SAB2223 Mechanics of Materials and Structures

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1 S2223 Mechanics of Materials and Structures TOPIC 2 SHER FORCE ND ENDING MOMENT Lecturer: Dr. Shek Poi Ngian

2 TOPIC 2 SHER FORCE ND ENDING MOMENT

3 Shear Force and ending Moment Introduction Types of beams Effects of loading on beams The force that cause shearing is known as shear force The force that results in bending is known as bending moment Draw the shear force and bending moment diagrams

4 Shear Force and ending Moment Members with support loadings applied perpendicular to their longitudinal axis are called beams. eams classified according to the way they are supported.

5 Shear Force and ending Moment Types of beam a) Determinate eam The force and moment of reactions at supports can be determined by using the 3 equilibrium equations of statics i.e F x =0, F y =0, and M=0 b) Indeterminate eam The force and moment of reactions at supports are more than the number of equilibrium equations of statics. (The extra reactions are called redundant and represent the amount of degrees of indeterminacy).

6 Shear Force and ending Moment In order to properly design a beam, it is important to know the variation of the shear and moment along its axis in order to find the points where these values are a maximum.

7 Principle of Moments The moment of a force indicates the tendency of a body to turn about an axis passing through a specific point O. The principle of moments, which is sometimes referred to as Varignon s Theorem (Varignon, ) states that the moment of a force about a point is equal to the sum of the moments of the force s components about the point.

8 Principle of Moments In the 2-D case, the magnitude of the moment is M o = F d

9 eam s Reactions If a support prevents translation of a body in a particular direction, then the support exerts a force on the body in that direction. Determined using F x =0, F y =0, and M=0 Example 1: The beam shown below is supported by a pin at and roller at. Calculate the reactions at both supports due to the loading. 20 kn 40 kn 2 m 3 m 4 m

10 Example 1 Solution Draw free body diagram 20 kn 40 kn R x 2 m 3 m 4 m R y taking the moment at, ΣM = 0 R = 0 9R = R = 33.3 kn ΣF y = 0 R + R = 0 R = R = 26.7 kn R ΣF x = 0 R x = 0

11 Example 2 Determine the reactions at support and for the overhanging beam subjected to the loading as shown. 15 kn/m 20 kn 4 m 3 m 2 m

12 Example 2 (cont.) Solution Draw free body diagram R x 15 kn/m 20 kn R 4 m 3 m 2 m R y taking the moment at, ΣM = 0 R (15 3) 5.5 = 0 7R = R = kn ΣF y = 0 R + R = 0 R = R = 3.93 kn ΣF x = 0 R x = 0

13 Types of Support s a general rule, if a support prevents translation of a body in a given direction, then a force is developed on the body in the opposite direction. Similarly, if rotation is prevented, a couple moment is exerted on the body.

14 Example 3 cantilever beam is loaded as shown. Determine all reactions at support. 5 kn/m 3 20 kn 4 15 knm 2 m 2 m 1 m

15 Example 3 (cont.) Solution Draw free body diagram R x M R y 5 kn/m 3 20 kn 4 15 knm ΣF x = 0 R x + 20 (4/5) = 0 R x = 16 kn 2 m 2 m 1 m ΣF y = 0 R y 0.5 (5)(2) 20(3/5) = 0 R y 5 12 = 0 R y = 17 kn ΣM = 0 M + 0.5(5)(2)(1/3)(2) + 20(3/5) (4) + 15 = 0 M = M = 66.3 knm

16 Shear Force and ending Moment a P R x a R R M V M P V V = shear force = the force that tends to separate the member = balances the reaction R R

17 Shear Force and ending Moment M = bending moment = the reaction moment at a particular point (section) = balances the moment, R.x From the equilibrium equations of statics, + F y =0; R V = 0 V = R + M a =0; M + R.x = 0 M = R.x

