Report Shear Force & Bending Moment in Influence Line

LS07

SHEAR FORCE INFLUENCE LINE

1.0 INTRODUCTION

Influence lines have important application for the design of structures that resist large

live loads. An influence line represents either 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 constructed, one can tell at glance where the moving load should be place on

the structures 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 be

calculated from the ordinates of the influence-line diagram. An influence line shows the

variation of an external or internal force as a unit load travels across the length of the

structure. For a beam-type bridge structure, the internal force may be either a transverse shear

force or bending moment acting at some discrete location along the length of the bridge.

Influence lines for truss bridges are commonly used to show the magnitude of the axial force

acting in a truss member as a unit load travels across the panel points of the bottom chord.

Furthermore, influence lines for the reaction forces at external support locations are employed

in all bridge types to determine the amount of load transferred to the bridge bearings. The two

common methods for constructing an influence line are through consideration of equilibrium

or by applying the concept of virtual work (i.e., the Müller- Breslau Principle).

1

2.0 PRINCIPLE

The principle line for bending moment at a section is the graph curves representing the variation of bending moment at a section of a various positions of the load of the span on the beam. The sign convension followed, is shown in the figure 1.

Consider a simply supported beam of span ‘L’ as shown in figure 1 and 2. It is required to draw influence line for bending moment at ‘C’ at a distance ‘a’ from the left support.

When the load ‘W’ is toward left of section ‘C’, at a distance ‘x’ from left support ‘A’ 0 < x < a

The shear force at C,

Vc = Va – W

M = +Rb ( L – a ) = ( L – a) ……………..[Equation 1]

(Considering right side of section c)When the load ‘W’ is towards right of section ‘D’ at a distance ‘x’ from the left support ‘A’ 0 < x < a

2

The bending moment at ‘D’ (considering left side of section C)

M = +Ra • a

M = + ……………..[Equation 2]

3.0 OBJECTIVES:

1) To determine the shear force influence line when the beam is subjected to a load

moving from left to right.

2) To plot the shear force influence line when the beam is subjected to a point load

moving from left to right.

3

4.0 APPARATUS

Sample of beam The support

Loads Beam with cut section

Indicator load hanger

4

5.0 PROCEDURES

The load cell is connected to the digital indicator.

The indicator is switched on. The indicator’s reading must be switch on 10 minutes before taking readings for stability of readings.

The two simple supports is fixed to the aluminium base at a distance equal to the span of the beam to be tested. The support is

screwed tightly to the base.

The load hanger is hanged at the 50 mm from the left support.

The indicator reading is noted. If it is not zero, the tare button on the indicator is pressed.

A load is placed on the load hanger

The indicator reading is recorded. This represent the shear force at cut section.

The load is removed from the hanger. The load hanger is moved 200 mm from the left support and step 7-11 is repeated.

The distance is increase each time by 50 mm.

Steps 7-11 repeated until the load which is end B.

5

6.0 RESULTS

Case 1:

Beam span = 1000mmDistance of the shear section from the left support = 390mmW1 = 2N

Table: case 1.

Distance from left support (mm)

Shear Force at x-xExperimental = Rb (N) Theory (N)

100 - 0.1 - 0.2200 - 0.1 - 0.4600 0.8 0.8800 0.4 0.4900 0.2 0.2

6

W1

L1

RA

RB

x

x

Case 2:

Beam span = 1000mmDistance of the shear section from the left support = 3900mmW1 = 2NW2 = 4N

a = 20mm

Table: case 2.

Distance from left support (mm)

Shear Force at x-xExperimental = Rb (N) Theory (N)

100 -0.4 -0.68200 -0.6 -1.28600 2.3 2.32800 1.2 1.12900 0.5 0.52

7

W1

RA

RB

x

x

W2

aL1

Case 3:

Beam span = 1000mmDistance of the shear section from the left support = 390mmW1 =2W2 = 4NW3 = 2N

a = b = 20mm

Table: case 3.

