P3.1 Report

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    ABSTRACT

    The engineers bending formula is only applicable to sections bent about its principal axes.

    For other cases, we have to use the generalized bending equation. This generalized bendingequation makes use of the assumption that plane section before bending will still be a plane

    even after bending. The other assumption is that the material obeys the idealized HookesLaw. This equation is applicable to all sections regardless their geometry and their load

    direction.

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    1. OBJECTVESThe objectives of this project are to verify the generalized bending theory by applying

    it in an experiment and to increase students knowledge on some features ofunsymmetrical bending.

    2. THEORYThe EulerBernoulli beam equation or better known as the Generalized BendingEquation is the means of calculating the load-carrying and deflection characteristics

    of beams. It is largely used when the section bending is not about its principle axes,

    such as unsymmetrical bending. The load-stress formula can be derive by assuming

    that the equilibrium equations, compatibility conditions and the stress-strain relations

    (Hooks law) be satisfied. And that the plane section before bending remains plane

    after bending.

    2.1Equilibrium EquationThe application of the equation of equilibrium to the free body in Figure 7.9b yields:

    0 = ZZ A

    Mx= yZZ A

    My= - xZZ AEq 1.1

    where

    ZZ is the normal stress in the beam due to bending A denotes an element of area in the cross section x = the perpendicular distance to the centroidaly-axis y = the perpendicular distance to the centroidalx-axis My = the bending moment about they-axis Mx = the bending moment about thex-axis

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    2.2Geometry Deformation(Strain)By considering two plane cross-section perpendicular to the bending axis before bending,

    with distance Lzz between their centroids. And after bending the extension LZZ betweenthe two planes can be represented as a linear function of x & y:

    LZZ = a' + b'x + c'y

    since ZZ = LZZ/LZZ , therefore:

    = + +

    2.3Stress-Strain RelationsAccording to Hooks law:

    ZZ = EZZ

    where E is young's modulus.

    Therefore

    =

    + +

    ZZ = a + bx + cyEq 1.2

    where

    a = Ea'/LZZ b = Eb'/LZZ c = Ec'/LZZ

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    2.4Generalized Bending EquationBy substituting 1.2 into 1.1

    0 = (a + bx + cy)A = aA + bxA+ cyA

    Mx= y(a + bx + cy)A = ayA + bxyA+ cy2A

    My= - x(a + bx + cy)A = -axA - bx2A - cyxA

    since xA = 0&yA = 0

    andy2A = Ix ,x2A = Iy & xyA = Ixy

    0 = aA

    Mx = bIxy + cIx

    My = - bIy - cIxy

    where Ix & Iy are the second moment of inertia with respect to the x and y axis

    respectively, and Ixy is the product moment of inertia.

    Solving the equation to obtain the constants:

    a = 0 (since A 0)

    = +

    = +

    Eq 1.3

    substituting Eq 1.3 into Eq 1.2

    = +

    + ( + )

    This is the generalized bending equation that will be used to calculate the normal stress.

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    2.5Principal Moment of Inertia

    Figure 1 Principal Axes

    2.5.1 Principal Axes

    In order to find the principal axis of inertia, let Ix & Iy are the second moment of inertia

    with respect to the x and y axis respectively, and Ixy is the product moment of inertia.

    Let (u,v) be the principal axes with the same origin and in the same plane as the (x,y)

    axes, and is the angle (x,y) must rotate to coincide with (u,v), is positive in thecounter clockwise direction.

    u =xcos +ysin

    v =ycos xsin

    to find the second and product moment of inertia

    Iu= v2A = (ycos xsin)2A

    Iv= u2A = (xcos +ysin)2A

    Iuv= uvA = (xcos +ysin)(ycos xsin)A

    after integrating and knowing that: y2A = Ix ,x2A = Iy & xyA = Ixy and using

    trigonometric identities. The following equation cans be obtain:

    Iu= 0.5[Ix

    + Iy] + 0.5[I

    xI

    y]cos2 - I

    xysin2

    Iv= 0.5[Ix + Iy] - 0.5[IxIy]cos2 + Ixysin2

    Iuv= 0.5[IxIy]cos2 + Ixysin2

    From the above equations, it can be shown that

    Iu +Iv= Ix+ Iy

    =

    +

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    3. EQUIPMENTSThe equipments used for this experiment are listed as follows.

    3.1

    Frame A

    Figure 2 Frame A

    3.2 Frame B

    Figure 3 Frame B

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    3.3 TDS 303 Data Logger

    Figure 4 TDS 303 Data Logger

    4. PROCEDUREThe procedures of the experiment are stated as follows:

    a)The power of the TDS 303 Data Logger and the frame readout is turned on.b)Then the test rigs are unloaded and the strain gauge circuits are balanced in order tozero the initial reading of each channel.

    c) It is followed by loading the beam according to case1 of Table 1.d)When the load is stabilised the strain reading is taken.e)The steps (c) and (d) are repeated for cases 2, 3 and 4.

