SM1 SHEAR FORCE AND BENDING MOMENT

Objectives

To show that the bending moment at a cut section of a beam is equal to the algebraic sum of the moment acting to the left or right of the section and the shear force at a cut section of a beam is equal to the algebraic sum of the acting to the left or right of the section.

Theory

In designing structural members such as beam, factors that are required to be considered are the shear force and the bending moment. The distribution of the forces on a member can be studied to design structures that can evenly distribute the loads, and prevent structural failure. When a transverse force is applied to any section of a beam, there will be internal forces that prevent the beam from breaking. At such condition, the shear force at any section in the beam is equal to the algebraic sum of the force taken on either side of the section. Similarly, for bending moment at any section is equal to the algebraic sum of the moments of the forces taken on either side of the section as well. Bending moment is induced when there is an external force that is causing the beam to bend.

Figure 1 show a schematic diagram of beam, and the shear force(SF) and bending moment(BM) of the beam at a section, x-x can be calculated by applying the equations below. In an experiment, to measure the bending moment at the section x-x, the section need to be cut.

Figure 1: Schematic diagram of beam is single point load

When the load is to the left of x-x;

Shear force at section x-x: SFx-x=W1-RA OR SFx-x=RBWhen the load is to the right of x-x;

SFx-x=RAWhen the load is to the left of x-x;

Bending Moment at section x-x is: BMx-x=W1L1-RALx OR BMx-x=RBL2When the load is to the right of x-x;

OR BMx-x=Lx

Figure 2: Bending moment sign convention

Force that cause beam to bend downward will induce positive moment, while force that cause beam to bend upward will induce negative moment, as shown in figure 2.

Figure 3: Schematic diagram of beam with multiple point load

For multiple load,

Shear force at section x-x is:

SFx-x=W1+W2+W3-RA OR SFx-x=RBBending moment at section x-x is:

BMx-x=W1Lx-L1+W2Lx-L2+W3Lx-L3-RA(Lx) OR BMx-x=RBL4Apparatus

Figure 4: Shear force and bending moment apparatus

Procedure

The switch of the indicator is turned on. Then, two simple support is fixed to the aluminium base at the distance equal to the span of the beam to be tested. The support is screwed tightly to the base. A load hanger is hung to the beam and the beam is placed on the supports. The load hanger is placed to the desired location. Readings on the indicators are noted, and the tare button is pressed when the values are not equal to zero. Next, the load is placed on the load hanger. The beam is gently tapped to ensure hysteresis error is reduced. The indicator readings are recorded once the values showed no changes when tap. After that, load is added to the load hanger and the indicators readings are recorded. Loads are then subsequently added and recorded for 4 more sets of readings. After five sets of readings are obtained, the location of the load is moved closer to the cut section and the experiment is repeated. Lastly, experiment with two load hangers and three load hangers is carried out.

L1 (mm) Beam span (mm) L2 (mm)

Case 1 200 900 305

Case 2 400 900 305

Table 1: Experiment 1 conditions in different cases

L1= Distance of the load from the left support (mm)

L2= Distance of the cut section from the right support (mm)

L1 (mm) L2 (mm) L3 (mm) Beam span (mm) L4 (mm)

Case 3 120 360 – 900 305

Case 4 200 400 – 900 305

Case 5 90 180 270 900 305

Case 6 80 180 280 900 305

Case 7 70 170 470 900 305

Table 2: Experiment 2 conditions in different cases

L1= Distance of the load from the left support (mm)

L2= Distance of the second load from the left support (mm)

L3= Distance of the third load from the left support (mm)

Distance of the load cell from the top of the beam at the cut section= 185mm

Results:

To calculate theoretical value for L1=200mm:

According to Figure 1, L1 = 0.2m, L2 = 0.305m, L = Lx+L2 = 0.9m, Take moment at point A,

752475-4445?MA+ = 0

W1 (L1) – RB (L) = 0

RB = W?L?L Equation (7)

Resolve forces vertically,

?FY ?+ = 0

RA + RB = W1

RA = W1 – RB Equation (8)

