TABLE OF CONTENT
Number Description Page
Student Ethic Code
Group Members 2
1.0 Objective of the experiment 3
2.0 Learning Outcomes 3 3.0 Introduction 3 4.0 Theory 3 5.0 Apparatus 6 6.0 Procedures 7 7.0 Result 8 8.0 Calculation 10 9.0 Discussion 20 10.0 Conclusion 22 11.0 References 23 Appendix 24
1.0 OBJECTIVE
1.1 Part 1 : To plot moment influence line
1.2 Part 2 : To apply the use of a moment influence on a simply supported beam
2.0 LEARNING OUTCOMES
2.1 Application of engineering knowledge in practical application
2.2 To enhance the technical competency in civil engineering through laboratory application.
2.3 Communicate effectively in group.
2.4 To identify problem, solving and finding out appropriate solution through laboratory application.
3.0 INTRODUCTION
Moving loads on beam are common features of design. Many road bridges are constructed from beam, and as such have to be designed to carry a knife edge load, or a string of wheel loads, or a uniformly distributed load, or perhaps the worst combination of all three. To find the critical moment in section, influence line is used.
4.0 THEORY
Part 1: This experiment examines how moment varies at a cut section as a unit
load moves from one end another (see diagram 1). From the diagram, moment influence equation can be written.
For a unit load between 0 < x < a ,
Mx = ( L – x ) a - 1 (a – x )……….(1)
L
For unit load between a < x < b ,
Mx = xb / L – ( x – a )…………..(2) ‘ cut ‘ 1 ( unit load ) Mx x Mx RA = (1-x/L) RB = x/L a b L Figure 1
Part 2: If the beam is loaded as shown below, the moment at the ‘cut’ can be
calculated using the influence line. (See diagram 2).
Moment at the ‘cut’ section = F1y1 + F2y2 + F3y3 ……….(3)
( y1, y2, and y3 are coordinates derived from the influence line in terms of x1, x2, x3, a, b and L ) a+b = L x1 x2 x3 y1 y2 y3
Moment influence line for cut section
5.0 APPARATUS
Moment Influence Line
Hanger with load
6.0 PROCEDURES
Part 1:
1. The digital forces meter reads zero with no load.
2. Hanger with any mass between 150 – 300 g was place at the first grooved hanger support at the left support and the digital forces reading were recorded in Table 1.
3. The procedure repeated to the next grooved hanger until to the last groove hanger at the right hand support.
4. Calculation in Table 1 completed.
Part 2:
1. Three load hangers with any load between 50 – 400 g was placed on it and placed it at any position between the supports. The position and the digital forces display reading recorded in Table 2.
2. The procedure repeated with three other location. 3. The calculation in Table 2 completed.
7.0 RESULT
Load = 200 g = 1.962 N
Location of load from left hand support (m) Digital Force Display Reading (N) Moment at cut section (N) Experimental influence line value (N) Theoretical Influence lines value (Nm) 0.04 0.2 0.025 0.013 0.013 0.06 0.3 0.038 0.019 0.019 0.08 0.4 0.050 0.025 0.026 0.10 0.5 0.063 0.032 0.032 0.12 0.6 0.075 0.038 0.038 0.14 0.7 0.088 0.045 0.045 0.16 0.8 0.100 0.051 0.051 0.18 0.9 0.113 0.057 0.057 0.20 1.0 0.125 0.064 0.064 0.22 1.1 0.138 0.070 0.07 0.24 1.2 0.150 0.076 0.076 0.26 1.3 0.163 0.083 0.083 0.30 1.4 0.175 0.089 0.096 0.32 1.2 0.150 0.076 0.082 0.34 1.0 0.125 0.064 0.07 0.36 0.8 0.100 0.051 0.055 0.40 0.4 0.050 0.025 0.027 Table 1 Notes :
2. Experimental Influence line values = Moment (Nm) Load (N)
3. Calculate the theoretical value using the equation 1 for load position 40 – 260 mm and equation 2 for load position 320mm and 400mm.
