Design Report
Questions
Solution
GIVEN DATA
- Outside Diameter of shaft Do =100 mm
- Length of shaft L = 1.25 m or 1250 mm
- Force acting on center of shaft F= 850 N
- Maximum deflection allowed (mid-point) = 0.10 mm
Consider a case in which shaft is simply supported at its ends and load of 850 N is applied at its center.
Bending stiffness formula
s= F/δ=(C_1 EI)/L^3 s= F/δ=(C_1 EI)/L^3
- F = Force acting on shaft
- δ= Permissible deflection
- E = Elastic modulus
- I = Moment of inertia of shaft
- L = Length of shaft
- C_1 = constant value equals to 48
a) All the three materials can be used as a shaft material in combination of resin. Calculate inside diameter of all 3 cases.
Consider the case of carbon fiber standard module.
- E = 230 GPa
F/δ = (C_1 EI)/L^3
850/0.1 = (48*230*1000*I)/〖1250〗^3
- I = 1.5*〖10〗^6 〖mm〗^4
- I = π/64 (〖D_O〗^4-〖D_I〗^4 )
Similarly inside diameters for other 2 materials calculated as same as above , the values of insider diameters are shown in table below .
From the above table, the required cost is minimum for Carbon fiber standard module. So carbon fiber standard module would be least expansive i.e252 .8 AUD.
