Project 3

Firstly we solve the equations using strength of materials formula:

deflection formula is given by: PL^3/3EI, where P is the load, L is the span length , E is youngs modulus and I is moment of interia of the cross section.

solving for

1)rectangle cross section:

I=BD^3/12.

=> I= (50x50^3)/12

=> I= 520833.33mm^4.

therefore deflection D is given by,

D = (1000x1000^3)/3x210x10^3x520833.33

=> D = 3.0476mm.

2)circular cross section

I = (PIxd^4)/64

I = (3.14x100^4)/64

I = 4908738.521mm^4.

therefore deflection D is,

D = (1000x1000^3)/3x210x10^3x4908738.521

=> D = 0.3233mm.

2) I section

given that area of rectangle and area of I section are equal. hence,

2500/3 = Lxb

833.33=50xb

=>b=16.66

moment of inertia I is given by

I=(bd^3/12) - (b1d1^3/12)

=> I = 2062816.543mm^4.

therefore the deflection is given by,

D = 1000x1000^3/3x210x10^3x2062816.543

=> D = 0.7694mm.

Now we verify the results using mechanical apdl and workbench:

1) rectangular cross section:

MAX deformation : 3.05203mm.

2)circular cross section:

max deformation :0.325315mm

3) for I section :

max deformation :0.00077871m= 0.77871mm.

From the above obtained results we can see that the values obtain are very much closer to the values obtained by using the SOM formula. Hence verified.