Date of Experiment:
Report due date: Report submission date:
Checked by:
Item/ Marks
Format/ 10
Abstract & Introduction/ 10 Figures & Diagrams/ 15 Materials & Method/ 10 Results Discussions/ 45
References/ 10
Total
SHEAR FORCE AND BENDING MOMENT
GEORGE KENJI PUTRA
0304559
Justin Moo Xian Yuen Leong Yok Ben Manish Kumar Sing Domun
Ng Yi Ming
School of Engineering
Taylor’s University
Malaysia
12 May 2014
TABLE OF CONTENTS
ABSTRACT 2 1.0 INTRODUCTION 2 2.0 EXPERIMENTAL DESIGN 2 2.1 Apparatus 3 2.2 Methods 3 2.3 Procedures 33.0 RESULTS AND CALCULATIONS 3
4.0 DISCUSSIONS 15
5.0 CONCLUSION 16
ABSTRACT
The objective of this lab session is to acquire the experimental values of bending moment and shear force that acted on simply supported beam that has some points to hang the load. After conducting this experiment and analyzing the data, the results of experimental calculation will be compared with theoretical calculation.
1.0 INTRODUCTION
Shear force is the internal resistance created in beam cross sections, in order to balance transverse external load acting on beam, while bending moment is bending effect due to the forces that act on the beam. There are several types of beams, such as: cantilever beam, simply supported beam, overhanging beam, rigidly fixed beam and continuous beam. In this particular experiment, we were dealing with simply supported beam.
Having better understanding about these two things are very important in engineering field, since nowadays development of construction is growing rapidly.
In theory, when the loads apply to the beam, the beam hold those loads by giving off internal stresses and strains inside its interior. The internal force that acts vertically to the longitudinal axis is the shear force. The bending moment is the internal couple forces of the beam. The summation of those internal forces need to be in equilibrium state in order to hold the external force that applied to the beam.
2.0 EXPERIMENTAL DESIGN
Figure 1. Rough Design for Shear Force and Bending Moment Experiment
. . . . .
Aluminum Profile Frame
Simply Supported Beam
Console Box Loading Point
Holder
2.1 Apparatus
Aluminum profile frame
Simply supported beam (0.8 m length)
Loading Point Holder Loads
Hangers
Console 2.2 Methods
Calibrate the console box and the beam before taking the reading. Put the loads to the hanger, and hang it on the point holder. Wait until the values shown on console box are stable, then only take the reading. For the next reading, make sure the console box is recalibrated back to zero and make sure the beam is flat by using water level ruler.
2.3 Procedures
1. Level the beam by using water level ruler. 2. Set the console box to zero.
3. Add 2 N load to the hanger, and hang it at loading point holder A (0.095 m from left). 4. Write down the reading that shown on console box.
5. Repeat the step 1–4 by using load of 4 N and 6 N.
6. Repeat the step 1-5 by hanging the load at loading point holder B (0.245 m from left). 7. Hang the loads at both point A and B, with combination load of 2-2 N, 4-4 N, and 6-6
N, and don’t forget to do step 1 and 2 before start the experiment.
8. Write down the reading separately according to which loading point holder the loads were.
3.0 RESULTS AND CALCULATIONS
Table 1. Experimental Values for Loading Point Holder A (Distance: 0.095 m)
Load (N) Shear Force (N) Bending Moment (Nm)
2 0.202 0.100
4 0.405 0.200
6 0.603 0.160
Table 2. Experimental Values for Loading Point Holder B (Distance: 0.245 m)
Load (N) Shear Force (N) Bending Moment (Nm)
2 0.505 0.090
4 1.013 0.280
Table 3. Experimental Values for Double Loading Point Holder (A and B)
Load 1 (N) Load 2 (N) Shear Force (N) Bending Moment (Nm)
2 2 0.708 0.360
4 4 1.422 0.700
6 6 2.135 0.800
Drawing the free body diagram will make our life easier to calculate the theoretical value of shear force and bending moment. Look at next page for free body diagram and also calculation for the theoretical values of shear force and bending moment. In this case, 4 N is used to be the example. 4 N Load Calculation Example at Point A
Figure 2. Free Body Diagram for 4 N Load at Point A (0.095 m) Applying equilibrium conditions:
∑Fy = 0, taking up as positive
Ra + Rb – 4 = 0
4 = Ra + Rb
∑MA = 0, taking clockwise as positive
4(0.095) – Rb(L) = 0
0.38 – 0.8Rb = 0
Rb = 0.475 N
4 = Ra + 0.475
Section the beam right before the force (4 N).
