Shear Force Influence Line

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FACULTY OF CIVIL AND ENVIRONMENTAL ENGINEERING

DEPARTMENT OF STRUCTURE AND MATERIAL

ENGINEERING

LAB MATERIAL

REPORT

Subject Code BFC 31901

Code & Experiment Title OPEN ENDED – SHEAR FORCE INFLUENCE LINES

Course Code 2 BFF

Date 04/03/2012

Section / Group SECTION 9 / GROUP 7

Name MUHAMMAD IKHWAN BIN ZAINUDDIN (DF100018)Members of Group 1.NUR EZRYNNA BINTI MOHD ZAINAL (DF100118)

2.MUHAMMAD NUH BIN AHMAD ZAIRI (DF100093)

3.NUR EEZRA ATHIRLIA BINTI GHAZALI (DF100147)

4.MUHAMMAD HUZAIR BIN ZULKIFLI (DF100040)

5.ZIRWATUL FAUZANA BINTI CHE JEMANI (DF100027)

Lecturer/Instructor/Tutor EN. MOHD KHAIRY BIN BURHANUDIN

Received Date 20/04/2012

Comment by examiner Received



STUDENT CODE OF ETHIC

(SCE)

DEPARTMENT OF STRUCTURE AND MATERIAL

ENGINEERING

FACULTY OF CIVIL & ENVIRONMENTAL ENGINEERINGUTHM

We, hereby confess that we have prepared this report on our effort. We also admit not to receive

or give any help during the preparation of this report and pledge that everything mentioned in the

report is true.

___________________________

Student Signature

Name : MUHAMMAD IKHWAN BIN ZAINUDDIN

Matric No. : DF100018

Date : 20/04/2012

_______________________

Student Signature

Name : MUHAMMAD NUH BIN AHMAD ZAIRI


Date : 20/04/2012

___________________________

Student Signature

Name : NUR EEZRA ATHIRLIA BINTI GHAZALI


Date : 20/04/2012

___________________________

Student Signature

Name : MUHAMMAD HUZAIR BIN ZULKIFLI


Date : 20/04/2011

___________________________

Student Signature

Name : NUR EZRYNNA BINTI MOHD ZAINAL


Date : 20/04/2012

_______________________

Student Signature

Name : ZIRWATUL FAUZANA BINTI CHE JEMANI


Date : 20/04/2012



SCOPE OF WORK GROUP 7

NO NAME SCOPE OF WORK

1 Nur Ezrynna binti Mohd Zainal Calculation of the laboratory report

2 Muhammad Ikhwan in Zainuddin Calculation of the laboratory report

3 Nur Eezra Athirlia binti Ghazali Discussion of the laboratory report

4 Muhammad Nuh bin Ahmad Zairi Theory and record data of the laboratory

report

5 Zirwatul Fauzana binti Che Jemani Conclusion of the laboratory report

6 Muhammad Huzair bin Zulkifli Procedure and result of the laboratory report



1.0 INTRODUCTION

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, i.e. 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.

2.0 OBJECTIVE

2.1 To plot Shear force influence line.

2.2 To verify the use of a shear force influence on a simply supported beam

2.3 Understanding about the envelopes of maximum influence line values

2.4 To construct influence line for maximum end shear in a beam supporting a series of

moving concentrated loads

3.0 LEARNING OUTCOME

3.1 The application the engineering knowledge in practical application

3.2 To enhance technical competency in structural engineering through laboratory

application.

3.3 To communicate effectively in group

3.4 To identify problem, solving and finding out appropriate solution through laboratory

application



4.0 THEORY

Defination: An influence line is a plot of the magnitude of the resulting reaction/axial

force/shear/moment generated in a beam or structure as a unit load travels across its length. Influence

lines can be generated for any of these actions (reactions, axial forces, shears, or moments) in a

structure. Influence lines are used to determine where to place moving loads on a structure to obtain

maximum results (reactions, shears, moments, axial forces), and to compute these reactions or other

actions (shears/moments/axial forces) once the loads are placed in critical positions.

Part 1: This Experiment examines how shear force varies at a cut section as a unit load moves from

one end to another (see Figure 1). From the diagram, shear force influence line equation can be

writen.

R A = load (a-x) + (digital force x 0.125) (before cut)

a

R A = (digital force x 0.125) (after cut)

a

‘Cut’

x 1 (unit load) M x

M x

R A= (1 – x/L)

R B=x/L

a b

L

Figure 1 - shear force varies at a cut section



Part 2: If the beam are loaded as shown in Figure 2, the shear force at the ‘cut’ can be calculated

using the influence line. (See Figure 2).

