8/13/2019 177195895 Shear Force Influence Line

http://slidepdf.com/reader/full/177195895-shear-force-influence-line 1/33

FACULTY OF CIVIL AND ENVIRONMENTAL ENGINEERING DEPARTMENT OF STRUCTURE AND MATERIAL ENGINEERING LAB MATERIAL

REPORTSubject Code Code & Experiment Title Course Code Date Section / Group Name Members of Group BFC 31901 OPEN ENDED ± SHEAR FORCE INFLUENCE LINES 2 BFF 04/03/2012 SECTION 9 / GROUP 7 MUHAMMAD IKHWAN BIN ZAINUDDIN (DF100018) 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) EN. MOHD KHAIRY BIN BURHANUDIN 20/04/2012

Lecturer/Instructor/Tutor Received Date

Comment by examiner

Received

8/13/2019 177195895 Shear Force Influence Line

http://slidepdf.com/reader/full/177195895-shear-force-influence-line 2/33

STUDENT CODE OF ETHIC (SCE)DEPARTMENT OF STRUCTURE AND MATERIAL ENGINEERING FACULTY OF CIVIL & ENVIRONMENTAL ENGINEERING UTHM

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 Name ___________________________ Student Signature : NUR EZRYNNA BINTI MOHD ZAINAL

Matric No. : DF100018 Date : 20/04/2012

Matric No. : DF100118 Date : 20/04/2012

 ___________________________ Student Signature Name : NUR EEZRA ATHIRLIA BINTI GHAZALI Name

 ___________________________ Student Signature : MUHAMMAD HUZAIR BIN ZULKIFLI

Matric No. : DF100147 Date : 20/04/2012

Matric No. : DF100040 Date : 20/04/2011

 _______________________ Student Signature Name : MUHAMMAD NUH BIN AHMAD ZAIRI Name

 _______________________ Student Signature : ZIRWATUL FAUZANA BINTI CHE JEMANI

Matric No. : DF100093 Date : 20/04/2012

Matric No. : DF100027 Date : 20/04/2012

8/13/2019 177195895 Shear Force Influence Line

http://slidepdf.com/reader/full/177195895-shear-force-influence-line 3/33

SCOPE OF WORK GROUP 7

NO 1 2 3 4 5 6

NAME Nur Ezrynna binti Mohd Zainal Muhammad Ikhwan in Zainuddin Nur Eezra Athirlia binti Ghazali Muhammad Nuh bin Ahmad Zairi Zirwatul Fauzana binti Che JemaniMuhammad Huzair bin Zulkifli

SCOPE OF WORK Calculation of the laboratory report Calculation of the laboratory report Discussion of the laboratory report Theory and record data of the laboratory report Conclusion of the laboratory report Procedure and result of the laboratory report

8/13/2019 177195895 Shear Force Influence Line

http://slidepdf.com/reader/full/177195895-shear-force-influence-line 4/33

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 pointon 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 andthen connected together to generate the influence line for the function.

2.0

OBJECTIVE 2.1 2.2 2.3 2.4 To plot Shear force influence line. To verify the useof a shear force influence on a simply supported beam Understanding about the envelopes of maximum influence line values To construct influence line for maximum end shear in a beam supporting a series of moving concentrated loads

3.0

LEARNING OUTCOME 3.1 3.2 3.3 3.4 The application the engineering knowledge in pr

actical application To enhance technical competency in structural engineering through laboratory application. To communicate effectively in group To identify problem, solving and finding out appropriate solution through laboratory application

8/13/2019 177195895 Shear Force Influence Line

http://slidepdf.com/reader/full/177195895-shear-force-influence-line 5/33

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 ata cut section as a unit load moves from one end to another (see Figure 1). Fromthe diagram, shear force influence line equation can be writen.

