EXPERIMENT 1 : REACTION ON BEAM 1. OBJECTIVE To test the reaction on simply supported beam under point load 1. INTRODUCTION The reaction on beam reacted as what Newton Third Law stated. “Action and reaction is the same but opposite”. Total reaction component depends on the types of supports. The structure under reaction is in equilibrium. This situation must occured so that the structure will not move. For that to happened, this structure must obey the rule of equilibrium. The load must be equalized by the reaction on the supports. The magnitude of action can be calculated using static equation. If this structure is externally determinacy, static equation was enough to solve the calculation. The steps to calculate the reaction are as below: 1. drew a image of the structure 1. identified the supports 1. Marked the reaction that will occur at all support, such as vertical reaction, horizontal reaction, and moment reaction. Assume that component react at all directions. 1. Write and solve all static equation which suitable to identify all the reaction magnitudes. 1. If the answer calculated was negative, this show that the direction pointed was opposite. If the value show positive value, this show that the direction was correct. These steps should be followed in order to solve the reaction of structure which have not more than three reactions. W R1 R2 1. APPARATUS Apparatus is set up as show in the picture below. Diagram 1: One point load M1 M3 m Diagram 2: Two point load M2 M 1050 mm x M1 M3 m m n y Balance spring Diagram 2: Two point load M2 1. PROCEDURES Apparatus was set up as shown in Diagram 1 and 2. The beam is to be making sure that it was flatted by using adjustable screw. The initial reading is on balance spring is jot down 1. Beam was loaded with one point load at different distance (x) from support (A). Action value on beam was observed and jot down at different weight (M) at different distance. 1. Two point loads (M1 and M2) was applied as shown in Diagram 2 and the reaction at difference was jot down carefully. 1. One point load, M3 was placed at distance y. Reaction on beam by the balance spring was jot down carefully. Values of M3 and y are unknown and have to be identifying from this experiment. The reaction R1, R2, R3, and R4 id calculated using principle of moment and value have to be obtained by comparing the reading from balance spring. 1. RESULT For one point load, M =1 kg Distance, x (mm) Reaction R1 on A(N) Reaction R2 on B(N) Experiment Calculation Experiment Calculation 375 18 0.64 16 0.36 500 17 0.52 18 0.48 750 15 0.29 20 0.71 1000 12 0.05 22 0.95 Table 1 For two point load, M1 =1 kg , M2 =2 kg Distance, m Distance, n Reaction R3 on A(N) Reaction R4 on B(N) (mm) (mm) Experiment Calculation Experiment Calculation 125 875 24 1.21 31 2.48 250 750 25 1.33 30 1.67 375 625 27 1.50 28 1.50 Table 2 For point load, M3, Distance, y (mm) Reaction R5 on B(N) Reaction R6 on B(N) M3 (N) 300 2.0 N 0.5 N 2.5 N CALCULATION F = m x g Where, F = force created m = weight of object g = gravitation force (10ms־² ( Moment for force on certain point was equal to the total of object’s weight multiple with the distance of the object. M = Force x Distance Where, M = moment 1. EXAMPLE OF CALCULATION Table 1 X M A B R1 R1 1050mm For one point load, M1 = 1 KG Taking moment about B Clockwise moment = anticlockwise moment Therefore, R x 1050 = 1 x 675 Resolving vertically, R1 = 675 / 1050 force up = force down R1 = 0.64 N R1 + R2 = 1 3.21 + R2 = 1 R2 = 1-0.64 R2= 0.36 N Table 2 n m M2 M1 B A R3 R4 1050 mm For two point load, M1= 1 kg , M2 = 2 kg Taking moment about B, Clockwise = anticlockwise Therefore, R3 x 1050 mm = (1x925) + (2x175) R3 = 1275/ 1050 R3= 1.21 N Resolving vertically, Force up = force down R3+ R4 = 3 R4= 3/1.21 R4=2.48N Taking moment about B Clockwise = anticlockwise Therefore, R 3x x1050 = 1(800) + 2 (300) R 3 = 1400/1050 R 3 = 1.33 N Resolving vertically, Force up = force down , R 3 + R4 = 3 1.33 + R4 = 3 R4 =1.67N Taking moment about B, Resolving vertically, Clockwise = anticlockwise Force up = force down Therefore, R 3 + R4 = 3 R 3 (1050) = 1 (675) + 2 (452) R 3= 1.50 N R4 = 1.50 N TABLE 3 Taking moment about B Force up = force down Therefore , clockwise=anticlockwise R5 = 2.0 N R5 + R6 = M3 R6 = 0.5 N 2.0 + 0.5= 2.5 N 1. DISCUSSION Explain the reasons of occurring errors and give your conclusion. In conducting experiment a person encounters one or more errors, systematics error and random error. In accuracy occurs in the reading due to : 1. The imbalance of the equalizer spring. 1. Environmental factors such as wind and vibration. 1. The flaw of the apparatus. 1. Human error (parallax error) when set up experiment misread an instrument or mistake in calculation. Facts that causes the error of experiment’s conclusion. In accuracy in the experiment can be reduces by taking several precautions method including : 1. Taking several readings and taking and average as the final result 1. Reading and eye level must be perpendicular to reduce parallax error. 1. The experiment should be done in secure and closed environment to reduce error caused by the environment surrounding. 1. CONCLUSION 1. Beam is equilibrium condition when total force is equal. 1. Uncertainly in result can be avoid by taking more than one readings and calculate the average reading and replace the old apparatus to new one.