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
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
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.