RA = load (a-x) + (digital force x 0.125) a RA = (digital force x 0.125) a

(before cut)

(after cut)

`Cut' x 1 (unit load) Mx RA = (1 ± x/L) a L Figure 1 - shear force varies at a cut section b RB=x/L Mx

8/13/2019 177195895 Shear Force Influence Line

http://slidepdf.com/reader/full/177195895-shear-force-influence-line 6/33

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 influenceline in terms of x3 ,a , b and L) x1, x2,

F1

F2

F3

ab  L

x1 x2 x3Shear Force Influence line for cut y1 y2 y3

Figure 2 - beam are loaded 5.0 APPARATUS

8/13/2019 177195895 Shear Force Influence Line

http://slidepdf.com/reader/full/177195895-shear-force-influence-line 7/33

6.0

PROCEDURES Part 1: i. ii. iii. iv. 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.iii.

Check the Digital Force Meter reads zero with no load. Place hanger with 200g of mass at support and locate it at the left support and record the Digital Forcereading in Table 1. Repeat the procedure number 2 with different distance Complete the calculation in Table 1.

Repeat the procedure with three other locations. 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 RA (kN) RB (kN) Theoretical Influence Line Value RA (kN) RB (kN)

0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20 0.22 0.24 0.26 0.34 0.36 0.38 0.40

0.1 0.2 0.2 0.3 0.3 0.3 0.4 0.4 0.5 0.5 0.6 0.6 -0.2 -0.2 -0.1 -0.1

0.1 0.2 0.2 0.3 0.3 0.3 0.4 0.4 0.5 0.5 0.6 0.6 -0.2 -0.2 -0.1 -0.1

0.892 0.868 0.803 0.779 0.714 0.648 0.624 0.559 0.535 0.470 0.446 0.381 0.083 0.083 0.042 0.042 Table 1

0.089 0.113 0.178 0.202 0.267 0.333 0.357 0.422 0.446 0.511 0.535 0.600 1.064 1.064 1.023 1.023

0.891 0.848 0.802 0.759 0.714 0.668 0.625 0.580 0.534 0.491 0.445 0.402 0.223 0.

177 0.134 0.089

0.090 0.133 0.179 0.222 0.267 0.313 0.356 0.401 0.447 0.490 0.536 0.579 0.758 0.804 0.847 0.892

8/13/2019 177195895 Shear Force Influence Line

http://slidepdf.com/reader/full/177195895-shear-force-influence-line 8/33

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 =

ShearForce ( N ) Load ( N )

3. Calculate the theoretical value using the equation 1 for load position 40 to260 mm and equation 2 for load position 320mm to 380mm.

Part B : Position of hanger from left hand support (m) 100g 1 2 3 4 0.04 0.38 0.26 0.34 200g 0.20 0.08 0.38 0.22 300g 0.36 0.14 0.04 0.06 Digital Force Reading(N) 0.5 1.2 0.6 1.1 Exp. Moment (N) RA 1.921 4.009 3.182 3.795 RB 3.924 1.877 2.704 2.091 Theoretical Moment (N) RA RB

Location

2.727 3.389 3.748 2.141 3.345 2.541 3.745 2.141

Table 2

8/13/2019 177195895 Shear Force Influence Line

http://slidepdf.com/reader/full/177195895-shear-force-influence-line 9/33

GRAPH OF INFLUENCE LINE VALUE (REACTION A) VERSUS LOCATION OF LOAD FROM LEFT HAND SUPPORT1 Theoritical Value Experimental Value 0.9

0.8

INFLUENCE LINE VALUE FOR REACTION A

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0 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

LOCATION OF LOAD FROM LEFT HAND SUPPORT

8/13/2019 177195895 Shear Force Influence Line

http://slidepdf.com/reader/full/177195895-shear-force-influence-line 10/33

7.0

DATA ANALYSIS 7.1 PART A (Experimental):