Shear force at ‘cut’ section = F1y1 + F2y2 + F3y3

(y1, y2 and y3 are ordinates derived from the influence line in terms of x1, x2,

x3 ,a , b and L)

F1 F2 F3 Lba

x1

x2

x3

Shear Force Influence

line for cut

y1 y2 y3

Figure 2 - beam are loaded

5.0 APPARATUS



6.0 PROCEDURES

Part 1:

i. Check the Digital Force Meter reads zero with no load.

ii. Place hanger with 200g of mass at support and locate it at the left support and

record the Digital Force reading in Table 1.

iii. Repeat the procedure number 2 with different distance

iv. Complete the calculation in Table 1.

Part 2:

i. Place three load hangers with different mass at same position between the

supports. Record the positions and the Digital Force Display reading in Table

2.

ii. Repeat the procedure with three other locations.

iii. Complete the calculation in Table 2

7.0 RESULT

Part A :

Location Of

Load From Left

Hand Support

(m)

Digital

Force

Display

Reading (N)

Shear

Force At

Cut

Section (N)

Experimental Influence

Line Value

Theoretical Influence Line

Value

R A (kN) R B (kN) R A (kN) R B (kN)

0.04 0.1 0.1 0.892 0.089 0.891 0.090

0.06 0.2 0.2 0.868 0.113 0.848 0.133

0.08 0.2 0.2 0.803 0.178 0.802 0.179

0.10 0.3 0.3 0.779 0.202 0.759 0.222

0.12 0.3 0.3 0.714 0.267 0.714 0.267

0.14 0.3 0.3 0.648 0.333 0.668 0.3130.16 0.4 0.4 0.624 0.357 0.625 0.356

0.18 0.4 0.4 0.559 0.422 0.580 0.401

0.20 0.5 0.5 0.535 0.446 0.534 0.447

0.22 0.5 0.5 0.470 0.511 0.491 0.490

0.24 0.6 0.6 0.446 0.535 0.445 0.536

0.26 0.6 0.6 0.381 0.600 0.402 0.579

0.34 -0.2 -0.2 0.083 1.064 0.223 0.758

0.36 -0.2 -0.2 0.083 1.064 0.177 0.804

0.38 -0.1 -0.1 0.042 1.023 0.134 0.847

0.40 -0.1 -0.1 0.042 1.023 0.089 0.892

Table 1



Notes:

1. Shear force at cut section is the same value given by Digital force reading. Add – ve sign to the

value for positions 320mm to 380mm.

2.

Experimental Influence line values = )(

)(

N Load

N ShearForce

3. Calculate the theoretical value using the equation 1 for load position 40 to 260 mm and equation 2

for load position 320mm to 380mm.

Part B :

Location

Position of hanger from

left hand support (m)Digital

Force

Reading(N)

Exp.

Moment (N)

Theoretical

Moment (N)

100g 200g 300g R A R B R A R B

10.04 0.20 0.36 0.5

1.921 3.924 2.727 3.389

20.38 0.08 0.14 1.2

4.009 1.877 3.748 2.141

30.26 0.38 0.04 0.6

3.182 2.704 3.345 2.541

40.34 0.22 0.06 1.1

3.795 2.091 3.745 2.141

Table 2



0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 0.22 0.24 0.26 0.34 0.36 0.38 0.4

I N F L U E N C E L I N E V A L U E F O R R E A C T I O N A

LOCATION OF LOAD FROM LEFT HAND SUPPORT

Theoritical Value

Experimental Value

GRAPH OF INFLUENCE LINE VALUE (REACTION A) VERSUS LOCATION OF

LOAD FROM LEFT HAND SUPPORT



7.0 DATA ANALYSIS

7.1 PART A (Experimental):

7.1.1 Before Cut

Load = 100g 100g x 9.81 = 0.981 N

1000

R A = 0.981 (a-x) + (digital force x 0.125)

