1) Tarek Hegazy, Univ. of Waterloo tarek@uwaterloo.ca

  8. + Y=0 All Down All Up Now, then, 2 t 2 t / m 1m 1 2m 2m 2m 2m Ya Yc 4 Ya + 2 Yc = 16 . . . (1) +Ya +Yc +1 -2x4 -2 = 0 Ya + Yc = 9 . . . (2) Solve (1) & (2), Ya = -1 , Yc = 10 Opposite direction Same direction Tarek Hegazy, Univ. of Waterloo tarek@uwaterloo.ca

  9. Final Reactions: Shear is Paralle to the section & Perpendicular to the beam Axial (i.e., Normal) force is parallel to the beam Section 2 Section 3 7 10 8 6 9 1 2 5 3 4 2 t 4 4 2 t/m 1m 1 2m 2m 2m 2m 1 10 Let’s first define the beam sections with changes in load or beam’s Shape. We are now ready to draw the B.M.D. and the S.F.D. Now, we analyze each section separately, considering only one side of the structure. Tarek Hegazy, Univ. of Waterloo tarek@uwaterloo.ca

  10. B.M.D. + 2 t 4 4 2 t/m 1m 1 2m 2m 2m 2m 1 10 Section Analysis: Analyze the right side or the left side, whichever has less calculation. 7 10 8 6 9 2 1 5 3 4 Tarek Hegazy, Univ. of Waterloo tarek@uwaterloo.ca

  11. B.M.D. + 2 t 4 4 2 t/m 1m Section 1 Left S 1 2m 2m 2m 2m 1 10 Section 1 B.M. = -1 . 0 = 0 7 10 8 6 9 2 1 5 3 4 0 Tarek Hegazy, Univ. of Waterloo tarek@uwaterloo.ca

  12. B.M.D. + 0 -6 2 t 4 4 2 t/m 1m S Section 2 Left 1 2m 2m 2m 2m 1 10 Section 2 B.M. = -1 . 2 - 4 . 1 = - 6 7 10 8 6 9 2 1 5 3 4 Tarek Hegazy, Univ. of Waterloo tarek@uwaterloo.ca

  13. B.M.D. + 0 -6 2 t 4 4 2 t/m 1m S Section 3 Left 1 2m 2m 2m 2m 1 10 Section 3 B.M. = - 1 . 2 - 4 . 1 +10 . 0 = -6 7 10 8 6 9 2 1 5 3 4 Tarek Hegazy, Univ. of Waterloo tarek@uwaterloo.ca

  14. B.M.D. + 0 0 -6 2 t 4 4 2 t/m 1m S Section 4 Left 1 2m 2m 2m 2m 1 10 Section 4 B.M. = -1 . 4 -4 . 3-4 . 1 +10 . 2 = 0 7 10 8 6 9 2 1 5 3 4 Tarek Hegazy, Univ. of Waterloo tarek@uwaterloo.ca

  15. B.M.D. + 0 0 -6 2 t 4 4 2 t/m S Section 5 Left 1m 1 2m 2m 2m 2m 1 10 Section 5 B.M. = 0 (same as section 4) 7 10 8 6 9 2 1 5 3 4 Tarek Hegazy, Univ. of Waterloo tarek@uwaterloo.ca

  16. B.M.D. + 0 0 0 -6 2 t 4 4 S Section 10 Right 2 t/m 1m 1 2m 2m 2m 2m 1 10 Section 10 B.M. = +1 . 0 = 0 10 7 8 6 9 2 1 5 3 4 Tarek Hegazy, Univ. of Waterloo tarek@uwaterloo.ca

  17. B.M.D. + +2 0 0 0 -6 2 t 4 4 S 2 t/m Section 9 Right 1m 1 2m 2m 2m 2m 1 10 Section 9 B.M. = +1 . 2 = +2 10 7 8 6 9 2 1 5 3 4 Tarek Hegazy, Univ. of Waterloo tarek@uwaterloo.ca

  18. B.M.D. + +2 0 0 0 -6 2 t 4 4 S 2 t/m Section 8 Right 1m 1 2m 2m 2m 2m 1 10 Section 8 B.M. = +1 . 2 = +2 10 7 8 6 9 2 1 5 3 4 Tarek Hegazy, Univ. of Waterloo tarek@uwaterloo.ca

  19. B.M.D. + +2 0 0 0 0 -6 2 t 4 4 S S 2 t/m Sections 7 & 6 Right 1m 1 2m 2m 2m 2m 1 10 Sections 7 & 6 B.M. = 0 10 7 8 6 9 2 1 5 3 4 Tarek Hegazy, Univ. of Waterloo tarek@uwaterloo.ca