7.1.1 Before Cut Load = 100g 100g x 9.81 1000 = 0.981 N

RA = 0.981 (a-x) + (digital force x 0.125) a At 0.04 m

RA = 0.981 (0.3 ± 0.04) + (0.1 x 0.125) 0.3 RA = 0.892 kN

åfy = åfy ¯ RA + RB = 0.981 RB = 0.981 ± 0.892 RB = 0.089 kN

At 0.06 m

RA = 0.981 (0.3 ± 0.06) + (0.2 x 0.125) 0.3 RA = 0.868 kN

åfy = åfy ¯ RA + RB = 0.981 RB = 0.981 ± 0.868 RB = 0.113 kN

8/13/2019 177195895 Shear Force Influence Line

http://slidepdf.com/reader/full/177195895-shear-force-influence-line 11/33

At 0.08 m

RA = 0.981 (0.3 ± 0.08) + (0.2 x 0.125) 0.3 RA = 0.803 kN

åfy = åfy ¯ RA + RB = 0.981 RB = 0.981 ± 0.803 RB = 0.178 kN

At 0.10 m

RA = 0.981 (0.3 ± 0.10) + (0.3 x 0.125) 0.3 RA = 0.779 kN At 0.12 m

åfy = åfy ¯ RA + RB = 0.981 RB = 0.981 ± 0.779 RB = 0.202 kN

RA = 0.981 (0.3 ± 0.12) + (0.3 x 0.125) 0.3 RA = 0.714 kN

åfy = åfy ¯ RA + RB = 0.981 RB = 0.981 ± 0.714 RB = 0.267 Kn

8/13/2019 177195895 Shear Force Influence Line

http://slidepdf.com/reader/full/177195895-shear-force-influence-line 12/33

At 0.14 m

RA = 0.981 (0.3 ± 0.14) + (0.3 x 0.125) 0.3 RA = 0.648 kN

åfy = åfy ¯ RA + RB = 0.981 RB = 0.981 ± 0.648 RB = 0.333 kN

At 0.16 m

RA = 0.981 (0.3 ± 0.16) + (0.4 x 0.125) 0.3 RA = 0.624 kN At 0.18 m

åfy = åfy ¯ RA + RB = 0.981 RB = 0.981 ± 0.624 RB = 0.357 kN

RA = 0.981 (0.3 ± 0.18) + (0.4 x 0.125) 0.3 RA = 0.559 kN

åfy = åfy ¯ RA + RB = 0.981 RB = 0.981 ± 0.559 RB = 0.422 kN

8/13/2019 177195895 Shear Force Influence Line

http://slidepdf.com/reader/full/177195895-shear-force-influence-line 13/33

At 0.20 m

RA = 0.981 (0.3 ± 0.14) + (0.5 x 0.125) 0.3 RA = 0.535 kN At 0.22 m

åfy = åfy ¯ RA + RB = 0.981 RB = 0.981 ± 0.535 RB = 0.446 kN

RA = 0.981 (0.3 ± 0.22) + (0.5 x 0.125) 0.3 RA = 0.470 kN At 0.24 m

åfy = åfy ¯ RA + RB = 0.981 RB = 0.981 ± 0.470 RB = 0.511 kN

RA = 0.981 (0.3 ± 0.24) + (0.6 x 0.125) 0.3 RA = 0.446 kN

åfy = åfy ¯ RA + RB = 0.981 RB = 0.981 ± 0.446 RB = 0.535 kN

8/13/2019 177195895 Shear Force Influence Line

http://slidepdf.com/reader/full/177195895-shear-force-influence-line 14/33

At 0.26 m

RA = 0.981 (0.3 ± 0.26) + (0.6 x 0.125) 0.3 RA = 0.381 kN

åfy = åfy ¯ RA + RB = 0.981 RB = 0.981 ± 0.381 RB = 0.60 kN

7.1.2 After Cut Load = 100g RA = (digital force x 0.125) a At 0.34 m 100g x 9.81 1000 = 0.981 N

RA = (-0.2 x 0.125) 0.3 RA = - 0.083 kN RA = 0.083 kN (¯)