Distance from left support (mm)

Shear Force at x-xExperimental = Rb (N) Theory (N)

100 -0.5 -1.16200 -1.1 -1.76600 3.1 3.04800 1.4 1.44900 0.7 0.64

8

W1

L1

RA

RB

x

x

W2

a b

W3

7.0 CALCULATIONS

Case 1 :

P1

100mm

390mm C 610mm

b/L = 0.61

+-

i

a/L=-0.39

Load, P1 = 2kN

At L1 = 100mm

i/100 = -0.39/390

i = -0.1

Vc = - 0.1 x 2N = - 0.2 N

At L2 = 200mm

i/200 = -0.39/390

i = -0.2

Vc = - 0.2 x 2N = - 0.4 N

At L3= 600mm

i/400 = 0.61/610

i = 0.4

Vc = 0.4 x 2N = 0.8 N

At L1 = 800mm

i/200 = 0.61/610

i = 0.2

Vc = 0.2 x 2N = 0.4 N

9

At L1 = 900mm

i/100 = 0.61/610

i = 0.1

Vc =0.1 x 2N = 0.2 N

Case 2 :

P1 P 2

100mm 20mm

390mm C 610mm

b/L = 0.61

i j a/L=-0.39

Load, P1 = 2kN P2 = 4 kN

At L1 = 100mm

j/120 = -0.39/390

j = -0.12

Vj = - 0.12 x 4 = - 0.48 N

Vc = -0.48 – 0.2 = -0.68 N

At L2 = 200mm

j/220 = -0.39/390

j = -0.22

Vj = - 0.22 x 4 = - 0.88 N

Vc = -0.8 – 0.4 = -1.28 N

At L3= 600mm

j/380 = 0.61/610

j = 0.38

10

Vj = 0.38 x 4 = 1.52 N

Vc = 0.8 – 1.52 = 2.32 N

At L4 = 800mm

j/180 = 0.61/610

j = 0.18

Vj = 0.18 x 4 = 0.72N

Vc = 0.72+0.4 = 1.12N

At L5 = 900mm

j/80 = 0.61/610

j = 0.08

Vj = 0.08 x 4 = 0.32N

Vc = 0.32+0.2 = 0.52N

Case 3 :

P1 P 2

100mm20mm20mm

390mm C 610mm

b/L = 0.61

i j k a/L=-0.39

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Load, P1 = 2kN P2 = 4 kN P3 = 2 kN

At L1 = 100mm

k/140 = -0.39/390

k = -0.14

Vk = - 0.14 x 2 = - 0.28 N

Vc = -0.28 – 0.68= -0.96 N

At L2 = 200mm

k/240 = -0.39/390

k = -0.24

Vk = - 0.24 x 2 = - 0.48 N

Vc = -0.48 – 1.28= -1.76 N

At L3= 600mm

k/360 = 0.61/610

k = 0.36

Vk = 0.36 x 2 = 0.72 N

Vc = 0.72+2.32 = 3.04 N

At L4 = 800mm

k/160 = 0.61/610

k = 0.16

Vk = 0.16 x 2 = 0.32 N

Vc = 0.32+1.12 = 1.44 N

At L5 = 900mm

k/60 = 0.61/610

k = 0.06

Vk = 0.06 x 2 = 0.12 N

Vc = 0.12+0.52 = 0.64 N

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8.0 DISCUSSION AND ANALYSIS

Refer to the graph of Load vs. Shear force.

In this experiment, we used the load 100g, 200g, 600g, 800g, and

900g . From this experiment, the value for the experimental and

theoretical are decrease first and then increase after the cut section.

The value for the experimental and theoretical value is nearly same.

The value is depend on the location of the load. Its shows that the

location is one of the causes for the differences between the value.

We should know that, influence lines can be used to calculate the

shear force at the cut section.

The calculation of percentage of error:

Percentage of error = Theory – Experiment X 100% Theory

Table: case 1.

Distance from left support

(mm)

Shear Force at x-x Percentages of Error, %Experimental = Rb

(N)Theory (N)

100 - 0.1 - 0.2 50.0200 - 0.1 - 0.4 75.0600 0.8 0.8 0.0800 0.4 0.4 0.0900 0.2 0.2 0.0

Table: case 2.

Distance from left support

(mm)

Shear Force at x-x Percentages of Error, %Experimental = Rb

(N)Theory (N)

100 -0.4 -0.68 41.18200 -0.6 -1.28 53.13600 2.3 2.32 0.86800 1.2 1.12 7.14900 0.5 0.52 3.85

13

Table: case 3.

