moment influence line (part 2)
TRANSCRIPT
STRUCTURE AND MATERIAL ENGINEERING
TITLE : MOMENT INFLUENCE LINES (U2)
1.0 OBJECTIVE
1.1 To plot moment influence line.
1.2 To apply the use of a moment influence on a simply supported beam.
2.0 LEARNING OUTCOMES
2.1 Application the engineering knowledge in practical application.
2.2 To enhance technical competency in structural engineering through
laboratory application.
2.3 To 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 such have to be designed to carry a knife edge load,
or a string of wheel loads, or a uniformity distributed load, or perhaps the worst
combination of all three. To find the critical moment at a section, influence line is
used.
4.0 THEORY
Definition: Influence line is defined 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.
An influence line for a given function, such as a reaction, axial force, shear force, or
bending moment, is a graph that shows the variation of that function at any given
point on a structure due to the application of a unit load at any point on the structure.
An influence line for a function differs from a shear, axial, or bending moment
diagram. Influence lines can be generated by independently applying a unit load at
several points on a structure and determining the value of the function due to this
load, for example shear, axial, and moment at the desired location. The calculated
values for each function are then plotted where the load was applied and then
connected together to generate the influence line for the function.
For example, the influence line for the support reaction at A of the structure shown in
Figure 1, is found by applying a unit load at several points (See Figure 2) on the
structure and determining what the resulting reaction will be at A. This can be done
by solving the support reaction YA as a function of the position of a downward acting
unit load. One such equation can be found by summing moments at Support B.
Figure 1 - Beam structure for influence line example
Figure 2 - Beam structure showing application of unit load
MB = 0 (Assume counter-clockwise positive moment)
-YA(L)+1(L-x) = 0
YA = (L-x)/L = 1 - (x/L)
The graph of this equation is the influence line for the support reaction at A (See
Figure 3). The graph illustrates that if the unit load was applied at A, the reaction at A
would be equal to unity. Similarly, if the unit load was applied at B, the reaction at A
would be equal to 0, and if the unit load was applied at C, the reaction at A would be
equal to -e/L.
Figure 3 - Influence line for the support reaction at A
Once an understanding is gained on how these equations and the influence lines they
produce are developed, some general properties of influence lines for statically
determinate structures can be stated.
a
‘cut’
L
b
x1 (unit load) Mx
Mx
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 a unit load between 0 ≤ x ≤ b ,
Mx = _xb_ - (x-a) …………(2) L
Figure 1
RA = (1-x/L) RB = x/L
y1 y3y2
x2
x1
x3
F1 F2 F3a + b = L
Moment influence line for cut section
Part 2: If the beam is loaded as shown below, the moment at the ‘cut’ can be
calculated using the influence line. (See Figure 2).
Moment at ‘cut’ section = F1y1 + F2y2 + F3y3……………(3)
(y1, y2 and y3 are ordinates derived from the influence line in terms of x1, x2, x3,
a, b and L).
Figure 2
5.0 APPARATUS
5.1 Bending moment Machine
5.2 Weights (Loadings)
6.0 PROCEDURES
6.1 Part 1
6.1.1 The Digital Force Display meter is checked that it reads zero with
no loading that the structure is subjected to.
6.1.2 A hanger with a 200g mass is placed on the left of the cut.
6.1.3 The Digital Force Display reading is recorded in Table 1.
6.1.4 Repeat steps to the next grooved hanger until to the last grooved
hanger at the right hand support.
6.1.5 Calculation in Table 1 is completed.
6.2 Part 2
6.2.1 The Digital Force Display meter is checked that it reads zero with
no loading that the structure is subjected to.
6.2.2 Three loads were placed with 100g, 200g and 300g and place them
at positions between the supports. The positions and the Digital
Force Display were recorded and the force reading is converted
into bending moment (Nm) using:
Bending moment at a cut (Nm) = Displayed Force x 0.125
6.2.3 The theoretical bending moment at the cut and also the support
reaction, RA and RB are calculated and entered into Table 2.