Figure 3. Sectioned Beam for 0 ≤ x ≤ 0.095 m ∑Fy = 0, taking up as positive
Ra – V = 0
V = 3.525 N
∑M = 0, taking clockwise as positive (at cut-off section) Ra(x) - M = 0
M = 3.525(x) At x = 0 m; M = 0
At x = 0.095 m; M = 0.335 Nm
Then make a section again, this time 0.095 ≤ x ≤ 0.8 m.
Figure 4. Sectioned Beam for 0.095 ≤ x ≤ 0.800 m ∑Fy = 0, taking up as positive
Ra – 4 –V = 0
∑M = 0, taking clockwise as positive (at cut-off section) Ra(x) – 4(x – 0.095) – M = 0 At x = 0.095 m; 3.525(0.095) – 4(0.095-0.095) = M M = 0.335 Nm At x = 0.800 m; 3.525(0.800) – 4(0.800-0.095) = M M = 0 Nm
The shear force and bending moment diagram for calculation above would be like figures below:
Figure 6. Bending Moment Diagram for 2 N Load at Point A (0.095 m)
4 N Load Calculation Example at Point B
Figure 7. Free Body Diagram for 4 N Load at Point B (0.245 m) Applying equilibrium conditions:
∑Fy = 0, taking up as positive
4 = Ra + Rb
∑MA = 0, taking clockwise as positive
4(0.245) – Rb(L) = 0
0.98 – 0.8Rb = 0
Rb = 1.225 N
4 = Ra + 1.225
Ra = 2.775 N
Section the beam right before the force (4 N).
Figure 8. Sectioned Beam for 0 ≤ x ≤ 0.245 m ∑Fy = 0, taking up as positive
Ra – V = 0
V = 2.775 N
∑M = 0, taking clockwise as positive (at cut-off section) Ra(x) - M = 0
M = 2.775(x) At x = 0 m; M = 0
Then make a section again, this time 0.245 ≤ x ≤ 0.8 m.
Figure 9. Sectioned Beam for 0.245 ≤ x ≤ 0.800 m ∑Fy = 0, taking up as positive
Ra – 4 –V = 0
V = -1.225 N
∑M = 0, taking clockwise as positive (at cut-off section) Ra(x) – 4(x – 0.245) – M = 0
At x = 0.245 m; 2.775(0.245) – 4(0.245-0.245) = M M = 0.680 Nm
At x = 0.800 m; 2.775(0.800) – 4(0.800-0.245) = M M = 0 Nm
The shear force and bending moment diagram for calculation above would be like figures on the next page:
Figure 10. Shear Force Diagram for 4 N Load at Point B (0.245 m)
4 N Load Calculation Example at Point A and B (Double Point)
Figure 12. Free Body Diagram for 4-4 N Load at Point A and B (Double Point) Applying equilibrium conditions:
∑Fy = 0, taking up as positive
Ra + Rb – 4 – 4 = 0
8 = Ra + Rb
∑MA = 0, taking clockwise as positive
4(0.095) + 4(0.245) – Rb(L) = 0
0.38 + 0.98 – 0.8Rb = 0
Rb = 1.700 N
8 = Ra + 1.700
Ra = 6.300 N
Section the beam right before the force (4 N).
∑Fy = 0, taking up as positive Ra – V = 0
V = 6.300 N
∑M = 0, taking clockwise as positive (at cut-off section) Ra(x) - M = 0
M = 6.300(x) At x = 0 m; M = 0
At x = 0.095 m; M = 0.599Nm
Then make a section again, this time 0.095 ≤ x ≤ 0.245 m.
