Shear Forces and Bending Moments

Transcription

1 Shear Forces and ending oments lanar (-D) Structures: ll loads act in the same plane and all deflections occurs in the same plane (-y plane) ssociated with the shear forces and bending moments are normal stresses and shear stresses.

2 Types of eams, oads, and Supports

3 rocedure for determining shear force and bending moment Determine the reactions using the euilibrium conditions of the overall structure Cut the beam at the cross section at which shear force and bending moment are to be determined. Draw a free-body diagram Set up euilibrium euations of the F..D. to determine shear force and bending moment at the cross section Draw the shear force and bending moment diagrams

4 Sign Convention Sign conventions for shear force and bending moment.

5 Find the shear force and the bending moment at cross sections to the left and to the right of the midpoint. eactions ( ) F O O O To the left of the midpoint: 8 O O F

6 To the right of the midpoint: 8 O O O F The shear force does not change (because the vertical force acting on the free body do not change) but the bending moment increases by an amount eual to O. To the left of the load O O F

7 O O O 8 O 8 O

8 Find the shear force and bending moment at a distance from the free end of the beam. eactions at (concentrate the distributed load) ( ) F O O O O O t a distance O Intensity of the distributed load.

9 () () O O

10 Find the shear force and bending moment at a distance 5ft from. eactions (concentrate the distributed load) F 9lb lb ( lb / ft)( ft ft) ( ft) ( lb)( 9 ft) ( lb / ft)( ft)( 5 ft) lb lb Shear Forces and bending moments: From to the concentrated load : <<9ft F

11 ( ) ( ) From the concentrated load to : 9<<ft F ( ) ( 9) ( ) Then for 5ft 5 ( 5) lb ( ) ( 5) 585lb ft

12 elationships between oads, Shear Stresses and ending oments Distributed oads F ertical ( d ) d ending oment (ccw as positive) d d d d ( d ) d ( d ) F ertical ( ) d d The concentrated loads cause abrupt changes in the shear force wherever they are located. d d d d ( ) d ( ) s the differentials are small, the bending moment does not change as we pass through the point of application of a concentrated load.

13 oads in the form of couples: F ertical ( ) The shear force does not change at the point of application of a couple. O O ( ) d ( ) The bending moment changes abruptly at the point of application of a couple.

14 evisit the following eample : Find the shear force and bending moment at a distance from the free end of the beam. O

15 C then For C d d d d o o O O O () C then For C d d o o O _ ()

16 Shear Force and ending oments Diagrams ial Forces Torsion

17 Shear Force and ending oment Diagrams Finding the reactions:

18

19 Shear-force and bending moment diagrams for a simple beam with several concentrated loads. eactions

20

21 Simple beam with a uniform load over part of the span.

22 The maimum bending moments occurs at the location where the shear force is zero

23 Cantilever beam with two concentrated loads.

24 Cantilever beam with a uniform load. d d d d d C For then () d C ( ) For then d For then ()

25 d d d C For then () d d d d C For then () For / then /8

26 C d d d a a a a D D C a a a a D D C a a D D C a ( ) D C d C d d d C a a C a a C a a C a

27 ( ) ( ) D D C a a

28 eam with an overhang. eactions C F ertical C 5.5k.5k C (.)( ) (.)( )( 8) ( ).

29 Simple beam F 7.5kN.5kN eactions ( )( 5.kN / m )( m ) ( 5.)( )( 7) ( )( ) ( 5.)( )( ) ( 8) d d For << then For then () ( ) d 5 C ( ) d d 7.5kN 5.kN-m kn-m -.5kN 5.kN-m For << then () -.5kN () kn-m.kn-m ( ).kn-m 5.kN-m -.5kN.5 C kn m C ( ) C C ( ) ( ) For << then () -.5kN ( ).5 C 5kN m C ( ) ( ).5 ( ) ( ) ( ).5 ( ) 5kN m ( ) () for.5 then a 5.5kN-m 5.5kN m (.5) (.) 5.kN m For <<8 5 C ( ) 5 ( ) ( ) C C ( 8) ( )

30 ethod of Superposition This can be treated as the superposition of four simple cases. The total shear force diagram is the sum of the individual shear force diagrams for each case. The total bending moment diagram is the sum of the individual bending moment diagrams for each case.

