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
Definition : Influence line is define as a line representing the changes in either moment, shear force, reaction or displacement at a section of a beam when a unit load moves on the beam.
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
Figure 2
Figure 3
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.
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.
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 0.05 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 0.10 0.051 0.051 0.18 0.9 0.113 0.058 0.057 0.20 1.0 0.125 0.064 0.064 0.22 1.1 0.138 0.07 0.07 0.24 1.2 0.15 0.077 0.076 0.26 1.3 0.163 0.083 0.083 0.30 1.5 0.188 0.10 0.096 0.32 1.3 0.163 0.083 0.082 0.34 1.1 0.138 0.07 0.07 0.36 0.8 0.10 0.051 0.055 0.40 0.4 0.05 0.025 0.027 Table 1 Notes :
1. Moment at cut section = Digital force reading x 0.125 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 40 100 200 2.1 0.263 0.261 2 80 160 260 2.9 0.363 0.366
3 360 340 80 2.1 0.263 0.260
4 260 400 60 1.6 0.200 0.190
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
Load (N) = 0.025
1.962 = 0.013 m
Theoretical Influence lines value;
Equation 1 for load position 40 to 260 mm
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
When x = 0.32 m
Mx = (0.32) (0.14) – (0.32 – 0.3) 0.44
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 1.962 N 2.943 N x1 x2 x3
y1 y2 y3
Moment influence line for cut section
*For location 1,
Experimental moment at cut section (Nm) = Digital force reading x 0.125 = 2.1 x 0.125 = 0.263 Nm Moment at cut : ∑Mx = 0 Mx = 1(0.3)- x (0.3) – 1 (0.3-x) 0.44 = 0.3 - 0.3x – 0.3 + x 0.44 Mx = 0.318x When x = 0.3 Mx = 0.318x = 0.318 (0.3) = 0.095 Nm
Use interpolation to get y1,y2 and y3
y1, 0.095 = y1 0.3 0.04 0.3y = 0.0038
y1 = 0.013 m y2, 0.095 = y2 0.3 0.1 y2 = 0.032 m y3, 0.095 = y3 0.3 0.2 y3 = 0.063 m
Theoritical moment at cut section (Nm) = F1y1 + F2y2 + F3y3 = 0.981 (0.013) + 1.962 (0.032) + 2.943 (0.063) = 0.261 Nm 0.981 N 1.962 N 2.943 N x1 x2 x3 y1 y2 y3 *For location 2,
Experimental moment (Nm) = 0.363 Nm When y1 = 0.025 m , y2 = 0.051 m , y3 = 0.082m Theoritical moment (Nm) = 0.366 Nm 2.943 N 1.962 N 0.981 N x1 x2 x3 y3 y2 y1 *For location 3, Experimental moment (Nm) = 0.263 Nm When y1 = 0.054m , y2 = 0.068m , y3 = 0.025m Theoritical moment (Nm) = 0.260 Nm
2.943 N 0.981 N 1.962 N x1 x2 x3 y3 y1 y2 *For location 4, Experimental moment (Nm) = 0.4125 Nm When y1 = 0.082m , y2 = 0.027m , y3 = 0.019m Theoritical moment (Nm) = 0.190 Nm
F 1 a b cut RA = = RB x L 9.0 DISCUSSIONS PART 1
1. Derive equation 1 and 2.
ΣFx = 0 ΣFy = RA + RB – 1 = 0 RA + RB = 1 RA( L ) – 1( L – x ) = 0 RAL = 1(L- x)
RA = 1( L – x ) L = 1 - x L RB = 1 – (1 – x) = x L L Equation 1 ; 0 ≤ x ≤ a -Mx + RA(a) – 1(a - x) = 0 Mx = (1 – x/L)a – 1(a - x) = (L – x)a – 1(a - x) L Equation 2 ; a≤ x ≤ b Mx – RB(b) + 1(x - a) = 0 Mx = RB (b) – 1(x - a) = x/L (b) – 1(x -a) = xb/L – 1(x -a)
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?
GRAPH EXPERIMENTAL VALUE (Nm) VERSUS THEORETICAL VALUE (Nm) VERSUS DISTANCE (m)
0 0.05 0.1 0.15 0.2 0.25 0 .0 4 0 .0 6 0 .0 8 0 .1 0 .1 2 0 .1 4 0 .1 6 0 .1 8 0 .2 0 .2 2 0 .2 4 0 .2 6 0 .3 0 .3 2 0 .3 4 0 .3 6 0 .4 DISTANCE (m) M O M E N T ( N m ) Theoretical Value Experimental Value
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.
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 (%)
0.263 0.261 0.77
0.363 0.366 0.82
0.263 0.26 1.15
0.2 0.19 5.26
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 behaviour of a beam in terms of its moment reaction value. This method is useful to check every cross section for a particular beam.