7.0 RESULT
PART 1
Location Of Load
From Left Hand
Support (M)
Digital ForceDisplay
Reading (N)
Moment AtCut Section
(Nm)
ExperimentalInfluence Line
Value (Nm)
TheoreticalInfluence
Lines Value(Nm)
0.04 0.2 0.025 0.013 0.0130.06 0.3 0.038 0.019 0.0190.08 0.4 0.050 0.025 0.0250.10 0.5 0.063 0.032 0.0320.12 0.6 0.075 0.038 0.0380.14 0.7 0.088 0.045 0.0450.16 0.8 0.100 0.051 0.0510.18 0.9 0.113 0.058 0.0570.20 1.0 0.125 0.064 0.0640.22 1.2 0.150 0.076 0.0700.24 1.3 0.163 0.083 0.0760.26 1.6 0.200 0.102 0.0830.30 1.2 0.150 0.076 0.0960.32 0.9 0.113 0.058 0.0820.34 0.7 0.088 0.045 0.0680.36 0.5 0.063 0.032 0.0550.40 0.1 0.013 0.007 0.027
Moment at cut section = Digital force reading x 0.125
a. Moment at cut section= 0.2 x 0.125= 0.025 Nm
b. Moment at cut section= 0.3 x 0.125= 0.038 Nm
c. Moment at cut section= 0.4 x 0.125= 0.050 Nm
d. Moment at cut section= 0.5 x 0.125= 0.063 Nm
e. Moment at cut section= 0.6 x 0.125= 0.075 Nm
f. Moment at cut section= 0.7 x 0.125= 0.088 Nm
g. Moment at cut section= 0.8 x 0.125= 0.100 Nm
h. Moment at cut section= 0.9 x 0.125= 0.113 Nm
i. Moment at cut section= 1.0 x 0.125= 0.125 Nm
j. Moment at cut section= 1.2 x 0.125= 0.150 Nm
k. Moment at cut section= 1.3 x 0.125= 0.163 Nm
l. Moment at cut section= 1.6 x 0.125= 0.200 Nm
m. Moment at cut section= 1.2 x 0.125= 0.150 Nm
n. Moment at cut section= 0.9 x 0.125= 0.113 Nm
o. Moment at cut section= 0.7 x 0.125= 0.088 Nm
p. Moment at cut section= 0.5 x 0.125= 0.063 Nm
q. Moment at cut section= 0.1 x 0.125= 0.013 Nm
Experimental Influence line values = Moment (Nm)
Load (N)
a. Experimental Influence line values = 0.025 1.962
= 0.013 Nmb. Experimental Influence line values = 0.038
1.962= 0.019 Nm
c. Experimental Influence line values = 0.050 1.962
= 0.025 Nmd. Experimental Influence line values = 0.063
1.962= 0.032 Nm
e. Experimental Influence line values = 0.075 1.962
= 0.038 Nmf. Experimental Influence line values = 0.088
1.962= 0.045 Nm
g. Experimental Influence line values = 0.100 1.962
= 0.051 Nmh. Experimental Influence line values = 0.113
1.962= 0.058 Nm
i. Experimental Influence line values = 0.125 1.962
= 0.064 Nmj. Experimental Influence line values = 0.150
1.962= 0.076 Nm
k. Experimental Influence line values = 0.163 1.962
= 0.083 Nml. Experimental Influence line values = 0.200
1.962= 0.102 Nm
m. Experimental Influence line values = 0.150 1.962
= 0.076 Nmn. Experimental Influence line values = 0.113
1.962= 0.058 Nm
o. Experimental Influence line values = 0.088 1.962
= 0.045 Nmp. Experimental Influence line values = 0.063
1.962= 0.032 Nm
q. Experimental Influence line values= 0.013 1.962
