The Reaction Forces on a Supported Beam - Assignment Example

ASSESSMENT ON THE BENDING MOMENT AND TENSILE TESTING Assessment on the Bending Moment and Tensile Testing test to identify the reaction forces on a simply supported beam For a simply supported beam forces and torque develop. The beam is usually horizontal and supports loads acting vertically downwards. At the points where the beam is supported there are reaction forces that act vertically upwards. These forces are refereed to as reaction forces. There are two main types of forces applied vertically to a loaded beam. These forces are Point loads: these forces act at a point (Boresi, 1993). Distributed loads. These loads are distributed over a given length of the beam. This type of load is referred to as a uniformly distributed load (UDL) Fig 1 showing the elements of a simply supported beam Materials required for the test 1. Beam (the beam should be rectangular in cross-section ) length of the beam = 1m 2. A pair of simple supports (wedge shaped) 3. Spirit level 4. A set of weights 5. A hanger 6. Steel rule 7. Force gauge 8. Data logger Procedure 1) Set the two support on a test table that is flat 2) Set up the beam on a simply supported stand. Use the spirit level to ensure that the beam is laid totally flat. 3) Hang the loads on the beam at points equal to 0.4 m and 0.6 m from one end of the beam. 4) Simple support stands. 5) Inserts the force gauge on the two ends of the simply supported beam. 6) Connect the force gauge on the data logger in order to measure and record the force measurement. 7) Take the first reading 8) Move the weights to another location and take the measurement The diagram below shows the experimental setup for the determination of the reaction forces on a simply supported beam. Figure 2 showing the experimental set up for determination of the reaction forces on a simply supported beam. (2)Sources of errors In the experiment described above, the sources of error are 1. Error of measurement when using the steel rule. The loads may not be placed at their exact location 2. Error when measuring the forces 3. The beam selected may not have uniform internal properties 4. It is difficult to ensure that the beam is not tilted at a small angle. (3) For a simply supported beam with points load as shown below; The reaction at point (B) is calculated by calculating moment from point (A). Let the distance from point load (C) be X1 from point A and the distance from point load (D) be X2 from the support (A) Anticlockwise moment = clockwise moment Or Anticlockwise moment- clockwise moment = 0 After calculation of the reaction at point B, the force at point A is computed by; Sum of the upward forces = sum of the downward forces C+D = A+ B in this case B is known and hence the force is calculated by; A= (C+D) –B Where A, B, C and D are the forces acting on the beam. Question 4 For the given beam problem below, and using the mathematical equations in Question 3, calculate the reaction forces of RA and RB. Take that the span of the beam is exactly 1m measured from point A. Taking moments from point A Reaction force Calculating the reaction at point A Sum of upward forces = sum of downward forces Question 5 For the given beam problem below, and using the mathematical equations in Question 3, calculate the reaction forces of RA and RB. Take that the span of the beam is exactly 1m measured from point A. Taking moments from point RA Anticlockwise moments = clockwise moments The reaction force RB = 117N Calculating the forces RA Upward Forces = Downward Forces Reaction force RA = 133N Question 6 For the given beam problem below, and using the mathematical equations in Question 3, calculate the reaction forces of RA and RB. Take that the span of the beam is exactly 1m measured from point A. The beam must be in balance. This means that the anticlockwise force must equal the clockwise moments. Taking moments from point A Part B – Tensile Testing TENSILE TESTING Experiment for testing the mechanical property of a material when subjected to tensile load: In the tensile load testing, a carefully prepared specimen on the material is subjected to pulling force or a tensile load. This employs the use of a tensile testing machine. With this machine, a load is applied on the prepared specimen. The elongation on the test piece is recorded using a dial gauge. This gauge is a small gauge that senses very small elongation per applied load. The procedure for conducting the experiment is outlined below; 1) Take a specially prepared specimen of specified length and cross-section area. 2) From materials tables determine the yield load for the material. This helps in preventing the snapping of the test piece when the dial gauge is still attached on the test piece 3) Mount the test piece on the tensile testing machine 4) Attach the dial gauge on the test piece 5) Gradually increase the load in equal increments say 10 KN, after every increment note the elongation on the test piece. Record the data in a table as shown below; Table 1 Load applied Elongation 0 KN 10 KN 20KN 30KN 40KN 50KN On nearing the yield load remove the dial gauge, continue recording the load applied. Apply the load till the material snaps/ breaks. Note the applied load before the material breaks. From the collected data a graph of load against extension can be plotted. 