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OBJECTIVE 1.1 Part 1: To plot Shear force influence line. 1.2 Part 2: To verify the use of a shear force influence on a simply supported beam

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INTRODUCTION Moving loads on beams are common features of design. Many road bridges are constructed from beam, and as such have to be designed to carry a knife edge load, or a string of wheel loads, or a uniformly distributed load, or perhaps the worst combination of all three. The method of solving the problem is to use influence lines.

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THEORY Defination: Shear influence line is defined as a line representing the changes in shear force at a section of a beam when a unit load moves on the beam 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 written. For 0 ≤ x ≤ a a shear line is given by: Sy = − x/ L…………… (1) For a ≤ x ≤ b shear line is given by: Sy = 1− x L. ………… (2)

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 diagram 2). Shear force at ‘cut’ section = F1 y1 + F2 y2 + F3 y3 … (3) (y1, y2 and y3 are ordinates derived from the influence line in terms of x1, x2, x3, a, b and L)

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APPARATUS 4.1

Shear Force machine

4.2

Weight (Loadings)

Beam

Digital Force Display

Load

5.0

PROCEDURES Part 1 1.

Digital Force Display meter reads zero with no load is checking.

2.

Hanger with any mass range between 100g to 300g was placed at the first grooved hanger support at the left support and the Digital Force reading recorded in Table 1.

3.

The procedure to the next grooved hanger until to the last grooved hanger at the right hand support was repeated.

4.

The calculation in Table 1 was completed.

Part 2 1.

Three load hangers with 100g. 200g and 300g mass respectively placed at any position between the supports. The positions and the Digital Force Display reading recorded in Table 2.

2.

The produce with three other locations was repeated.

3.

The calculation in Table 2 was completed.

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RESULT

Part 1: 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 lines value

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

0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.0 1.1 1.2 -0.5 -0.4 -0.3 -0.2

0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.0 1.1 1.2 -0.5 -0.4 -0.3 -0.2

0.102 0.153 0.204 0.255 0.306 0.357 0.408 0.459 0.510 0.510 0.561 0.612 -0.255 -0.204 -0.153 -0.051

-0.091 -0.136 -0.182 -0.227 -0.273 -0.318 -0.364 -0.409 -0.455 -0.500 -0.545 -0.591 0.227 0.182 0.136 0.091

Part 2: Position of hanger from left hand Location 1 2 3 4

6.1

support ( m ) 100g 0.22 0.16 0.26 0.10

CALCULATION

200g 0.24 0.08 0.04 0.24

300g 0.06 0.24 0.14 0.38

Shear force Digital Reading (N)

Theoretical Shear ( Nm )

2.0 2.4 1.8 0.9

1.962 1.581 1.156 1.000

Part 1: 1) Experimental Influence line values = Shear Force (N) Load (N) Eg.

Experimental Influence line values =

0.2 N 200 x 9.81/1000

2) Theoretical Influence lines value; 0.04 ≤ x ≤ 0.26m Theoretical value, Sy = -x/L Eg.

Theoretical value, Sy = -0.04/ 0.44 = - 0.091

3) Theoretical Influence lines value; 0.32 ≤ x ≤ 0.38m Theoretical value, Sy =1 -x/L Eg.

Theoretical value, Sy = 1 -0.34/ 0.44 = 0.227

Part 2: Location 1

= 0.102

Y3

Y1

Y2

300g

100g

200g

60mm 220mm

140mm 240mm 300mm

a/L 1. a/L

2. b/L

3. y1 / 220

y1

4. y2 / 240

y2

5. y3 / 60 =

b/L

=

300/440

=

0.682

=

140/440

=

0.318

=

0.682 / 300

=

0.500

=

0.682 / 300

=

0.546

0.682 / 300

y3 Theoretical Shear

=

0.136

= = =

F1y1 + F2y2 + F3y3…. [(0.1 x 0.5) + (0.2 x 0.546) + (0.3 x 0.136)] x 9.81 1.962 Nm

Location 2 Y2 200g

Y1

Y3

100g

300g

80mm 160mm

140mm 240mm 300mm

a/L

1. y1 / 160

y1

2. y2 / 80 =

y2

3. y3 / 240

y3

b/L

=

0.682 / 300

=

0.248

0.682 / 300 =

0.124

=

0.682 / 300

=

0.372

Theoretical Shear

= = =

F1y1 + F2y2 + F3y3…. [(0.1 x 0.248) + (0.2 x 0.124) + (0.3 x 0.372)] x 9.81 1.581 Nm

Location 3 Y2 200g

Y3

Y1

300g

100g

40mm 140mm

140mm 260mm 300mm

a/L

1. y1 / 200

y1

2. y2 / 40 =

b/L

=

0.682 / 300

=

0.403

0.682 / 300

y2

3. y3 / 140

y3

Theoretical Shear

=

0.062

=

0.682 / 300

=

0.217

= = =

F1y1 + F2y2 + F3y3…. [(0.1 x 0.403) + (0.2 x 0.062) + (0.3 x 0.217)] x 9.81 1.156 Nm