åfy = åfy ¯ RA + RB = 0.981 RB = 0.981 + 0.083 RB = 1.064 kN

8/13/2019 177195895 Shear Force Influence Line

http://slidepdf.com/reader/full/177195895-shear-force-influence-line 15/33

At 0.36 m

RA = (-0.2 x 0.125) 0.3 RA = - 0.083 kN RA = 0.083 kN (¯) At 0.38 m

åfy = åfy ¯ RA + RB = 0.981 RB = 0.981 + 0.083 RB = 1.064 kN

RA = (-0.1 x 0.125) 0.3 RA = - 0.042 kN RA = 0.042 kN (¯) At 0.40 m

åfy = åfy ¯ RA + RB = 0.981 RB = 0.981 + 0.042 RB = 1.023 kN

RA = (-0.1 x 0.125) 0.3 RA = - 0.042 kN RA = 0.042 kN (¯)

åfy = åfy ¯ RA + RB = 0.981 RB = 0.981 + 0.042 RB = 1.023 Kn

8/13/2019 177195895 Shear Force Influence Line

http://slidepdf.com/reader/full/177195895-shear-force-influence-line 16/33

7.2

PART A (Theoritical):

Distance : 0.04mmMB = 0 RA (0.44) 0.981 (0.40) = 0 RA (0.44) = 0.392 RA = 0.891 kN Fy =Fy RA + RB 81 RB = 0.981 ± 0.891 RB = 0.090 kN

Distance : 0.06mmMB = 0 RA (0.44) 0.981 (0.38) = 0 RA (0.44) = 0.373 RA = 0.848 kN Fy =Fy RA + RB 81 RA = 1.784 RB = 0.981 ± 0.848 RB = 0.133 kN

Distance : 0.08mmMB = 0 RA (0.44) 0.981 (0.36) = 0 RA (0.44) = 0.353 RA = 0.802 kN

Fy =Fy RA + RB = 0.981 RA = 1.784 RB = 0.981 ± 0.802 RB = 0.179 kN

Distance : 0.10mmMB = 0 RA (0.44) 0.981 (0.34) = 0 RA (0.44) = 0.334 RA = 0.759 kN Fy =Fy RA + RB 81 RA = 1.784 RB = 0.981 ± 0.759 RB = 0.222 kN

Distance : 0.12mmMB = 0 RA (0.44) 0.981 (0.32) = 0 RA (0.44) = 0.314 RA = 0.714 kN Fy =Fy RA + RB 81 RA = 1.784 RB = 0.981 ± 0.714 RB = 0.267 kN

RA = 1.784

8/13/2019 177195895 Shear Force Influence Line

http://slidepdf.com/reader/full/177195895-shear-force-influence-line 17/33

Distance : 0.14mmMB = 0 RA (0.44) 0.981 (0.30) = 0 RA (0.44) = 0.294 RA = 0.668 kN Fy =Fy RA + RB 81 RB = 0.981 ± 0.668 RB = 0.313 kN

Distance : 0.16mmMB = 0 RA (0.44) 0.981 (0.28) = 0 RA (0.44) = 0.275 RA = 0.625 kN Fy =Fy RA + RB 81 RA = 1.784 RB = 0.981 ± 0.625 RB = 0.356 kN

Distance : 0.18mmMB = 0 RA (0.44) 0.981 (0.26) = 0 RA (0.44) = 0.255 RA = 0.580 kN Fy =Fy RA + RB 81 RA = 1.784 RB = 0.981 ± 0.580 RB = 0.401 kN

Distance : 0.20mmMB = 0 RA (0.44) 0.981 (0.24) = 0 RA (0.44) = 0.235 RA = 0.534 kN Fy =Fy RA + RB 81 RA = 1.784 RB = 0.981 ± 0.534 RB = 0.447 kN

Distance : 0.22mmMB = 0 RA (0.44) 0.981 (0.22) = 0 RA (0.44) = 0.216 RA = 0.491 kN Fy =Fy RA + RB 81 RA = 1.784 RB = 0.981 ± 0.491 RB = 0.490 kN