a

At 0.04 m

R A = 0.981 (0.3 – 0.04) + (0.1 x 0.125) ∑fy ↑ = ∑fy ↓

0.3 R A + R B = 0.981R A = 0.892 kN R B = 0.981 – 0.892

R B = 0.089 kN

At 0.06 m

R A = 0.981 (0.3 – 0.06) + (0.2 x 0.125) ∑fy ↑ = ∑fy ↓

0.3 R A + R B = 0.981

R A = 0.868 kN R B = 0.981 – 0.868

R B = 0.113 kN



At 0.08 m

R A = 0.981 (0.3 – 0.08) + (0.2 x 0.125) ∑fy ↑ = ∑fy ↓

0.3 R A + R B = 0.981

R A = 0.803 kN R B = 0.981 – 0.803

R B = 0.178 kN

At 0.10 m

R A = 0.981 (0.3 – 0.10) + (0.3 x 0.125) ∑fy ↑ = ∑fy ↓

0.3 R A + R B = 0.981

R A = 0.779 kN R B = 0.981 – 0.779

R B = 0.202 kN

At 0.12 m

R A = 0.981 (0.3 – 0.12) + (0.3 x 0.125) ∑fy ↑ = ∑fy ↓

0.3 R A + R B = 0.981

R A = 0.714 kN R B = 0.981 – 0.714R B = 0.267 Kn



At 0.14 m

R A = 0.981 (0.3 – 0.14) + (0.3 x 0.125) ∑fy ↑ = ∑fy ↓

0.3 R A + R B = 0.981

R A = 0.648 kN R B = 0.981 – 0.648

R B = 0.333 kN

At 0.16 m

R A = 0.981 (0.3 – 0.16) + (0.4 x 0.125) ∑fy ↑ = ∑fy ↓

0.3 R A + R B = 0.981

R A = 0.624 kN R B = 0.981 – 0.624

R B = 0.357 kN

At 0.18 m

R A = 0.981 (0.3 – 0.18) + (0.4 x 0.125) ∑fy ↑ = ∑fy ↓

0.3 R A + R B = 0.981

R A = 0.559 kN R B = 0.981 – 0.559

R B = 0.422 kN



At 0.20 m

R A = 0.981 (0.3 – 0.14) + (0.5 x 0.125) ∑fy ↑ = ∑fy ↓

0.3 R A + R B = 0.981

R A = 0.535 kN R B = 0.981 – 0.535

R B = 0.446 kN

At 0.22 m

R A = 0.981 (0.3 – 0.22) + (0.5 x 0.125) ∑fy ↑ = ∑fy ↓

0.3 R A + R B = 0.981

R A = 0.470 kN R B = 0.981 – 0.470

R B = 0.511 kN

At 0.24 m

R A = 0.981 (0.3 – 0.24) + (0.6 x 0.125) ∑fy ↑ = ∑fy ↓

0.3 R A + R B = 0.981

R A = 0.446 kN R B = 0.981 – 0.446

R B = 0.535 kN



At 0.26 m

R A = 0.981 (0.3 – 0.26) + (0.6 x 0.125) ∑fy ↑ = ∑fy ↓

0.3 R A + R B = 0.981

R A = 0.381 kN R B = 0.981 – 0.381

R B = 0.60 kN

7.1.2 After Cut

Load = 100g 100g x 9.81 = 0.981 N

1000

R A = (digital force x 0.125)