= 0.007 Nm
Theoretical Influence lines valueEquation 1 for load position 40 to 260 mm
MX = (L-x)a – 1(a-x) L
a. MX = (0.44 – 0.04)0.30 – 1(0.30-0.04) 0.44
= 0.013 Nmb. MX = (0.44 – 0.06)0.30 – 1(0.30-0.06)
0.44= 0.019 Nm
c. MX = (0.44 – 0.08)0.30 – 1(0.30-0.08) 0.44
= 0.025 Nmd. MX = (0.44 – 0.10)0.30 – 1(0.30-0.10)
0.44= 0.032 Nm
e. MX = (0.44 – 0.12)0.30 – 1(0.30-0.12) 0.44
= 0.038 Nmf. MX = (0.44 – 0.14)0.30 – 1(0.30-0.14)
0.44= 0.045 Nm
g. MX = (0.44 – 0.16)0.30 – 1(0.30-0.16) 0.44
= 0.051 Nmh. MX = (0.44 – 0.18)0.30 – 1(0.30-0.18)
0.44= 0.057 Nm
i. MX = (0.44 – 0.20)0.30 – 1(0.30-0.20) 0.44
= 0.064 Nmj. MX = (0.44 – 0.22)0.30 – 1(0.30-0.22)
0.44= 0.070 Nm
k. MX = (0.44 – 0.24)0.30 – 1(0.30-0.24) 0.44
= 0.076 Nml. MX = (0.44 – 0.26)0.30 – 1(0.30-0.26)
0.44= 0.083 Nm
Theoretical Influence lines valueEquation 2 for load position 320 to 400 mm
MX = xb – (x-a) L
a. MX = (0.32)(0.14) – (0.32-0.30) 0.44
= 0.082 Nmb. MX = (0.34)(0.14) – (0.34-0.30)
0.44= 0.068 Nm
c. MX = (0.36)(0.14) – (0.36-0.30) 0.44
= 0.055 Nmd. MX = (0.40)(0.14) – (0.40-0.30)
0.44= 0.027 Nm
Part 2
LocationPosition of hanger from left hand support (m)
Digital Force
Reading (N)
ExperimentMoment
Theoretical Moment
(Nm)100g 200g 300g
1 0.06 0.28 0.36 1.6 0.2000 0.21502 0.40 0.28 0.10 2.5 0.3130 0.31313 0.28 0.80 0.36 1.9 0.2380 0.23194 0.12 0.24 0.38 2.2 0.2750 0.2615
Experimental Moment = Digital Force Reading x 0.125
1. 1.6N x 0.125m = 0.2000Nm
2. 2.5N x 0.125m = 0.3130Nm
3. 1.9N x 0.125m = 0.2380Nm
4. 2.2N x 0.125m = 0.2750Nm
Load = 100g
100g x 1kg x 9.81 1000g
= 0.981N
Load = 200g
200g x 1kg x 9.81 1000g
= 1.962N
Load = 300g
300g x 1kg x 9.81 1000g
= 2.943N
Theoretical Moment
Location 1
x1
0.14m
Moment influence line
Mc = length before cut – length before cut x length before cut Length beam
Mc = 0.3 – 0.3 x 0.3 0.44
Mc = 0.0955m
x1 = 0.08mx2 = 0.16mx3 = 0.38m
‘Cut’
2.943N
y1
0.0955
1.962N 0.981N
x3 x2
y2 y3
0.3m
Influence lines x
0.0955 = _y1_0.3 0.08
y1 = 0.0255m
0.0955 = _y2_0.3 0.16
y2 = 0.0509m
0.0955 = y3 __ 0.14 (0.44-0.38)
y3 = 0.0409m
Moment at ‘cut’ section = F1y1+ F2y2 + F3y3
= 2.943(0.0255) + 1.962(0.0509) + 0.981(0.0409)= 0.0750 + 0.0999 + 0.0401= 0.2150Nm
Location 2
0.981N 1.962N 2.943N
x1
0.3m 0.14m
Moment influence line
Mc = length before cut – length before cut x length before cut Length beam
Mc = 0.3 – 0.3 x 0.3 0.44
Mc = 0.0955m
x1 = 0.04mx2 = 0.16mx3 = 0.34m
‘Cut’
y1
0.0955
x3
x2
y2
y3
Influence lines x
0.0955 = _y1_0.3 0.04
y1 = 0.0127m
0.0955 = _y2_0.3 0.16
y2 = 0.0509m
0.0955 = y3 __ 0.14 (0.44-0.34)
y3 = 0.0682m
Moment at ‘cut’ section = F1y1+ F2y2 + F3y3