RA = 1.784

8/13/2019 177195895 Shear Force Influence Line

http://slidepdf.com/reader/full/177195895-shear-force-influence-line 18/33

Distance : 0.24mmMB = 0 RA (0.44) 0.981 (0.20) = 0 RA (0.44) = 0.196 RA = 0.445 kN Fy =Fy RA + RB 81 RB = 0.981 ± 0.445 RB = 0.536 kN

Distance : 0.26mmMB = 0 RA (0.44) 0.981 (0.18) = 0 RA (0.44) = 0.177 RA = 0.402 kN Fy =Fy RA + RB 81 RA = 1.784 RB = 0.981 ± 0.402 RB = 0.579 kN

Distance : 0.34mmMB = 0 RA (0.44) 0.981 (0.10) = 0 RA (0.44) = 0.098 RA = 0.223 kN Fy =Fy RA = + 1RB = 0.981 RB = 0.981 ± 0.223 RB = 0.758 kN

Distance : 0.36mmMB = 0 RA (0.44) 0.981 (0.08) = 0 RA (0.44) = 0.078 RA = 0.177 kN Fy =Fy RA + RB 81 RA = 1.784 RB = 0.981 ± 0.177 RB = 0.804 kN

Distance : 0.38mmMB = 0 RA (0.44) 0.981 (0.06) = 0 RA (0.44) = 0.059 RA = 0.134 kN Fy =Fy RA + RB 81 RA = 1.784 RB = 0.981 ± 0.134 RB = 0.847 kN

RA = 1.784

8/13/2019 177195895 Shear Force Influence Line

http://slidepdf.com/reader/full/177195895-shear-force-influence-line 19/33

Distance : 0.40mmMB = 0 RA (0.44) 0.981 (0.04) = 0 RA (0.44) = 0.039 RA = 0.089 kN Fy =Fy RA + RB 81 RB = 0.981 ± 0.089 RB = 0.892 kN

7.3

PART B (Experimental):RA = 1.784

Location 10.981 N 1.962 N

2.943 N

0.04m 1.921 N

0.16m

0.10m

0.06m

0.08m 3.924N

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 ± 1.921N RB = 3.924 N

8/13/2019 177195895 Shear Force Influence Line

http://slidepdf.com/reader/full/177195895-shear-force-influence-line 20/33

Location 21.962 N 2.943 N 0.981 N

0.08m 4.009 N

0.06m

0.16m

0.08m

0.06m 1.877N

RA = Load x Length + (Digital Force x 0.125 ) 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 = 1.877N

Location 32.943 N 0.981 N 1.962 N

0.04m 3.182N

0.22m

0.04m

0.08m

0.06m 2.704 N

RA = Load x Length + (Digital Force x 0.125 ) 0.3 RA = 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 + 2.9

43N + 1.962N RA + RB = 0.981N + 2.943N + 1.962N RB = 5.883N ± 3.182 N RB = 2.704 N

8/13/2019 177195895 Shear Force Influence Line

http://slidepdf.com/reader/full/177195895-shear-force-influence-line 21/33

Location 42.943 N 1.962 N

0.981 N

0.06m 3.795 N

0.16m

0.08m

0.04m

0.1m 2.091 N

RA = Load x Length + (Digital Force x 0.125 ) 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 = 2.091N

7.4

PART B (Theoritical):

Location 10.981 N 1.962 N 2.943 N

RA0.04 m 0.16 m 0.1 m 0.06 m 0.08 m

RB

åMA = 0 0.981 (0.04) + 1.962 (0.2) + 2.943 (0.36) ± RB (0.44) = 0 1.491 ± 0.44 RB = 0RB = 3.389 N åMB = 0 RA (0.44) ± 0.981 (0.4) ± 1.962 (0.24) ± 2.943 (0.08) = 0 0.44 RA .20 = 0 RA = 2.727 N

