LAB Report - TWO & Three Hinged ARCH ( Group 1) PDF

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UNIVERSITI TEKNOLOGI MARA KAMPUS PULAU PINANGnullSEMESTER : MAR - AUG 2021 DATE OF LAB : 7 APRIL 2021 GROUP : PEC2215D1 LAB LEVEL OEL : 1LECTURER : PUAN NOOR SYAFEEKHA BINTI MOHAMAD SAKDUN CO2 : Organize laboratory work on structural elements and materials. PO5 : Ability to utilize appropriate techn...

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CES511 –STRUCTURAL ENGINEERING LABORATORY LAB EXPERIMENT: TWO & THREE HINGED ARCH (CO2:PO5) SEMESTER GROUP LECTURER

: MAR - AUG 2021 DATE OF LAB : 7 APRIL 2021 : PEC2215D1 LAB LEVEL OEL : 1 : PUAN NOOR SYAFEEKHA BINTI MOHAMAD SAKDUN

CO2: Organize laboratory work on structural elements and materials. PO5: Ability to utilize appropriate techniques, resources and modern engineering and IT tools in predicting and modelling of complex civil engineering problems with an understanding of the limitations. NO

STUDENT ID

NAME

1.

2020982463

ABANG NAZIRUL IZZAT BIN ABANG YUSUF

2.

2020991447

ANNUR JASMIN BINTI LUKMAN

3.

2020982413

MUHAMMAD AKMAL DANISH BIN MUDZAFFAR

4.

2020970511

NUR AFIQAH IZZATI BINTI OTHMAN

5.

2020986097

SITI NUR AIN BINTI MOHD FAIZAN

1

2

CRITERIA 3 4

TOTAL

PSYCHOMOTOR PERFORMANCE RUBRIC

NO.

PERFORMANCE SCALE

CRITERIA Developing 0

1

2

Ability to design and conducted a research- based experiment (P1-P2)

Successfully performs experiment without guidance based on level of Openness (P3-P4)

3

Ability to manipulate the data leading to findings (P5-P6)

4

Ability to adapt the obtained result with logical justification (P5-P6)

Functional 1

2

Proficient 3

Advanced 4

Often requires instructor to design / identify basic idea / task of the experiment.

Generally able to design / identify basic idea / task of the experiment.

Helps are required to conduct the whole experiment.

Helps are required with refinement in several major details and conduct the experiment.

Able to conduct the experiment with minimal guide from the instructor.

Cannot complete tasks and standard procedures.

Low ability to complete tasks and standard procedures.

Successfully complete experiment procedures with moderate supervision.

Successfully complete experiment procedures with minimal supervision.

Incorrect interpretation of trends and comparison of data indicating a lack of understanding of results.

Incomplete interpretation of trends and comparison of data indicating a lack of understanding of results.

Minimal with 1 sentence describing the main finding of the experiment.

Almost all of the results have been correctly interpreted but without sufficient support of important trends or data comparisons.

All of the results have been correctly interpreted, with only 1 sufficient support of important trends or data comparisons.

Failed to show any idea at all

No ability to adapt the result with any suitable justification.

Show effort to give reasonable justification but incorrect

Display effort to understand the finding with major flawed interpretation

Display effort to understand the finding with minimum flawed interpretation

Unable to design/ identify the basic idea / task of the experiment. Helps are required to conduct the whole experiment.

Has a very low ability to design/ identify the basic idea / task of the experiment.

Instruction to Students 1. Two (2) weeks duration is given for each lab report submission. 2. Any plagiarism found or not properly cited, the group will be penalized and marks will be deducted.

Independently designed / identified idea / task of the experiment. Able to conduct a research-based experiment.

Successfully complete experiment independently.

5 Independently designed / identified the idea / task of the experiment with additional supporting references. Show an outstanding ability in conducting research-based experiment. Successfully complete experiment independently.

