Unit II Shear and Bending in Beams

Size: px

Start display at page:

Download "Unit II Shear and Bending in Beams"

Sylvia Stokes
6 years ago
Views:

1 Unit II Shear and Bending in Beams Syllabus: Beams and Bending- Types of loads, supports - Shear Force and Bending Moment Diagrams for statically determinate beam with concentrated load, UDL, uniformly varying load. Theory of Simple Bending nalysis of Beams for Stresses Stress Distribution at a cross Section due to bending moment and shear force for Cantilever, simply supported and overhanging beams with different loading conditions Flitched Beams 1

2 Unit II Shear and Bending in Beams Objective: To know the mechanism of load transfer in beams and the induced stress resultants. 2

3 References: Unit II Shear and Bending in Beams 1. Rajput.R.K. Strength of Materials, S.Chand and Co, New Delhi, Bhavikatti. S., "Solid Mechanics", Vikas publishing house Pvt. Ltd, New Delhi, Junnarkar.S.B. andshah.h.j, Mechanics of Structures, Vol I, Charotar Publishing House, New Delhi,

4 Types of Loads: 1. Point loads: Unit II Shear and Bending in Beams 2 kn 2. Uniformly distributed load (UDL): The loads are uniformly applied over the entire length of the beam. It can be shown as follows: 2 kn/m 3. Uniformly varying load (UVL): Triangular or trapezoidal loads fall under this category. The variation in intensities of such loads is constant. It can be shown as follows: 2 kn/m 2 kn/m 2 kn/m 1 kn/m 4

5 Unit II Shear and Bending in Beams Types of supports: 1. Roller support: 2. Hinged support: 3. Fixed support: 5

6 Unit II Shear and Bending in Beams : 1. cantilever beams: cantilever beam which is fixed at one end and free at the other end. 2. Simply supported beams: simply supported beam rests freely on hinged support at one end and roller support at the other end. 3. Overhanging beams: If a beam extends beyond its supports it is called an overhanging beam. Over hanging portion could be either any one of the sides or both the sides. 6

7 Unit II Shear and Bending in Beams Statically indeterminate beams: 1. Propped cantilever beam: cantilever beam with propped support at the free end. 2. Fixed beam: beam with both supports fixed. 3. Continuous beam: beam with more than two supports. 7

8 Shear force: Unit II Shear and Bending in Beams Shear force at a section of a loaded beam may be defined as the algebraic sum of all vertical forces acting on any one side of the section. Sign Convention: Section line Section line + ve - ve 8

9 Unit II Shear and Bending in Beams Bending Moment: Bending moment at a section of a loaded beam may be defined as the algebraic sum of all moments of forces acting on any one side of the section. Sign Convention: Sagging moment: Moments which bend the beam upwards and cause compression in the top fibre and tension in the bottom fibre are taken as positive. Hogging Moment: Moments which bend the beam downwards and cause compression in the bottom and tension in the top fibre are taken as negative. +ve Sagging moment -ve Hogging moment 9

10 1. Cantilever beam subjected to a point load at the free end: (i) Shear force (S.F.) Calculations: Sign convention: Section line + ve Section line - ve X X L m x W B B= + W XX= +W = + W W W +ve S.F.D. W 10

11 1. Cantilever beam subjected to a point load at the free end: (ii) Bending Moment(B.M.) Calculations: Sign convention: +ve Sagging moment -ve Hogging moment X X L m x W B B= 0 XX= -Wx = - WL WL -ve B.M.D. Wx 0 11

12 1. Cantilever beam subjected to a point load at the free end: (iii) S.F.D. & B.M.D. Diagrams: W B W L m +ve S.F.D. W WL -ve B.M.D 12

13 2. Cantilever beam subjected to point loads as shown in Fig. Draw S.F.D. & B.M.D.: (i) Shear force (S.F.) Calculations: Sign convention: Section line + ve Section line - ve B= 0 C= +2 kn D (without Pt. Load at D)=+ 2 kn D (with Pt. Load at D)=+ 5 kn = + 5 kn 5 kn 1 m 3 kn D 2 m 5 kn 2 kn +ve S.F.D. 2 kn C 1 m 2 kn B 0 13