Location 4 Y1

Y2

100g

200g

Y3 300g

100mm 140mm 240mm 60mm 300mm

a/L

4. y1 / 100

=

b/L

0.682 / 300

y1

5. y2 / 240

y2

6. y3 / 60 =

y3

Theoretical Shear

7.0

=

0.155

=

0.682 / 300

=

0.372

0.682 / 300 =

0.043

= = =

F1y1 + F2y2 + F3y3…. [(0.1 x 0.155) + (0.2 x 0.372) + (0.3 x 0.043)] x 9.81 1.008 Nm

DISCUSSIONS

Part 1: 1. Derive equation 1 and 2. Equation 1 ∑ Mcut = 0 ∑ Fy = 0 (L-x)/L -1 – Sy = 0 Sy = -x/L Equation 2 ∑ Mcut = 0 ∑ Fy = 0

(L-x)/L – Sy = 0 Sy = (L-x)/L Sy = 1 – x/L 2. On the same graph paper, plot the theoretical and experimental values against distance from left hand support.

3. Comment on the shape of graph. What does it tell you about how shear force varies at the cut section as a load moved on the beam? The experimental result increases with the increasing of the distance of load from the left hand support at the left side of the cut. Based on the result, the values of shear force at cut section (N) increases when a load moves nearer towards the cut. 4. Comment on the experimental result compared to the theoretical result. Based on the results that we got, shows a totally different result between the theoretical and experimental values. For the experimental influence line value, there are a big different between those experimental and theoretical. Overall,

based on the procedure, we followed the right instruction. It might be the error of the machine itself and not in the good condition. Part 2 : 1. Comment the experimental result and the theoretical result in Table 2. In this part, we used the load 100g, 200g and 300g. From this experiment, the value for the location 1 to 4, the value for the experimental is bigger than the theoretical value. The value is depend on the location but the value for both results is not so much differences.

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CONCLUSION

Part 1 : From the experiment, we know that the value for the experimental and theoretical values is totally difference. From the graph it shows totally difference result between theoretical and experimental result. Based on the result, the values of shear force at cut section (N) increases when a load moves nearer towards the cut. Part 2 :

From the experiment, 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.

View more...
OBJECTIVE 1.1 Part 1: To plot Shear force influence line. 1.2 Part 2: To verify the use of a shear force influence on a simply supported beam

2.0

INTRODUCTION Moving loads on beams are common features of design. Many road bridges are constructed from beam, and as such have to be designed to carry a knife edge load, or a string of wheel loads, or a uniformly distributed load, or perhaps the worst combination of all three. The method of solving the problem is to use influence lines.

3.0

THEORY Defination: Shear influence line is defined as a line representing the changes in shear force at a section of a beam when a unit load moves on the beam 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 written. For 0 ≤ x ≤ a a shear line is given by: Sy = − x/ L…………… (1) For a ≤ x ≤ b shear line is given by: Sy = 1− x L. ………… (2)

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 diagram 2). Shear force at ‘cut’ section = F1 y1 + F2 y2 + F3 y3 … (3) (y1, y2 and y3 are ordinates derived from the influence line in terms of x1, x2, x3, a, b and L)

4.0

APPARATUS 4.1

Shear Force machine

4.2

Weight (Loadings)

Beam

Digital Force Display

Load

5.0

PROCEDURES Part 1 1.

Digital Force Display meter reads zero with no load is checking.

2.

Hanger with any mass range between 100g to 300g was placed at the first grooved hanger support at the left support and the Digital Force reading recorded in Table 1.

3.

The procedure to the next grooved hanger until to the last grooved hanger at the right hand support was repeated.

4.

The calculation in Table 1 was completed.

Part 2 1.

Three load hangers with 100g. 200g and 300g mass respectively placed at any position between the supports. The positions and the Digital Force Display reading recorded in Table 2.

2.

The produce with three other locations was repeated.

3.

The calculation in Table 2 was completed.

6.0

RESULT

Part 1: 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 lines value

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

0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.0 1.1 1.2 -0.5 -0.4 -0.3 -0.2

0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.0 1.1 1.2 -0.5 -0.4 -0.3 -0.2

0.102 0.153 0.204 0.255 0.306 0.357 0.408 0.459 0.510 0.510 0.561 0.612 -0.255 -0.204 -0.153 -0.051

-0.091 -0.136 -0.182 -0.227 -0.273 -0.318 -0.364 -0.409 -0.455 -0.500 -0.545 -0.591 0.227 0.182 0.136 0.091

Part 2: Position of hanger from left hand Location 1 2 3 4

6.1

support ( m ) 100g 0.22 0.16 0.26 0.10

CALCULATION

200g 0.24 0.08 0.04 0.24

300g 0.06 0.24 0.14 0.38

Shear force Digital Reading (N)

Theoretical Shear ( Nm )

2.0 2.4 1.8 0.9

1.962 1.581 1.156 1.000

Part 1: 1) Experimental Influence line values = Shear Force (N) Load (N) Eg.