2. What do the terms “elastic” and “uniform plastic” limits mean in the context of tensile test results? Elastic limit When a material is subjected to tensile load it length changes. The increase in length is directly proportional to the applied load. This means that after application of load, the material returns to its original length. After continued increment of the tensile load, A point is reached where the extension due to the applied load is not directly proportional to the extension. This point is called the elastic limit Plastic limit After exceeding the elastic limit, the material becomes plastic. This means that the extension caused by applied load is not proportional and the material does not return to its original point when the applied load is removed. A point is reached where the material breaks, this is the plastic limit. 3. Describe what is meant by the Hooke’s law and how this information can be found on a typical Force (N) Vs Extension (mm) curve for common non-ferrous alloy. Hooks law Hooks law states that the extension on a material is directly proportional to the applied load as long as the elastic limit is not exceeded. For a material subjected to tensile load and the graph plotted, the straight part of the graph represents the region under which the hooks law applies. This is the elastic region. Where the straight line begins to curve the point is called the elastic limit. Load Figure 2: showing the graph of load against extension 4 Using the test data plotted from such a tensile experiment on two different metals, namely copper and brass shown in the Figure below answer the following: (A)What is the average “yield” and “ultimate” forces for the materials? Yield load Yield load: this is the maximum stress in which the material begins to deform plastically. Before yield point the material behaves elastically. From the graph provided the yield force for brass is equal to 7000N this is equivalent to 7KN From the graph the ultimate load for copper is 3500 N this is equivalent to 3.5 KN Ultimate load This is the maximum load that is applied before the material finally fails From the graph the maximum ultimate stress for brass is given by 8300 N this is equivalent to 8.3 KN From the graph, the ultimate load for copper is given by 4500 (B) Suggest a measure of “ductility” and subsequently calculate the average “ductility” of the materials given that the original length of ALL the tensile specimens was 25mm Ductility can be defined as the ability of a material to withstand plastic deformation without rapture. Ductile materials deform easily under tension. When a material is subjected to tensional loads the length of the material increases while the cross sectional area reduces. These two indicators are used for the measurement of ductility. Ductility measurement using change in length Ductility measurement using change in cross-sectional area Percent elongation = (final gauge length –initial gauge length)/initial gauge length Percentage increase in area (Original cross section area- minimum final area)/original cross-sectional area Ductility for brass specimen Original length of brass specimen = 25 mm Extension on the brass specimen = 7 mm Final length of the brass specimen = 25mm+7mm= 32mm Ductility = percent elongation = Ductility for copper specimen Original length of copper specimen = 25 mm Extension of the copper specimen = 13 mm Final length = 38 mm Ductility = percent elongation = (C)Suggest an alternative measure of ductility in engineering applications where continuous drawing or rolling process is employed? I.e. when the measure of elongation is more difficult. When the measure of elongation is more difficult because there is no uniformity of length and elongation. The change in crossectional area can be used to compute the ductility. The use of change in cross-sectional area is more reliable as compared to elongation since the crossectional area changes uniformly Percentage increase in area (Original cross section area- minimum final area)/original cross-sectional area Where (A0 )is the original crossectional area, (Amin ) is the final reduced area (D) Which material absorbs greater amount of energy up to the point of fracture? Justify your choice of answer. Brass material absorbs more energy up to the point of fracture. This is because the material resists elongation and hence more energy must be applied to pull the material. (E)In your opinion, what are the possible contributing factors to the variation in the acquired test data? The variation in data is due to the impurities in copper and brass elements. These impurities alter the materials properties including ductility and its yield strength. The variation can also be caused by alloying elements. Also heat treatment can also alter the properties of the material References Boresi, A. P, Schmidt, R. J. and Sidebottom, O. M. 1993. Advanced Mechanics of Materials. New York: Wiley Read More

