Moment Influence Line (Part 2)

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.