a

At 0.34 m

R A = (-0.2 x 0.125) ∑fy ↑ = ∑fy ↓

0.3 R A + R B = 0.981

R A = - 0.083 kN R B = 0.981 + 0.083

R A = 0.083 kN (↓) R B = 1.064 kN



At 0.36 m

R A = (-0.2 x 0.125) ∑fy ↑ = ∑fy ↓

0.3 R A + R B = 0.981

R A = - 0.083 kN R B = 0.981 + 0.083

R A = 0.083 kN (↓) R B = 1.064 kN

At 0.38 m

R A = (-0.1 x 0.125) ∑fy ↑ = ∑fy ↓

0.3 R A + R B = 0.981

R A = - 0.042 kN R B = 0.981 + 0.042

R A = 0.042 kN (↓) R B = 1.023 kN

At 0.40 m

R A = (-0.1 x 0.125) ∑fy ↑ = ∑fy ↓

0.3 R A + R B = 0.981

R A

= - 0.042 kN R B

= 0.981 + 0.042

R A = 0.042 kN (↓) R B = 1.023 Kn



7.2 PART A (Theoritical):

Distance : 0.04mm

Distance : 0.06mm

Distance : 0.08mm

Distance : 0.10mm

Distance : 0.12mm

MB = 0

R A (0.44) 0.981 (0.40) = 0

R A (0.44) = 0.392

R A = 0.891 kN

Fy =Fy

R A + R B = 0.981

R B = 0.981 – 0.891

R B = 0.090 kN

MB = 0

R A (0.44) 0.981 (0.38) = 0

R A (0.44) = 0.373

R A = 0.848 kN

Fy =Fy

R A + R B = 0.981

R B = 0.981 – 0.848

R B = 0.133 kN

Fy =Fy

R A + R B = 0.981

R B = 0.981 – 0.802

R B = 0.179 kN

MB = 0

R A (0.44) 0.981 (0.36) = 0

R A (0.44) = 0.353

R A = 0.802 kN

MB = 0

R A (0.44) 0.981 (0.34) = 0

R A (0.44) = 0.334

R A = 0.759 kN

Fy =Fy

R A + R B = 0.981

R B = 0.981 – 0.759

R B = 0.222 kN

MB = 0

R A (0.44) 0.981 (0.32) = 0

R A (0.44) = 0.314

R A = 0.714 kN

Fy =Fy

R A + R B = 0.981

R B = 0.981 – 0.714

R B = 0.267 kN



Distance : 0.14mm

Distance : 0.16mm

Distance : 0.18mm

Distance : 0.20mm

Distance : 0.22mm

MB = 0

R A (0.44) 0.981 (0.30) = 0

R A (0.44) = 0.294

R A = 0.668 kN

Fy =Fy

R A + R B = 0.981

R B = 0.981 – 0.668

R B = 0.313 kN

MB = 0

R A (0.44) 0.981 (0.28) = 0

R A (0.44) = 0.275

R A = 0.625 kN

Fy =Fy

R A + R B = 0.981

R B = 0.981 – 0.625

R B = 0.356 kN

MB = 0

R A (0.44) 0.981 (0.26) = 0

R A (0.44) = 0.255

R A = 0.580 kN

Fy =Fy

R A + R B = 0.981

R B = 0.981 – 0.580

R B = 0.401 kN

MB = 0

R A (0.44) 0.981 (0.24) = 0

R A (0.44) = 0.235

R A = 0.534 kN

Fy =Fy

R A + R B = 0.981

R B = 0.981 – 0.534

R B = 0.447 kN

MB = 0

R A (0.44) 0.981 (0.22) = 0

R A (0.44) = 0.216

R A = 0.491 kN

Fy =Fy

R A + R B = 0.981

R B = 0.981 – 0.491

R B = 0.490 kN



Distance : 0.24mm

Distance : 0.26mm

Distance : 0.34mm

Distance : 0.36mm

Distance : 0.38mm

MB = 0

R A (0.44) 0.981 (0.20) = 0

R A (0.44) = 0.196

R A = 0.445 kN

Fy =Fy

R A + R B = 0.981

R B = 0.981 – 0.445

R B = 0.536 kN

MB = 0

R A (0.44) 0.981 (0.18) = 0

R A (0.44) = 0.177

R A = 0.402 kN

Fy =Fy

R A + R B = 0.981

R B = 0.981 – 0.402

R B = 0.579 kN

MB = 0

R A (0.44) 0.981 (0.10) = 0

R A (0.44) = 0.098

R A = 0.223 kN

Fy =Fy

R A + R B = 0.981

R B = 0.981 – 0.223

R B = 0.758 kN

MB = 0

R A (0.44) 0.981 (0.08) = 0

R A (0.44) = 0.078

R A = 0.177 kN

Fy =Fy

R A + R B = 0.981

R B = 0.981 – 0.177

R B = 0.804 kN

MB

= 0

R A (0.44) 0.981 (0.06) = 0

R A (0.44) = 0.059

R A = 0.134 kN

Fy =Fy

R A + R B = 0.981

R B = 0.981 – 0.134

R B = 0.847 kN



Distance : 0.40mm

7.3 PART B (Experimental):

Location 1

RA = Load x Length + (Digital Force x 0.125 )

0.3

RA = 0.981N x 0.26 + (0.5 x 0.125 ) + 1.962x 0.1 + (0.5x 0.125 )

0.3 0.3

RA = 1.921 N

RA + RB = 0.981N + 2.943N + 1.962N

RA + RB = 0.981N + 2.943N + 1.962N

RB = 5.883N –

1.921NRB = 3.924 N

MB = 0

R A (0.44) 0.981 (0.04) = 0

R A (0.44) = 0.039

R A = 0.089 kN

Fy =Fy

R A + R B = 0.981

R B = 0.981 – 0.089

R B = 0.892 kN

0.981 N 1.962 N

1.921 N 3.924N

0.04m 0.16m 0.10m 0.06m 0.08m

2.943 N



Location 2


0.3

RA = 1.962N x 0.22 + (1.2 x 0.125 ) + 2.943 x 0.16 + (1.2 x 0.125 )