= 0.981(0.0127) + 1.962(0.0509) + 2.943(0.0682)= 0.0125 + 0.0999 + 0.2007= 0.3131Nm
Location 3
1.962N0.981N2.943N
x1
0.14m
Moment influence line
Mc = length before cut – length before cut x length before cut Length beam
Mc = 0.3 – 0.3 x 0.3 0.44
Mc = 0.0955m
x1 = 0.08mx2 = 0.16mx3 = 0.36m
‘Cut’
y1
0.0955
x3 x2
y2 y3
0.3m
Influence line x
0.0955 = _y1_0.3 0.08
y1 = 0.0255m
0.0955 = _y2_0.3 0.16
y2 = 0.0509m
0.0955 = y3 __ 0.14 (0.44-0.36)
y3 = 0.0546m
Moment at ‘cut’ section = F1y1+ F2y2 + F3y3
= 2.943(0.0255) + 0.981(0.0509) + 1.962(0.0546)= 0.0750 + 0.0499 + 0.107= 0.2319Nm
Location 4
0.981N1.962N2.943N
x1
0.14m
Moment influence line
Mc = length before cut – length before cut x length before cut Length beam
Mc = 0.3 – 0.3 x 0.3 0.44
Mc = 0.0955m
x1 = 0.06mx2 = 0.20mx3 = 0.32m
‘Cut’
y1
0.0955
x3 x2
y2 y3
0.3m
Influence line x
0.0955 = _y1_0.3 0.06
y1 = 0.0191m
0.0955 = _y2_0.3 0.20
y2 = 0.0637m
0.0955 = y3 __ 0.14 (0.44-0.32)
y3 = 0.0819m
Moment at ‘cut’ section = F1y1+ F2y2 + F3y3
= 2.943(0.0191) + 1.962(0.0637) + 0.981(0.0819)= 0.0562 + 0.1250 + 0.0803= 0.2615Nm
8.0 DISCUSSIONS
Part 2
1. Calculate the percentage difference between experimental and theoretical results in Table 2. Comment on why results differ.
Percentage difference :
Different of experiment moment and theoretical moment x 100%Experiment Moment
Location 1
0.2150-0.200 x 100% 0.2000
= 7.50%
Location 2
0.3131-0.3130 x 100% 0.3130
= 0.03%
Location 3
0.2380-0.2319 x 100% 0.2380
= 2.56%
Location 4
0.2750-0.2615 x 100% 0.2750
= 4.90%
Comment :
Like every other experiment, there is no doubt that the experiment can
always improved in all aspects according to the passing of time. In this
experiment, it can be seen from the results that there is differences between the
experimental and theoretical results. It should be noticed that in the experiment, a
procedure states that the Force Display has to be checked to be zero, and if not, it
is calibrated until it is zero. In this procedure, this promotes the possibility that
there exist loadings even before the experimental loadings are subjected to the
beam. This loading may not be from any weights but could be external factors
such as environment.
It is possible that this had effect during the calibration to zero of the Meter
and this has caused error during the readings of values from the Force Display.
Not only that, the weights may have variation in their weights and therefore a
number of them might have more or less than the required weight and this either
reduces or increases the loading than she value that was assumed it to be.