Distance from left support (mm)

Shear Force at x-x Percentages of Error, %Experimental = Rb

(N)Theory (N)

100 -0.5 -1.16 56.89200 -1.1 -1.76 37.50600 3.1 3.04 1.97800 1.4 1.44 2.78900 0.7 0.64 9.38

From the calculation of error in this experiment, it shows that the error at the distance

of 100mm and 200mm is high which is more than 50%. But for the distances of

600mm, 800mm and 900mm, the percentages of error are small. This experimental

results are sometimes 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

From the result that we get, there are some errors that make our result not accurate

and contribute the error between the experiment and theory.

i. Digital indicator is not too accurateAlthough the value of experiment quite near with the value of theory a there are still have error. The digital indicator is not too accurate.

ii. The digital indicator is too sensitiveWhen we taking the reading, the screen show that the reading not in static. That mean the digital indicator is too sensitive with the wind and the surrounding movement.

iii. The load hanger is shakingWhen we taking the reading, we put the load to the hanger. When the load is putting to the hanger, the hanger is shaking and the reading of digital indicator is change. So it effect the reading.

iv. Parallax error: Reading the ruler scale The ruler scale is in centimeter (cm). So, when the reading process, we can’t get the accurate value, because the scale are not suitable for our eye to read with accurately

v. The beam is sensitive

14

When we do the experiment, the beam is moving when we try to put the load. When we want to change the holder of hanger to right side, the beam is not in the original position yet.

9.0 CONCLUSION

While doing this experiment, we get the value of the theoretical is exactly the same

from the experiment value. Hence, the objective of this experiment is proven. So, we

know that our experiment was archived the objective. After the experiment, we have

learned how to determine the shear force influence line when the beam is subjected to

a load moving from left to right.

We also learn how to plot the shear force influence line when the beam is subjected to

a point load moving from left to right.

15

LS08

BENDING MOMENT INFLUENCE LINE

1.0 INTRODUCTION

The definition of bending moment which is the internal load generated within a bending

element whenever a pure moment is reacted, or a shear load is transferred by beam action

from the point of application to distant points of reaction. 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 such as shear, axial, and moment at the desired location. 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.

16

2.0 PRINCIPLES

The principle line for bending moment at a section is the graph curves representing the variation of bending moment at a section of a various positions of the load of the span on the beam. The sign convension followed, is shown in the figure 1.

Consider a simply supported beam of span ‘L’ as shown in figure 1 and 2. It is required to draw influence line for bending moment at ‘C’ at a distance ‘a’ from the left support.

When the load ‘W’ is toward left of section ‘C’, at a distance ‘x’ from left support ‘A’ 0 < x < a

The shear force at C,

Vc = Va – W

M = +Rb ( L – a ) = ( L – a) ……………..[Equation 1]

(Considering right side of section c)

17

When the load ‘W’ is towards right of section ‘D’ at a distance ‘x’ from the left support ‘A’ 0 < x < a

The bending moment at ‘D’ (considering left side of section C)

M = +Ra • a

M = + ……………..[Equation 2]

3.0 OBJECTIVE

1) To determine the bending moment influence line when the beam is subjected to a

load moving from left to right

2) To determine the value of the function to this load of influence line.

3) To calculate values for each function are then plotted where the load was applied

and then connected together to generate the influence line for the function.

18

4.0 APPARATUS

Sample of beam The support

Loads Beam with cut section

Indicator load hanger

19

5.0 PROCEDURE

The load cell is connected to the digital indicator.

The indicator is switched on. The indicator’s reading must be switch on 10 minutes before taking readings for stability of readings.

The two simple supports is fixed to the aluminium base at a distance equal to the span of the beam to be tested. The support is

screwed tightly to the base.

The load hanger is hanged at the 50 mm from the left support.

The indicator reading is noted. If it is not zero, the tare button on the indicator is pressed.

A load is placed on the load hanger

The indicator reading is recorded. This represent the shear force at cut section.

The load is removed from the hanger. The load hanger is moved 200 mm from the left support and step 7-11 is repeated.

The distance is increase each time by 50 mm.

20

Steps 7-11 repeated until the load which is end B.

