1. Analysis of Beams and Frames Theory of Structure - I

2. Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan 2 Lecture Outlines Shear and Moment Diagrams for Beams Shear and Moment Diagrams for a Frames Moment Diagrams Constructed by the Method of Superposition Deflected Curves

3. Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan 3.. BC A D F1F1 F3F3 F2F2 w = w(x) M1M1. M2M2. w x xx w(x)xw(x)x xx xx w(x)w(x) O. M +  M V +  V  F y = 0: + +  M O  = 0: M V Shear and Moment Diagrams for a Beam

4. Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan 4 ----------(4-2) ----------(4-1) Dividing by  x and taking the limit as  x 0, these equation become Slope of Moment Diagram = Shear Slope of Shear Diagram = -Intensity of Distributed Load

5. Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan 5 ----------(4-4)----------(4-3) Equations (4-1) and (4-2) can be “integrated” from one point to another between concentrated forces or couples, in which case Change in Shear = -Area under Distributed Loading Diagram and Change in Moment = Area under Shear Diagram

6. Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan 6 xx M V M +  M V +  V. M´M´ O F xx M V M +  M V +  V +  M O  = 0:  F y = 0: + Thus, when F acts downward on the beam,  V is negative so that the shear diagram shows a “jump” downward. Likewise, if F acts upward, the jump (  V) is upward. In this case, if an external couple moment M ´ is applied clockwise,  M is positive, so that the moment diagram jumps upward, and when M acts counterclockwise, the jump (  M) must be downward.

7. Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan 7 MLML MRMR M´M´ 0 P VLVL VRVR MLML MRMR VLVL VRVR MLML MRMR w0w0 VLVL VRVR Slope = V L Slope = V R 0 0 MLML MRMR MLML MRMR 0 0 VLVL VRVR MRMR -w o Slope = V L Slope = V R MLML

8. Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan 8 VLVL VRVR MLML MRMR w1w1 w2w2 VLVL VRVR MLML MRMR w1w1 w2w2 VRVR VLVL MLML MRMR Slope = -w 1 Slope = -w 2 Slope = V R Slope = V L MLML VLVL VRVR Slope = w 1 Slope = -w 2 MRMR Slope = V R Slope = V L

9. Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan 9 Example 4-7 Draw the shear and moment diagrams for the beam shown in the figure. 9 m 20 kN/m

10. Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan 10 9 m 20 kN/m + SOLUTION (2/3)9 = 6 m (1/2)(9)(20) = 90 kN 30 kN 60 kN x V (kN) 30 + 60 - x M (kNm) V = 0 M  F y = 0: + x = 5.20 m x +  M x = 0: M = 104 kNm 104 V = 0 = 5.20 m x 30 kN

11. Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan 11 Example 4-8 Draw the shear and moment diagrams for each of the beam shown in the figure. P L L MOMO L wowo L wowo

12. Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan 12 + L MoMo SOLUTION P L P PL 0 0 MoMo 0 x V P + x M -PL - x V x M MoMo

13. Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan 13 L wowo wo Lwo L 0 wo L2wo L2 2 L wowo (w o L)/2 0 wo L2wo L2 6 x V wo Lwo L + x M -w o L 2 2 - x V (w o L)/2 + x M -w o L 2 6 -

14. Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan 14 Example 4-9 Draw the shear and moment diagrams for the beam shown in the figure. 3 kN 5 kNm AB CD 3 m1.5 m

15. Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan 15 3 kN 5 kNm AB CD 3 m1.5 m SOLUTION 0.67 kN 2.33 kN V (N) x (m) 0.67 + -2.33 - M (kNm) x (m) 2.01 + -1.49 3.52 - +

16. Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan 16 Example 4-10 Draw the shear and moment diagrams for the compound beam shown in the figure. Assume the supports at A is fix C is roller and B is pin connections. 12 m 15 m 8 kN 30 kNm A B C hinge

17. Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan 17 AyAy AxAx MAMA SOLUTION CyCy ByBy BxBx BxBx ByBy 0 = = 2 kN = 0 = 6 kN 8 kN 30 kNm = 48 kNm 12 m 15 m 8 kN 30 kNm A B C hinge

18. Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan 18 V (kN) x (m) 6 6 -2 x (m) M (kNm) -48 24 -30 8 m - + - 12 m 15 m 8 kN A B C 30 kNm 6 kN 48 kNm 2 kN

19. Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan 19 Example 4-11 Draw the shear and moment diagrams for the compound beam shown in the figure. Assume the supports at A and C are rollers and B and D are pin connections. 5 kN 3 kN/m 2 kN/m 60 kN m Hinge 10 m6 m4 m6 m A B C D

20. Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan 20 ByBy BxBx BxBx ByBy AyAy CyCy DyDy DxDx = 0 kN = 16 kN = 4 kN = 16 kN 0 = = 0 kN = 45 kN= -6 kN SOLUTION 5 kN 9 kN 4 m 60 kN m 20 kN 5 kN 3 kN/m 2 kN/m 60 kN m Hinge 10 m6 m4 m6 m A B C D

21. Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan 21 V (kN) x (m) -16 -21 24 6 M (kN m) x (m) 2 m 60 64 -96 -180 + - 4 5 kN 2 kN/m 60 kN m Hinge 10 m6 m 4 m 6 m B C D 3 kN/m A 4 kN 45 kN 6 kN

22. Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan 22 P1P1 B P2P2 AyAy CxCx CyCy BxBx ByBy MBMB BxBx ByBy MBMB BxBx ByBy MBMB BxBx ByBy MBMB P1P1 P2P2 AyAy CxCx CyCy P1P1 P2P2 A B C Shear and Moment Diagrams for a Frame

23. Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan 23 CxCx CyCy BxBx ByBy MBMB MBMB P2P2 P 2 = B x P2P2 AyAy BxBx ByBy MBMB ByBy B y = C y MBMB MBMB MBMB A B C

24. Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan 24 Example 4-12 Draw the shear and moment diagrams for the frame shown. Assume A, C and D are pinned and B is a fixed joint. 3 kN/m 15 kN 4 m 12 m 60 kNm A B C D

25. Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan 25 Find the Reaction 15 kN 4 m B 36 kN 60 kNm AxAx AyAy DxDx DyDy = 5 kN = 42 kN = 41 kN = -27 kN CxCx CyCy CxCx CyCy = 5 kN = 42 kN 3 kN/m 15 kN 4 m 12 m 60 kNm A B C D

26. Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan 26 V (kN) x (m) BxBx ByBy MBMB Member AB = 276 kNm 12 m 3 kN/m A B A x =41 kN A y = 27 kN = 5 kN = 27 kN 41 5 M (kNm) x (m) 276

27. Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan 27 - - V (kN) x (m) 15 kN 4 m B 5 kN 42 kN C Member BC 41 kN 27 kN BxBx ByBy MBMB 12 m 3 kN/m A B = 5 kN = 27 kN = 276 kNm 5 kN 27 kN 276 kNm -27 -42 M (kNm) x (m) 276 168

28. Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan 28 - V (kN) x (m) 60 kNm 5 kN 42 kN D C 12 m Member CD -5 M (kNm) x (m) 60 +

29. Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan 29 Bending moment diagram of frame 276 A B + 168 C + 60 D + 3 kN/m 15 kN 4 m 12 m 60 kNm A B C D

30. Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan 30 Example 4-13 Draw the moment diagram for the frame shown. Assume A is pin, C is a roller, and B is a fixed joint. 40 kN/m 80 kN 4 m 2 m 3 m A B C

31. Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan 31 120 kN 1.5 m 36.87 o 80 kN 4 m 2 m 3 m B AyAy AxAx CyCy +  M A = 0; +  F x = 0; +  F y = 0; 82.5 kN = - (120)(1.5) - (80)(6) + 8C y = 0 C y = 82.5 kN, ญ 120 kN = -A x + 120 = 0; A x = 120 kN, ฌ = 2.5 kN - A y - 80 + 82.5 = 0; A y = 2.5 kN, ฏ

32. Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan 32 BxBx ByBy MBMB BxBx ByBy MBMB B 80 kN 82.5 kN B C 2 m 2.5 kN 120 kN A B 36.87 o 1.5 m By´By´ Bx´Bx´ MB´MB´ By´By´ Bx´Bx´ MB´MB´ +  M B = 0: Member BC +  F y = 0: 2.5 kN = -B y - 80 + 82.5 = 0, B y = 2.5 kN, ฏ 36.87 o B y´ cos 36.87 B y´ sin 36.87 -B x´ cos 36.87 + B y´ sin 36.87 + 0 = 0-----(1) Joint B +  F x = 0; 36.87 o B x´ cos 36.87 B x´ sin 36.87 -B x´ sin 36.87 - B y´ cos 36.87 + 2.5 = 0-----(2) +  F y = 0; =2.5 kN = 0 kN =170 kNm 170 kNm= From eq. (1) and (2): B x´ = 1.5 kN B y´ = 2 kN 1.5 kN = = 2 kN =1.5 kN =170 kNm 2 kN= 0 kN = 170 kNm = -M B -80(2) + 82.5(4) = 0, M B = 170 kNm

33. Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan 33 1.5 kN 170 kNm 2 kN B 120 kN 36.87 o 2.5 kN 120 kN 36.87 o 53.13 o 5 m 120sin36.87 o 1.5 kN 170 kNm 2 kN 70 kN 97.5 kN 120 sin 36.87/5 =14.4 kN/m 70 -2 x (m) M (kNm) 4.86m + - 170.1 170 x (m) V (kN)

34. Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan 34 40 kN/m 80 kN A B C - - 82.5 kN B C 170 kNm 2.5 kN V (kN)x (m) -2.5 -82.5 M (kNm)x (m) 170 165 170 + 165 + A C B M (kNm)

35. Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan 35 L P M x -PL L wowo M x Parabolic curve Most loading on beams in structural analysis will be a combination of the loadings shown in the figure below: Moment Diagrams Constructed by the Method of Superposition

36. Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan 36 M x MoMo L MoMo L wowo M x Cubic curve

37. Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan 37 M (kNm) x (m) -20 70 = M (kNm) x (m) 90 M (kNm) x (m) -20 + M (kNm) x (m) -20 + 12 m 20 kNm 5 kN/m 12 m 5 kN/m = 12 m 20 kNm + 12 m 20 kNm +

38. Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan 38 Example 4-14 Draw the moment diagrams for the beam shown at the top of the figure below using the method of superposition. Consider the beam to be cantilevered from the support at B. 6 m 20 kNm 5 kN/m 2 m

39. Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan 39 6 m 20 kNm 5 kN/m 2 m SOLUTION 8.33 kN 6.67 kN 6 m 8.33 kN + 6 m 5 kN/m + 20 kNm = M (kNm) x (m) 49.98 + M (kNm) x (m) -20 M (kNm) x (m) 4.84 = M (kNm) x (m) -30 +

40. Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan 40 + M positive moment, concave upward - M negative moment, concave downward P1P1 P2P2 M x inflection point M x P1P1 P2P2 Deflected Curve

41. Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan 41