8/13/2019 177195895 Shear Force Influence Line

http://slidepdf.com/reader/full/177195895-shear-force-influence-line 22/33

Location 21.962 N 2.943 N 0.981 N

RA0.08 m 0.06 m 0.16 m 0.08 m 0.06 m

RB

åMA = 0 1.962 (0.08) + 2.943 (0.14) + 0.981 (0.38) ± RB (0.44) = 0 0.942 ± 0.44 RB = 0 RB = 2.141 N åMB = 0 RA (0.44) ± 1.962 (0.36) ± 2.943 (0.3) ± 0.981 (0.06) = 0 0.44 RA1.649 = 0 RA = 3.748 N  Location 32.943 N 0.981 N 1.962 N

RA0.04 m 0.22 m 0.04 m 0.08 m 0.06 m

RB

åMA = 0 2.943 (0.04) + 0.981 (0.26) + 1.962 (0.38) ± RB (0.44) = 0 1.118 ± 0.44 RB = 0 RB = 2.541 N åMB = 0 RA (0.44) ± 2.943 (0.4) ± 0.981 (0.18) ± 1.962 (0.06) = 0 0.44 RA1.472 = 0 RA = 3.345 N

8/13/2019 177195895 Shear Force Influence Line

http://slidepdf.com/reader/full/177195895-shear-force-influence-line 23/33

Location 42.943 N 1.962 N 0.981 N

RA0.06 m 0.16 m 0.08 m 0.04 m 0.1 m

RB

åMA = 0 2.943 (0.06) + 1.962 (0.22) + 0.981 (0.34) ± RB (0.44) = 0 0.942 ± 0.44 RB = 0 RB = 2.141 N åMB = 0 RA (0.44) ± 2.943 (0.38) ± 1.962 (0.22) ± 0.981 (0.1) = 0 0.44 RA1.648 = 0 RA = 3.745 N

8/13/2019 177195895 Shear Force Influence Line

http://slidepdf.com/reader/full/177195895-shear-force-influence-line 24/33

Influence Line: Theoretical (N)

 Location 10.981 N 1.962 N 2.943 N

RA0.04 m 0.16 m 0.1 m 0.06 m 0.08 m

RB

1

Y1 Y20

RB

Y3

Y3 RA Y1 Y2

0

1

8/13/2019 177195895 Shear Force Influence Line

http://slidepdf.com/reader/full/177195895-shear-force-influence-line 25/33

Reaction at Support A Y1 0.36 Y1 Y2 0.2 Y2 Y3 0.04 Y3 RA = = = = = = = = 1.0 0.44 0.82 m 1.0 0.44 0.45 m 1.0 0.44 0.09 m 0.981 (0.09) + 1.962 (0.45) + 2.943 (0.82) 3.40 KN

Reaction at Support B Y1 0.4 Y1 Y2 0.24 Y2 Y3 0.08 Y3 RB = = = = = = = = 1.0 0.44 0.91 m 1.0 0.44 0.55 m 1.0 0.44 0.18 m 0.981 (0.91) + 1.962(0.55) + 2.943 (0.18) 2.50 KN

Checking Force å FY F = = 0 F

RA + RB= 1.962 + 2.943 + 0.981 RA + RB = 5.89 KN 3.40 + 2.50 = 5.89 5.90 = 5.89

8/13/2019 177195895 Shear Force Influence Line

http://slidepdf.com/reader/full/177195895-shear-force-influence-line 26/33

 Location 21.962 N 2.943 N 0.981 N

RA0.08 m 0.06 m 0.16 m 0.08 m 0.06 m

RB

1

Y1 Y20

RB

Y3

Y3 RA Y1 Y2

0

1

8/13/2019 177195895 Shear Force Influence Line

http://slidepdf.com/reader/full/177195895-shear-force-influence-line 27/33

Reaction at Support A Y1 0.38 Y1 Y2 0.14 Y2 Y3 0.08 Y3 RA = = = = = = = = 1.0 0.44 0.86 m 1.0 0.44 0.32 m 1.0 0.44 0.18 m 1.962 (0.18) + 2.943 (0.32) + 0.981 (0.86) 2.14 KN

