Supports Reactions of a Loaded Beam
1) To experimentally verify the equations for the equilibrium of non-concurrent forces by measuring the supports reactions of a loaded beam
A uniform metal beam, two triangular supports, two flat-top scales, weights, and a calculator
The conditions for the static equilibrium of a loaded beam are:
ΣFx = 0 ; ΣFy = 0 ; ΣT = 0.
For a horizontal beam that is supported at two points and loaded as shown, the external forces acting on it are all vertical. Since there are no horizontal forces acting on it, the condition ΣFx = 0 is automatically satisfied. By applying the equations: ΣFy = 0 and ΣT = 0, two unknown reactions at its two supports can be calculated and then compared with their corresponding measured values. This verifies the validity of the net torque equilibrium equation ΣT = 0. Note that although no rotation takes place in this experiment; however, since the sum of torques caused by forces is set equal to zero for static equilibrium, we say that rotational equilibrium is achieved. If a rotating object is rotating at a constant angular velocity, its angular acceleration α = 0. This makes the net torque acting on it to be zero that means ΣT = Iα = 0. If an object is not rotating and is at rest, the equation ΣT = Iα = 0 still holds true because α = 0 due to no rotation.
The point about which the sum of torques or moments is to be calculated is arbitrary. Any point may be selected for this purpose. The point may be chosen on the beam or out of it. The final values that one calculates for the reactions will come out the same regardless of the point chosen.
Note: Since this experiment involves weights of up to 5-kg, it is recommended that the experimental set up be at the center of the table and far from the edges to avoid weights accidentally fall on the floor. Although all students must have covered-toe-shoes on, those with weaker shoes must specifically keep distance from the table edges.
The beam to be used is divided into 8 equal sections by 8 lines, as shown. The supports are triangular prisms, as shown. When placed on the table, each support has a relatively sharp top edge for the beam to be placed on top of it. This guarantees the measurement of the position of the reaction forces on the beam with good precision.
For the purpose of force measurement, two identical flat-top scales must be used. On the top of each scale, one triangular support must be placed and then zeroed. This way, the weights of the supports will be excluded from calculations. The beam should then be placed on the top edges of the supports and its position adjusted as shown.
Following are the specific steps to be taken:
1) Determine the weight of the aluminum beam by using an appropriate scale provided for this experiment.
2) Place the supports at B and H by adjusting the positions of the scale, supports, and beam.
3) Place a 5-kilogram weight centered at B. Read the reaction forces (from the scales) and record them in the table provided under the Data section.
4) Move the 5-kg weight to C. Record the new reactions.
5) Repeat Step 4 with the 5 kg weight moved to positions D, E, F, G and H respectively.
6) Calculate the expected or accepted values of the reactions for each case by applying the two equilibrium equations discussed above. Determine a %error on each reaction in each case.
7) Plot the experimental values of reaction at B versus the location of 5 kg weight on the beam.
1) Place the supports at E and H.
2) Place a 5-kg weight at G. Read scale reactions and record them in a similar table.
3) Let the 5-kg weight remain at G. Place a 1-kg weight at D. Record the new reactions.
4) Repeat Step 3 with the one-kilogram weight moved to the C and B positions, respectively.
5) Calculate the accepted values of reactions and determine a %error on each reaction as you did in Part 1.
6) Plot the analytical reactions at location H versus the location of the 1- kg weight on the beam.
From the graph in Step 6, determine the location of the 1-kg weight when the reaction at point H is zero. (It will be off the beam and therefore it is somewhat hypothetical).
g = 9.81 m/s2.
|Measured Reactions (N)||Calculated Reactions (N)||%error on RB||%error on RH|
|1||5 at A|
|2||5 at B|
|3||5 at C|
|4||5 at D|
|5||5 at E|
|6||5 at F|
|7||5 at G|
|8||5 at H|
|Note: Observe significant figures|
|Measured Reactions (N)||Calculated Reactions (N)||%error on RE||%error on RH|
|1||5kg at G|
|2||5 at G, 1 at D|
|3||5 at G, 1 at C|
|4||5 at G, 1 at B|
|Note: Observe significant figures|
Comparison of the results:
Provide the percent error formula used and the calculated percent errors in each case.
State your conclusions of the experiment.
Provide a discussion if necessary.