All of the results have been interpreted correctly, with sufficient support of important trends or data comparisons

Show excellent understanding between the results to adapt with logical justification

LabReport: CES511 – STRUCTURAL ENGINEERING LABORATORY

UNIVERSITI TEKNOLOGI MARA KAMPUS PULAU PINANG

INTRODUCTION (TWO HINGED ARCH) The two hinged arch is a statically indeterminate structure of the first degree. Atypical two-hinged arch is shown in Fig.6 (b).The horizontal thrust is the redundant reaction and is obtained using strain energy methods. Two hinged arch is made determinate by treating it as a simply supported curved beam and horizontal thrust as a redundant reaction. The arch spreads out under external load. Horizontal thrust is the redundant reactionism obtained using strain energy method.

The horizontal thrust is given by: H = 5 W L (k4 – 2k3 + k) / ( 8h )

OBJECTIVES To study two hinged arch for the horizontal thrust of the roller end for a given system of loading and to compare the same with those obtained analytically.

APPARATUS: Support frame, two hinged arch assembly, simple support, roller support, ruler, load hanger, digital indicator, vernier caliper, set of point load weight.

PROCEDURE: 1. The indicator is switched on. The indicator must be switched on 10 minutes before taking readings to ensure the stability of the reading. 2. The two supports are fixed tightly to the support frame. The frame of the arch is measured. 3. The ‘Tare’ button is pressed to set the dial indicator reading to zero. 4. The load is placed on the load hanger at the arch. 5. The indicator reading is recorded. 6. The applied load is increased and step 5 and 5 is repeated. 7. The experiment is repeated to get an average of two readings.

INTRODUCTION (THREE HINGED ARCH) An arch may be looked upon as a curved girder, either a solid rib or braced, supported at its ends, and carrying transverse loads which are frequently vertical. Since the transverse loading at any section normal to the axis of the girder is at an angle to the normal face, an arch is subjected to three restraining forces: thrust, shear force and bending moment. Depending upon the number of hinges, arches may be divided into four classes which is three hinged arch, two hinged arch, single hinged arch, and fixed arch (hinge less arch). A three-hinge arch is statically determinate structures while the rest three hinged arch is statically determinate structures while the rest three arches are statically indeterminate. In bridge construction, especially in railroad bridges, the more used arches are two-hinged and the fixed end ones.

The horizontal thrust is given by:

OBJECTIVE •

To study three hinged arches for the horizontal thrust of the roller end for a given system of loading and to compare the same with those obtained analytically.

•

To determine the relationship between applied load and the horizontal support of a three hinged parabolic arch.

PROBLEM STATEMENT In the case of three-hinged arch, we have three hinges: two at the support and one at the crown thus making it statically determinate structure. There are four reaction components in the three-hinged arch. One more equation is required in addition to three equations of static equilibrium for evaluating the four reaction components. Taking moment of the hinge of all the forces acting on either side of the hinge can set up the required equation.

APPARATUS 1. Three-Hinged arch apparatus 2. Load hanger 3. Dial gauge 4. Vernier calliper 5. Ruler

PROCEDURE 1.

The roller support and simple support were fixed to the support frame at distance equals to span of arch.

2.

The roller support was anchored, and the pulley was located on the inside of the arch.

3.

The wire rope was taking at the roller end of the arch and passed over a pulley attached to the roller support.

4.

Load hanger was placed at the end of wire rope and at selected location of arch.

5.

The dead load was offset by placing sufficient load on the load hanger at the end of wire rope, so the arch was leveled.

6.

Load was placed on load hanger at selected location of arch.

7.

Sufficient loads were added on load hanger to level the arch.

8.

Applied load and load were recorded.

9.

Applied load was increased, and step 6 to 9 were repeated.

10.

Test was made twice to obtain average readings.