18 Shear Force and ending Moment P F Q a R a a R b R a x 3 P x 2 F x 1 V M F y = 0 R a P F V = 0 V = R a P F M a = 0 M F.x 1 P.x 2 + R a.x 3 = 0 M = R a.x 3 F.x 1 P.x 2

19 Shear Force and ending Moment Shape deformation due to shear force V V Shape deformation due to bending moment M M Positive shear force diagram drawn above the beam Positive bending moment diagram drawn below the beam

20 Example 4 a) Calculate the shear force and bending moment for the beam subjected to a concentrated load as shown in the figure, and draw the shear force diagram (SFD) and bending moment diagram (MD). b) If P = 20 kn and L = 6 m, draw the SFD and MD for the beam. P kn L/2 L/2

21 Example 4 (cont.) Solution a) P kn R x R y L/2 L/2 R y y taking the moment at, ΣM = 0 R L + P L/2 = 0 R = P/2 kn ΣF y = 0 R + R = P R = P P/2 R = P/2 kn ΣF x = 0 R x = 0

22 Example 4 (cont.) Solution (cont.) P kn P/2 L/2 L/2 P/2 P/2 P/2 SFD MD (+) (-) P/2 P/2 0 (+) 0 PL / 4

23 Example 4 (cont.) Solution (cont.) b) 20 kn 10 kn 3 m 3 m 10 kn 10 kn 10 kn SFD MD (+) (-) 10 kn 10 kn 0 (+) 0 30 knm

24 Example 5 Calculate the shear force and bending moment for the beam subjected to a concentrated load as shown in the figure, then draw the shear force diagram (SFD) and bending moment diagram (MD). 15 kn 3 m 2 m

25 Example 5 (cont.) Solution a) 15 kn R x R y 3 m 2 m R y y taking the moment at, ΣM = 0 R = 0 R = 9 kn ΣF y = 0 R + R = 15 R = 15 9 R = 6 kn ΣF x = 0 R x = 0

26 Example 5 (cont.) Solution (cont.) b) 15 kn 6 kn 3 m 2 m 9 kn 6 kn 6 kn SFD (+) (-) MD 9 kn 9 kn 0 (+) 0 18 knm

27 Example 6 Calculate the shear force and bending moment for the beam subjected to an uniformly distributed load as shown in the figure, then draw the shear force diagram (SFD) and bending moment diagram (MD). 5 kn/m 3 m

28 Example 6 (cont.) Solution a) 5 kn/m R x R y 3 m R y y taking the moment at, ΣM = 0 R /2 = 0 R = 7.5 kn ΣF y = 0 R + R = 5 3 R = R = 7.5 kn ΣF x = 0 R x = 0

29 Example 6 (cont.) Solution (cont.) a) 5 kn/m 3 m 7.5 kn 7.5 kn 7.5 kn SFD (+) (-) MD 7.5 kn 0 (+) knm

30 Example 7 Calculate the shear force and bending moment for the beam subjected to the loads as shown in the figure, then draw the shear force diagram (SFD) and bending moment diagram (MD). 2 kn/m 3 m

31 Example 7 (cont.) Solution 2 kn/m R y 3 m R y y taking the moment at, ΣM = 0 2 3/2 3 2/3 R 3 = 0 R = 2 kn ΣF y = 0 R + R = 2 3/2 R = 3 2 R = 1 kn ΣF x = 0 R x = 0

32 Example 7 (cont.) Solution (cont.) 2 kn/m R y 3 m R y 1 kn SFD (+) (-) MD 0 (+) 0 2 kn knm

33 Example 7 (cont.) Solution (cont.) x 1 kn 1 2x/3(x)(1/2) V = 0 V = 1 2x 2 /6 If x = 0, V = 1 kn and x = 3, V = 2 kn M + 1 x 2x/3(x)(1/2) (x/3) = 0 M = x x 3 /9 dm M = maximum when 0 dx dm dx 2x/3 V M 2 3x x x m 3 Therefore, M maximum M = (1.732) (1.732) 3 /9 M = knm