6.0 RESULT

6.1 CASE 1

Beam span = 1000 mmDistance of the shear section from the left support = 390 mmW1 = 2 NDistance of load cell from the beam cross section = 175mm

Table: case 1

Distance from left support (mm)

Bending Moment at x-xExperimental = ( F*175) Theory (Nmm)

100 0.6 x 175 = 105 122200 1.4 x 175 = 245 244600 1.8 x 175 = 315 312800 0.9 x 175 = 157.5 156900 0.4 x 175 = 70 78

6.2 CASE 2

21

W1

L1

RA RB

x

x

Bea

m span = 1000 mm

Distance of the shear section from the left support = 390 mm

W1 = 2 N

W1 = 4 NDistance of load cell from the beam cross section = 175mm

Table: case 2

Distance from left support (mm)

Bending Moment at x-xExperimental = ( F*175) Theory (Nmm)

100 2.5 x 175 = 437.5 414.8200 4.6 x 175 = 805 780.8600 5.3 x 175 = 927.5 904.8800 2.5 x 175 = 437.5 436.8950 1.2 x 175 = 210 202.8

6.3 CASE 3

Beam span = 1000 mmDistance of the shear section from the left support = 390 mmW1 = 2 NW2 = 4 NW3 = 2 NDistance of load cell from the beam cross section = 175 mm

22

W1

RA RB

x

x

W2

aL1

W1

L1

RA RB

x

x

W2

a b

W3

Table: case 3.

Distance from left support (mm)

Bending Moment at x-xExperimental = ( F*175) Theory (Nmm)

100 3.5 x 175 = 612.5 585.6200 6.4 x 175 = 1120 1073.6600 7.0 x 175 = 1225 1185.6800 3.1 x 175 = 542.5 561.6900 1.5 x 175 = 262.5 249.6

7.0 CALCULATION

For case 1:

P1

100mm

390mm C 610mm

q

ab/L = 390 x 610 / 1000

= 237.9

Load, P1 = 2kN

At L1 = 100mm

q/100 = 237.9/390

23

q = 61

Mc = 61 x 2N = 122 Nm

At L2 = 200

q/200 = 237.9/390

q = 122

Mc = 122 x 2 = 244 Nm

At L3 = 600

q/400 = 237.9/610

q = 156

Mc = 156 x 2 = 312 Nm

At L4 = 800

q/200 = 237.9/610

q= 78

Mc = 78 x 2 =156 Nm

At L4 = 800

q/100 = 237.9/610

q= 39

Mc = 39 x 2 = 78 Nm

For case 2:

P1 P2

100mm 20mm

390mm 610mm

q

24

r

Load, P1 = 2kN P2 = 4kN

At L1 = 100mm

r/120 = 237.9/390

r = 73.2

Mr = 73.2 x 4N = 292.8Nm

Mc =292.8+122 =414.8 Nm

At L2 = 200

r/220 = 237.9/390

r = 134.2

Mr = 134.2x 4N = 536.8Nm

Mc =536.8+122 = 244 = 780.8 Nm

At L3 = 600

r/380 = 237.9/610

r = 148.2

Mr = 148.2 x 4N = 592.8Nm

Mc =592.8+312 =904.8 Nm

At L4 = 800

r/180 = 237.9/610

r = 70.2

Mr = 70.2 x 4N = 280.8Nm

Mc =280.8+156 =436.8 Nm

At L5 = 900

r/80 = 237.9/610

r = 31.2

Mr = 31.2 x 4N = 124.8Nm

25

Mc =124.8+78 =202.8 Nm

For case 3:

P1 P2 P3

100mm 20mm20mm

390mm 610mm

q

r s

Load, P1 = 2kN P2 = 4kN P3 = 2 kN

At L1 = 100mm

s/140 = 237.9/390

s = 85.4

Ms = 85.4 x 2N = 170.8Nm

Mc =170.8+414.8 =585.6 Nm

26

At L2 = 200

s/240 = 237.9/390

s = 146.4

Ms = 146.4 x 2N = 292.8Nm

Mc =292.8+780.8 =1073.6 Nm

At L3 = 600

s/360 = 237.9/610

s = 140.4

Ms = 140.4 x 2N = 280.8Nm

Mc =280.8+904.8 =1185.6 Nm

At L4 = 800

s/160 = 237.9/610

s = 62.4

Ms = 62.4 x 2N = 124.8Nm

Mc =124.8+436.8 =561.6 Nm

At L5 = 900

s/60 = 237.9/610

s = 23.4

Ms = 23.4 x 2N = 46.8Nm

Mc =46.8+202.8 =249.6 Nm

27

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8.0 DISCUSSSION AND ANALYSIS