  20. B.M.D. -3 -3 w.L2/8 = 1 w.L2/8 = 1 2 t 4 4 2 t/m 1m 1 1 10 2m 2m 2m 2m Now, we connect the moment values +2 0 0 0 0 -6 Tarek Hegazy, Univ. of Waterloo tarek@uwaterloo.ca

  21. S.F.D. + -1 2 t 4 4 Section 1 Left 2 t/m 1m S 1 2m 2m 2m 2m 1 10 Shear is a force perpendicular to the beam Section 1 S.F. = -1 7 10 8 6 9 2 1 5 3 4 Tarek Hegazy, Univ. of Waterloo tarek@uwaterloo.ca

  22. S.F.D. + -1 -5 2 t 4 4 2 t/m 1m Section 2 Left S 1 2m 2m 2m 2m 1 10 Section 2 S.F. = -1 - 4 = - 5 7 10 8 6 9 2 1 5 3 4 Tarek Hegazy, Univ. of Waterloo tarek@uwaterloo.ca

  23. S.F.D. + +5 -1 -5 2 t 4 4 2 t/m Section 3 Left 1m S 1 2m 2m 2m 2m 1 10 Section 3 S.F. = - 1 - 4 +10 = +5 7 10 8 6 9 2 1 5 3 4 Tarek Hegazy, Univ. of Waterloo tarek@uwaterloo.ca

  24. S.F.D. + +5 +1 -1 -5 2 t 4 4 2 t/m Section 4 Left 1m S 1 2m 2m 2m 2m 1 10 Section 4 S.F. = - 1 - 4 +10 - 4 = +1 7 10 8 6 9 2 1 5 3 4 Tarek Hegazy, Univ. of Waterloo tarek@uwaterloo.ca

  25. S.F.D. + +5 +1 -1 -5 2 t 4 4 2 t/m S Section 5 Left 1m 1 2m 2m 2m 2m 1 10 Section 5 S.F. = 0 7 10 8 6 9 2 1 5 3 4 Tarek Hegazy, Univ. of Waterloo tarek@uwaterloo.ca

  26. S.F.D. + +5 +1 -1 -5 2 t 4 4 S Section 6 Left 2 t/m 1m 1 2m 2m 2m 2m 1 10 Section 6 S.F. = 0 7 10 8 6 9 2 1 5 3 4 Tarek Hegazy, Univ. of Waterloo tarek@uwaterloo.ca

  27. S.F.D. + +5 +1 -1 -1 -5 2 t 4 4 S Section 10 Right 2 t/m 1m 1 2m 2m 2m 2m 1 10 Section 10 S.F. = -1 7 10 8 9 6 2 1 5 3 4 Tarek Hegazy, Univ. of Waterloo tarek@uwaterloo.ca

  28. S.F.D. + +5 +1 -1 -1 -1 -5 2 t 4 4 S 2 t/m Section 9 Right 1m 1 2m 2m 2m 2m 1 10 Section 9 S.F. = -1 10 8 9 7 6 2 1 5 3 4 Tarek Hegazy, Univ. of Waterloo tarek@uwaterloo.ca

  29. S.F.D. + +1 +5 +1 -1 -1 -1 -5 2 t 4 4 S 2 t/m Section 8 Right 1m 1 2m 2m 2m 2m 1 10 Section 8 S.F. = -1 +2 = +1 10 8 9 7 6 2 1 5 3 4 Tarek Hegazy, Univ. of Waterloo tarek@uwaterloo.ca

  30. S.F.D. + +1 +1 +5 +1 -1 -1 -1 -5 2 t 4 4 S 2 t/m Sections 7 Right 1m 1 2m 2m 2m 2m 1 10 Sections 7 S.F. = -1 +2 = +1 10 8 9 7 6 2 1 5 3 4 Tarek Hegazy, Univ. of Waterloo tarek@uwaterloo.ca

  31. S.F.D. 2 t 4 4 2 t/m 1m 1 1 10 2m 2m 2m 2m +1 +1 +5 Now, we connect the shear values +1 -1 -1 -1 -5 Tarek Hegazy, Univ. of Waterloo tarek@uwaterloo.ca

  32. +2 0 0 0 0 -3 -3 w.L2/8 = 1 w.L2/8 = 1 -6 2 t Final Answer 4 4 2 t/m 1m 1 1 10 2m 2m 2m 2m +1 +1 +5 +1 -1 -1 S.F.D. -1 -5 B.M.D Tarek Hegazy, Univ. of Waterloo tarek@uwaterloo.ca

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