Bending Stress and Strain

Transcription

1 Bending Stress and Strain DEFLECTIONS OF BEAMS When a beam with a straight longitudinal ais is loaded by lateral forces, the ais is deformed into a curve, called the deflection curve of the beam. We will determine the equations for finding the deflection curve and also find the deflections at specific points along the ais of the beam.

2 Differential equations of the deflection curve Consider a cantilever beam with a concentrated load acting upward at the free end. Under the action of the load, the ais of the beam deforms into a curve. The reference aes have their origin at the fied end of the beam. is positive to the right and y is positive upwards. The deflection ν is the displacement in the y direction of any point on the ais of the beam. We must epress ν as a function of the coordinate.

3 When the beam is bent, there is not only a deflection at each point along the ais but also a rotation. The angle of rotation (also known as angle of inclination and angle of slope) of the ais of the beam is the angle between the -ais and the tangent to the deflection curve. Consider the points m and m located at a distance and δ from the origin respectively. Point m has a deflection equal to ν and point m has a deflection equal to ν δν, where δν is the increment in deflection as we move from m to m.

4 The angle of rotation is for point m and δ for point m, i.e., the angle between the lines normal to the tangents at points m and m is δ. 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. δ ρ δ δσ Where δs is the distance along the deflection curve between m and m. Therefore the curvature κ ρ δ δs The slope of the deflection curve is the first derivative δν/δ and is equal to tan. Similarly, sin δν/δs and cos δ/δs

5 If the angle of rotation is very small then δs ~ δ ; the curvature k /ρ δ /δ ; and tan dν /d. Thus, if the rotations of the beam are very small, we can assume that the angle of rotation (in radians) and the slope δν /δ are equal. The bending moment causes tension at the bottom and compression at the top. At the neutral surface there is no tension or compression, then y c Ma σ y σ c Ma

6 Taking the derivative of with respect to : If the material is linearly elastic and it obeys Hooke s law, the curvature is: M is the bending moment and EI is the fleural rigidity of the beam. This equation is known as the differential equation of the deflection curve. It can be integrated in each particular case to find the deflection ν, provided the bending moment M and fleural rigidity EI are known as functions of. δ δ υ δ δ M κ ρ EI δ υ δ Sign Conventions The and y aes are positive to the right and upwards, respectively. The deflection ν is positive upwards. The slope δν /δ and angle of rotation are positive when CCW with respect to the positive ais. The curvature κ is positive when the beam is bent concave upward The bending moment M is positive when it produces compression in the upper part of the beam. M EI δ δ υ

7 Additional equations can be obtained from the relations between bending moment M, shear force V, and intensity q of distributed load V M q V EI M δ δ δ δ δ υ δ We will consider two different cases: Non prismatic beams (EI changes with ) Prismatic beams (EI constant) Nonprismatic Beams q V M EI V M EI δ δ δ δ δ υ δ δ δ δ δ δ υ δ δ δ

8 Prismatic Beams These equations will be referred to as the bending-moment equation, the shear force equation and the load equation, respectively. δ υ δm EI δ δ 4 δ υ δ M EI 4 δ δ V δv δ q Deflections by Integration of the Bending-Moment Equation Regardless of the number of bending-moment epressions, the general procedure for solving the differential equations is as follows:. For each region of the beam we substitute the epression for M into the differential equation and integrate to obtain the slope ν δν /δ.. Each such integration produces one constant of integration.. Net, we integrate the slope equation to obtain the corresponding deflection ν.

9 . Again each integration produces a new constant. 4. The two constants of integration for each region of the beam are evaluated from known conditions pertaining to the slopes and deflections. The conditions fall into three categories: () Boundary Conditions () Continuity Conditions () Symmetry Conditions. Boundary conditions: Pertain to the deflections and slopes at the supports of a beam. Eg. At a simple support (either pin or roller) the deflection is zero and at a fied support both the deflection and the slope are zero.