For different loads, by using equation (7),

When the load, W1 is 10N, RB = (10)(0.2) / 0.9 = 2.2N

When the load, W1 is 20N, RB = (20)(0.2) / 0.9 = 4.4N

When the load, W1 is 30N, RB = (30)(0.2) / 0.9 = 6.7N

When the load, W1 is 40N, RB = (40)(0.2) / 0.9 = 8.9N

When the load, W1 is 50N, RB = (50)(0.2) / 0.9 = 11.1N

Finding force RA, by using equation (8),

When the load, W1 is 10N, RA = 10 – 2.2 = 7.8N

When the load, W1 is 20N, RA = 20 – 4.4 = 15.6N

When the load, W1 is 30N, RA = 30 – 6.7 = 23.3N

When the load, W1 is 40N, RA = 40 – 8.9 = 31.11N

When the load, W1 is 50N, RA = 50 – 11.1 = 38.9N

Calculating shear force at section x-x by using equation (1),

When the load, W1 is 10N, SFx-x = RB = 2.2N

When the load, W1 is 20N, SFx-x = RB = 4.4N

When the load, W1 is 30N, SFx-x = RB = 6.7N

When the load, W1 is 40N, SFx-x = RB = 8.9N

When the load, W1 is 50N, SFx-x = RB = 11.1N

Calculating bending moment at section x-x by using equation (2),

When the load, W1 is 10N, BMx-x = RBL2 = (2.2)(0.305) = 0.7Nm

When the load, W1 is 20N, BMx-x = RBL2 = (4.4)(0.305) = 1.3Nm

When the load, W1 is 30N, BMx-x = RBL2 = (6.7)(0.305) = 2.0Nm

When the load, W1 is 40N, BMx-x = RBL2 = (8.9)(0.305) = 2.7Nm

When the load, W1 is 50N, BMx-x = RBL2 = (11.1)(0.305) = 3.4Nm

Load, W (N) Shear force (N) Bending moment (Nm) L1 (mm)Theoretical Experimental Theoretical Experimental 10 2.2 2.2 0.7 2.3 x 0.185= 0.43 200

20 4.4 4.6 1.3 4.8x 0.185 = 0.89 200

30 6.7 7.2 2.0 7.6x 0.185 = 1.41 200

40 8.9 9.2 2.7 10.2 x 0.185 = 1.89 200

50 11.1 12.6 3.4 13.9 x 0.185 = 2.57 200

Table 3: Experimental and theoretical results under single point load when L1=200mmLoad, W (N) Shear force (N) Bending moment (Nm) L1 (mm)Theoretical Experimental Theoretical Experimental 10 4.4 3.8 1.3 4.1 x 0.185= 0.76 400

20 8.9 8.5 2.7 9.4x 0.185 = 1.74 400

30 13.3 13.9 4.1 15.9x 0.185 = 2.94 400

40 17.7 18.2 5.4 21.8 x 0.185 = 4.03 400

50 22.2 23.1 6.8 29.1 x 0.185 = 5.38 400

Table 4: Experimental and theoretical results under single point load when L1=400mm.

Figure 5: Graph of shear force versus load

Figure 6: Graph of bending moment versus load

Experiment 1: L1=200mmBending Moment=1×L2×L1L =0.305×0.20.9 =0.0678 =Slopetheory, BMShear Force=1×L1L =0.20.9 =0.2222

=Slopetheory, SFFor average percentage error,

Percentage error, SF=Slopetheory,SF-Slopeexp,SF×100Slopetheory,SF%Percentage error, SF=0.2222-12.6-2.250-10×1000.2222% =17.01%Percentage error, BM=Slopetheory,BM-Slopeexp,BM×100Slopetheory,BM%Percentage error, BM=0.0678-2.57-0.4350-10×1000.0678% =6.44%Experiment 1: L1=400mmBending Moment=1×L2×L1L =0.305×0.40.9 =0.1356 =Slopetheory, BMShear Force=1×L1L =0.40.9 =0.4444 =Slopetheory, SFFor average percentage error,