    Case Frame Px Py

    1 A 300 N 0 N

    2 A 300 N 200 N

    3 B 20.5 kg 0 kg

    4 B 20.5 kg 20.5 kg

    Table1: Loading values for different cases

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    5. EXPERIMENTAL RESULTS

    Ix Ixy Iy E

    OR

    208.8 x 103

    mm4

    72.74 x 103

    mm4

    73.64 x 103

    mm4

    70 x 109

    Pa

    (Aluminium)208.8 x 10

    -9m

    472.74 x 10

    -9m

    473.64 x 10

    -9m

    4

    Table 1 Properties of Material

    *Note: Different load direction compared with Frame ATable 2 Loading of individual cases

    Table 3 Moment Calculation

    Table 4 Experimental Data for Strain

    Case Frame Px Py

    1 A 300N 0N

    2 A 300N 200N

    3 B 20.5kg * 0kg

    4 B 20.5kg* 20.5kg

    Case Formula Mx/Nm My/Nm

    1 Mx = Py X

    d

    My= Px X

    d

    d=0.33m

    0 -49.5

    2 -33 -49.5

    3 0 33.18

    4 -33.18 33.18

    Case 1 Case 2 Case 3 Case 4

    Gauge

    No

    (horizontal

    frame)

    (vertical

    frame)

    (horizontal

    frame)

    (vertical

    frame)

    (horizontal

    frame)

    (vertical

    frame)

    (horizontal

    frame)

    (vertical

    frame)

    1 -190 -217 -281 -300 109 127 25 53

    2 -120 -179 -195 -237 68 106 -1 55

    3 -53 -141 -115 -177 25 82 -32 58

    4 12 -104 -33 -115 -13 60 -61 59

    5 79 -67 49 -53 -54 37 -90 60

    6 149 -29 133 5 -95 14 -119 62

    7 219 5 219 67 -135 -6 -147 64

    8 281 43 296 130 -175 -28 -175 66

    9 357 80 386 189 -217 -52 -205 67

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    Table 5 Distances X and Y for calculation of theoretical strain

    Case 1 Case 2 Case 3 Case 4

    Formula

    = ( + ) ( + )( )

    (horizontal

    frame)

    (vertical

    frame)

    (horizontal

    frame)

    (vertical

    frame)

    (horizontal

    frame)

    (vertical

    frame)

    (horizontal

    frame)

    (vertical

    frame)

    Gauge

    No 9 411 94 454 218 -275 -63 -232 61

    Corner

    C -254 -254 -366 -366 171 171 58 58

    Table 6 Theoretical Strain

    Experimental () Theoretical () Difference ()Formula tan = [ Ixtan + Ixy ]/[ Iy + Ixytan ]Case 1 70 71 1

    Case 2 63 64 1

    Case 3 71 71 0

    Case 4 97 91 6

    Table 7 Experimental and Theoretical Neutral Angle,

    Table 9 Theoretical Strain vs Experimental Strain Error Percentage

    Case 1 - 4

    Horizontal

    Frame

    Vertical Frame

    Gauge No

    9

    x1= 36.32 x 10-3

    m

    x2=-9.29 x 10-3

    m

    y1 = 23.23 x 10-3

    my2= -45.17 x 10

    -3

    m

    Corner C

    x3=-9.29 x 10-3

    m

    y3= 23.23 x 10-3

    m

    Gauge

    No 9 13.1% 14.9% 15.0% 13.3% 21.1% 17.5% 11.6% -9.8%

    Corner

    C 25.2% 14.6% 23.2% 18.0% 36.3% 25.7% 56.9% 8.6%

    x3 x1C

    x2

    y3

    y2

    y1CENTROID

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    6. DISCUSSION AND CONCLUSION6.1 Bok Chian Check

    a) Without using the generalized bending equation, describe briefly an alternative method

    of calculating the normal stresses at any point on an unsymmetrical beam in bending,

    From the principal moment of inertia, the angle of the neutral axis can be found using

    tan2 = -(2Ixy)/( Ix + Iy). Therefore the principal axis can be determine with respect to thex & y axis. Hence the symmetric loading equation can be used:

    ZZ = Mv V+ Mv U = 0

    Iv Iu

    the equation can be simplified to:

    V = Iutan = tan

    U Iv

    b) For the existing experimental setup for Frame A and B, determine the expressions for

    Mxand My at the loading point, in terms of the applied loads.

    Moment = Force x Perpendicular Distance

    Mx = Px/2 x 0.33 (Nm)

    My = Py/2 x 0.33 (Nm)

    c) For Frame A only: When Px is loaded first, observe the load reading of Py is applied.

    Explain why?

    When Pyis applied, the load reading of Px decreases. This is because when Pyis applied,

    due to the unsymmetrical section of the beam, it cause a increase in moment about the x-

    axis, in the positive directions, which is the direction Px is also acting. Therefore it results

    in the sensor detecting a lower net force acting in the Px direction.

    d) Briefly explain why the NA is not 90 degrees to the applied load.

    Neutral axis is an axis in the cross section of a beam or shaft along that have zero

    longitudinal str