Reaction at Support B Y1 0.36 Y1 Y2 0.3 Y2 Y3 0.06 Y3 RB = = = = = = = = 1.0 0.44 0.82 m 1.0 0.44 0.68 m 1.0 0.44 0.14 m 1.962 (0.82) + 2.943 (0.68) + 0.981 (0.14) 3.75 KN

Checking Force å FY F RA + RB 3.75 + 2.14 5.89 = 5.89 = = 0 F = 5.89 KN = 5.89

RA + RB= 1.962 + 2.943 + 0.981

8/13/2019 177195895 Shear Force Influence Line

http://slidepdf.com/reader/full/177195895-shear-force-influence-line 28/33

 Location 32.943 N 0.981 N 1.962 N

RA0.04 m 0.22 m 0.04 m 0.08 m 0.06 m

RB

1

Y1 Y20

RB

Y3

Y2 RA Y1

Y3

0

1

8/13/2019 177195895 Shear Force Influence Line

http://slidepdf.com/reader/full/177195895-shear-force-influence-line 29/33

Reaction at Support A Y1 0.38 Y1 Y2 0.26 Y2 Y3 0.04 Y3 RA = = = = = = = = 1.0 0.44 0.86 m 1.0 0.44 0.59 m 1.0 0.44 0.09 m 2.943 (0.09) + 0.981 (0.59) + 1.962 (0.86) 2.53 KN

Reaction at Support B Y1 0.4 Y1 Y2 0.18 Y2 Y3 0.06 Y3 RB = = = = = = = = 1.0 0.44 0.91 m 1.0 0.44 0.41 m 1.0 0.44 0.14 m 2.943 (0.91) + 0.981 (0.41) + 1.962 (0.14) 3.36 KN

Checking Force å FY F RA + RB = = 0 F = 5.89 KN

RA + RB= 1.962 + 2.943 + 0.981 2.53 + 3.36 = 5.89 5.89 = 5.89

8/13/2019 177195895 Shear Force Influence Line

http://slidepdf.com/reader/full/177195895-shear-force-influence-line 30/33

 Location 42.943 N 1.962 N 0.981 N

RA0.06 m 0.16 m 0.08 m 0.04 m 0.1 m

RB

1

RB Y20

Y1

Y3

Y1 RA

Y2

Y3

0

1

8/13/2019 177195895 Shear Force Influence Line

http://slidepdf.com/reader/full/177195895-shear-force-influence-line 31/33

Reaction at Support A Y1 0.34 Y1 Y2 0.22 Y2 Y3 0.06 Y3 RA = = = = = = = = 1.0 0.44 0.77 m 1.0 0.44 0.50 m 1.0 0.44 0.14 m 2.943 (0.14) + 1.962 (0.50) + 0.981 (0.77) 2.15 KN

Reaction at Support B Y1 0.38 Y1 Y2 0.22 Y2 Y3 0.1 Y3 RB = = = = = = = = 1.0 0.44 0.86 m 1.0 0.44 0.50 m 1.0 0.44 0.23 m 2.943 (0.86) + 1.962 (0.50) + 0.981 (0.23) 3.74 KN

Checking Force å FY F RA + RB = = 0 F = 5.89 KN

RA + RB= 1.962 + 2.943 + 0.981 2.15 + 3.74 = 5.89 5.89 = 5.89

8/13/2019 177195895 Shear Force Influence Line

http://slidepdf.com/reader/full/177195895-shear-force-influence-line 32/33

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. ii. 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. 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 takingthe 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 itaffects 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 fromleft to right.

10.0i. ii.

REFERENCESSTRUCTURAL ANALYSIS (2009), Bambang Prihartanto MECHANICS OF MATERIALS, James M. Gere, Barry J. Goodno

8/13/2019 177195895 Shear Force Influence Line

http://slidepdf.com/reader/full/177195895-shear-force-influence-line 33/33