Abstract
In this lab experiment the beam design was evaluated for a given bending load so that the best engineering design can be used to design the beam for a given load. In order to evaluate the given design of bean against a given load stress and strain produce in beam when load is applied in it was calculated. Stresses in beam were calculated using theoretical formulas and the deflection was measured using foil strain gauges. Similarly the buckling of columns was also studied experimentally for different ends conditions to evaluate the load bearing capacity of columns made of aluminum and brass.
STRAIN MEASUREMENT IN BEAM BENDING
Introduction
For any load bearing system beams are its main supporting structure which provides necessary strength to the structure. Beam provides the required strength to the structure due to strength of its material and it shape. Shape of the beam defines the second moment of inertia which when combine with the load and point of application of load gives the value of strength a beam can provide to its structure. There are many different types of beams like simply supported beam, continuous beam and over hanging beam were each type can have different types of shapes like square beam, rectangular beam or I beam. From the above mention shapes of beam I beam is the most used beam due to its greater second moment of inertia for given height and width.
In construction work application one of the most important elements used is beam. Beam is used to withstand load which is applied laterally along the length of the beam. The applied load can be a point load which is applied at one single point or it can be distributed load. The uniformly distributed load expands all over the length of beam and acts with same magnitude at all points on the beam. Due to bending load the beam can be deflected downwards under the applied load [1]. This deflection must be measured and analyzed to evaluate the effectiveness of beam towards its load bearing capacity. The deflection in beams can be measured using strain measurement and this is done by using strain sensors [2]. The examples of such strain gauges are foil gauges, vibrating wire gauges, micro electromechanical strain gauges and optical sensors etc. in this experiment the foil gauges were used because they are easy to use and are more economical and produce accurate results.
Experimental apparatus
Wooden plank
Foil gauges (attached to wooden plank by a glue with good strain transference)
Wheatstone bridge circuit
Oscilloscope or multimeter
Loads
CALCULATIONS
The following dimensions of beam are given
Load applied; W = 100 kN/m2
Beam length; L = 100 cm = 1 m
Beam width; w = 3.5 cm
Beam Breadth, h = 14.3 cm
From beam dimensions the second moment of area, ‘I’ can be calculated as
Second moment of area
I =(wh^3)/12
Putting in values
I =(0.035* (0.143)^3)/12 = 8.53 x 〖10〗^(6) m^4
The beam is assumed to be simply supported and the load is applied at the middle point
For a maximum load of 100 kg
W = 100* 9.81 = 981 N
For simply supported beam the maximum bending moment will be at the centre of beam where the deflection will be the maximum but the shear force will be the minimum.
The maximum moment would be
M =WL/4 =(981*1)/4 = 245.25 Nm
The maximum bending stress can be calculated as
Bending stress
σ =My/I
Where ‘y’ is the distance from neutral axis to the edge of beam and this is equal to half of beam breadth (height)
so y =14.3/2 = 7.15 cm
Bending stress
σ =(245.25* 0.0715)/(8.53 * 〖10〗^(6) )
σ = 2.05* 〖10〗^6 Pa = 2.05 MPa
For three point test, the flexural strength can be calculated using the following formula
σ =3LF/(2wh^2 )
Putting in values
σ =(3*1*981)/[2*0.035*(0.143)^2 ]
σ = 2.05* 〖10〗^6 Pa = 2.05 MPa
This is exactly same as the maximum bending stress in the beam
For uniform distributed load
For this particular case when the load is 100 k N/m2 which is uniformly distributed and the beam length is 2.6 m. Let’s assume the width of beam is 10 cm and height is 20 cm. Then the load on the beam will be
100 kN/m2* 0.1 m = 10 k N/m
Therefore
W =10 k N/m
This is uniformly distributed load and in this case the second moment of area would be
Second moment of area
I =(wh^3)/12
Putting in values
Second moment of area
I =(0.1* (0.2)^3)/12 = 6.67 x 〖10〗^(5) m^4
The distance from neutral axis to the beam edge
y =20/2 = 10 cm = 0.1 m
The maximum bending stress can be calculated as
Bending stress
σ =My/I
The maximum moment for uniformly distributed load would be
M =(WL^2)/4 =(10*1000* 2.62)/4 = 16900 Nm
Now
Bending stress
σ =(16900 x* 0.1)/ 6.67* 〖10〗^(5)
σ = 25.33*〖10〗^6 Pa = 25.33 MPa
This is the maximum stress the beam can take before failure. With a factor of safety of 3 the ultimate stress would be 25.33 x 3 = 76 MPa
W

I

M

Y

stresses

strain

196

0.00000853

49

411145.96

2.74097E05


245

0.00000853

61

0.0715

513932.44

3.42622E05

491

0.00000853

123

0.0715

1027864.9

6.85243E05

736

0.00000853

184

0.0715

1541797.3

0.000102786

981

0.00000853

245

0.0715

2055729.8

0.000137049

Discussion
A square shape simply supported beam was used during this experiment and material used for the beam was wood. As per instruction for experiment load was applied on the beam and stresses were calculated using formulas for simply supported beam. Strain was recorded using the strain gauge and Wheatstone bridge. Five different loads were applied on beam and stresses and strain for each of the load were calculated. Graphs for stress and strain was made against the force applied. According to the graph one which is between stress and force (load) the relation between stress and load is linear and directly proportional to each other. This means that increase in load increase the stress produce in beam and reduction in load reduces the stress produce. This change in stress is equal to change in load as the graph is linear. According to the graph two which is between strain and force (load) the relation between strain and load is linear and directly proportional to each other. This means that increase in load increase the strain produce in beam and reduction in load reduces the s strain produce. This change in strain is equal to change in load as the graph is linear.
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