Part 2,
Location
Position of hanger from left
hand support (m) Digital force reading (N) Experimental Moment (Nm) Theoretical moment (Nm) 100 gram 200 gram 300 gram 1 80 160 340 2.6 0.325 0.325 2 340 100 220 2.6 0.325 0.336 3 120 300 60 2.2 0.275 0.280 4 200 380 120 2.0 0.250 0.254 Table 2 Notes:
1. Experimental moment = Digital force reading x 0.125 2. Theoretical moment is calculated using equation (3)
8.0 CALCULATION EXAMPLE CALCULATION PART 1
Moment at cut section = 0.2 x 0.125 = 0.025 N
Experimental Influence line values = Moment (Nm) Load (N) = 0.025
1.962 = 0.013 m
Theoretical Influence lines value;
Equation 1 for load position 40 to 260 mm ( ) ( )……….(1)
Mx = (0.44 – 0.04) (0.3) – 1(0.3 – 0.04) 0.44
= 0.013 Nm
Equation 2 for load position 320mm to 400mm ( )………(2) When x = 0.32 m Mx = (0.32) (0.14) – (0.32 – 0.3) 0.44 = 0.082 Nm
PART 2 F1 = 100g = 100 x 9.81 1000 = 0.981N F2 = 200g = 200 x 9.81 1000 = 1.962N F3 = 300g = 300 x 9.81 1000 = 2.943N 0.981 N 1.962 N 2.943 N x1 x2 x3 y2
*For location 1,
Experimental moment at cut section (Nm) = Digital force reading x 0.125 = 2.6 x 0.125 = 0.325 Nm Moment at cut: For ∑ ( ) ( ) ( ) For ∑ ( ) ( ) When x = 0.3 Mx = 0.318x = 0.318 (0.3) = 0.095 Nm = - 0.682 (0.3) + 0.3 = - 0.095 Nm
Use interpolation to get y1,y2 and y3 y1, y2, y3,
Theoretical moment at cut section (Nm)
=
= 0.981 (0.025) + 1.962 (0.051) + 2.943 (0.068) = 0.325 Nm
Moment influence line for cut section *For location 2, Experimental moment (Nm) = 0.325 Nm Moment at cut: For ∑ ( ) ( ) ( ) For ∑ ( ) ( ) y1 y3 1.962 N 2.943 N 0.981 N x1 x2 x3 y2
When x = 0.3 Mx = 0.318x = 0.318 (0.3) = 0.095 Nm = - 0.682 (0.3) + 0.3 = - 0.095 Nm
Use interpolation to get y1,y2 and y3
y1, y2, y3,
Theoretical moment at cut section (Nm)
=
= 1.962 (0.032) + 2.943 (0.070) + 0.981 (0.068) = 0.336 Nm
Moment influence line for cut section *For location 3, Experimental moment (Nm) = 0.275 Nm Moment at cut : For ∑ ( ) ( ) ( ) For ∑ ( ) ( ) y1 y3 2.941 N 0.981 N 1.962 N x1 x2 x3 y2
When x = 0.3 Mx = 0.318x = 0.318 (0.3) = 0.095 Nm = - 0.682 (0.3) + 0.3 = - 0.095 Nm
Use interpolation to get y1,y2 and y3
y1, y2, y3,
Theoretical moment at cut section (Nm)
=
= 2.943 (0.019) + 0.981 (0.038) + 1.962 (0.095) = 0.280 Nm
Moment influence line for cut section *For location 4, Experimental moment (Nm) = 0.250 Nm Moment at cut: For ∑ ( ) ( ) ( ) For ∑ ( ) ( ) y1 y3 2.941 N 0.981 N 1.962 N x1 x2 x3 y2
When x = 0.3 Mx = 0.318x = 0.318 (0.3) = 0.095 Nm = - 0.682 (0.3) + 0.3 = - 0.095 Nm
Use interpolation to get y1,y2 and y3
y1, y2, y3,
Theoritical moment at cut section (Nm)
=
= 2.943 (0.038) + 0.981 (0.063) + 1.962 (0.041) = 0.254 Nm
F 1 a b cut RA = = RB x L 9.0 DISCUSSIONS PART 1
1. Derive equation 1 and 2.
∑ ∑ ∑ ( ) ( ) ( ) ( ) ( ) Equation 1 ; ( ) – ( ) ( – ) – ( ) ( – ) – ( ) Equation 2 ; – ( ) ( ) ( ) – ( ) ( ) – ( ) – ( )
2. On the graph, plot the theoretical and experimental value against distance from left and support. Comment on the shape of graph. What does it tell u about how moment varies at the cut section as a load moved on the beam?
From the graph, a peak shaped graph can be obtained. The peak is the weakest point of the beam where there is a hinge in the beam. As load is being moved on the beam, the influence line which was constructed can be used to obtain the value of the moment. As load is moved across near to it, the moment will increase. So does the other way round when load is moving further than the hinge, the value of moment will decrease as the load is moving towards the support at the end. As the load is moving along towards the hinge from both side of support, it will come to a peak where the value of moment is the same.
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 0.0 4 0.0 6 0.0 8 0.1 0.12 0.14 0.16 0.18 0.2 0.22 0.24 0.26 0.3 0.32 0.34 0.36 0.4 M o m e n t (N m ) Distance (m)
Graph of Theoretical Value (Nm)versus
Experimental Value (Nm) versus Distance (m)
Theoretical Influence lines value (Nm)
Experimental influence line value (N)
3. Comment on the experimental results and compare it to the theoretical results.
The experimental results that we obtained are quite accurate and compare to the theoretical results, the experimental results are only slightly different with theoretical results. When we were conducted the experiment, we tried to minimize the error by ensuring the Digital Force Meter reads zero with no load before we place the hangers.
PART 2
1. Calculate the percentage difference between experimental and theoretical results in table 2. Comment on why the results differ.