Question 6

For the given beam problem below, and using the mathematical equations in Question 3, calculate the reaction forces of RA and RB. Take that the span of the beam is exactly 1m measured from point A.   

The beam must be in balance. This means that the anticlockwise force must equal the clockwise moments.

Taking moments from point A

Part B – Tensile Testing

TENSILE TESTING

Experiment for testing the mechanical property of a material when subjected to tensile load:

In the tensile load testing, a carefully prepared specimen on the material is subjected to pulling force or a tensile load. This employs the use of a tensile testing machine. With this machine, a load is applied to the prepared specimen. The elongation on the test piece is recorded using a dial gauge. This gauge is a small gauge that senses very small elongation per applied load. The procedure for conducting the experiment is outlined below;

• Take a specially prepared specimen of specified length and cross-section area.
• From materials tables determine the yield load for the material. This helps in preventing the snapping of the test piece when the dial gauge is still attached to the test piece
• Mount the test piece on the tensile testing machine
• Attach the dial gauge on the test piece
• Gradually increase the load in equal increments say 10 KN, after every increment note the elongation on the test piece. Record the data in a table as shown below;

Table 1

Load applied

Elongation

0 KN

 

10 KN

 

20KN

 

30KN

 

40KN

 

50KN

 

 

 

 

On nearing the yield load remove the dial gauge, continue recording the load applied. Apply the load till the material snaps/ breaks. Note the applied load before the material breaks.

From the collected data a graph of load against extension can be plotted.

1. What do the terms “elastic” and “uniform plastic” limits mean in the context of tensile test results?

 

Elastic limit

When a material is subjected to tensile load it length changes.  The increase in length is directly proportional to the applied load. This means that after application of load, the material returns to its original length. After continued increment of the tensile load, A point is reached where the extension due to the applied load is not directly proportional to the extension. This point is called the elastic limit 

Plastic limit

After exceeding the elastic limit, the material becomes plastic. This means that the extension caused by the applied load is not proportional and the material does not return to its original point when the applied load is removed. A point is reached where the material breaks, this is the plastic limit.

3. Describe what is meant by Hooke’s law and how this information can be found on a typical Force (N) Vs Extension (mm) curve for common non-ferrous alloy.
   

   Hooks law

   Hooks law states that the extension on the material is directly proportional to the applied load as long as the elastic limit is not exceeded.

   For a material subjected to tensile load and the graph plotted, the straight part of the graph represents the region under which the hooks law applies. This is the elastic region. Where the straight line begins to curve the point is called the elastic limit.  

    

                  

   Extension

      
   

                                    Load

   Figure 2: showing the graph of load against the extension

    

   4 Using the test data plotted from such a tensile experiment on two different metals, namely copper and brass shown in the Figure below answer the following:

   (A)What are the average “yield” and “ultimate” forces for the materials?

    

   Yield load

   Yield load: this is the maximum stress in which the material begins to deform plastically. Before the yield point, the material behaves elastically.

   From the graph provided the yield force for brass is equal to 7000N, this is equivalent to 7KN

   From the graph the ultimate load for copper is 3500 N this is equivalent to 3.5 KN

   Ultimate load

   This is the maximum load that is applied before the material finally fails

   From the graph the maximum ultimate stress for brass is given by 8300 N this is equivalent to 8.3 KN

   From the graph, the ultimate load for copper is given by 4500

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