34 Example 8 Calculate the shear force and bending moment for the beam subjected to the loads as shown in the figure, then draw the shear force diagram (SFD) and bending moment diagram (MD). 3 kn/m 4 m

35 Example 8 (cont.) Solution 3 kn/m M 4 m y taking the moment at, ΣM = 0 M = 3 4/2 4/3 M = 8 knm ΣF y = 0 R = 3 4/2 R = 6 kn R ΣF x = 0 R x = 0

36 Example 8 (cont.) Solution (cont.) 3 kn/m 8 knm 4 m 6 kn SFD (-) 6 kn MD 8 knm (-) 0

37 Example 9 Calculate the shear force and bending moment for the beam subjected to the loads as shown in the figure, then draw the shear force diagram (SFD) and bending moment diagram (MD). 3 knm 10 kn/m 12 kn 2 m 4 m 1 m

38 Example 9 (cont.) Solution 3 knm 10 kn/m 12 kn 2 m 4 m 1 m R y R y y taking the moment at, ΣM = 0 R / = 0 R = kn ΣF y = 0 R + R = R = R = kn ΣF x = 0 R x = 0

39 Example 9 (cont.) Solution (cont.) 3 knm 10 kn/m 12 kn SFD 2 m 4 m 1 m kn kn kn 12 kn (+) (+) (-) kn MD 3 knm 0 0 (+) 12 knm knm

40 Example 10 Calculate the shear force and bending moment for the beam subjected to the loads as shown in the figure, then draw the shear force diagram (SFD) and bending moment diagram (MD). 4 kn/m 5 kn 3 m 2 m

41 Example 10 (cont.) Solution 4 kn/m 5 kn M R y 3 m 2 m y taking the moment at, ΣM = 0 M / = 0 M = 43 knm ΣF y = 0 R = R = R = 17 kn ΣF x = 0 R x = 0

42 Example 10 (cont.) Solution (cont.) 43 knm 4 kn/m 5 kn 17 kn 3 m 2 m SFD 17 kn (+) 5 kn 43 knm MD (-) 10 knm 0

43 Example 11 Calculate the shear force and bending moment for the beam subjected to the loads as shown in the figure, then draw the shear force diagram (SFD) and bending moment diagram (MD). 25 kn 5 kn 20 kn C D E 1 m 1 m 2 m 3 m

44 Example 11 (cont.) Solution 25 kn 5 kn 20 kn R x C D E R y R y y taking the moment at, 1 m 1 m 2 m 3 m ΣM = 0 ΣF y = R y 7 = 0 R y + R y = R y = kn R y = R y = kn ΣF x = 0 R x = 0

45 Example 11 (cont.) Solution (cont.) SFD 25 kn 5 kn 20 kn C D E 0 kn 1 m 1 m 2 m 3 m kn kn MD

46 Example 12 Calculate the shear force and bending moment for the beam subjected to the loads as shown in the figure, then draw the shear force diagram (SFD) and bending moment diagram (MD). 10 kn/m 20 kn C D 4 m 4 m 2 m

47 Example 12 (cont.) Solution 10 kn/m 20 kn R x R y 4 m C 4 m R y 2 m D y taking the moment at, ΣM = R y 8 = 0 R y = 35 kn ΣF y = 0 R y + R y = R y = R y = 25 kn ΣF x = 0 R x = 0

48 Example 12 (cont.) Solution (cont.) 10 kn/m 20 kn SFD 0 C 25 kn 4 m 4 m 35 kn 25 4 x 20 x m D x 4 x x x 40x 100 x 2.5 MD

49 Example 13 Calculate the shear force and bending moment for the beam subjected to the loads as shown in the figure, then draw the shear force diagram (SFD) and bending moment diagram (MD). 10 kn/m C D 1 m 3 m 1.5 m