Percentage error, SF=Slopetheory,SF-Slopeexp,SF×100Slopetheory,SF%Percentage error, SF=0.4444-23.1-3.850-10×1000.4444% =8.57%Percentage error, BM=Slopetheory,BM-Slopeexp,BM×100Slopetheory,BM%Percentage error, BM=0.1356-5.38-0.7650-10×1000.1356% Percentage error, BM=14.82%Percentage of error between theory and experiment (%)

Shear force Bending moment

Case 1 17.01 6.44

Case 2 8.57 14.82

Table 5: Percentage of error between theory and experiment

Experiment 2: Multiple point load

Load (N) Shear force (N) Bending moment (Nm) Length of load from left support (mm)

W1W2W3Theoretical Experimental Theoretical Experimental L1L2L310 10 – 5.3 5.3 1.6 1.0 120 360 –

20 30 – 17.8 18.3 5.4 3.8 200 400 –

10 10 10 6.0 6.2 1.8 1.2 90 180 270

10 20 30 14.2 14.7 4.3 3.3 80 180 280

10 20 30 22.4 26.1 6.8 5.4 70 270 470

Table 6: Experimental and theoretical results under multiple point load.

Figure 7: Shear force comparison between theory and experiment

Figure 8: Bending moment comparison between theory and experiment

For percentage error,

Percentage error=theorectical value-experimental valuetheoretical value×100%Table 7: Percentage of error between theory and experiment

Percentage of error between theory and experiment (%)

Shear force Bending moment

Case 3 5.3-5.35.3×100=01.6-1.01.6×100=37.5Case 4 2.81 29.63

Case 5 3..33 33.3

Case 6 3.52 23.26

Case 7 16.52 20.59

Discussion:

Under single point loading case 1(when L1=200mm) and case 2(when L1=400mm), the result obtained are quite accurate. From the graph of shear force versus load plotted in figure 5 it can be observed that the theoretical and the experimental lines are almost in line together and most of the time the experimental results tend to have lesser values than the theoretical values. The largest difference between theoretical and experimental value for case 1 is 0.6N when the load applied is 40N, which contribute to 6.74% error. Meanwhile, for case 2 the largest percentage of error is 12.36% when load is 20N. However, the average percentage error (table 5) calculated from the slope are 2.39% for both cases which is acceptable. Next, for multiple point loading experiment, the percentage of error for each cases are tabulated in table 7. Based on the bar chart in figure 7, the percentage of error for case 3 to 6 are quite less, but case 7 is exceptionally higher with a 16.52% error. This large degree of error in case 7 may be due to many factors, and will be discussed in later part. Overall, both experiment 1 and 2 are quite accurate.

As for the results of bending moment for case 1 and case 2, the difference between the theoretical and experimental values for both cases tend to get larger as the load applied is increased, indicating inaccuracy as the load increases. The difference can clearly be seen on the graphs plotted in figure 6 and similar with shear force, the experimental results are lesser than the theoretical results. The average percentage of error for case 1 bending moment is 18.88%, while 29.94% for case 2, resulting in a theoretical and experimental line with large degree of difference in steepness as seen in figure 6. Next, for the bending moment of multiple point loading cases, the differences can be seen in figure 8. The largest percentage of error for bending moment is 37.5%, as tabulated in table 7. On the other hand, it turns out that the lowest percentage error for bending moment is case 7, that is 20.59%. Like shear force, the graph of bending moment has an increasing trend. Although the experimental bending moment lines in figure 6 are not a smooth line, they can be represented by a trend line. Then, if the percentage of error is low, it is expected that the trend line will become more in line with the theoretical lines. Overall, the bending moment for experiment 1 and 2 can be considered as inaccurate and this may be due to some probable causes.