10 . Continuity conditions: They occur at points where the regions of integration meet, (point C in the beam shown below). The deflection curve for this beam is physically continuous at point C. Therefore the deflection of point C as determined for the left and right hand part of the beam must be equal. Similarly, the slopes found for each part of the beam must be equal at point C.

11 . Symmetry conditions: They may also be available. Eg. If a simple beam supports a uniform load throughout its length, we know in advance that the slope of the deflection curve at the midpoint must be zero. Each boundary, continuity and symmetry condition leads to an equation containing one or more of the constants of integration. The number of independent conditions always matches the number of constants of integration, we can always solve these equations for the constants. This method is sometimes referred to as the method of successive integrations.

12 Eample Determine the equation of a deflection curve for a simple beam AB supporting a uniform load of intensity q acting throughout the span of the beam, as shown in the figure. Also, determine the maimum deflection δ ma at the midpoint of the beam and the angles of rotation A and B at the supports. (Note: the beam has length L and constant fleural rigidity EI).

13 Solution Bending moment in the beam. The bending moment at the cross section distance from the left-hand support is obtained from the free-body diagram. M ql δ υ δ ( ) q EI This equation can now be integrated to obtain the slope of the beam. Slope of the beam: EI υ δ ql δ q q EIυ ql C 4 6 δ C is a constant of integration

14 Symmetry Condition: the slope of the deflection curve at midspan is equal to zero. ( ) L ( L ) q EIυ L ql q υ 4EI 4 6 ( L 6L 4 ) C 0 C ql 4 As epected, the slope is negative (i.e. clockwise) at the left-hand end of the beam (0), positive( i.e. anticlockwise) at the right-hand end (L), and equal to zero at the midpoint ( ½ L). Deflection of the beam : The deflection is obtained by integrating the equation for the slope. q EI υ ql ql C

15 Boundary Condition: the deflection of the beam at the left-hand support is equal to zero (ν 0 when 0) or ν(0) 0 Applying this condition yields C 0; hence the equation of the deflection curve is q υ 4EI ( L L ) Maimum deflection :From symmetry we know that the maimum deflection occurs at the midpoint of the span. Thus, for L/ we obtain Angles of Rotation υ L 5qL 4 84EI Downward deflection. The maimum angles of rotation occur at the supports of the beam. Then for 0 and for L ql ql A υ ( 0) cw B υ( L) ccw 4EI 4EI

16 Eample Determine the equation of the deflection curve for a cantilever beam AB subjected to a uniform load of intensity q (see figure). Also, determine the angle of rotation B and the deflection δ B at the free end. (The beam has a length L and a constant fleural rigidity EI). Solution Equation of the slope of the beam Bending moment equation ql q M EIυ ql ql ql q EIυ C 6

17 Boundary Condition: slope of the beam is zero at the support ν (0) 0, then C 0 Deflection of the Beam: Integration of the slope ql EIυ q υ 6EI ql ql q EIυ boundary _ conditions υ q υ 4EI ( 6L 4L ) ql ( L L ) 4 0 C Angle of rotation and deflection at the free-end of the beam: 0 then C q 6 0 B ql 6EI υ L δ B υ L 4 ql 8EI

18 Deflections by Integration of the Shear-Force and Load Equations The equations EIν V and EIν - q in terms of the shear load V and the distributed load q may also be integrated to obtain the slopes and deflections. Eample Determine the equation of the deflection curve for a cantilever beam AB supporting a triangularly distributed load of maimum intensity q o (see figure below). (Note: the beam has length L and constant fleural rigidity EI)

19 Solution: Differential equation of the deflection curve: The intensity of the distributed load is given by the following equation: qo( L ) q The fourth - order differential equation becomes L qo( L ) qo ( L ) EI '''' q V EIν ''' C ν L L Shear Force in the beam: The first integration of the above equation gives Because the shear force is zero at L then ν (L) 0 and C 0 The equation simplifies to : V EIν ''' q ( L ) o L