50 Example 13 (cont.) Solution 10 kn/m C D R x R y 1 m 3 m 1.5 m R y y taking the moment at, ΣM = R y 5.5 = 0 R y = kn ΣF y = 0 R y + R y = 10 3 R y = R y = kn ΣF x = 0 R x = 0

51 Example 13 (cont.) Solution (cont.) 10 kn/m 0 1 m kn C 3 m D 1.5 m kn SFD x 3 x x 3 x x x 30x x MD

52 Example 14 Calculate the shear force and bending moment for the beam subjected to the loads as shown in the figure, then draw the shear force diagram (SFD) and bending moment diagram (MD). 50 kn 50 kn 25 kn/m C D 5 m 5 m 4 m

53 Example 14 (cont.) Solution 50 kn 50 kn 25 kn/m R x C D R y 5 m 5 m R y 4 m y taking the moment at, ΣM = R y 10 = 0 R y = 220 kn ΣF x = 0 R x = 0 ΣF y = 0 R y + R y = R y = = 130 kn

54 Example 14 (cont.) Solution (cont.) 50 kn 50 kn 25 kn/m 0 5 m C 5 m 4 m D 130 kn 220 kn SFD MD 337.5

55 Contra Point of SF and M Contra point is a place where positive shear force/bending moment shifting to the negative region or vice versa. Contra point for shear: V = 0 Contra point for moment: M = 0 When shear force is zero, the moment is maximum. Maximum shear force usually occur at the support / concentrated load.

56 Contra Point of SF and M 10 kn/m 20 kn 0 C D 25 kn 4 m 4 m 35 kn 2 m SFD 25 4 x 20 Contra Point of Shear Force x (maximum negative moment) MD Contra Point of ending Moment (maximum positive moment)

57 Statically Determinate Frames For a frame to be statically determinate, the number of unknown (reactions) must be able to solved using the equations of equilibrium. 2 kn 4 kn/m ΣM = 0 ΣF y = 0 ΣF x = 0 R x R y R y

58 Example 15 Calculate the shear force and bending moment for the frame subjected to the loads as shown in the figure, then draw the shear force diagram (SFD) and bending moment diagram (MD). 4 kn/m 3 m 2 kn 3 m C R x D 4 m R y R y 5 m

59 Example 15 (cont.) Solution C 3 m 2 kn 3 m R x 4 kn/m D 4 m ΣM = R y 5 = 0 R y = 11.2 kn ΣF y = 0 R y + R y = 4 5 R y = = 8.8 kn R y 5 m R y ΣF x = 0 R x = 2 kn

60 Example 15 (cont.) Solution (cont.) M C 4 kn/m M D R Cy C D C M C 3 m 2 kn 3 m 2 kn M D = 0 knm 8.8 kn ΣM = 0: 2 3 M C = 0 M C = 6 knm ΣF y = 0: R cy = 8.8 kn R Dy R 5 m Cy R Dy ΣF y = 0: R Cy + R Dy = 4 5 R Dy = = 11.2 kn ΣM C = 0: M C R Dy 5 M D = 0 D 4 m 11.2 kn ΣF y = 0: R Dy = 11.2 kn

61 Example 15 (cont.) Solution 8.8 kn C x 2 kn 5 x 11.2 kn D C 6 knm 6 knm 15.7 knm 6 knm D x 5 x x x 20x 44 x 2.2 M max = = 15.7 knm

62 References 1. Hibbeler, R.C., Mechanics Of Materials, 8th Edition in SI units, Prentice Hall, Gere dan Timoshenko, Mechanics of Materials, 3rd Edition, Chapman & Hall. 3. Yusof hmad, Mekanik ahan dan Struktur Penerbit UTM 2001

Internal Internal Forces Forces

Internal Internal Forces Forces Internal Forces ENGR 221 March 19, 2003 Lecture Goals Internal Force in Structures Shear Forces Bending Moment Shear and Bending moment Diagrams Internal Forces and Bending The bending moment, M. Moment