Figure 14. Sectioned Beam for 0.095 ≤ x ≤ 0.245 m ∑Fy = 0, taking up as positive
Ra – 4 –V = 0
V = 2.300 N
∑M = 0, taking clockwise as positive (at cut-off section) Ra(x) – 4(x – 0.095) – M = 0
At x = 0.095 m; 6.300(0.095) – 4(0.095-0.095) = M M = 0.599 Nm
At x = 0.245 m; 6.300(0.245) – 4(0.245-0.095) = M M = 0.944 Nm
Once again make a section, this time 0.245 ≤ x ≤ 0.800 m.
Figure 15. Sectioned Beam for 0.245 ≤ x ≤ 0.800m ∑Fy = 0, taking up as positive
Ra – 4 – 4 – V = 0
6.300 – 8 = V V = -1.700 N
∑M = 0, taking clockwise as positive Ra .x - 4(x – 0.095) – 4(x – 0.245) – M = 0 6.300(x) – 4(x – 0.095) – 4(x – 0.245) = M At x = 0.245 m; 6.300(0.245) – 4(0.245 – 0.095) – 4(0.245 – 0.245) M = 0.944 Nm At x = 0.800 m; 6.300(0.800) – 4(0.800 – 0.095) – 4(0.800 – 0.245) = M M = 0 Nm
The shear force and bending moment diagram for calculation above would be like figures on the next page:
Figure 15. Shear Force Diagram for 4-4 N Load at Point A and B (Double Point)
Table 4. Theoretical Values for Loading Point Holder A (Distance: 0.095 m)
Load (N) Shear Force (N) Bending Moment (Nm)
2 -0.238 0.167
4 -0.475 0.335
6 -0.712 0.502
Table 5. Theoretical Values for Loading Point Holder B (Distance: 0.245 m)
Load (N) Shear Force (N) Bending Moment (Nm)
2 1.388 0.340
4 2.775 0.680
6 4.162 1.020
Table 6. Theoretical Values for Double Loading Point Holder (A and B) with 2-2 N Load Distance (m) Shear Force (N) Bending Moment (Nm)
0 3.150 0
0.095 1.150 0.299
0.245 -0.850 0.472
0.800 -0.850 0
Table 7. Theoretical Values for Double Loading Point Holder (A and B) with 4-4 N Load Distance (m) Shear Force (N) Bending Moment (Nm)
0 6.300 0
0.095 2.300 0.599
0.245 -1.700 0.944
0.800 -1.700 0
Table 8. Theoretical Values for Double Loading Point Holder (A and B) with 6-6 N Load Distance (m) Shear Force (N) Bending Moment (Nm)
0 9.450 0
0.095 3.450 0.898
0.245 -2.550 1.415
0.800 -2.550 0
4.0 DISCUSSIONS
The tables and the graphs above show that the values of the load is proportional to the values of shear force as well as bending moment. Which means, the bigger the load applied, the bigger shear
force and bending moment will produced. This statement is valid to both conditions, one point loaded and also double point loaded.
Nonetheless, the experimental results are not really accurate since the console box somehow were only showing one direction (in this case is positive). As we all can see on theoretical values, there are some negative forces which means the force acting downwards. If only the console box worked how it supposed to be, the experimental values would not have big different compared to the theoretical ones.
Other than that, the shocks from another table that we sharing with affected the number that coming out of the console box.
5.0 CONCLUSION
In conclusion, this experiment show us the values of theoretical calculation are larger than the actual values that measured by a console box. Both single loaded and double loaded experienced the same condition. In outline, from the experiment, when the load is added, the number of shear force and bending moment will get higher too.
REFERENCES
1. Analysis of Beams | Shear Force & Bending Moment Diagram ~ Learn Engineering. [ONLINE] Available at: http://www.learnengineering.org/2013/08/shear-force-bending-moment-diagram.html. [Accessed 11 May 2014].
2. Shear Force and bending diagrams. [ONLINE] Available at: http://www.roymech.co.uk/Useful_Tables/Beams/Shear_Bending.html. [Accessed 11 May 2014].
3. Civil Engineering: TYPES OF BEAMS & TYPES OF LOADINGS. [ONLINE] Available at: http://civilengineerworks.blogspot.com/2011/12/types-of-beams-types-of-loadings.html. [Accessed 11 May 2014].