Experimental Influence line values =

0.2 N 200 x 9.81/1000

2) Theoretical Influence lines value; 0.04 ≤ x ≤ 0.26m Theoretical value, Sy = -x/L Eg.

Theoretical value, Sy = -0.04/ 0.44 = - 0.091

3) Theoretical Influence lines value; 0.32 ≤ x ≤ 0.38m Theoretical value, Sy =1 -x/L Eg.

Theoretical value, Sy = 1 -0.34/ 0.44 = 0.227

Part 2: Location 1

= 0.102

Y3

Y1

Y2

300g

100g

200g

60mm 220mm

140mm 240mm 300mm

a/L 1. a/L

2. b/L

3. y1 / 220

y1

4. y2 / 240

y2

5. y3 / 60 =

b/L

=

300/440

=

0.682

=

140/440

=

0.318

=

0.682 / 300

=

0.500

=

0.682 / 300

=

0.546

0.682 / 300

y3 Theoretical Shear

=

0.136

= = =

F1y1 + F2y2 + F3y3…. [(0.1 x 0.5) + (0.2 x 0.546) + (0.3 x 0.136)] x 9.81 1.962 Nm

Location 2 Y2 200g

Y1

Y3

100g

300g

80mm 160mm

140mm 240mm 300mm

a/L

1. y1 / 160

y1

2. y2 / 80 =

y2

3. y3 / 240

y3

b/L

=

0.682 / 300

=

0.248

0.682 / 300 =

0.124

=

0.682 / 300

=

0.372

Theoretical Shear

= = =

F1y1 + F2y2 + F3y3…. [(0.1 x 0.248) + (0.2 x 0.124) + (0.3 x 0.372)] x 9.81 1.581 Nm

Location 3 Y2 200g

Y3

Y1

300g

100g

40mm 140mm

140mm 260mm 300mm

a/L

1. y1 / 200

y1

2. y2 / 40 =

b/L

=

0.682 / 300

=

0.403

0.682 / 300

y2

3. y3 / 140

y3

Theoretical Shear

=

0.062

=

0.682 / 300

=

0.217

= = =

F1y1 + F2y2 + F3y3…. [(0.1 x 0.403) + (0.2 x 0.062) + (0.3 x 0.217)] x 9.81 1.156 Nm

Location 4 Y1

Y2

100g

200g

Y3 300g

100mm 140mm 240mm 60mm 300mm

a/L

4. y1 / 100

=

b/L

0.682 / 300

y1

5. y2 / 240

y2

6. y3 / 60 =

y3

Theoretical Shear

7.0

=

0.155

=

0.682 / 300

=

0.372

0.682 / 300 =

0.043

= = =

F1y1 + F2y2 + F3y3…. [(0.1 x 0.155) + (0.2 x 0.372) + (0.3 x 0.043)] x 9.81 1.008 Nm

DISCUSSIONS

Part 1: 1. Derive equation 1 and 2. Equation 1 ∑ Mcut = 0 ∑ Fy = 0 (L-x)/L -1 – Sy = 0 Sy = -x/L Equation 2 ∑ Mcut = 0 ∑ Fy = 0

(L-x)/L – Sy = 0 Sy = (L-x)/L Sy = 1 – x/L 2. On the same graph paper, plot the theoretical and experimental values against distance from left hand support.

3. Comment on the shape of graph. What does it tell you about how shear force varies at the cut section as a load moved on the beam? The experimental result increases with the increasing of the distance of load from the left hand support at the left side of the cut. Based on the result, the values of shear force at cut section (N) increases when a load moves nearer towards the cut. 4. Comment on the experimental result compared to the theoretical result. Based on the results that we got, shows a totally different result between the theoretical and experimental values. For the experimental influence line value, there are a big different between those experimental and theoretical. Overall,

based on the procedure, we followed the right instruction. It might be the error of the machine itself and not in the good condition. Part 2 : 1. Comment the experimental result and the theoretical result in Table 2. In this part, we used the load 100g, 200g and 300g. From this experiment, the value for the location 1 to 4, the value for the experimental is bigger than the theoretical value. The value is depend on the location but the value for both results is not so much differences.

8.0

CONCLUSION

Part 1 : From the experiment, we know that the value for the experimental and theoretical values is totally difference. From the graph it shows totally difference result between theoretical and experimental result. Based on the result, the values of shear force at cut section (N) increases when a load moves nearer towards the cut. Part 2 :

From the experiment, 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.

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