GROUP MEMBERS INFORMATION
NUR SYAZILA BT AZLIN
AF080286
NUR ZAKINA ELMY BT AHMAD ZAKI
AF080282
(Discussion and conclusion)
(Recorder) Introduction and result
TAN LEE SIANG
TEH CING CING
(Photographer)
AF080296
Appendix
TUN WAI FEI
AF080324
Group leader (compile and divide job)
AF080292
(data analyze) Calculation
1.0
OBJECTIVE
1.1
To find the form factor
1.2
To investigate the load deflection relationship for a beam to the point of plastic collapse
2.0
LEARNING OUTCOME 2.1
Able to apply the structural knowledge in practical application.
2.2
To improve the technical efficiency through the laboratory work.
2.3
Able to communicate effectively in team work.
2.4
Able to recognize the problem, solving and getting the solution through experimental work
3.0
INTRODUCTION During the design process for beams it would not be unreasonable for one to assume that no part of the beam should experience a stress greater than that allowable for the working material. However, it can be found that a beam will withstand much larger forces before collapse than simple elastic theory predict.
4.0
THEORY
When a beam is bent around the neutral axis, the stress through the beam section varies with the distance from the neutral axis, form the greatest att the extreme fibres ( y= maximum) to zero at the neutral axis ( y = 0)
If the beam is subjected to an increasing bending moment, the stress will build up through the section to a maximum at the extreme fibres. This means that although the outer parts of the beam may well have yielded and are behaving plastically, the inner parts may still be behaving elastically and resisting load. If the bending moment continues to increse, the plastic portion will move further into the beam leaving a smaller elastic core. This called the partially plastic condition. The beam will continue to resist the bending moment although with an incresing rate of deflection as the plastic portion moves further toward the neutral axis. ( See diagram below )
Neutral axis σ y
σ y
Plastic portion
Eventually the elastic portion will far enough into the beam andthe beam will be “fully” plastic. It will form a plastic hinge and be unable ti resist any further bending moment ( shown below)
The ratio of the “fully’ plastic bending to the “just” plastic moment is call FORM FACTOR. The form factor is entirely dependent on the shape of the beam and not on the size, material or fixing condition. For cantilever beam Yield stress σ y =
4WL …………….equation 1 bd
For the simply supported beam Maximum bending moment, M p =
WL ……….. equation 2 4
Bending moment at yield of the extreme fibre M y = σ y I y Experiment form factor = M p M y where the text book value is 1.5
5.0
PROCEDURE
Experiment (Simply Supported Beam)
1. Take the specimen beam and measure the cross section, calculate the second moment of area for the specimen. 2. Ensure the clamp plates are removed and place the specimen beam across the chucks of the unit. 3. Push the roller mechanism outwards to its stop. 4. Put the pin through the load cell fork and wind the load cell down until the pin just touches the specimen beam zero both the load cell and the indicator. 5. Wind the load cell down to cause a measured deflection of 3mm and the reading of the force required. 6. Continue to wind the load cell down in 3mm step until there is no or very little increase in load for each increment of deflection. 7. Enter the result into Table 1.
6.0
APPARATUS
Sketch the apparatus used in this experiment and named the essential components.
PLASTIC BENDING OF THE BEAM
MEASUREMENT DEFLECTION DIGITAL FORCE DISPLAY
FORCE OUTPUT
7.0 RESULT
Deflection (mm)
Force ( N )
0
0
3
41
6
73
9
91
12
107
15
123
18
131
21
138
24
141
27
143
30
144
33
145
36
142
Table 1 : Result for experiment ( Simply Supported Beam)
8.0
DISCUSSION Experiment 1
1. Plot the graph Force vs Deflection and from your result comment on the shape of the resulting plot. 2. From Table 1, note the collapse load, and using the bending moment diagram calculate the plastic moment ( Mp). The maximum deflection is 33mm, when the force reaches to 145 N.