31

32 ( ) ()()( ) ()()( ) ()()( ) ()( ) ( ) eactions kn kn.5 7.5

33 ( ) ()()( ) ()()( ) ()()( ) ()( ) ( ) ( )

34 kn kn ( ) ( ) eactions k k C ( ) ( ) ( )( ) ( )( ) 8 8

35 kn kn ( ) ( ) ( ) ( ) ( )( ) ( )( )

36 eactions F kn ( )( kn ) kn ( )( ) ( )( ) ( )( ) kn m oading kn m kn m O O kn / m kn kn kn kn ( kn / m)( ) kn m kn ( )... ( )... ( )... ()... () ( )( ) 8 8.5( ) ( )... ()... ()... ().5

37 d d ( ) ( ) C C.5 C C The structure shown is constructed of a W rolled-steel beam. Draw the shear and bending-moment diagrams for the beam and the given loading..5 Solution eplace the kip load with an euivalent force-couple system at D. Find the reactions at by considering the beam as a rigid body. F ()( 8 ) kips 8kips d d ( 8) 9C ()()( 8 ) ( 5)( ) 8 d d C C ( ) d d ( 8) C ( ) C C 5 ( ) C 8 C 8 ft d d ( ) ( )

38 8 8 () 8 ( )... (8)... ( )... ( )... () ( ) ( ) ( ) (8) 9... ( ) 8... ( ) 8... ( ) 8

39

40 Draw the shear and bending moment diagrams for the beam shown. d d d d F C woa woa C C a C woa C woa a a w w o wo a C d d ( ) ( ) w C w w C o wo a wo a C w o w w o wo a wo a wo a C woa woa

41 ( ) ( ) wa wa wa C wa C wa wa wa C wa C d d o o o o a o o o o a a w a w a w a o o C o C

42 Shear and ending Stresses in eams d d constant Four oint ending (ure bending in the central region) eams under pure bending

43 Curvature of a eam ρ is the radius of curvature ds κ ρdθ ρ dθ ds Sign convention for the curvature For very small deflections dsd dθ κ ρ d ρ

44 Deformations of a beam in pure bending: (a) a side view dθ of beam, (b) cross section beam, and (c) deformed ρ d ρ beam. To evaluate normal strains, consider the line ef and assumed that the - ais is the neutral ais of the undeformed beam. The length of the line ef after bending takes place: dθ d y ρ ( ρ y) dθ d d s the original length of ef is d, then the elongation is y d ε d κy ρ For y positive there is a shortening of the length, that is a negative eleongation.

45 Normal Stresses in eams (linearly elastic materials) ε y d ρ σ Eε Ey ρ The stresses acting on the cross section vary linearly with the distance. y y ( ve) σ ( ve) ( ve) σ ( ve) It is necessary to locate the origin of the coordinates, that is the neutral ais of the cross section. Note: (a)the resultant force in the -direction is zero (pure bending). (b)the resultant moment is eual to the bending moment. (a) The resultant force in the -direction is zero (pure bending). Eyd ρ E ρ σ d yd yd This means that the first moment of the area of the cross section (evaluated with respect to the z-ais) is zero. The z-ais must pass through the centroid of the cross section.

46 (b) The resultant moment is eual to the bending moment. d yσ d y d I EI ρ yσ d y d E ρ E ρ oment-curvature Euation y d The element of moment acts opposite in direction to the positive bending moment. EI is called the Fleural igidity (measure of the resistance of a beam to bending) σ Ey ρ ρ σ Ey EI σ ρ Ey σ y I Fleure formula. It calculate the bending stresses or fleural stresses elationships between signs of bending moments and signs of curvatures.

47 lease Note: Consider the points m and m located at a distance and δ from the origin respectively. oint m has a deflection eual to ν and point m has a deflection eual to ν δν, where δν is the increment in deflection as we move from m to m. The angle of rotation is θ for point m and θ δθ for point m, i.e., the angle between the lines normal to the tangents at points m and m is δθ. The point of intersection of these normals is the center of curvature O and the distance O to m is the radius of curvature ρ.

48 ds Slope of Sinθ dυ ds If the angle of rotation θ is very small then δs ~ δ ; the curvature κ /ρ δθ / δ ; and tan θ ~ θ δν /δ. δθ δ υ δ δ δ υ δ EI EI ρ dθ κ ρ Cosθ d ds dθ ds the Deflection Curve tanθ dυ θ d ρ σ δ υ κ ρ Ey ρ EI δ Where is the bending moment and EI is the fleural rigidity of the beam. This euation is known as the differential euation of the deflection curve. This euation can be integrated in each particular case to find the deflection ν, provided the bending moment and fleural rigidity EI are known as functions of. dυ d dθ d

49 aimum stresses Doubly Symmetric Shapes σ σ c I c I S S bh I S bh S and S are known as the section moduli. asic ssumptions I πd S πd eam is slender; plane y is a plane of symmetry where the load is applied and the deflections takes place. The thickness remains constant. The ais of the beam passes through the centroid.

50