discussion and conclusion

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Discussion In this structural analysis of gymnasium equipment we have analysis the barbell. The material for this equipment is A-36 steel. We have obtained the references value of this material from the textbook. The ultimate stress for this material is max= 400MPa. The modulus elasticity, E of this material is 200GPa and modulus rigidity, G is 75GPa. Next, the yield strength is y= 250Mpa. We have analysis the material with the weight 5 kg at barbell which is at structure 1, 10kg at structure 4, and 2.5 at structure 5. We have analysis the equipment by using the appropriate calculation. The main part of this equipment is the barbell which is structure 1. The barbell is exerted by three forces which are P, Ra and Fbar. From the equilibrium equation we use load, P (5kg) and Fbar (2kg) to get reaction force, Ra. Then, we have calculate three forces to find the shear force and bending moment of the beam to obtain shear and bending moment diagram. Firstly we draw the shear diagram and then from the shear diagram we can draw the bending diagram. Area under the shear curve between two points is equal to the change in bending moment between the same two points. Next, we use the bending moment diagram to get the maximum moment so we can find the normal stress by using the equation: σ = M S By using the equation we get the normal stress equal to 115.87 MPa. The reference value for maximum stress of the A-36 steel is 400 MPa. Since the value obtained for the normal stress is smaller than the reference value for the steel we used, we conclude that the barbell safely support the load to which it will be subjected. We assume that the barbell is in static condition. If the force is exerted at the end of the barbell, deflection will occurs. The deflection of the barbell can be finding by using the formula: δ= PL AE

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Page 1: Discussion and Conclusion

Discussion

In this structural analysis of gymnasium equipment we have analysis the barbell. The material for this equipment is A-36 steel. We have obtained the references value of this material from the textbook. The ultimate stress for this material is max= 400MPa. The modulus elasticity, E of this material is 200GPa and modulus rigidity, G is 75GPa. Next, the yield strength is y= 250Mpa. We have analysis the material with the weight 5 kg at barbell which is at structure 1, 10kg at structure 4, and 2.5 at structure 5.

We have analysis the equipment by using the appropriate calculation. The main part of this equipment is the barbell which is structure 1. The barbell is exerted by three forces which are P, Ra and Fbar. From the equilibrium equation we use load, P (5kg) and Fbar (2kg) to get reaction force, Ra. Then, we have calculate three forces to find the shear force and bending moment of the beam to obtain shear and bending moment diagram. Firstly we draw the shear diagram and then from the shear diagram we can draw the bending diagram. Area under the shear curve between two points is equal to the change in bending moment between the same two points.

Next, we use the bending moment diagram to get the maximum moment so we can find the normal stress by using the equation:

σ=MS

By using the equation we get the normal stress equal to 115.87 MPa. The reference value for maximum stress of the A-36 steel is 400 MPa. Since the value obtained for the normal stress is smaller than the reference value for the steel we used, we conclude that the barbell safely support the load to which it will be subjected.

We assume that the barbell is in static condition. If the force is exerted at the end of the barbell, deflection will occurs. The deflection of the barbell can be finding by using the formula:

δ= PLAE

Next, we use the value of deflection to get the value of strain using the equation:

ε= δL

Furthermore, we can calculate the deflection of the member by using the formula of deflection.

EId2 yd x2

=−Px

Page 2: Discussion and Conclusion

The formula must be integrated twice to obtain the slope and the deflection of the member. The first the integration is to find the slope and the second integration is to find the deflection.

At structure 2, we can find the stress that act at the upper of barbell stand by using the equation:

σ= PA

So, the stress exerted at this position is equal to 20.44 kPa.

At structure 3, there will be buckling act upon the barbell stand. Using E=200 GPa and max=400MPa for A-36 steel material. Using formula:

Pcr= π EI(KL)

σcr=PcrA

The critical stress is equal to 415.61 MPa.

At structure 4 and structure 5, all force can be finding by using the equilibrium equation:

F=0

M=0

Structure 4 also has load that exerted at barbell stand. The load is supported by a pin, so we consider all the stress in the pin. By using the concept of equilibrium in particle, we can find the resultant force at the pin.

We consider the pin is in single shear. So, to calculate the shear stress we use the single shear stress formula as below and we get the value of shear stress τ=138.76MPa.

τ= PA

Page 3: Discussion and Conclusion

Conclusion

Based on this analysis we deduced that the load that exerted to the material must be appropriate. From this assignment we can conclude that strength of material is very important in everyday life. Everything surrounding us is related to equipment that we use. In this structural analysis of gymnasium equipment we have analysed, we can resolve that this gymnasium equipment is safe to be use. This is proven after we calculated all the structure, the value obtained is less than the reference value of this material from the text book.

The machine or this gymnasium equipment can be broke and damaged if some part of the structure or the supporting holder can’t withstand the exert force from the barbell or other force included. This situation will also may cause injury to the user.

The main objective of the study of the mechanics of materials is to provide the future engineer with the means of analysing and designing various machines and structures. Both the analysis and design of the gymnasium equipment involve the entire basic of mechanics.

Overall, the fundamental principles learned in the class help us to relate with the real life occurrences. Besides, we are now able to do analysis and apply theories to real life problem and situations. Lastly, good teamwork skills among the member are developed after going through this assignment.