DATA AND ANALYSIS Data collection Span of arch (m)

: 1.00

Height of arch (m) : 0.32 Set 1 Distance from pinned support to Point load 1, W1: 0.26m Distance from pinned support to Point load 4 , W4: 0.63m Horizontal Thrust (N) Theoretical Experimental W1 W4 4.8 2.24 2.81 9.3 4.48 5.63 13.8 6.72 8.44 18.3 8.96 11.25 22.0 11.20 14.06 Total 68.2 33.6 42.19

Load, W1 (N)

Load, W4 (N)

5 10 15 20 25

5 10 15 20 25

Test 1 Test 2 Test 3 Test 3 Test 4

Percentage error

= =

𝐸𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡𝑎𝑙− ∑𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙

68.2− 75.79 75.79

= 10.01%

x 100

x 100

Set 2 Distance from pinned support to Point load 1, W1: 0.26m Distance from pinned support to Point load 2 , W2: 0.39m Load, W1 (N) Test 1 Test 2 Test 3 Test 3 Test 4

Percentage error

5 10 15 20 25

= =

Horizontal Thrust (N) Load, W2 (N) Theoretical Experimental W1 W2 5 4.8 2.24 2.88 10 9.3 4.48 5.76 15 13.7 6.72 8.64 20 18.4 8.96 11.52 25 22.1 11.20 14.40 Total 68.3 33.6 43.20

𝐸𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡𝑎𝑙− ∑𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 68.3−76.8 76.8

= 11.07%

x 100

x 100

Analysis Theoretical formula:

H = 5WL (k4–2k3+ k)/(8h) Where; H = horizontal thrust W = load L = span k = distance h = height Given:

Span of arch, L (m) : 1.00 Height of arch, h (m)

: 0.32

Set 1 Distance from pinned support to Point load 1, W1: 0.26m Distance from pinned support to Point load 4 , W4: 0.63m Test 1 Load, W1: H = 5(5)(1) [0.264 – 2(0.26)3 + 0.26)]/ (8x0.32) = 2.24N

Load, W4: H = 5(5)(1) [0.634 – 2(0.63)3 + 0.63)]/ (8x0.32) = 2.81N

Test 2 Load, W1: H = 5(10)(1) [0.264 – 2(0.26)3 + 0.26)]/ (8x0.32) = 4.48N

Load, W4: H = 5(10)(1) [0.634 – 2(0.63)3 + 0.63)]/ (8x0.32) = 5.63N

Test 3 Load, W1: H = 5(15)(1) [0.264 – 2(0.26)3 + 0.26)]/ (8x0.32) = 6.72N

Load, W4: H = 5(15)(1) [0.634 – 2(0.63)3 + 0.63)]/ (8x0.32) = 8.44N

Test 4 Load, W1: H = 5(20)(1) [0.264 – 2(0.26)3 + 0.26)]/ (8x0.32) = 8.96N

Load, W4: H = 5(20)(1) [0.634 – 2(0.63)3 + 0.63)]/ (8x0.32) = 11.25N

Test 5 Load, W1: H = 5(25)(1) [0.264 – 2(0.26)3 + 0.26)]/ (8x0.32) = 11.20N

Load, W4: H = 5(25)(1) [0.634 – 2(0.63)3 + 0.63)]/ (8x0.32) = 14.06N

Set 2 Distance from pinned support to Point load 1, W1: 0.26m Distance from pinned support to Point load 2, W2: 0.39m Test 1 Load, W1: H = 5(5)(1) [0.264 – 2(0.26)3 + 0.26)]/ (8x0.32) = 2.24N