Experimental Moment (Nm) Theoretical moment (Nm) Percentage Different (%) 0.325 0.325 0 0.325 0.336 3.27 0.275 0.280 1.79 0.250 0.254 1.57
The experimental results are slightly different from theoretical results are due to human error and instrument sensitivity as the reading of the instrument keep changing when we conducted the experiment.
10.0 CONCLUSION
As a conclusion, both objectives were achieved. Moment influence line could be plot and the influence line can be use to determine the moment. We were able to identify the reaction and behavior of a beam in terms of its moment reaction value. This method is useful to check every cross section for a particular beam.
11.0 REFERENCES
Ando, J., Komatsuda, T., and Kamiya, A. 1988. Cytoplasmic calcium responses to fluid shear stress in cultured vascular endothelial cells. In Vitro 24: 871-877.
Ando, J., Nomura, H., and Kamiya, A. 1987. The effects of fluid shear stress on the migration and proliferation of cultured endothelial cells. Microvascular Research 33: 6270.
Caplan, B.A. and Schwartz, C.J. 1973. Increased endothelial cell turnover in areas of in vivo Evans blue uptake in the pig aorta. Atherosclerosis 17: 401-417.
Davies, P.F., Remuzzi, A., Gordon, EJ., Dewey, C.F., and Gimbrone, M.A. 1986. Turbulent fluid shear stress induces vascular endothelial cell turnover in vitro. Proceedings of the National Academy of Sciences U.S.A. 83: 2114-2117.
deSouza, P.A., Levesque, M.J., and Nerem, R.M. 1986. Electrophysiological response of endothelial cells to fluid- imposed shear stress. Federation Proceedings 45: 471.
Dewey, C.F., Jr., Bussolari, S.R., Gimbrone, M.A., Jr., and Davis, P.F. 1981. The dynamic response of vascular endothelial cells to fluid shear stress. Journal of Biomechanical Engineering 103: 177-185.
Diamond, S.L., Eskin, S.G., and McIntire, L.V. 1989. Fluid flow stimulates tissue plasminogen activator. Science 243: 1483-1485.
Frangos, J.A., McIntire, L.V., Eskin, S.G., and Ives, C.L. 1985. Flow effects on prostacyclin production by cultured human endothelial cells. Science 227: 1477-1479.
Leung, D.Y.M., Glagov, S., and Mathews, M.B. 1976. Cyclic stretching stimulates synthesis of matrix components by arterial smooth muscle cells in vitro. Science 191: 475-477.
Levesque, M.J., Liepsch, D., Moravec, S., and Nerem, R.M. 1986. Correlation of endothelial cell shape and wall shear stress in a stenosed dog aorta. Arteriosclerosis 6: 220229.
APPENDIX
Influence lines play an important part in the design of bridges, industrial crane
rails, conveyor belts, and other structures where loads move across their span.
An influence line represents the variation of the reaction, shear, moment, or deflection at a specific point in a member as a concentrated force moves over the member. Once this line is constructed, one can tell at a glance where the moving load should be placed on a structure so that it creates the greatest influence at the specified point. Furthermore, the magnitude of the associated reaction, shear, moment, or deflection at the point can than be calculated from the ordinates of the influence line diagram.
A bending moment exists in a structural element when a moment is applied to the element so that the element bends. Moments and torques are measured as a force multiplied by a distance so they have as unit newton-metres (N·m) , or foot-pounds force (ft·lbf). The concept of bending moment is very important in engineering (particularly
in civil and mechanical engineering) and physics.
Tensile stresses and compressive stresses increase proportionally with bending moment, but are also dependent on the second moment of area of the cross-section of the structural element. Failure in bending will occur when the bending moment is sufficient to induce tensile stresses greater than the yield stress of the material throughout the entire cross-section. It is possible that failure of a structural element in shear may occur before failure in bending; however the mechanics of failure in shear and in bending are different.
The bending moment at a section through a structural element may be defined as "the sum of the moments about that section of all external forces acting to one side of that section". The forces and moments on either side of the section must be equal in order to counteract each other and maintain a state of equilibrium so the same bending moment will result from summing the moments, regardless of which side of the section is selected.
Moments are calculated by multiplying the external vector forces (loads or reactions) by the vector distance at which they are applied. When analyzing an entire element, it is sensible to calculate moments at both ends of the element, at the beginning, centre and end of any uniformly distributed loads, and directly underneath any point loads. Of course any "pin-joints" within a structure allow free rotation, and so zero moment occurs at these points as there is no way of transmitting turning forces from one side to the other.
If clockwise bending moments are taken as negative, then a negative bending moment within an element will cause "sagging", and a positive moment will cause "hogging". It is therefore clear that a point of zero bending moment within a beam is a point of contra flexure—that is the point of transition from hogging to sagging or vice versa.
It is more common to use the convention that a clockwise bending moment to the left of the point under consideration is taken as positive. This then corresponds to the second
derivative of a function which, when positive, indicates a curvature that is 'lower at the centre' i.e. sagging. When defining moments and curvatures in this way calculus can be more readily used to find slopes and deflections.
