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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