Load, W4: H = 5(5)(1) [0.394 – 2(0.39)3 + 0.39)]/ (8x0.32) = 2.88N

Test 2 Load, W1: H = 5(10)(1) [0.264 – 2(0.26)3 + 0.26)]/ (8x0.32) = 4.48N

Load, W4: H = 5(10)(1) [0.394 – 2(0.39)3 + 0.39)]/ (8x0.32) = 5.76N

Test 3 Load, W1: H = 5(15)(1) [0.264 – 2(0.26)3 + 0.26)]/ (8x0.32) = 6.72N

Load, W4: H = 5(15)(1) [0.394 – 2(0.39)3 + 0.39)]/ (8x0.32) = 8.64N

Test 4 Load, W1: H = 5(20)(1) [0.264 – 2(0.26)3 + 0.26)]/ (8x0.32) = 8.96N

Load, W4: H = 5(20)(1) [0.394 – 2(0.39)3 + 0.39)]/ (8x0.32) = 11.52N

Test 5 Load, W1: H = 5(25)(1) [0.264 – 2(0.26)3 + 0.26)]/ (8x0.32) = 11.20N

Load, W4: H = 5(25)(1) [0.394 – 2(0.39)3 + 0.39)]/ (8x0.32) = 14.40N

Graph analysis Set 1

Horizontal Thrust (N) vs Load (N) 50 45

Horizontal thrust (N)

40 35 30 25

Theoretical, W4

20

Theoretical, W1

15

Experimental

10 5 0

5

10

15

20

25

Load (N)

Set 2

Horizontal Thrust (N) vs Load (N) 50 45

Horizontal thrust (N)

40 35 30 25

Theoretical, W2

20

Theoretical, W1

15

Experimental

10

5 0 5

10

15

Load (N)

20

25

APPARATUS MEASUREMENTS

L

W3 W2

W5 W4

W6

h

Roller Support

Pinned Support W1

W7

Measurements

Value (m)

Span of arch, L

1.00

Height of arch, h

0.23

Distance from pinned support to Point Load 1

0.125

Distance from pinned support to Point Load 2

0.250

Distance from pinned support to Point Load 3

0.375

Distance from pinned support to Point Load 4

0.500

Distance from pinned support to Point Load 5

0.625

Distance from pinned support to Point Load 6

0.750

Distance from pinned support to Point Load 7

0.875

Data For Point Load Set 2 Point Test 1 Test 2 Test 3 Test 4 Test 5

Load, W2 (N) 5 10 15 20 25

Load, W6 (N) 5 10 15 20 25

Horizontal Thrust (N) 5.2 10.7 16.6 21.9 27.1

Data For Combination Load – Uniformly Distributed Load And Point Load Set 1 Point

UDL, (N)

Load, W3 (N)

Load, W5 (N)

Test 1

9.6

Test 2

9.6 + 9.4

10

10

Test 3

9.6 + 9.4 + 9.8

15

15

Horizontal Thrust (N) 15.4 30.7 41.1

Data For Point Load - SET 1 Span of arch, L (m)

= 1.000

Distance from pinned support to Point Load 2 to W1, 𝑘1 (m)

= 0.125

Distance from pinned support to Point Load 5 to W5, 𝑘5 (m)

= 0.625

Height of arch, h (m)

= 0.230

𝐋𝐨𝐚𝐝, 𝐖𝟏 (N)

𝐋𝐨𝐚𝐝, 𝐖𝟓 (N)

Test 1

5

Test 2

Horizontal Thrust (N) Theoretical

Experimental

𝐖𝟏

𝐖𝟓

5

5.2

1.649

3.931

10

10

10.7

3.297

7.862

Test 3

15

15

16.6

4.946

11.792

Test 4

20

20

21.9

6.594

15.720

Test 5

25

25

27.1

8.243

19.654

81.5

24.729

58.959

∑

Percentage error

=

=

∑Experimental −∑Theoretical ∑Theoretical

81.5 −83.688 83.688

= 2.614%

× 100

× 100

Data Analysis 5WL(k4 − 2k 3 + k) Horizontal Thrust, H = 8h

Test 1 At W1 ,

At W5 ,

H = 1.649

H = 3.931

At W1 ,

At W5 ,

H = 3.297

H = 7.862

H=

5(5)(1.0)[(0.125)4 − 2(0.125)3 + 0.125] 8(0.230)

Test 2

H=

5(10)(1.0)[(0.125)4 − 2(0.125)3 + 0.125] 8(0.230)

H=

H=

5(5)(1.0)[(0.625)4 − 2(0.625)3 + 0.625] 8(0.230)

5(10)(1.0)[(0.625)4 − 2(0.625)3 + 0.625] 8(0.230)