Refer to the graph of Load vs. Bending Moment

P1 P2 P3

100mm20mm20mm

390mm C 610mm

Shear foerce diagram +

-

i 1

k1 j1

Bending moment diagram

+

q1

r1 s1

The accuracy:

29

The calculation of percentage of error:

Percentage of error = Theory – Experiment X 100% Theory

Table: case 1

Distance from left support

(mm)

Bending Moment at x-x Percentage of Error, %Experimental =

( F*175)Theory (Nmm)

100 0.6 x 175 = 105 122 13.93200 1.4 x 175 = 245 244 0.41600 1.8 x 175 = 315 312 0.96800 0.9 x 175 = 157.5 156 0.96900 0.4 x 175 = 70 78 10.26

Table: case 2

Distance from left support

(mm)

Bending Moment at x-x Percentage of Error, %Experimental =

( F*175)Theory (Nmm)

100 2.5 x 175 = 437.5 414.8 5.47200 4.6 x 175 = 805 780.8 3.10600 5.3 x 175 = 927.5 904.8 2.51800 2.5 x 175 = 437.5 436.8 0.16950 1.2 x 175 = 210 202.8 3.55

Table: case 3.

Distance from left support

(mm)

Bending Moment at x-x Percentage of Error, %Experimental =

( F*175)Theory (Nmm)

100 3.5 x 175 = 612.5 585.6 4.59200 6.4 x 175 = 1120 1073.6 4.32600 7.0 x 175 = 1225 1185.6 3.32800 3.1 x 175 = 542.5 561.6 3.40900 1.5 x 175 = 262.5 249.6 5.17

The percentages of the error is small or less than half. It shows here that the accuraccy of the

experiment is high. So, we conclude that this experiment is success because the experimental

and theoritical value are nearly same.

Sometimes, the value of theoritical is too much difference. This is maybe because of the error

that occur during the experiment. There are some errors that make result not accurate and

contribute the error between the experiment and theory.

30

a. Digital indicator is not too accurateAlthough the value of experiment quite near with the value of theory a there are still have error. The digital indicator is not too accurate.

b. The digital indicator is too sensitiveWhen we taking the reading, the screen show that the reading not in static. That mean the digital indicator is too sensitive with the wind and the surrounding movement.

c. The load hanger is shakingWhen we taking the reading, we put the load to the hanger. When the load is putting to the hanger, the hanger is shaking and the reading of digital indicator is change. So it effect the reading.

d. Parallax error: Reading the ruler scale The ruler scale is in centimeter (cm). So, when the reading process, we can’t get the accurate value, because the scale are not suitable for our eye to read with accurately

e. The beam is sensitive When we do the experiment, the beam is moving when we try to put the load. When we want to change the holder of hanger to right side, the beam is not in the original position yet.

The ways to overcomes his problem:

a. Check the apparatus condition whether it good or not.

b. While putting the load at the hanger, we must put it slowly so that it won’t shake and affect the reading

c. When we take the reading, we had to wait until the digital indicator constant and the reading not move anymore.

d. We also must wait until the reading shown by the indicator is totally constant before record it.

9.0 CONCLUSION

From our discussion, the influence line concept and its application is to specific

structural systems reached a certain conclusion at the end of the establishment phrase

of structural theory . However , the general influences lines theory actually belongs to

31

the classical phase of structural theory. the result and the theory of left support, middle

support and right support. This is may be because of experimental error done during

the experiment session. The errors can be the condition during experiment, wind or

error in handling the equipments during the experiment.

10.0 REFERENCES

1) http://composite.about.com/library/glossary/b/bldef-b604.htm

2) http://en.wikipedia.org/wiki/Influence_line

32

3) http://www.engr.uky.edu

4) www.bending and shear/htm

5) www.bending/influenceline.com

6) Structural analysis- RC Hibbeler..sixth edition

11.0 APPENDICES

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