20 Bending Moment in The Beam: qo ( L ) M EIν '' C Integrating a second time: 6L The bending moment is zero at the free end of the beam ν (L) 0 Therefore C 0 and the equation simplifies to qo ( L ) M EIν '' Slope and Deflection of the Beam: 6L The third and fourth integration yield EIν ' q 4L o 4 ( L ) C EIν qo 0L ( L The boundary conditions at the fied support, where the slope and the deflection equal zero, are ν (0) 0 ; ν(0) 0, therefore ) 5 C C 4 C qol 4 C 4 q o L 4 0

21 qo ν ' (4L 4LEI qo ν (0L 0LEI 6L 0L 4L 5L ) ) Angle of Rotation and Deflection at the Free end of the Beam: The angle of rotation B and the deflection δ B at the free end of the beam are obtained by substituting L in the above equations: B qol ν '( L) δ B ν ( L) 4EI q o L 4 0EI

22 Eample A simple beam AB with an overhang BC supports a concentrated load P at the end of the overhang. The main span of the beam has a length L and the overhang has a length ½L. Determine the equations of the deflection curve and the deflection δ C at the end of the overhang (see figure). (Note: the beam has constant fleural rigidity EI) Solution : We must write separate differential equations for parts AB and BC of the beam.

23 Reaction at support A ½ P Reaction at support B P / The shear forces in parts AB and BC are V -½P (0 L) V P (L L/) In which is measured from end A of the beam. The third-order differential equations for the beam now become EIν V-½P (0 L) EIν VP (L L/) Bending Moments of the Beam: Integration of the preceding two equations yields the bendingmoment equations: M EIν -½P C (0 L) M EIν P C (L L/) The bending moments at points A and C are zero ν (0) 0 ν (L/) 0

24 Using these conditions we get C 0 and C -PL / therefore M EIν -½P (0 L) M EIν -½P(L ) (L L/) These equations can be verified by determining the bending moments from free-body diagrams and equations of equilibrium. Slopes and Deflections of the Beam The net integration yield the slopes: EIν -P / 4 C (0 L) EIν -½P(L ) C 4 (L L/) The only condition on the slopes is the continuity condition at point B (slope as found for part AB slope as found for part BC) -P / 4 C -½P(L ) C 4 for L then C 4 C ¾PL

25 The third and last integration give EIν - (/)P C C 5 (0 L) EIν -(/)P (9L ) C 4 C 6 (L L/) The boundary conditions are that the deflections in A and B are zero. ν(0)0 ; ν(l)0 Hence, we obtain: C 5 0 ; C PL / and substituting in previous equations C 4 5PL / 6 and C 6 - PL / 4 All constants of integrations have now been evaluated. Substituting in the above equations ν (/EI) P (L ) (0 L) ν -(/EI) P (L 0L 9L - ) (L L/) Note: the deflection is always positive (upward) in part AB and negative (downward) in BC. Deflection at the end of the overhang: L/ δ C -ν (L/) PL /(8EI)

26 Plain Strain We will derive the transformation equations that relate the strains in inclined directions to the strain in the reference directions. State of plain strain - the only deformations are those in the y plane, i.e. it has only three strain components, y and y. Plain stress is analogous to plane stress, but under ordinary conditions they do not occur simultaneously, ecept when σ -σ y and when ν 0 Strain components, y, and y in the y plane (plane strain).

27 Comparison of plane stress and plane strain.

28 Transformation Equations for Plain Strain Assume that the strain, y and y associated with the y plane are known. We need to determine the normal and shear strains ( and y ) associated with the y ais. y can be obtained from the equation of by substituting 90 for.