14 2. Cantilever beam subjected to point loads as shown in Fig. Draw S.F.D. & B.M.D.: (ii) Bending Moment(B.M.) Calculations: Sign convention: +ve Sagging moment B= 0 C= 0 D = 2 2 = 4 knm -ve Hogging moment 1 m 3 kn D -ve 2 m 4 knm 2 kn 0 C 1 m B 0 = = 9 knm 9 knm B.M.D 14

15 2. Cantilever beam subjected to point loads as shown in Fig. Draw S.F.D. & B.M.D.: 3 kn 2 kn (i) S.F.D. & B.M.D. D C B 1 m 2 m 1 m 5 kn 9 knm 5 kn 2 kn +ve S.F.D. -ve 2 kn 4 knm B.M.D. 0 15

16 3. Cantilever beam subjected to u.d.l as shown in Fig. Draw S.F.D. & B.M.D.: (i) Shear force (S.F.) Calculations: Sign = + : Section line Section line w/m L m X X x B + ve B= 0 XX= +wx = +wl - ve wl +ve S.F.D. wx 16

17 3. Cantilever beam subjected to u.d.l as shown in Fig. Draw S.F.D. & B.M.D.: (i) Bending Moment(B.M.) Calculations: W /m X Sign convention: -ve L m +ve X Hogging moment Sagging moment x B B= 0 XX= --wx x = wx2 2 2 = wl L = wl ve wx 2 2 wl 2 2 B.M.D. 17

18 3. Cantilever beam subjected to u.d.l as shown in Fig. Draw S.F.D. & B.M.D.: (iii) S.F.D & B.M.D: wl W /m L m X X x B wl 2 2 +ve S.F.D. -ve B.M.D. wx wx

19 3. Cantilever beam subjected to u.d.l as shown in Fig. Draw S.F.D. & B.M.D.: 1.5 m 1.5 m 4 kn/m B 19

20 3. Cantilever beam subjected to u.d.l as shown in Fig. Draw S.F.D. & B.M.D.: 6 kn 1.5 m -ve 6 kn +ve S.F.D. 1.5 m 4.5 knm 4 kn/m B 13.5 knm B.M.D. 20

21 4. simply supported beam of span L carries a central concentrated load W. Draw S.F.D. & B.M.D.: W L m 21

22 4. simply supported beam of span L carries a central concentrated load W. Draw S.F.D. & B.M.D.: W W/2 L m W/2 W/2 +ve -ve S.F.D. WL/4 +ve W/2 B.M.D. 22

23 4. simply supported beam of span L carries a eccentric concentrated load W as shown in Fig. Draw S.F.D. & B.M.D.: W a b L m 23

24 Solution: (i) Calculation of reactions: M = 0 +ve W C a b L m B R B L + W a = 0 R B = Wa L R R B F V = 0 +ve R + R B W = 0 R + Wa L W = 0 R = Wb L 24

25 Solution (contd ): (ii) Shear force (S.F.) Calculations: Sign Convention: Section line + ve Section line - ve S.F. at = Wb L S.F. at C (between and C)= Wb L S.F. at C(between C and B)= Wa L S.F. at B= Wa L Wb L Wb L a +ve W C b L m S.F.D. -ve B Wa L Wa L 25

26 Solution: (ii) Bending Moment(B.M.) Calculations: Sign Convention: M =0 +ve Sagging -ve Hogging Wb L a W C b L m Wab L +ve B Wa L M C = Wb L Wab a = L B.M.D. M B = 0 26

27 5. simply supported beam of span L carries two point loads as shown in Fig. Draw S.F.D. & B.M.D.: 4 kn 3 kn 1.5 m 3.5 m 1 m 27

28 Solution: (i) Calculation of reactions: M = 0 +ve R (R B 6) + (3 5) + (4 1.5) = 0 R B = 3. 5 kn 4 kn 3 kn 1.5 m 3.5 m 1 m B R B F V = 0 +ve R + R B 4 3 = 0 R = 0 R = 3. 5 kn 28