Test 3 At W1 ,

At W5 ,

H = 4.946

H = 11.792

H=

5(15)(1.0)[(0.125)4 − 2(0.125)3 + 0.125] 8(0.230)

H=

5(15)(1.0)[(0.625)4 − 2(0.625)3 + 0.625] 8(0.230)

Test 4 At W1 ,

At W5 ,

H = 6.594

H = 15.72

H=

5(20)(1.0)[(0.125)4 − 2(0.125)3 + 0.125] 8(0.230)

H=

5(20)(1.0)[(0.625)4 − 2(0.625)3 + 0.625] 8(0.230)

H=

5(25)(1.0)[(0.625)4 − 2(0.625)3 + 0.625] 8(0.230)

Test 5 At W1 , H=

5(25)(1.0)[(0.125)4 − 2(0.125)3 + 0.125] 8(0.230)

H = 8.243

At W5 ,

H = 19.654

Data For Combination Load – SET 1 Span of arch, L (m)

= 1.000

Distance from pinned support to Point Load 3 (W3), 𝑘1 (m)

= 0.375

Distance from pinned support to Point Load 5 (W5), 𝑘5 (m)

= 0.625

Height of arch, h (m)

= 0.230

UDL (N)

𝐋𝐨𝐚𝐝, 𝐖𝟑 (N)

𝐋𝐨𝐚𝐝, 𝐖𝟓 (N)

Experimental

Theoretical

Horizontal Thrust (N)

Test 1

9.6

5

5

15.4

49.327

Test 2

19.0

10

10

30.7

181.947

Test 3

28.8

15

15

41.1

403.178

Test 4

38.7

20

20

52.6

713.020

Test 5

38.7

25

25

60.2

891.275

200.0

2238.747

∑

Percentage error

=

=

∑Experimental −∑Theoretical ∑Theoretical

200.0 −2238.747 2238.747

= 0.91%

× 100

× 100

Data Analysis Test 1 (W = 5) 1.0 (9.6)( )(1.0) (5)(0.375)(1.0) (5)(0.625)(1.0) 2 ][ H=[ ]+ 2(0.23) 2(0.23) 2(0.23) H = 49.327

Test 2 (W = 10) 1.0 (19.0)( 2 )(1.0) (10)(0.375)(1.0) (10)(0.625)(1.0) ][ H=[ ]+ 2(0.23) 2(0.23) 2(0.23)

H = 181.947

Test 3 (W = 15) 1.0 (28.8)( 2 )(1.0) (15)(0.375)(1.0) (15)(0.625)(1.0) ][ ]+ H=[ 2(0.23) 2(0.23) 2(0.23)

H = 403.178

Test 4 (W = 20) 1.0 (38.7)( )(1.0) (20)(0.375)(1.0) (20)(0.625)(1.0) 2 ][ H=[ ]+ 2(0.23) 2(0.23) 2(0.23)

H = 713.020

Test 5 (W = 25) 1.0 (38.7)( )(1.0) (25)(0.375)(1.0) (25)(0.625)(1.0) 2 ][ H=[ ]+ 2(0.23) 2(0.23) 2(0.23)

H = 891.275

DISCUSSION (TWO HINGED) In this Two Hinged Arch experiment, students need to conduct a two hinged arch experiment with pinned-end support and pinned-end support. This experiment is to study two-hinged arch for the horizontal thrust of the pinned-end by given system of loading and to compare the same with those obtain analytically. There are two set of data which differ in distance of the load from pinned support. For the first set, the distance of point load 1, W1, from pinned support is 0.26m while for point load 4, W4, from pinned support is 0.63m. For the second set, the distance of point load 1, W1, from pinned support is 0.26m while fore point load 2, W2, from pinned support is 0.39m. The point loads used are 5N, 10N, 15N, 20N and 25N. This experiment shows the reading of horizontal thrust using system load of the indicator which are shown in table, but the horizontal value was calculated by using the given formula: H = 5WL(k4 – 2k3 – k / (8h) According to the data, the graph plot...