0.3 0.3

RA = 4.009 N

RA + RB = 0.981N + 2.943N + 1.962N

RA + RB = 0.981N + 2.943N + 1.962N

RB = 5.883N – 4.009 N

RB = 1.877 N

Location 3


0.3RA = 2.946N x 0.26 + (0.6 x 0.125 ) + 0.981 x 0.04 + (0.6x 0.125 )

0.3 0.3

RA = 3.182 N

RA + RB = 0.981N + 2.943N + 1.962N

RA + RB = 0.981N + 2.943N + 1.962N

RB = 5.883N – 3.182 N

RB = 2.704 N

2.943 N 0.981 N

3.182N 2.704 N

0.04m 0.22m 0.04m 0.08m 0.06m

1.962 N

1.962 N 2.943 N

4.009 N 1.877N

0.08m 0.06m 0.16m 0.08m 0.06m

0.981 N



Location 4


0.3

RA = 2.943N x 0.24 + (1.1 x 0.125 ) + 1.962 x 0.08 + (1.1 x 0.125 )

0.3 0.3

RA = 3.795 N

RA + RB = 0.981N + 2.943N + 1.962N

RA + RB = 0.981N + 2.943N + 1.962N

RB = 5.883N – 3.795 N

RB = 2.091 N

7.4 PART B (Theoritical):

Location 1

∑MA = 0

0.981 (0.04) + 1.962 (0.2) + 2.943 (0.36) – R B (0.44) = 0

1.491 – 0.44 R B = 0

R B = 3.389 N

∑MB = 0

R A (0.44) – 0.981 (0.4) – 1.962 (0.24) – 2.943 (0.08) = 0

0.44 R A – 1.20 = 0

R A = 2.727 N

2.943 N 1.962 N

3.795 N 2.091 N

0.06m 0.16m 0.08m 0.04m 0.1m

0.981 N

0.04 m 0.16 m 0.1 m 0.06 m 0.08 m

0.981 N 1.962 N 2.943 N

R A R B



Location 2

∑MA = 0

1.962 (0.08) + 2.943 (0.14) + 0.981 (0.38) – R B (0.44) = 0

0.942 – 0.44 R B = 0

R B = 2.141 N

∑MB = 0

R A (0.44) – 1.962 (0.36) – 2.943 (0.3) – 0.981 (0.06) = 0

0.44 R A – 1.649 = 0

R A = 3.748 N

Location 3

∑MA = 0

2.943 (0.04) + 0.981 (0.26) + 1.962 (0.38) – R B (0.44) = 0

1.118 – 0.44 R B = 0

R B = 2.541 N

∑MB = 0

R A (0.44) – 2.943 (0.4) – 0.981 (0.18) – 1.962 (0.06) = 0

0.44 R A – 1.472 = 0

R A = 3.345 N

1.962 N 2.943 N 0.981 N

R A R B

0.08 m 0.06 m 0.16 m 0.08 m 0.06 m

0.04 m 0.22 m 0.04 m 0.08 m 0.06 m

2.943 N 0.981 N 1.962 N

R A R B



Location 4

∑MA = 0

2.943 (0.06) + 1.962 (0.22) + 0.981 (0.34) – R B (0.44) = 0

0.942 – 0.44 R B = 0

R B = 2.141 N

∑MB = 0

R A (0.44) – 2.943 (0.38) – 1.962 (0.22) – 0.981 (0.1) = 0

0.44 R A – 1.648 = 0

R A = 3.745 N

0.06 m 0.16 m 0.08 m 0.04 m 0.1 m

2.943 N 1.962 N 0.981 N

R A R B



Influence Line: Theoretical (N)

Location 1

0.04 m 0.16 m 0.1 m 0.06 m 0.08 m

0.981 N 1.962 N 2.943 N

R A R B

RB

Y3

Y2

0

1

1

0

Y1

Y3

Y2

Y1RA



Reaction at Support A

Y1 = 1.0

0.36 0.44

Y1 = 0.82 m

Y2 = 1.0

0.2 0.44

Y2 = 0.45 m

Y3 = 1.0

0.04 0.44

Y3 = 0.09 m

R A = 0.981 (0.09) + 1.962 (0.45) + 2.943 (0.82)

= 3.40 KN

Reaction at Support B

Y1 = 1.0

0.4 0.44

Y1 = 0.91 m

Y2 = 1.0

0.24 0.44

Y2

= 0.55 mY3 = 1.0

0.08 0.44

Y3 = 0.18 m

R B = 0.981 (0.91) + 1.962(0.55) + 2.943 (0.18)

