Experiment 2

Shear force in beams

Objective:

Part 1: To comprehend the action of shear force in a beam for different load position.

Part 2: To comprehend the action of shear force in a beam for different loading at the selected position.

SET OF APPARATUS: Shear Force Machine HFC 4.

Procedure:

Part 1:

1. The beam to be set up so that the face of the normal section is 300mm from the support A and 600mm to the right support B as shown in Diagram (2a).

2. The first load hanger (W 1) to be positioned at 100mm from support A of the beam, the second load hanger (W 2) at part C (in the groove provided), and the third load hanger (W 3) at 400mm from right support B of the beam.

3. Adjustment to be used on the underslung spring and spring balance to align the two parts of the beam and the spring reading is to be recorded.

4. 10N weight is to be loaded on the third load hanger (W 3), realign the beam and the new reading in the spring balance is to be recorded.

5. Another 10N weight to be added to the third load hanger (W 3), realign the beam and the new reading on the spring balance is to be recorded.

6. Procedure (3), (4), and (5) are to be repeated using the second load hanger (W 2) which is 300mm from support A of the beam. All spring balance readings are to be recorded.

7. Procedure (3), (4), and (5) are to be repeated using the first load hanger (W 1) which is 100mm from left support A of the beam. All spring balance readings are to be recorded.

Part 2:

1. One load hanger is to be positioned at the middle of part X of the beam, second load hanger at part C, and third load hanger at the middle of part Y as shown in Diagram (2b).

2. The adjustment on the underslung spring and spring balance to align the two parts of the beam. The readings of the spring balance to be recorded.

3. 5N weight load is to be loaded to the first load hanger (W 1) and a 12N weight load to the third load hanger (W 3). The readings of the spring balance to be recorded.

4. Each load hanger to be unloaded.5. Procedure (2), (3), and (4) are to be repeated with different weight

loads in accordance with the table in part 2.

Measurements

Part 1

a) Load on third hanger ( 400mm from B) (N)

0 10 20

Spring balance force (N) 3.0 7.0 12.0Exp. shear (VE) (N) - 4.0 9.0Theo. Shear (VT) (N) - 4.4 8.9

Ratio VE/VT - 0.91 1.01

a) Load on second hanger ( 300mm from

A) (N)

0 10 20

Spring balance force (N) 2.5 9.0 15.0Exp. shear (VE) (N) - 6.5 12.5Theo. Shear (VT) (N) - 6.7 13.3

Ratio VE/VT - 0.97 0.94

a) Load on first hanger ( 100mm from A) (N)

0 10 20

Spring balance force (N) 3 1.5 1.0Exp. shear (VE) (N) - -1.5 -2.0Theo. Shear (VT) (N) - -1.1 -2.2

Ratio VE/VT - 1.09 0.91

Part 2

Load (N) at positions Balance force (N)

Shear V (N) RatioW1 W2 W3 Exp. Theory0 0 0 2.5 - 0 05 0 12 6.0 4.2 4.8 0.885 2 10 6.75 4.3 5.2 0.835 10 10 12.0 9.5 10.6 0.90

Average of Ratio Exp. / Theory = 0.87

Notations:

VE = Experimental shear force, N

VT = Theoretical shear force, N

W1 = Weight on the first load hanger, N

W2 = Weight on the second load hanger, N

W3 = Weight on the third load hanger, N

Calculations & Results

Load on third hanger (400mm from B)

For 10N

Exp. shear (N) = 6.75-3

= 3.75 N

For 20N

Exp. Shear (N) = 12-3

= 9 N

Load on second hanger (300mm from A)

For 10N

Exp. shear (N) = 9.0-2.5

= 6.5 N

For 20N

Exp. Shear (N) = 15.0-2.5

= 12.5 N

Load on first hanger (100mm from A)

For 10N

Exp. shear (N) = 1.5-3 = -2 N

For 20N

Exp. Shear (N) = 1-3

= -1.5N

Theoretical shear

Part 1 (a)

W 3 = 10N W 3 = 20N

∑ MA =0 → -10(0.5) +0.9By =0 ∑ MB =0→-0.9Ay+20(0.4) =0

By = 5.56N Ay = 8.89N

∑ Fy = 0 → Ay + 5.56-10=0

Ay = 4.44N

∑ Fy = 0 → 4.44 – 10-V=0 ∑ Fy = 0→8.89-20-V=0

V = -5.56N V= -11.11N

Part 1(b)

W 2 = 10N W 2 = 20N

∑ MB =0→-0.9Ay+10(0.6) =0 ∑ MB=0→-0.9Ay+20(0.6)=0

Ay = 6.67N Ay=13.33N

∑ Fy = 0→6.67-10-V=0 ∑ Fy = 0→13.33-20-V=0

V= -3.33N V= -6.67N

Part 1(c)

W 1 = 10N W 1 = 20N

∑ MB =0→-0.9Ay+10(0.8) =0 ∑ MB=0→-0.9 Ay 20(0.8) =0

Ay = 8.89N Ay =17.78N

∑ Fy = 0→8.89-10-V=0 ∑ Fy = 0→17.78-20-V=0

V=-1.11N V=-2.22N

Part 2

W 1 =5N, W 2 = 0, W 3 = 12N ∑ MB =0→-0.9Ay+5(0.8) +12(0.4)=0 Ay = 9.78N

∑ Fy = 0→9.78-5-V=0 ∑ Fy = 0→9.78-12-5-V=0

V= 4.78N V = -7.22N

W1=5N, W2= 2N, W3= 10N ∑ MB =0→-0.9Ay+5(0.8) +2(0.6)+10(0.4)=0

Ay = 10.2

∑ Fy = 0→10.2-5-V=0 ∑ Fy = 0→10.2-5-2-10-V=0

V = 5.2N V = -6.8N

∑ Fy = 0→10.2-5-2-V=0

V = 3.2N

W 1 =5N, W 2 = 10N, W 3 = 10N

∑MB =0→0.9Ay+5(0.8)+10(0.6)+10(0.4)=0

Ay = 15.6N

∑ Fy = 0→15.6-5-V=0 ∑ Fy = 0→15.6-5-10-10-V=0

V = 10.6N V = -9.4N

∑ Fy = 0→15.6-5-10-V=0

V = 0.6N

Discussion

The results of this experiment showed that the weights applied to the beam caused shear forces to occur within the beam. The amount of shear force is related the position of the weights. As the position of the weights got closer to the spring balance, so did the shearing force on the beam increase. It was also observed that the amount of weight was directly proportional to the shear force.

Certain values were negative because of the position of the weights were very close to the supports. This caused a negative shear force value.

This experiment was a challenging one because the spring balance and underslung spring had to be adjusted simultaneously and carefully in order to achieve correct results. Also it had to be ensured that the beam was stable and aligned before recording any values. For the improvement of any similar experiments, it is suggested that the spring balance be oiled before used. For during this experiment, the spring balance was very insensitive, and it could be felt that that was friction within the internal components of the spring balance.

Conclusion

This experiment proved to be a success because the difference between the experimental results and theoretical calculations were less than ten percent.

In the first part of this experiment, loads were applied on a beam in different positions to comprehend the action of shear force. During the second part, different loads were applied to the beam.

The results of this experiment showed that the closer the loads are to the center, the larger the shearing force. Also, greater amounts of load will cause higher shearing forces.