There could be many factors that affect the accuracy of the experiment as mentioned. One of the cause may be due to the restraint caused by the load cells mounted on the beam. This may be the reason that influence the accuracy of the bending moment reading in the experiment. The only way to improve the results is to minimize the restrain as much as possible by adjusting the connection between the load cell and the beam or use a more sophisticated load cell equipment. Another cause of error may be due to hysteresis error. Hysteresis error will cause the force distribution to get stuck at any part of the beam. That is why tapping is required to reduce this error. Next, error could be caused by non-uniform load distribution as well, if the load is not placed at the centre of the load hanger or is slanting. Nevertheless, error may occur if the nut of the load cell is not aligned properly. It is noticeable that the shear force experiment results are accurate enough but not for the bending moment experimental results. This inaccuracy can most probably be a result of the bad nut tuning. Since load cell is a sensitive measuring equipment, the nut that connect the load cell to the beam in this experiment need to be adjusted as good as possible to obtain an accurate result.

When load is applied to a beam, there will be internal shear force to counter the load. If there is no shear force to counteract the load, the beam will break apart. Nevertheless, when load is applied to a beam, the beam will tend to bend and induce bending moment. Shear force and bending moment are independent of the beam material properties. No matter what is the material’s elastic modulus, the bending moment and shear force will still remain the same. The difference between different material subjected under load will be the deflection, whether the material tend to bend more or less.

It can be seen in figure 5 and figure 6 that shear force and bending moment will increase as the load is closer to the cut section of the beam. The increase in L1 resulted in a steeper line in figure 5 and 6 as compared to those with the same load but shorter L1. As load moved closer to the cut section, it is farther from the support. And based on the equation:

SFx-x=W×L1LSFx-x?L1Hence, shear force is directly proportional to the distance of the load taken from the left support. Therefore, the farther the load is from the left support, the greater will the shear force be. And since bending moment is the product of shear force and length of the opposite side support to the cut section, bending moment increases as shear force increases (BMx-x?SFx-x) if L2 is constant.

Next, the horizontal load cell is used to measure the bending moment only. Therefore, changing the location of the horizontal load cell will only affect the bending moment measurements. In the experiment, positive bending moment is induced when the load is added, but the results obtained from the load cell provide a negative sign in front of each measurement. This is because, the beam experienced a tensile force. If the horizontal load cell is located at 200mm from the top of the beam instead of bottom, the load cell bending moment reading will have opposite sign, which is positive.

For this experiment, there are a few assumptions that need to be made. Firstly, the weight of the beam and the weight of the load hanger should be neglected. Hence, in calculation, the weight is not taken into account and during the experiment tare button is press after hanging the load hanger. Another assumption is that, the load cell is able to measure the shear force and bending moment completely and the load is not slanted, which provide a complete load distribution. On the other hand, since bending moment is induced when the beam is bent, the beam is assumed to be horizontal when the tare button on the bending moment indicator is pressed. Lastly, the shear force and bending moment is assumed to be the same regardless of the type of material of the beam.

Conclusion:

In conclusion, since the percentage error of shear force for most cases (case 1 to 6) are acceptable, with the highest error of 3.52%, the shear force measured at a cut section can be conclude to be equal to the algebraic sum of the forces acting to the left or right of the section. However, throughout this experiment there exist a high percent of error for bending moment. The highest percent of error is 37.5% for bending moment. These errors can be caused by many factors such as hysteresis error, non-uniform load distribution (load slanting), restraint caused by the load cells, and may be due to the imperfect tightness of the nut connecting the load cell to the beam. Therefore, more measures need to be taken to ensure the accuracy of the experiment. For, the measurement of the bending moment, if the location of the horizontal load cell is changed from the top to the beam the value measured will be positive due to the compressive force experience by the beam.

Reference

Beer, F ; Johnston, R ; DeWolf, J ; Mazurek, D. 2015. Mechanics of materials. New York: McGraw-Hill Education.

Civil Engineer. 2006. Bending moment at a section of beam. Viewed on 20 June 2017. Available from: ;http://civilengineer.webinfolist.com/mech/bm.htm;.

Applied measurement LTD: Aldermasten. 2004. What are hysteresis errors? Viewed on 21 June 2017. Available from: ;http://www.appmeas.co.uk/technical-notes/what-are-hysteresis-errors.html;.