29 For an element of size δ δy In the direction, the strain produces an elongation δ. The diagonal increases in length by δcos. In the y direction, the strain y produces an elongation y δy. The diagonal increases in length by y δysin. The shear strain y produces a distortion. The upper face moves y δy. This deformation results in an increase of the diagonal equal to: y δycos

30 The total increase Δd of the diagonal is the sum of the preceding three epressions, thus: Δd δcos y δysin y δycos The normal strain in the direction is equal to the increase in length divided by the initial length δs of the diagonal. Δd / ds cos δ/δs y sin δy/δs y cos δy/δs Observing that δ/δs cos and δy/δs sin cos cos sin sin sin cos sin cos

31 Shear Strain y associated with y aes. This strain is equal to the decrease in angle between lines in the material that were initially along the and y aes. Oa and Ob were the lines initially along the and y ais respectively. The deformation caused by the strains, y and y caused the Oa and Ob lines to rotate and angle α and β from the and y ais respectively. The shear strain y is the decrease in angle between the two lines that originally were at right angles, therefore, y αβ.

32 The angle α can be found from the deformations produced by the strains, y and y. The strains and y produce a cwrotation, while the strain y produces a ccw-rotation. Let us denote the angle of rotation produced by, y and y as α, α and α respectively. The angle α is equal to the distance δ sin divided by the length δs of the diagonal: α sin d/ds α y cos dy/ds α y sin dy/ds Observing that d/ds cos and dy/ds sin. The resulting ccwrotation of the diagonal is α - α α - α - ( y ) sin cos - y sin

33 The rotation of line Ob which initially was at at 90 o to the line Oa can be found by substituting 90 for in the epression for α. Because β is positive when clockwise. Thus β ( y ) sin( 90) cos( 90) y sin ( 90) β - ( y ) sin cos y cos Adding α and β gives the shear strain y y α β - ( y ) sin cos y (cos -sin ) To put the equation in a more useful form: sin cos sin cos sin cos sin sin cos sin cos sin cos ( cos )

34 ( ) sin cos cos sin cos sin cos sin cos sin cos sin sin cos [ ] [ ] T T [ ] ( ) sin cos cos sin cos sin cos sin cos sin cos sin sin cos T [] Tensor Strain zz yz z zy yy y z y _

35 Transformation Equations for Plain Strain Using known trigonometric identities, the transformation equations for plain strain becomes: ( ) ( ) ( ) sin cos cos These equations are counterpart of the equations for plane stress where,, y and y correspond to σ, σ, τ y and τ y respectively. There are also counterparts for principal stress and Mohr s circle. y y sin

36 Principal Strains The angle for the principal strains is The value for the principal strains are P tan ( ) ( ) Maimum Shear The maimum shear strains in the y plane are associated with aes at 45o to the directions of the principal strains: ( ) ( ) or MA MA MA

37 Mohr s Circle for Plane Strain

38 Eample An element of material in plane strain undergoes the following strains: y 00-6 y Determine the following: (a) the strains of an element oriented at an angle 0 o ; (b) the principal strains and (c) the maimum shear strains. Solution ( ) ( ) ( ) sin cos cos sin

39 Then 50-6 (50-6 ) cos 60 o (900-6 )sin 60 o ½ y - (50-6 ) (sin 60 o ) ( )(cos 60 o ) Therefore y The strain y can be obtained from the equation y y y ( ) ( ) ( ) (b) Principal Strains The principal strains are readily determine from the following equations:

40 (c) Maimum Shear Strain The maimum shear strain is calculated from the equation: ½ ma SQR[(( y )/) ( ½ y ) ] or ma ( ) Then ma The normal strains of this element is aver ½ ( y ) 50-6

41 For -D problems [ ] z zy z yz y y z y Which is a symmetrical matri. As in the case of stresses: z zy z yz y y z y z zy z yz y y z y m l k

42

43 Dilatation (Volume strain) Under pressure: the volume will change p p p p V-ΔV z y V V Δ Δ 0 y z z y yz yz z y z y yz z y z z y y z y I I I I I I

44 Strain Deviator z y Δ Mean strain It produces a volume change (not a shape change) [ ] Δ Δ Δ z zy z yz y y z y D [ ] Strain Deviator Matri Δ Δ Δ D

45 Application : Strain Gauge and Strain Rosette