29 (ii) Shear force (S.F.) Calculations: Sign Convention: Section line + ve S.F. at =3.5 kn - ve 3.5 kn S.F. at C (between and C)=3.5 kn S.F. at C(between C and D)= 0.5 kn S.F. at D (between D and B)=-3.5 kn S.F. at B= 3.5 kn Section line 4 kn 3 kn C D B 1.5 m 3.5 m 1 m 3.5 kn kn S.F.D kn 3.5 kn 29

30 (iii) B.M.Calculations: Sign Convention: +ve Sagging M =0 4 kn 3 kn C D B -ve 1.5 m 3.5 m 1 m Hogging 3.5 kn 5.5 knm 3.5 kn 3.5 knm + M C = = 5.5 knm B.M.D. M D = = 3.5 knm M B = 0 30

31 5. simply supported beam of span L carries u.d.l throughout the. Draw S.F.D. & B.M.D.: w kn/m L m 31

32 5. simply supported beam of span L carries u.d.l throughout the. Draw S.F.D. & B.M.D.: X w kn/m B x X L m XX=0 wl/2 wl/2 wl wl/2 2 wx = 0 x = L/2 L/2 = wl2 8 + S.F.D. wl wl/2 B.M.D. 32

33 Over hanging beams: Problem: n over hanging beam of length 10 m is loaded as shown in Fig. Draw the S.F.D. and B.M.D. Mark the values at salient points. 15 kn/m 25 kn 20 kn/m C D B E 3 m 1 m 4 m 2 m 33

34 (i) Support Reactions: 15 kn/m 25 kn 20 kn/m C D B E M = 0 +ve 3 m 1 m 4 m 2 m R R B R B = 0 F V = 0 R B = kn +ve R = 0 R = kn 34

35 (ii) S.F. calculations: S. = kn S. XX = x 5x 2 = x 2 C= =1.88 kn S.F. between C&D=1.88 kn X 3 m 15 kn/m 25 kn 20 kn/m C D B E X Parabola m 4 m 2 m kn kn 35

36 (ii) S.F. calculations: 15 kn/m 25 kn 20 kn/m S.F. between D&B= C D B E m 1 m 4 m 2 m = kn R =24.37 R B = (including reaction at B)= =+40 kn Parabola E= = S.F.D. 36

37 (iii) B.M. calculations: M = 0 M C = =50.68 knm M D = =52.5 knm 3 m 15 kn/m 25 kn 20 kn/m C D B E 1 m 4 m 2 m R =24.37 R B = knm 37

38 (iii) B.M. calculations: M B = (20 2 1) = 40 knm M E =0 M XX = x (x 2) =0 x = m 3 m 15 kn/m 25 kn X 20 kn/m C D B E 1 m 4 m 2 m R =24.37 R B = Cubic parabola knm + B.M.D. x X Point of contraflexure - 40 knm Parabola 38

39 Simply supported beam: Problem: simply supported beam is loaded as shown in Fig. Draw the S.F.D. and B.M.D. Mark the values at salient points. 2 kn/m 3 knm C D 2 kn E B 1.5 m 0.5 m 1m 1 m 39

40 S.F.Diagram: X 2 kn 2 kn/m 3 knm C D E B XX= x = 0 x = 1.09 m 2.18 kn 1.5 m 0.5 m 1m 1 m 2.82 kn x 2.18 X S.F.D

41 B.M.Diagram: 1.09 m from =1.18 knm 2 kn/m 2 kn 2.18 kn 1.5 m 0.5 m 1m 1 m 2.82 kn 1.09 m X 1.18 X 3 knm C D E B 3.62 knm 2.82 knm + B.M.D. 41

42 Over hanging beam: Problem: Draw the shear force and bending moment diagram for the overhanging beam shown in Fig. Indicate the salient values on them. 5 kn B 20 kn 5 kn/m C D 2 knm E 1 m 2 m 3m 2 m 42

Procedure for drawing shear force and bending moment diagram:

Procedure for drawing shear force and bending moment diagram: Preamble: The advantage of plotting a variation of shear force F and bending moment M in a beam as a function of x' measured from one end of