Mp = wL 4 = 145 (860) 4 = 31.175 x 103 Nmm
Figure (a) Mp + Mp = PL 2
4
LOAD
3Mp = PL L/2 = 430mm
2
4
P
= 6Mp
L/2 = 430mm
L =
6 (31.175 x 103) 860
=
217.5 N (collapse load)
a)
C
MOMENT
b)
c)
d)
3. Using yield stress of 325 Mpa* Calculate the bending moment (My) to just cause yielding of the extreme fibres.
h
=
7.45 + 6.91 + 6.85
3
b
=
7.07mm
=
7.22 + 6.92 + 6.85 3
=
h =7.07mm
3.54mmm
b = 7.00mm
7.00mm
My = σyI y
σy =
325 Mpa = 325 N/mm2 I
=
bh 3 12
= 7.00 mm ( 7.07 mm ) 3 12 = 206.14mm4
My
=
(325 N/mm2) (206.14mm4) 3.54mm
=
18.925 x 103Nmm
4. Calculate the form factor (Mp/My) . Compare to the text book value
Form Factor = Mp My
=
31.175 x 103 Nmm 18.925 x 103Nmm
=
1.65
Compared to the text book value, For the rectangular section, S = bh4/4 = 1.500 bh2/6 The form factor is 1.65 >1.500
The experimental form factor is higher than theoretical form factor, where the different is 0.15. This slightly different because human error. When the person wind the load cell was not wind in same direction which is anti-clockwise direction. Error occurs when we change the direction of wind when loading. This means we need continuous wind load cell down and cannot wind back. Besides, system error also occur in this experiment, this error occur due to equipment. Before we start experiment, the roller support of the equipment is not placed at the end of the beam, this will result different result. Beside of this, the load cell fork also not place perpendicular to the beam specimen.
5.
Discuss the advantages of considering the extra available strength due to the plastic beam theory when designing structures. There are many advantages of considering the extra strength due to the plastic beam theory when designing structures. For example, we can improve safety of the building to prevent collapse. This is because when we consider extra strength also mean increase the safety factor of the structure.
Rather than that, we can determine the ultimate load or collapse load by consider extra strength due to the plastic beam. Ultimate load is important and we need to know when we using plastic construction material, this is because once the load excess the ultimate load, the beam will not return to the original length or dimension. Moreover, consider extra strength to the plastic beam can reduce the risk of failure due to the calculation error or additional load to the building. Once we design a structure, calculation or mistake may be occur, or some unexpected load will act on the building, so we need consuder extra strength due to the plastic beam. Besides, consider extra strength due to the plastic beam also will prevent craking occur. Cracking occur when the the part of the structure can not resist the load. By consider extra strengh risk of cracking may reduce. 9.0 RECOMMENDATION 1. Before start experiment,check the equipment and make equipment is in good condition. 2. Make sure the end of the beam is clamped, and the roller is placed at the end of the beam. 3. Before start ensure that is no zero error of equipment. 4. Make sure the specimen beam touch both the load cell and the indicator. 5. Before start experiment make sure the load cell fork is perpendicular to the beam specimen.
10.0
CONCLUSION
After carry on this experiment, we found that the relationship between load deflection of plastic material and the load act on it. Before ultimate load archive, when the load is increase the deflection also increase. When excess ultimate load, the deflection will decrease when the load is increase.
There are two types of boundary conditions in this experiment, simply supported panels and panels clamped at their two ends. We can conclude that in a plastic analysis the stress distribution is nonlinear and is dependent on the beam’s material. Where, strain distribution is linear about the neutral axis. According to the result we obtained, we can conclude that experimental value is slightly different compare to the text book value. Where the form factor we obtained is 1.65 and theoretical value is 1.5. Experimental value is slightly higher than theoretical value. This happen because system error and human error happen in this experiment.
11.0
i.
REFERENCES
“MECHANICS OF MATERIALS – FOURTH EDITION IN SI UNITS” Ferdinand P. Beer E. Russell Johnston, Jr. John T. Dewolf MC GRAW HILL HIGHER EDUCATION, 2006
ii.
“STRUCTURAL ANALYSIS – SIXTH EDITION IN SI UNITS” R.C Hibbeler PEARSON PRENTICE HALL,2006
iii.
Module UTHM mechanic of material