= 2.50 KN

Checking Force

∑ FY = 0

F = F

R A + R B= 1.962 + 2.943 + 0.981

R A + R B = 5.89 KN

3.40 + 2.50 = 5.89

5.90 = 5.89



Location 2

0.08 m 0.06 m 0.16 m 0.08 m 0.06 m

1.962 N 2.943 N 0.981 N

R A R B

RB

Y3

Y2

0

1

1

0

Y1

Y3

Y2

Y1RA




Y1 = 1.0

0.38 0.44

Y1 = 0.86 m

Y2 = 1.0

0.14 0.44

Y2 = 0.32 m

Y3 = 1.0

0.08 0.44

Y3 = 0.18 m

R A = 1.962 (0.18) + 2.943 (0.32) + 0.981 (0.86)

= 2.14 KN


Y1 = 1.0

0.36 0.44

Y1 = 0.82 m

Y2 = 1.0

0.3 0.44

Y2

= 0.68 mY3 = 1.0

0.06 0.44

Y3 = 0.14 m

R B = 1.962 (0.82) + 2.943 (0.68) + 0.981 (0.14)

= 3.75 KN

Checking Force

∑ FY = 0

F = F

R A + R B= 1.962 + 2.943 + 0.981

R A + R B = 5.89 KN

3.75 + 2.14 = 5.89

5.89 = 5.89



Location 3

0.04 m 0.22 m 0.04 m 0.08 m 0.06 m

2.943 N 0.981 N 1.962 N

R A R B

RB

Y3Y2

0

1

1

0

Y1

Y3Y2

Y1RA




Y1 = 1.0

0.38 0.44

Y1 = 0.86 m

Y2 = 1.0

0.26 0.44

Y2 = 0.59 m

Y3 = 1.0

0.04 0.44

Y3 = 0.09 m

R A = 2.943 (0.09) + 0.981 (0.59) + 1.962 (0.86)

= 2.53 KN


Y1 = 1.0

0.4 0.44

Y1 = 0.91 m

Y2 = 1.0

0.18 0.44

Y2

= 0.41 mY3 = 1.0

0.06 0.44

Y3 = 0.14 m

R B = 2.943 (0.91) + 0.981 (0.41) + 1.962 (0.14)

= 3.36 KN

Checking Force

∑ FY = 0

F = F

R A + R B= 1.962 + 2.943 + 0.981

R A + R B = 5.89 KN

2.53 + 3.36 = 5.89

5.89 = 5.89



Location 4

0.06 m 0.16 m 0.08 m 0.04 m 0.1 m

2.943 N 1.962 N 0.981 N

R A R B

RB

Y3

Y2

0

1

1

0

Y1

Y3

Y2

Y1

RA




Y1 = 1.0

0.34 0.44

Y1 = 0.77 m

Y2 = 1.0

0.22 0.44

Y2 = 0.50 m

Y3 = 1.0

0.06 0.44

Y3 = 0.14 m

R A = 2.943 (0.14) + 1.962 (0.50) + 0.981 (0.77)

= 2.15 KN


Y1 = 1.0

0.38 0.44

Y1 = 0.86 m

Y2 = 1.0

0.22 0.44

Y2

= 0.50 mY3 = 1.0

0.1 0.44

Y3 = 0.23 m

R B = 2.943 (0.86) + 1.962 (0.50) + 0.981 (0.23)

= 3.74 KN

Checking Force

∑ FY = 0

F = F

R A + R B= 1.962 + 2.943 + 0.981

R A + R B = 5.89 KN

2.15 + 3.74 = 5.89

5.89 = 5.89



8.0 DISCUSSION

The graph shows, 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 accurate. Although the value of experiment quite near with the

value of theory a there arestill have error. The digital indicator is not too accurate.

ii. The digital indicator is too sensitive. When 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 shaking . When 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 affects the reading.

iv. Parallax error . Reading the ruler scale. The ruler scale is in centimetre (cm). So, when the

reading process, we can’tget the accurate value, because the scale are not suitable for our eye

to read with accurately

v. 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 inthe

original position yet.

9.0 CONCLUSION

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

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.

10.0 REFERENCES

i. STRUCTURAL ANALYSIS (2009), Bambang Prihartantoii MECHANICS OF MATERIALS James M Gere Barry J Goodno

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Shear Force Influence Line

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